Support email articles Harvard college search blogs

Yeah, that's the problem with politicians and politics-related people.  They don't know anything about math.  I don't know if they do those "John Kerry voted to raise taxes x times" ads in your area.  They're a horrible misuse of statistics.  I like the book [i]Innumeracy[/i], by John Paulos, I think.  It has a lot of fun statistical stuff.

I wonder who u support?

...well, fox isn't the only channel. There are alot of other channels that are biased for Kerry and against Bush. The economy really isn't the fault of ne of those two, especially not Bush. The president has no control over economy.

[quote="mathmonkey"]Every now and then when I want something to laugh at, a flip over to fox news. Seriously, I laugh at how blatently biased they are. Once on some economic show they had, there was a bunch of economy analysts. One said that Kerry was bad for the economy, because when his poll numbers went up the economy went down. Of course everyone else on the show agreed with him.
I'm taking a statistics course right now, and I already know that correlation doesn't prove causation. For all anyone knows, something else (like oil prices rising) could have made both Kerry's poll numbers go up and stock prices drop. I'm in 10th grade and I know this! Just one of so many examples of fox's bias towards the republicans.[/quote]

What's really the point of this, other than a flame against a major news network? I think we can find some piece of mind without flaming.

Also, I'm finding that most democrats I know (in day-to-day life, not AoPS or online) are pretty fanatical. For some reason, the republicans I know in daily life are less fanatical. I sure don't get emails each day saying about Kerry's swift boat stuff, but I do get a lot of emails each week or so about Bush's IQ and other stuff like that. Might be just a coincidence though.

On the subject of Fox News' bias, have you ever read the opinion section in ant major newspaper?  Notice the number of anti-Bush articles there are?

My point is that FOX News is one of the few news sources that does [u]not[/u] have a blatant liberal bias.  You don't hear people complaining about how the New York Times is liberal, why is Fox News being conservative any different?

Well, considering how well a lot of political figures do it, I can't imagine it would be that difficult...

[quote="kentliu"]the president has lots to do with the economy.. for example, if kerry pass laws that will discourage outsourcing of jobs, then the economy will get better.. and if bush began selling the reserved oil, oil prices would drop.[/quote]

I really don't think Kerry could pass any law. Laws are passed by the legislative branch which is dominated by Republicans. All Kerry can do about it is whine and give speeches about how he's going to make everything better.

[quote="towbomb"]On the subject of Fox News' bias, have you ever read the opinion section in ant major newspaper?  Notice the number of anti-Bush articles there are?[/quote]

This is, of course, a complete dodge.  When people accuse Fox of having a conservative bias, it isn't a complaint about their editorializing; it's a complaint about their [i]news[/i].  As a simple example, people for whom Fox is their main news source are far more likely than the average American to believe that Saddam Hussein was directly responsible for the attacks of September 11th, which is false.  When you purchase a newspaper, you know that it has an op-ed page and you can reasonably guess the politics of the editors.  That has nothing to do with the bias in question.  What's important is the effect on actual news coverage.

[quote="towbomb"]You don't hear people complaining about how the New York Times is liberal...[/quote]

Well, I hear you complaining about it.  The Harvard Salient's title piece this month is about it.  In fact, every real conservative I've ever had a coversation about this with has complained about it.  In fact, the Times' public editor has done several pieces concerning it, published in the times.  A quick search of conservative blogs finds it plastered all over.  If you haven't heard it, you haven't tried hard enough.

[quote="yifman"]Also, I'm finding that most democrats I know (in day-to-day life, not AoPS or online) are pretty fanatical. For some reason, the republicans I know in daily life are less fanatical. I sure don't get emails each day saying about Kerry's swift boat stuff, but I do get a lot of emails each week or so about Bush's IQ and other stuff like that. Might be just a coincidence though.[/quote]

You need some perspective -- you live in Conneticut, a very Democratic state.  The case would be highly different if you were in, say, rural Georgia or Texas.

[quote="filletwho"]The president has no control over economy.[/quote]
If that's the case, why has Bush sold all 4 of his tax cuts as economic stimuli?  Why has he claimed credit every time a job is created?  You can't have him both get credit for good things and not get blamed for the bad things!

[quote="filletwho"]I really don't think Kerry could pass any law. Laws are passed by the legislative branch which is dominated by Republicans.[/quote]

Come now, don't be naive.  A president, among other things, is responsible for sending a budget to congress and signing every bill congress passes.  He has a significant role in pressuring members of Congress to support legislation he supports and so on.  This is, for example, why it is reasonable to tie the non-renewal of the Assault Weapons Ban to Bush -- he did no lobbying in the House whatsoever to get it passed.  Granted, when a President has a Congress opposed to him, his hands are rather tied.  (For example, compare the number of court appointees that were blocked under Clinton to those blocked under Bush.)  But in the Senate, it would take only one cross-over Republican (with a Democratic vice-president) to get Democrat-supported legislation passed, and the margin in the House is also quite small.  Additionally, it's not by any means impossible (if not exactly likely) for the Democratic party to come out of this election in control of the Senate.

[quote="JBL"][quote="yifman"]Also, I'm finding that most democrats I know (in day-to-day life, not AoPS or online) are pretty fanatical. For some reason, the republicans I know in daily life are less fanatical. I sure don't get emails each day saying about Kerry's swift boat stuff, but I do get a lot of emails each week or so about Bush's IQ and other stuff like that. Might be just a coincidence though.[/quote]

You need some perspective -- you live in Conneticut, a very Democratic state.  The case would be highly different if you were in, say, rural Georgia or Texas.
[/quote]

True, true.

[quote]filletwho wrote:
The president has no control over economy.

If that's the case, why has Bush sold all 4 of his tax cuts as economic stimuli? Why has he claimed credit every time a job is created? You can't have him both get credit for good things and not get blamed for the bad things!

filletwho wrote:
I really don't think Kerry could pass any law. Laws are passed by the legislative branch which is dominated by Republicans.

Come now, don't be naive. A president, among other things, is responsible for sending a budget to congress and signing every bill congress passes. He has a significant role in pressuring members of Congress to support legislation he supports and so on. This is, for example, why it is reasonable to tie the non-renewal of the Assault Weapons Ban to Bush -- he did no lobbying in the House whatsoever to get it passed. Granted, when a President has a Congress opposed to him, his hands are rather tied. (For example, compare the number of court appointees that were blocked under Clinton to those blocked under Bush.) But in the Senate, it would take only one cross-over Republican (with a Democratic vice-president) to get Democrat-supported legislation passed, and the margin in the House is also quite small. Additionally, it's not by any means impossible (if not exactly likely) for the Democratic party to come out of this election in control of the Senate.[/quote]

True, the president can propose and approve laws, but he cannot pass the laws without a majority vote in Congress, which is what I was reffering to. I was merely trying to point out that Kerry can't accomplish all those things he is promising. I'm not taking the blame off bush for the things he's proposing, but the bad economy is not [i]enirirely[/i] his fault. Does not Alan Greenspan have control over the budget policies? The decline in the stock market came 6 months before Bush took office. I'm not being naive about things, ppl who think Kerry can improve the economy with his imaginary policies are the ones being naive.[/quote]

[quote="h_s_potter2002"]Has anyone ever read the book [i]How to Lie With Statistics[/i]?[/quote]

That book is a classic. I read it back in the 1970s when I was on my high school debate team. The inflammatory title is to sell the book, of course. It is actually about how to recognize statistical lies.

[quote]It is actually about how to recognize statistical lies.[/quote] and one of the most effective come-back line I have used is:[quote]Isn't 87.6% of all statistical data  is made up? :) [/quote]

[quote]And can you please say which of John Kerry's policies is "imaginary?"[/quote]

John Kerry barely has any detailed plans. If he does outline nething, he doesn't have the budget to pay for it. He says he'll get rid of the national defecit while claiming he'll provide government funding for things like health-care. He claims he will not tax the middle class, but do u not think that the term middle class is vague. Many of us consider ouselves to be middle class while some of the less fortunate would consider us rich.

He might say things like he'll provide such and such funding for something but every word he says contradicts every other word. It seems inevitable that he'll either raise taxes on the "middle class" or not follow through with the paltry few policies he has outlined.

You clearly suffer from selective information intake.  Rather than complaining that you don't know anything about Kerry's plans, you might go do research on them or even have watched the debates where he layed some of them out.  I notice that you speak only in generalities about what he says, rather than pointing out specific instances of places where he has contradicted himself in discussing his policies, for example.  This shows a complete lack of effort on your part -- if you were really correct and not just insulting him, you would provide some content to support your claims.  (If your response to this includes the number 87, I'm going to decide that this discussion isn't worth having with you.)
You complain that he hasn't defined the term "middle class."  In fact, he has repeatedly pointed out that he intends to raise taxes on people making over $200,000 and no one else.  That is a specific policy goal targetting a specific group of people.  So I don't know what you're talking about when you say he doesn't give a definition to the term \"middle class.\"
As for funding of health care, the actual cost of his proposed program is relatively small.  George Bush, in the debate, put it at something like $7,000 a family.  Taking George at his word, despite the fact that I have no idea where that number came from, that would be vastly cheaper than what Americans currently pay for insurance today (roughly $5,000 per capita).  Simple steps, like mass-purchasing of medicines through medicare and medicaid, could save billions of dollars.

As for your point that it is unrealistic for Kerry to claim to be able to halve the deficit in the next 4 years, you are almost certainly correct.  What's interesting about this point, of course, is that Bush has claimed exactly the same thing while simultaneously proposing spending almost exactly the same amount of money Kerry has.  So at the very best, you get a tie from that point.

I find it offensive that you would attack me personaly, saying that i'm complaining, while i was merely trying to give a point. I will give u one more chance, if u continue to turn this personal, i would be happy to note shortcomings in your views. I did not list specifics because i was under the impression that this was a friendly discussion on views and thoughts, not a full scale debate as u have made it.

Back to the point, because I have an amazing well of time(actually not really but it's either this or HW) on my hands, i checked out Kerry's policy from a few sites, mainly Johnkerry.com. You said I was complaining, well, someone is complaining and it's not me. Kerry's outline of plans consists mainly of complaints on how this government is ineffectually run. A plan starts out with how such-and-such is done wrong and something to the effect of John Kerry is going to make it better. Mostly his plan is spending X amount of money. The most specific he gets is rolling back tax on companies that keep their jobs in America. He also says he's giving tax cuts to 99% of Americans. Now, it seems to me, that he's giving tax cuts and spending an awful lot of government funding for his "plans" while claiming to cut the defecit in half in the first four years. Where does all this money hail from? I'm pretty sure that raising taxes on the rich isn't even gonna produce all this money. So, that is where he contradicts himself.
About the middle class, yes, he does give general parameters, but i'm not sure what your 7000 is. When he promised not to raise taxes for the middle class, did he specify?
I did watch the debate and i noticed that most of the plans are vague or he changes the subject. One time he even sent ppl to a website.

Some policies of Kerry's which are specific I agree with, but I see no way that he has the money or the backing in Congress to accomplish this stuff.

One more thing, he talks about not having a coalition when going to Iraq and how he's going to change that. Well, I was given the impression by the debate that on his policy for N. Korea, he wants to end Chinese negotiations. That's another area of contradiction. When the Country has armed nuclear weapons, it's probably(this is only my opinion) not wise to end diplomacy and go aggressive.

Joel, can you not attack [i]people[/i] in every political discussion?  I understand that you have strong political beliefs, but please limit yourself to attacking Mr. Bush's policies, actions, and so on -- not the supposed weaknesses of the people who support him.  You only make yourself look bad when you are arrogant enough to tell someone they "clearly suffer from selective information intake".  I think you owe someone an apology.

You're both right, and I apologize.  Allow me to re-phrase some of the things I said in my last post in a non-attacking fashion.

First, Kerry has repeatedly stated that his tax increases are targetted at those making over 200,000 USD.  It follows that his tax breaks are targetted at those making less than 200,000USD, although one may probably fairly assume that for people at the upper end of that lower region, taxes will likely stay the same rather than decrease significantly.  This 200,000USD number has been stated repeatedly and unambiguously.  The tax breaks he has suggested have largely been laid out in terms of specific programs -- such as the child tax credit, for example -- that already benefit a well-defined group of people (in particular, those whose income is high enough to qualify for it).  I don't happen to know every detail of the child tax credit, but it was originally fashioned as a "middle class tax break," and so it is certainly reasonable to describe expanding it in the same terms.  (The way the break works, by the way, is as a fixed-dollar amount, so while it benefits wealthier individuals as well it does not do so in a disproportional way.  In fact, it dispropotionately affects people who make less money, because you can't get a tax break if you don't pay taxes.  That's what makes it for the "middle class."  Also, corporations don't have children -- it's strictly a personal income tax issue.)

The 7,000USD number was stated by George Bush in the debate as the cost of Kerry's proposed plan per individual.  I have no idea if it's an accurate number, but it's pretty darn cheap!

You'd be amazed how much money raising taxes on the rich can raise.  The tax cuts of the last 4 years have cost several trillion dollars -- almost all of it coming from people making 6 or more figures.  This makes sense, because those people (even though there aren't very many of them) pay a higher percentage of their income and have such enormous incomes that the combined effect is that you can raise (or lose) a lot of money by tinkering with their tax rates (and, of course, with other portions of the tax code).  As for his health plan, I believe Kerry has laid out its major aspects quite clearly.  Among other things, it involves creating a seperate insurance system for those who are seriously ill, taking them out ofthe traditional healthcare system and consequently decreasing total costs, as well as opening the Congressional health plan up to all Americans, among other options.  [url=http://www.johnkerry.com/issues/health_care/health_care.html]this[/url] webpage lays it out in full, and I don't see any of the unspecificity that you are complaining about.

Many problems simply are money problems.  For example, the initiative mentioned on the Kerry website about afterschool programs is essentially an issue of funding -- there are lots of afterschool programs that schools would run, if they could fund them.  The under0funding of N.C.L.B. is a problem that is purely a money problem: the framework for spending the money is already in place, there just hasn't been money allocated.

I find the rest of your complaints* to be politician-specific rather than Kerry specific.  Bush has proposed the same amount of new spending for the next 4 years that Kerry has, so his inability to fund it is equally apparent.  Everyone tries to change topics in debates when faced with a question they don't like: remember how Bush, when asked about jobs, spent his 2 minutes or whatever talking about NCLB?  Often, it makes perfect sense to send people to a website; why restate your whole platform when you only have 2 minutes and the content is readily accessible to listeners?

Finally, for North Korea, when asked directly about his plan in the first debate he stated that he would like to continue the 6-nation talks while opening up simultaneous bilateral talks as well.  That is certainly not ending negotiations or going agressive towards either N. Korea or China.  (Even if it were true, I don't see how it's a "contradiction.")  It's worth remembering here as well that Bush was the president who came upon the presidency with serious discussions happening between the U.S. and North Korea and just dropped them for more than 2 years.

Again, I apologize for my earlier rudeness.  I hope we can continue this discussion in a non-agressive fashion :)

* Note that this use of the word "complaint" is non-derogotory: it refers to your dislike of the characteristics of a person, rather than "complaining" (in the sense of whining) about anything.

After a bit more research, i concede that som plans are detailed and some are even good. But if Kerry taxes the ppl who earn over 200,000USD for enough money to fund all his programs, the economy could very well collapse. Those ppl making that much money are usually businness owners or other ppl that have a large influence in the economy. With more money taken away, businesses could lay off employees or close out.

Now, Kerry does contradict himself an awful lot. On the first debate, Kerry talked about mothers having to buy armor for soldiers as a Christmas present. Well, I thought he made a pretty good point, until I found out that he had voted against funding for things such as body armor. About North Korea, I really haven't found anything about Kerry's policy on North Korea(though it might be somewhere obvious or mbe came up in a debate after their voices put me to sleep).  When I said he contradicted himself there, I was referring to the fact that he supports having a strong coalition while he wants to end any help China is giving in the negotiations, though if you are right on his policy, then it seems there's no contradiction. However, when listening to Kerry speak, it seems that he never has ne firm position on bilateral talks and negotiations. That's what I don't like about Kerry, it's hard to find his real position on issues such as these.

Kerry does(several times) state that ppl making over 200,000USD are getting more tax cuts, but when he pledges not raise taxes for the middle class or give tax breaks to the middle class, does he say what middle class? When he complains about ppl making 200,000USD getting large tax cuts, he knows ppl will automatically asume that when he refers to middle class it will be ppl making less than 200,000USD.

Bush was entirely justified in talking about education when asked about jobs. The fact is, education can give more job opportunities.

What the economy issue all comes down to is the basic difference between Republicans and Democrats. Rep. would give tax breaks so businesses will thrive while Democrats will take the money and create plans for businesses to thrive.

What state do you(everyone, anyone) live in? Who is winning in your state?

[quote="filletwho"]Bush was entirely justified in talking about education when asked about jobs. The fact is, education can give more job opportunities.[/quote]

Okay, I should have been more explicit.  When asked about the fact that there have been 1.5 million private-sector jobs lost during his tenure (net loss is 500,000, more or less -- a whole bunch of government jobs have been created), Bush talked for maybe 5 or 10 seconds before moving the discussion to the NCLB.  Now, I know he didn't mean it this way, but imagine you're a 50-year old laid-off worker and you hear the president, in response to a question about your old job, talk about what he's doing for teenagers and how he's expanding community college options.  It was a completely out-there response because it dodged the entire issue of what to do about people who have lost jobs now.

[quote]Rep. would give tax breaks so businesses will thrive while Democrats will take the money and create plans for businesses to thrive. [/quote]
That's a silly thing to say.  If I give someone making 6,000,000 USD an extra 100,000 USD, it won't hardly do anything.  I could spend that same 100,000 USD dollars giving two people good-paying jobs.  The tax cuts have been multi-trillion dollar things, and yet our economy is struggling.  We would be far better off if that money had been invested, by the government, into something useful.  Kerry's health plan, for example, would be an extreme boon to American business since it would allow them to lower the amount that they have to pay for insurance.  (Additionally, it would be a boon to emergency rooms everywhere since more people would have insurance -- the uninsured are incredibly expensive to treat for illness.)  Your suggestion, that lowering business taxes leads to an immediately better economic situation, simply doesn't hold up to empirical examination.  It's a theory that doesn't work.

Okay, I now just don't understand your objection to his comments about the middle class.  Should he name every individual who will have lower taxes?  I just don't get your point.

[quote]On the first debate, Kerry talked about mothers having to buy armor for soldiers as a Christmas present. Well, I thought he made a pretty good point, until I found out that he had voted against funding for things such as body armor.[/quote]

Okay, so this is one of the most often-repeated and most absurd charges against Kerry.  There was no "Body Armor Bill."  There was a massive bill (refered to as an "omnibus bill,") which included funding for just about everything.  Initially, while this bill was being put together, Kerry voted for the inclusion of the body armor monies.  When the bill was completed, he (and other sensible senators) recognized that the bill contained, among its many provisions, a number that were quite outrageous.  Key among those that people have objected to, for example, were the provisions allowing for no-bid contracts.  Now, Republicans, as we all "know," are the party of free markets and competition, and yet their bill was specifically designed to avoid said competition in funding various efforts to be carried out by contractors.  So, as a result, a number of these individuals voted against the omnibus bill, which included the body armor provision.  It's important to recognize how the system works when talking about issues of what congresspeople have or have not voted for, or supported or passed.

[url=http://www.msnbc.msn.com/id/6146353/]Here[/url] is the relevant snippet from the first debate (I googled up a transcript of it):
[quote]LEHRER:  I want to make sureyes, sirbut in this one minute, I want to make sure that we understandthe people watching understand the differences between the two of you on this.
You want to continue the multinational talks, correct?

BUSH:  Right.

LEHRER:  And youre willing to do it...

KERRY:  Both.  I want bilateral talks which put all of the issues, from the armistice of 1952, the economic issues, the human rights issues, the artillery disposal issues, the DMZ issues and the nuclear issues on the table. [/quote]

So, that's what he said.  I haven't gone to read about what else he has to say about it.

Finally, to your first point:  people making more that 200,000USD were taxed significantly higher during the 90's, a period of significant economic expansion, than they are today, a period of economic stagnancy.  So, again, to link taxes (especially reasonably modest taxes, given what they pay in Europe or America 10 years ago) to economic collapse is silly.

I am from New York City.  I go to school in Cambridge, Massachussetts.  They're two of the most Democratic states, except with Republican governors :).

[quote]That's a silly thing to say.[/quote]

Ofcourse it's silly, I meant it as a joke, a jab at the politicians. But as to the theory, it's really no the place of one person to say it doesn't work. This theory cannot be proven or diproven over a short period of time because the actual process of the theory requires a long time. There are thousands of ppl who believe in this theory, are they not smart enough to figure out that the theory doesn't work? During the 90's, ppl were taxed significantly and the economy did thrive; but there is also the possibility that it was the result of the Reagon tax breaks kicking in.

I did not say there was a "body-armor bill". What I said was Kerry voted against funding for body armor. I made no mention of a special bill for body armor. Kerry might say that he was voting against the bill because of the no-bid contracts, but what his real motives were are unknown. Republicans do believe in free-market, but the belief should not overrule situation. If in that situation the contract made sense, it wouldn't do to not pass it just because it goes against a belief that does not consider situation.

I live in Michigan btw, a DEMOCRATIC state.

Yay! Michigan power!

filletwho, where in MI do you live? Are you taking the mmpc contests?

Yea, i took the MMPC's. They were harder than most other years. I think I got around 32, I was aiming for 35 or 36 but o well. I got the easiest question wrong.

[quote="filletwho"]I did not say there was a "body-armor bill". What I said was Kerry voted against funding for body armor.[/quote]

Which is a direct implication that there was an opportunity to vote, just on body armor, and he voted against it.  In fact, you go on to imply that he might have voted against the appropriations bill [i]because[/i] of the body armor provision.

(Yes, I know you didn't say that, either.  However, look at your own words!  You note that he voted against a bill including the money for body armor; state his explanation as to why he voted against it; and then try to cast doubt on that explanation, without giving one of your own.  A reasonable person reading what you wrote there would take it to imply that Kerry was purposefully trying to vote against the funding for body armor.  A reasonable person, reading your previous post, would take it to mean exactly what I said.  When you say "Kerry voted against X," that carries with it the implicit assumption that either Kerry was voting only on X, or that the other issues he was voting on were far less significant than X.  In this case, both of those are clearly not true -- and, notably, Kerry had voted to have the body armor in that bill!  You keep saying "I didn't really say such and such," but the fact is that you [i]do[/i] keep implying the things you deny.)

As for your other point, there are millions of people in this country who believe that, 6,000 years ago (more or less), an omnipotent power created the world whole and placed a single man and woman upon it, and we are all their ancestors.  And, after letting the world go more or less for a few millenia (with a flood every now and then), he randomly decided to impregnate some lady.  Now, I have nothing against Christianity or anything, but that is not a [i]scientifically sound[/i] theory -- it's perfectly religiously sound, and the existence of god and whatnot is reasonably self-consistent, but it's not scientifically valid.  So I don't think the fact that people believe in Reaganomics is good evidence of its validity, no.  Does that mean people who believe in it are stupid?  No, it just means that they are wrong.

I hardly think I'm [b]flipflopping[/b]. You didn't say that, you [i]implied[/i] that. No, I was trying to imply that I do realize there is no special bill for body armor.

[quote]I did not say there was a "body-armor bill". What I said was Kerry voted against funding for body armor.[/quote]

[quote]Kerry might say that he was voting against the bill because of the no-bid contracts, but what his real motives were are unknown.[/quote]

As you can see by my words, I have always implied that he was voting against body armor. When I say he voted against funding, i'm saying that those are likely motives for voting against the bill.

On the second point, comparing the belief of god to reagonomics is absolutely wrong. My interpretation of what you are saying is that just because everybody believes in it, it doesn't mean it's right, right? Well, you're example is totally off. Right now, we know that God is scientifically unreasonable because we know more now than back then. Well, right now, unless you know something others don't(mbe you're from the future? mbe you're smarter than every single person that believes in reagonomics?), it is wrong to say that reagonomics doesn't work. Sure, it [i]could[/i] be wrong, but for now, you [b]cannot[/b] say it doesn't work. It could be like me saying that light does not travel at constant speeds.

[quote="JBL"][People making over six figures] pay a higher percentage of their income [in taxes][/quote]

I saw an article in the Washington Post that said last year, Kerry paid roughly 10% of his 6,000,000 USD income in taxes, while Bush paid roughly 30% of his 600,000 USD income in taxes.  Now, who's being taxed more: the richer or the poorer people?

I don't happen to have any idea where you got those figures, since Kerry doesn't have a personal income anywhere near six million dollars.  Even if that had a factual basis, however, I don't have any idea what it has to do with what we're talking about, or even how one specific example is supposed to disprove something that is generally true, or why you seen to think that weakens my argument.  It is true, incidentally, that if you count total tax burden, in some instances the poor end up paying a larger percentage of their income in taxes compared to the wealthy, just not through income taxes (which was what we were talking about, I believe).

filletwho, I'm not sure why I'm having such a difficult time communicating with you.  To say that he voted against body armor is short-hand for saying he voted against funding the body armor -- what else it the world could it possibly mean?

Then, you say [quote]When I say he voted against funding, i'm saying that those are likely motives for voting against the bill. [/quote], which was what I said you were implying.  But you don't respond to my comment that that doesn't make any sense.  The reason it doesn't make any sense is that Kerry voted to have the funding for body armor included in the larger bill.  (I also said this in my last post.)  Furthermore, it wouldn't make any sense at all to vote against supplying body armor to the army, so I don't even have the faintest idea why you'd think an improbable circumstance makes a likely explanation for his vote.

As for Reaganomics, you again state that, if I deny something others believe, I must believe that I am smarter than them.  I don't see why that's the case at all.  As I said in my last post, I just think they're wrong; our respective intelligences have nothing to do with it.  Lots of people believe lots of wrong things.  I mean, 74% of Bush supporters believe Bush favors including labor and environmental standards in agreements on trade.  That doesn't necessarily mean that they're all stupid, just that they're all wrong -- the same applies in this case.
There is only tenuous evidence to support the effectiveness of income-tax cuts as an efficient economic stimulus and a reasonable amount of evidence to contradict it, so in the worst case a real scientist would say the question was undecided.  It's never wrong for a scientist to say something hasn't worked; it's only wrong for scientists to claim that something does always.  There's plenty of experimental evidence that light travels at a constant speed in a vacuum -- there ain't none that giving a millionaire a couple hundred thousand dollars gives me a job.

Well then... What the heck do you think CNN is?

CNN: biased, lying, liberal hacks

I guess this is a communication issue. I have a hard time posting out my full ideas online, I find it easier talking to a person. OK, I'm going to try to make myself as clear as possible. First I suggest that Kerry has no right to complain about the body armor situation since he voted against a bill that contained funding for body armor. Then, you say that there is no special bill for body armor, implying that i had no idea what the bill was in the first place. I try to explain that I know there was no such "special" bill, but he did vote against the body armor. You then say that that implies there is a special body armor bill. Well, from my wording, i can see why u might think that, but that is not what I was implying, so there. You had suggested that no-bid contracts is the reason Kerry voted against the bill. I try to say that you don't know that those were his reasons. He might have voted against the bill because he didn't like the body armor funding. Now you ask why he wouldn't want to give funding to body armor. Well, to save money. Kerry has been complaing about the national defecit and how so much money has been invested into the war has he not?
About Reagonomics, i'm saying that you must know something that others don't. Reagonomics is backed by logical reasoning. If you make earning money unprofitable for businesses, why should they continue to expand? You say that Reagonomics does not work, yet you provide no evidence or reasoning. That is why I ask if you know more than all those ppl who do believe in Reagonomics.
Finally, about light, are you saying there is absolutely no possibility that light has travels at inconsistant speeds? Sure there is circumstancial evidence. But there have been laws backed by circumstancial evidence that have been proven wrong due to new ideas, new evidence, more advanced technology, and circumstances that have never been considered before.
Now, how about I state this: Kerry will raise taxes for everyone. Now, I assume you believe that that is wrong. But I say you would be wrong. That is just as effective as your argument.

I apreciate your writing your arguments out in a "slower" way.  I also prefer face-to-face communication, although I do grant that this format has its advantages (like quotability and enforced taking-of-turns).

Now, onto the meat:

I generally agree with your assesment of our discussion up until this point:
[quote]He might have voted against the bill because he didn't like the body armor funding. Now you ask why he wouldn't want to give funding to body armor. Well, to save money. Kerry has been complaing about the national defecit and how so much money has been invested into the war has he not? [/quote]
Here, you leave out one exceptionally important thing that I pointed out:  [i]Kerry voted to include the funding for body armor in that larger bill[/i].  Within the scheme that you have set up, this vote makes no sense whatsoever.  It was clear the larger bill was going to pass eventually and Kerry knew that, so there is no possibility he was trying to prevent the body armor from being included with this vote.  (It's also worth noting that the actual cost of the body armor was relatively insignificant compared with the size of the overall appropriations bill it was part of, which was itself relatively insignificant with respect to the size of the defecit.)  Your suggested rationale simply can't adapt to this fact.

[quote]About Reagonomics, i'm saying that you must know something that others don't. Reagonomics is backed by logical reasoning. If you make earning money unprofitable for businesses, why should they continue to expand? You say that Reagonomics does not work, yet you provide no evidence or reasoning. That is why I ask if you know more than all those ppl who do believe in Reagonomics. [/quote]

Well, you make it sound like I'm the only person around who doesn't believe in tax cutting as an end to all of our ills!  Lots of people don't believe in Reaganomics, does that mean you're smarter than all of us or know something we all don't?  Reaganomics is [i]not[/i] backed by logical reasoning, because we aren't talking about corporate taxes or buisness taxes, we're talking primarily about personal income taxes!  I will grant you that an exceedingly high tax rate could potentially provide an economic disincentive to grow (the logical reasoning you're talking about), but there is absolutely no reason to think that, for example, the difference between a 34% and 38% upper bracket is anywhere near large enough to make a significant blip on the economic radar screen.
Your suggestion that the 90's economic boom could have been a result of the Reagan tax cuts doesn't stand up to your own logical test -- a tax cut kicks in immediately, so if it were going to spur growth, it should do so in the short term when people get an extra few hundred dollars or whatever on their refund at the end of the year -- why would the growth suddenly kick in ten years later after several tax hikes?  (Remember -- after Reagan cut taxes, both he and Bush I raised them during their terms.)

Finally, you are asserting a positive fact: "Reaganomics does work."  I am asserting the contrary: "There is no evidence to support Reaganomics."  [i]You[/i] are the one who, in order to have his argument taken seriously, needs to supply evidence to support it, not me.  This is the same as your final comment: you are asserting a positive claim while I am asserting a negative one.  You are in the scientifically difficult position of needing to show that something will or must happen.  The burden is on you to find evidence to support either claim.

[quote="JBL"]we are all their ancestors[/quote]

You mean descendents, right?

Kerry/Edwards 2004 So full of crap they need two Johns...  :D .  LoL, no offense intended.  Just something I heard today.

With your reasoning on the bill, it would have to be assumed that Kerry had the interests of the people in mind(extinct yet?). The funding was low. You're also asserting that Kerry had put the funding in in the first place. Who's fault is it that the funding was low?

Okay, Okay, about Reagonomics. No, you're not the only one who doesn't believe in Reagonomics. But what you did was make it sound like there was sound evidence that Reagonomics doesn't work. Since that can't be said, I was asking whether or not you knew something other ppl didn't. You need to provide proof for the fact that it is absolutely wrong, otherwise, you would just be providing propaganda that no one wants to listen to. Now, Reagonomics does have logic to it, even if you believe the logic is wrong, there is still logic and there are those who believe it is sound. I notice that you put lots of money figures in trying to prove Reagonomics wrong, well you need to look at the whole picture. Humans are, by nature, ambitious beings. You have this rich, succesful business owner. You give him some money. What do you expect him to do with the money? Put it under the matress? Most likely, he'll want more money. That's the logic free enterprise is based on. So, he invests the money in his business, thereby creating new jobs. If you're openminded enough, you'll see that this arguement makes sense. You're right, Bush sr. raised taxes in his terms after the Reagon tax breaks. That is part of the reason that the tax breaks took so long to kick in.

Finally, to you're last attack on my little example, I ask you not to look too deep into little details and look at the big picture. Would it be any different if you said Kerry will not raise taxes for everyone and I say that is wrong? You would have me provide proof right?

[quote="filletwho"]With your reasoning on the bill, it would have to be assumed that Kerry had the interests of the people in mind(extinct yet?). The funding was low. You're also asserting that Kerry had put the funding in in the first place. Who's fault is it that the funding was low?[/quote]

I'm afraid that this again goes too fast for me, and it seems that you are avoiding rather than dealing with my argument.  We had troops going to Iraq.  A bill was going to be passed to appropriate money for supplies, contracts, etc. related to the war.  This bill was guaranteed to pass.  Kerry voted to include the funding for body armor in this bill.  This shows that Kerry must have wanted there to be funds for the body armor.  You have said nothing to refute this in any way whatsoever.  Additionally, you suggested that Kerry would have voted against the bill because of funding issues.  In that case, Kerry's explanation for his actions (concerning no-bid contracts) makes far more sense than what you are saying, since they are very expensive while the body armor was relatively inexpensive.  I fail to see how your comments addressed either of these points, and I also don't have any idea where the phrase "low funding" came from -- this is your addition to this segment of the conversation and I have no idea where it came from and thus I do not feel compelled to defend against it, at least until you explain what it means.  (I'd prefer if you responded to my earlier points first, however.)

Referring to your example of the greedy businessman: one thing he might do is put that money in a retirement account, say -- nothing that is particularly useful.  If we were talking about the difference between 80% and 20% tax rates, maybe, I would grant that the later perhaps offered an opportunity for greater economic growth.  We aren't talking about that kind of difference: most people got less than $500 from the last Bush tax cut -- that isn't nearly enough money to make any kind of real difference.  But, when you take some money away from people who are already making far more money than they will spend, the government can do really big, cool useful things with it, like pave roads and pay for daycare or health insurance or education (all things which will have a long- or short-term benefit on the economy).  In doing all of those things, it will have to hire companies or people to do the actual work -- so the money ends up being injected into the economy anyhow.  Your argument rests on at least one of two things: (1) the actual effectiveness, from an economic point of view, of having people have more money (that is, the benefits of tax cuts) and (2) the uselessness of having the government have money (that is, the disadvantage of higher taxes).  You have yet to raise a single reasonable piece of evidence of either of those propositions, while I have raised a solid piece of evidence contradicting them: in the recent past, our largest period of economic growth has come with relatively high taxes, while currently we have an exceptionally slow economic recovery during a period of extremely low taxes.

[quote]Finally, to you're last attack on my little example, I ask you not to look too deep into little details and look at the big picture. Would it be any different if you said Kerry will not raise taxes for everyone and I say that is wrong? You would have me provide proof right?[/quote]

Yes, it would be different -- as the person asserting a fact, I would have an obligation to defend it.  Because it is a negative fact, you would probably have a much harder time countering my evidence, I think, but that's not really relevant.  You still haven't defended the claim, originally yours, that lowered taxes are beneficial.


I have made 622 mistakes. I suppose over integer ring is right.

Math kid, come on!

Could you possibly explain what that means?

One can not prove FLT to be right or wrong by using the knowledge no more advanced than some ring such as integer ring.

I am sorry.

I've got another proof of FLT:

It follows from the definitions. $ \Box$

I hope you have done a great job!

Have you submitted or posted it?

[quote="JRav"][quote]One can not prove FLT to be right or wrong by using the knowledge no more advanced than some ring such as integer ring. [/quote]

Can you prove that?[/quote]

I have got some reasons but not a completed proof.


Dear reader,

The solution to the second example of an AMC 12 problem given in the Math Jam "AoPS Classes Math Jam" on May 27 2008 pursued involves the determination of the integer closest to 8pi. However, 3.14 is adopted as an approximation to pi, thus 25.12 an approximation to 8pi, and it is claimed that the required integer is 25, without completing the proof. Would not any more detailed solution depend on calculus for approximating pi, with a suitable error determined to prove that |25 - 8pi| < 1/2?

Sincerely,
Daniel Tsai

In most cases in the AMC, you will usually need an approximation of an irrational number.

$ 3.14 < \pi < 3.15$
$ \implies 25.12 < 8\pi < 25.2$
$ \implies 0.12 < 8\pi \minus{} 25 < .2$
$ \implies |8\pi \minus{} 25| < .2 < .5$.

The bound in the top line is obvious...you can prove it with calculus error bounds, but I think it is acceptable rigor to just state that.


topology
real analysis
real analysis unsolved

If A and B are in R then

the closure of (A union B) = the closure of (A) union the closure of (B)

I've been thinking about it and I really have no idea where to even start. I understand it intuitively...I'm just not sure how to prove this rigorously.

I ll give you one part.

By definition $ \bar{A}$ is the smallest closed set containing $ A$.

So $ \bar{A}\cup \bar{B}$ is a closed set containing $ A$ and $ B$.

But $ \bar{A\cup B}$ is the smallest closed containing $ A$ and $ B$.

So $ \bar{A\cup B}\subset \bar{A}\cup \bar{B}$

*The closure of A is a closed set but not only that it is the smallest closed set containing A. The smallest means that if there is another closed closed set containing A, then the closure must be contained in that set. It is important you keep this in mind when solving problems involving this concept.

For the other direction you will need this fact: If $ A \subset B$, then $ cl(A) \subset cl(B)$ --- you can prove this fact if you apply the * above.

You know that $ A \subset A \cup B$ and $ B \subset A \cup B$. Now apply the above fact and this should give you the other direction.


Unclickable username?

Note that the username is in black, different from the other blue links.

(username = superquark1123)

Yeah it means that blog has guest posting enabled, and that's a guest that named himself that and posted in his blog.


calculus
integration
analytic geometry
articles
floor function
vector
linear algebra

I hope the problem statement is correct, I'm trying to remember it off the top of my head.

23 people meet to play football. They want to split into 2 teams of 11, picking one referree. They also want each team to have the same total weight. They notice that any of the 23 people can be chosen to be the referree. Prove that all 23 people weigh the same amount.

[hide]1.  First it is easy to show that all the weights are either even or odd.

2.  Next, notice that if we subtract the same number a or divide by the same number b, where b divides all the weights, then this still satisfies the problem given.

3.  Now let c be the smallest weight.  Then we will have one weight equal to 0 and others will be positive.

4.  Now, since 0 is even the other elements according to 1 are also even.  Now we can divide it by 2.

5.  Now, 0 will still remain in the set so we can continually divide by 2, and the elements will still have to remain even according to 1.

6. Now take the weight with the largest power of 2 as its factor.  Now if we divide by this power of 2, we will get numbers that are odd or rational numbers if the numbers are all different.  But they must be all even, according to (1) and (2).  So, the only way possible for an integer to stay even when dividing by 2 is when it is 0, namely all the elements are 0 so all the weights would have to be equal and we are done.

I think that is allowed, correct me if I am wrong.[/hide]

I have seen this problem before and it said clearly integral weights, so yeah, if it was irrational, I would be screwed.

Perhaps a more interesting version of this problem would be to prove it for all reals. Not sure if it is true though.

It is true for real weights.  The proofs I've seen all work by reduction to the integer case, but the method is beyond intermediate.  First, suppose you have a valid situation with real weights, not all equal.  You have to show that there exists an integer $n$ such that multiplying everyone's weights by $n$ gives you a new situation in which the sum of the fractional parts of all 23 weights is less than 1 and such that there exist two of the new weights which differ by at least 1.  If follows that you can round everyone's weights to the nearest integer and preserve all solutions.  Then one applies the proof for the integer case to derive a contradiction.

Separate challenge: find an integer $n$ and a set of $n$ positive integers such that, removing any one of the $n$ integers, the remaining $n-1$ integers can be divided into two sets ([i]not[/i] necessarily of equal size) such that the sum of the numbers in each set is equal.  Why does the proof given above fail in this case?

Too lazy to go into more detail but clearly mathgeniuse's #2 fails.

[quote="JBL"]First, suppose you have a valid situation with real weights, not all equal.  You have to show that there exists an integer $n$ such that multiplying everyone's weights by $n$ gives you a new situation in which the sum of the fractional parts of all 23 weights is less than 1 and such that there exist two of the new weights which differ by at least 1.  If follows that you can round everyone's weights to the nearest integer and preserve all solutions.  Then one applies the proof for the integer case to derive a contradiction.

[/quote]
Can you elaborate on "it follows that you can round everyone's weights to the nearest integer and preserve all solutions?

Thanks.

From http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=539533 :

[quote="mavropnevma"]When they are generically real, consider the subvector space of $\mathbb R$ over $\mathbb Q$ they generate; clearly this will be of some finite dimension $1 \le m \le 2n+1$. Then considering a basis, they can be identified with $m$-plets of rational coordinates, and the problem holds by each coordinate (because of the unique representation), so we reduced to the rational case.[/quote]

what does "they can be identified with $m$-plets of rational coordinates" mean?

Also, does someone know of a linear algebra solution involving adjacency matrices? I think I found that in an article with the solution some time ago but didn't read it since I was unfamiliar (and still am) adjacency matrices.

Wait, can the OP be proven with the [url=http://en.wikipedia.org/wiki/Ham_sandwich_theorem]Ham Sandwich Theorem[/url]  in any way?

Simple answer - no. Not all theorems that state that some "dividing" may be done apply to all problems where some "dividing" is needed. Let's be serious.

[quote="JBL"]Suppose that we have a nontrivial solution $(w_1, \ldots, w_{2n+1})$ in real numbers.  Then I claim (proof omitted) that there is some integer $m$ such that
[list]
[*]$\{mw_1\} +  \cdots + \{mw_{2n + 1}\} < 1$, and
[*]the set $\{\lfloor mw_1\rfloor, \ldots, \lfloor mw_{2n + 1}\rfloor\}$ contains at least two different values.[/list]  (Here $\{\cdot\}$ represents the fractional part; the second part of this claim is easy, it's the first part that's nontrivial.)

Since $(w_i)$ is a solution to the problem, we have a bunch of equations like $w_{i_1} + \ldots + w_{i_n} = w_{j_1} + \ldots + w_{j_n}$.  Thus we also have a bunch of equations like $mw_{i_1} + \ldots + mw_{i_n} = mw_{j_1} + \ldots + mw_{j_n}$, i.e., $(mw_i)$ is a solution as well.

Since $\sum_i \{mw_i\} < 1$, we have for any subset $\{k_1, \ldots, k_n\} \subset \{1, \ldots, 2n + 1\}$ that $\lfloor mw_{k_1} + \ldots + mw_{k_n}\rfloor = \lfloor mw_{k_1}\rfloor + \ldots + \lfloor mw_{k_n}\rfloor$.  Thus, from any equation of the form $mw_{i_1} + \ldots + mw_{i_n} = mw_{j_1} + \ldots + mw_{j_n}$ it follows that $\lfloor mw_{i_1} \rfloor+ \ldots +\lfloor mw_{i_n}\rfloor =\lfloor mw_{j_1}\rfloor + \ldots +\lfloor mw_{j_n}\rfloor$.

Finally, the previous sentence implies that $(\lfloor mw_i \rfloor)$ is also a solution to the original problem, but now with integer weights.  Finally, the second bullet point above assures us that it is a solution in which not all weights are equal.  This is a contradiction, so actually there could not have existed a nontrivial real solution in the first place.

Also, for your first question of mavropnevma, do you understand what a vector space is?[/quote]

Why is it (that there exists another solution in which not all weights are equal) a contradiction?

Yes, I know what a vector space is. Does mavropnevma's phrase mean we can assign different bases to the subspace? And what does "the problem holds by each coordinate mean?"

Sorry, I seem to have trouble understanding solutions unless everything is spelled out clearly. My understanding of math and understanding of English are completely separate, so I try to apply my understanding of English to guess at the logic/reasoning/theorem implied in every english phrase and try to consider the wording, but sometimes I just don't understand.

[quote="mavropnevma"][quote="hurdler"]Why is it (that there exists another solution in which not all weights are equal) a contradiction?
Does mavropnevma's phrase mean we can assign different bases to the subspace? And what does "the problem holds by each coordinate mean?"[/quote]
It is a contradiction, because for integers we can easily prove all weights must be equal.

In my solution, when we split the weights, and write the relations about the equal sums, those relations hold ad litteram on each coordinate. There are no different bases to the subspace, just one, be it $\{b_1,b_2,\ldots,b_m\}$. Then any of our weights has a unique writing as $w_i = \sum_{k=1}^m \lambda_{i,k} b_k$, with rational coefficients, and it's usual to represent $w_i$ as the $m$-plet $(\lambda_{i,1},\lambda_{i,2},\ldots,\lambda_{i,m})$. When you now eliminate one weight, and write the fact the rest can be split in two groups of $n$ with equal sum, that will hold for each coordinate, yielding $m$ relations, but where now rational values are involved; and we know the answer in the rational case.[/quote]

Oh right I forgot.

Ok I see now. Thanks!

That's very nice. I'm still learning about things like abstract vectors spaces.

[quote=JBL]
$\{mw_1\} +  \cdots + \{mw_{2n + 1}\} < 1$ [/quote]
How can we assure this

Anyone? I mean, it's quite straightforward by taking a really large $m$ , but i have trouble putting them into words.


Prove that for any acute triangle $ ABC$,we have the ineq:
$ \frac{cosAcosB}{cosC}\plus{}\frac{cosBcosC}{cosA}\plus{}\frac{cosCcosA}{cosB}\plus{}4cosAcosBcosC \geq 2$

[quote="SUPERMAN2"]Prove that for any acute triangle $ ABC$,we have the ineq:
$ \frac {cosAcosB}{cosC} \plus{} \frac {cosBcosC}{cosA} \plus{} \frac {cosCcosA}{cosB} \plus{} 4cosAcosBcosC \geq 2$[/quote]
Let $ a^2 \plus{} b^2 \minus{} c^2 \equal{} z,$ $ a ^ 2 \plus{} c^2 \minus{} b^2 \equal{} y$ and $ b^2 \plus{} c^2 \minus{} a^2 \equal{} x.$
Hence, your inequality is equivalent to
$ \sum_{cyc}(x^3y^3 \minus{} x^3y^2z \minus{} x^3z^2y \plus{} x^2y^2z^2)\geq0,$ which is Schur.


The largest term in the sequence $ x_n \equal{} 1000^n/n! ,$
$ n \equal{} 1, 2, 3, . . .$

(A) is $ x_{999}.$
(B) is $ x_{1001}.$
(C) is $ x_1.$
(D) does not exist.

[quote="keyree10"][hide]$ x_{999}$[/hide][/quote]

wow nice answer man.

[hide]
Observe that $ x_n\equal{}x_{n\minus{}1} \cdot \frac{1000}{n}$. The result follows directly.[/hide]


Let {i_1, i_2, ..., i_n}, n > 1, be a collection of real numbers such that i_1 + i_2 + ... + i_n = 0 and i_1^2 + i_2^2 + ... + i_n^2 = 1. Show that there are two numbers in the set such that their product is less than or equal to -1/n.

[color=cyan]Well, I can prove that at least one pair has product less that or equal to -1/(n(n-1)), but I suppose that isn't so close . . .[/color]

Show how you did that. There may be some ideas there that be used to prove the other inequality.

Well, I squared the sum and subtracted the sum of squares and divided by 2 to get

i_1*i_2 + i_1*i_3 + . . . + i_(n-1)*i_n = -1/2

Now, the Left Hand Side is greater or equal to
n(n-1)/2 * min(i_a*i_b) over the pairs of distinct (a,b) between 1 and n, of which there are n(n-1)/2.  So
n(n-1)/2 * min <= -1/2
min <= -1/(n(n-1)).  So at least 1 pair has a product at most -1/(n(n-1)).  But that isn't really close enough to what we need, I don't think.


Let a,b,c be real numbers with $a+b+c=2$.Prove that the following inequality occurs: \[a^{4}+b^{4}+c^{4}+\frac{23}{2}(ab+bc+ac)\geq\frac{53}{4}+9abc\]

"You fool me once, shame on me! You fool me twice, shame on the ghost!"

$\, \Longleftrightarrow \, \, \sum_\textrm{cyc }\left( a-1 \right)^{2}\left( a-\frac12 \right)^{2}\geq 0$.


Αν αφαιρεθούν τα αυγά από ένα καλάθι ανά $ 2,3,4,5$ και $ 6$ τότε παραμένουν αντίστοιχα $ 1,2,3,4,5$ αυγά. Όταν όμως αφαιρεθούν τα αυγά ανά $ 7$ στο τέλος το καλάθι είναι άδειο. Ποιός είναι ο ελάχιστος αριθμός αυγών που θα μπορούσε να περιέχει το καλάθι?

Ευκολάκι!
Σωστός ο basted.
[hide]Έστω $ x$ ο αριθμός των αυγών. Tότε $ x \plus{} 1 \equal{} \pi o \lambda 2,3,4,5,6$ και $ x \equal{} \pi o \lambda 7$
$ LCM(2,3,4,5,6) \equal{} 60$ όμως $ 7\nmid 59$ οπότε $ 120 \minus{} 1 \equal{} 119 \equal{} \pi o \lambda 7$
οπότε $ \boxed{x \equal{} 119}$
[/hide]

Ορφέας


calculus
derivative
algebra
polynomial
algebra unsolved

n is a degree of W(x) and n > 1 . Prove that:
$\frac{1}{w'(x_1)}+\frac{1}{w'(x_2)}+...+\frac{1}{w'(x_n)}=0$
where $x_1, x_2, ... , x_n$ are all real roots of W(x).

we may assume that $W$ is monic. then $W'(x_i)$ is a polynomial in $x_1,...,x_n$ of degree $n-1$ and only depends on them, not on the polynomial $W$.

we want to prove that $W'(x_1)\cdots W'(x_{n-1})+...+W'(x_2)\cdots W'(x_n)=0$. the LHS is a polynomial of degree $\leq (n-1)^2$. on the other hand, it is clear that it is symmetric in all variables(in fact, exchanging $x_i$ with $x_j$ does not change anything about $W$) and it gets zero if $x_i=x_j$ for some $i\neq j$. therefore all terms of the term $(x_i-x_j)^2$ divide our LHS; thus if it wasn't zero, it would have degree $\geq n(n-1)$, contradiction.

Peter

Using Lagrangdre:
We need to proof:
\[ \sum \frac{1}{x_i-x_1)(x_i-x_2)\cdots (x_i-x_{i-1})(x_i-x_{i+1})\cdots (x_i-x_n)}=0\]
using with $f(x)=1$ we have:
\[f(x)=\sum f(x_i)\frac{(x-x_1)(x-x_2)\cdots (x-x_{i-1})(x-x_{i+1})\cdots (x-x_n)}{x_i-x_1)(x_i-x_2)\cdots (x_i-x_{i-1})(x_i-x_{i+1})\cdots (x_i-x_n)}\]
And we have result (the coffience degree n-1 of f(x) equal 0)


calculus
integration
analytic geometry
function
trigonometry
vector
calculus computations

I have no idea how to integrate this:
$\iiint_{\mathbb{R}^3\setminus \{(0,0,0)\}} \frac{e^{-i(k_1x+k_2y+k_3z)}}{(x^2+y^2+z^2)^a}dx\,dy\, dz$ as $a,k_1,k_2,k_3\in\mathbb{R}$.

Could that be integrated in cylinder coordinates? I believe I have made some mistake in my calculation and I would like to see any other answers. :?

Note that this is a Fourier transform.

I learned a method for this (and put it into the notes for the undergraduate Fourier analysis course I teach.) It would be a long argument. The outline is something like this:

To start with, note that this integral cannot converge absolutely no matter what $a$ is. There's some hope that it means something classically when $0<a<\frac n2.$ It's probably best to think of its convergence in some very weak sense, such as in distributions.

We write the radial function $f(x,y,z)=\frac1{(x^2+y^2+z^2)^a}$ as a "linear combination" of Guassians:

$f(x,y,z)=\int_0^{\infty}g(s)e^{-s(x^2+y^2+z^2)}\,ds,$ determining just what $g(s)$ we need to do this.

Exploit the fact that we know the Fourier transform of $e^{-s(x^2+y^2+z^2)}$ (another Gaussian.) Take the Fourier transform of our integral, and interchange the act of taking the transform with the integral. Then do that integral.

When you are done you should have the function $\frac{C}{(k_1^2+k_2^2+k_3^2)^{\frac n2-a}}$ for some constant $C$ which I don't immediately remember.

Use spherical coodinate, when $\frac{1}{2}<a<\frac{3}{2}$
$\iiint_{\mathbb{R}^3\setminus \{(0,0,0)\}} \frac{e^{-i(k_1x+k_2y+k_3z)}}{(x^2+y^2+z^2)^a}dx\,dy\,dz$
$=\int_{0}^{2\pi}d\varphi\int_{0}^{\infty}r^2 dr\int_{0}^{\pi}\sin\theta d\theta r^{-2a}e^{-ikr\cos\theta}$
$=2\pi \int_{0}^{\infty}\frac{2\sin kr}{kr}r^{2-2a}dr$
$=\frac{4\pi}{k^{3-2a}}\int_{0}^{\infty}r^{1-2a}\sin r\,dr$
$=\frac{4\pi}{k^{3-2a}}\Gamma\left(1-a\right)\sin\frac{\pi\left(1-a\right)}{2}$
$=\frac{2\pi^2}{k^{3-2a}\Gamma\left(a\right)\sin \frac{\pi a}{2}}$.

Clearly, Buffalo is using $k$ for the norm of the vector: $k=\sqrt{k_1^2+k_2^2+k_3^2}.$ Also, where I use $n$ in my post, I probably should particularlize that to $n=3.$

Buffalo's result contains mine; we have the same power of $k$ and he worked out the specific value of the constant.


AMC
AIME
probability
geometry
geometric transformation
rotation

Three numbers, a1, a2, a3, are drawn randomly and without replacement from the set {1, 2, 3, ....., 1000}.  Three other numbers, b1, b2, b3, are then drawn randomly and without replacement from the remaining set of 997 numbers.  Let p be the probabilitiy that, after a suitable rotation, a brick of dimensions a1 X a2 X a3 can be enclosed in a box of dimensions b1 X b2 X b3, with the sides of the brick parallel to the sides of the box.  If p is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

[hide]
Given any sequence $c_1 \le c_2 \le c_3 \le c_4 \le c_5 \le c_6$, we can find the number of ways to arrange it into $a$'s and $b$'s such that the brick fits inside the box. That means every dimension of $a$ has a corresponding dimension of $b$ that is larger than it.

Since $c_1$ is the smallest dimension, it must go in $a$ because if it was in $b$ there would be no dimension smaller than it in $a$. Similarly, $c_6$ is in $b$. Now we must find the number of arrangements of $c_2$ through $c_5$ satisfying the conditions.

Listing them, we find:
$c_2c_3$ in $a$, $c_4c_5$ in $b$
$c_2c_4$ in $a$, $c_3c_5$ in $b$
$c_2c_5$ in $a$, $c_3c_4$ in $b$
$c_3c_4$ in $a$, $c_2c_5$ in $b$
$c_3c_5$ in $a$, $c_2c_4$ in $b$

So there are five arrangements in which the brick will fit inside the box. There are $3!$ ways to pick the first three (the $a$'s) and $3!$ ways to pick the second three (the $b$'s). So the probability is $\frac{5(3!)(3!)}{6!} = \frac{1}{4}$.[/hide]


limit
induction
real analysis
real analysis unsolved

What is the value of  $ 3\minus{}\frac{2}{3\minus{}\frac{2}{3\minus{}\frac{2}{3\minus{}\dots}}}\equal{}?$   I solve like that; $ 3\minus{}\frac{2}{a}\equal{}a$ and I found $ a\equal{}1$ and $ a\equal{}2$ .  But what is the correct answer? $ 1$ or $ 2$ ?

The given value is not actually well-defined.  We can define this number as the limit of the sequence

$ a_{n\plus{}1} \equal{} 3 \minus{} \frac{2}{a_n}$

but the value of the limit depends (as you have seen) on the initial value.  It makes sense to start with $ a_1 \equal{} 3$, in which case we will get $ \lim_{n \to \infty} a_n \equal{} 2$, but if we start with different values of $ a_1$ we may get the other limit instead.

If we do take that initial value, for example: let $ a_1 \equal{} 3$, $ a_{n \plus{} 1} \equal{} 3 \minus{} {2\over{a_n}}$.  Show, by induction, that $ 2 < a_{n \plus{} 1} < a_n$.


function
real analysis
real analysis unsolved

Is there a continuous real-valued function $ f(x)$ on the real line such that for any real number $ y$ there are precisely two real numbers $ x_{1}$ and $ x_{2}$ such that $ f(x_{1})\equal{}f(x_{2})\equal{}y$?

I'm thinking the answer is no, because a function have a minimum or a maximum would be violating this property. So the function doesn't have a minimum or a maximum, and hence would be constant, but again a constant function violates the given properties of $ f$. So...how do I prove this problem rigorously?

What's the minimum/maximum of $ e^{x}$ on $ \mathbb{R}$?

[quote="lzk320"]Oh wait, yeah $ e^{x}$ has no max/min on $ \mathbb{R}$. But then how would I go about doing this problem?[/quote]

Suppose there is such a function. Choose $ a < b$ with $ f(a) \equal{} f(b)$. In $ [a,b]$, $ f$ has a maximum or minimum that is not equal to the value at the end points. That max or min is taken on at some other point, leading to trouble. Draw a picture.


Jose has ten numbers. The average of these is 20. We remove one of the numbers, chaning the average to 19. What number was removed?

A) 20
B) 21
C) 39
D) 40
E) None of above

[hide]The sum of the original 10 numbers is 200.  The sum of the 9 numbers is 171.  That means that 29 was removed, or (e).[/hide]

That is the correct answer.

But try to put them in spoiler.  ;)


i got

[hide="probably wrong"]

180 soldiers, because the number of soldiers is cut in half because the weeks double, so its less than 300, then, they're getting 3/5 of the amount they would get so 3/5 of 300 is 180.....my logic is off it seems, it doesnt make much sense......
[/hide]

[hide="Answer"]$5\cdot (5\cdot 7)\cdot 600=105,000$ ounces of protein.
$3\cdot (10\cdot 7)\cdot x=105,000 \Rightarrow x=\boxed{500}$[/hide]

[quote="anirudh"]Allowing each soldier five ounces of protein each day, an army of 600 soldiers has enough provisions for five weeks. With the same total amount of provisions, how many soldiers could be fed for 10 weeks if each soldier receives only three ounces of protein per day?[/quote]

[hide="Solution"]
3000 ounces of protein can be used per day for five weeks.

1500 ounces of protein could be used per day for ten weeks.

Three ounces of protein per person per day, and 1500 ounces of protein per day translates to 500 soldiers (1500/3).
[/hide]

[quote="anirudh"]Allowing each soldier five ounces of protein each day, an army of 600 soldiers has enough provisions for five weeks. With the same total amount of provisions, how many soldiers could be fed for 10 weeks if each soldier receives only three ounces of protein per day?[/quote]

[hide="myanswer"]25*600=15000/10=1500/3=500[/hide]

[quote="236factorial"][quote="ajai"] because the number of soldiers is cut in half [/quote]

where did it say that?

[hide="answer"]600*5*5=15000
x*10*3=15000
x=500[/hide][/quote]

i assumed that because there were twice the amount of weeks, the people feeding the soldiers would have half the supplies, therefore cutting the number of soldiers in half, because they sent them home.....it doesnt make much sense :blush: I understand the real way......thanks  :D

That was actually correct, but you went the wrong way with the food change. They cut the amount of food to $\frac{3}{5}$, so they should have multiplied the number of soldiers by $\frac{5}{3}$, giving $500$, the correct answer.

[quote="anirudh"]Allowing each soldier five ounces of protein each day, an army of 600 soldiers has enough provisions for five weeks. With the same total amount of provisions, how many soldiers could be fed for 10 weeks if each soldier receives only three ounces of protein per day?[/quote]

OMy Solution:

600*5=3000. 3000/3=1000. 1000 for 5 weeks, so 500 for 10 weeks.

[hide="answer"]

600 * 5 = 3000 ounces.
Which means that there are 3000 ounces for 5 weeks. Which means that there can be 1500 ounces for 10 weeks. 1500/3(each soldier gets 3 ounces) = $\framebox{500}$
[/hide]


Hi there!

"If $ N$ is an odd integer, then $ N$ and $ 2N+1$  are relatively prime."   Proof or counter-example!

Suppose $ p$ is any prime dividing $ N>1$, then can $ p$ divide $ 2N+1$?
No!  This is just like the proof that there are an infinite amount of prime numbers.  So, they have no common prime divisor, in other words, they are relatively prime.  If $ N=1$, then $ \gcd(N,2N+1)=\gcd(1,3)=1$, so the statement is true here too.

Hello, baz.

[quote]
This is just like the proof that there are an infinite amount of prime numbers
[/quote]

Do you mean the "traditional one":

Assume there are a finite number, M, of primes, the largest being p_M. Consider the number that is the product of these, plus one: p_1...p_M +1. By construction, this number is not divisible by any of the p_i. Hence it is either prime itself, or divisible by another prime greater than p_M, contradicting the assumption.

If so, any other reader could be most kind and explain to us why baz is correct?    :P
What about if $ N$ is even?  Does it make any difference?

Answer

Let the prime factorisation of $ N$ be:

$ N = p_{1}^{k_{1}}p_{2}^{k_{2}}\ldots p_{m}^{k_{m}}$

$ \Rightarrow 2N+1 = 2p_{1}^{k_{1}}p_{2}^{k_{2}}\ldots p_{m}^{k_{m}}+1$

We can see that $ 2N+1$ leaves remainder 1 on division by each of $ p_{1},\,p_{2}\ldots ,p_{m}$

Hence $ N$ and $ 2N+1$ share no common prime factors making them relatively prime.

[hide="Quick"] $ gcd(N, 2N+1) = gcd(N, 1) = 1$ by Euclidean Algorithm. [/hide]
[hide="In other words..."] If $ d | N, d | 2N+1$, then $ d | (2N+1)-2(N) = 1$. [/hide]
[hide="Generalization"] $ gcd(a, b) = 1 \Leftrightarrow \exists x, y \in \mathbb{Z}: ax+by = 1$ [/hide]

[quote="fskilnik"]Precisely, sludgethrower. Thanks for joining.

Being more explicit: if $ p>1$ is any prime number that divides $ N$, $ p$ divides $ 2N$ therefore $ p$ does NOT divide $ 2N+1$. In other words, $ N$ and $ 2N+1$ do not have any positive divisor in common except for $ 1$ (because if they did they would have a prime factor in common and they don´t!) and that garantees they are relatively prime.

As far as my last question is concerned, the same argument follows for every integer $ N>0$ ; even or odd is simply irrelevant![/quote]
Precisely and well put!  Sorry that I did not get back to you earlier, but we are fortunate for all those extraordinary members on this site  :) .


This PDF Contains Inequalities Questions Discussed at the Delhi INMO TC

Thanks Aman. The problems are very easy too.
My camp is only on the 3 days before the exam . So I wont be able to help u now. And is ur camp over? How was it?

Well It is not over yet !! There would be 2-3 More Sessions , Of Geometry , Functional Equations , Pell's Equation and Invariance Principle

This Camp is Really Helpful , I Learnt a Lot new Tricks and Techniques useful in Problem Solving!!

Which Region are you from , Aravind??

Well I am from Tamil Nadu.
Its good that you people have 6 days camp. Enjoy. Invariance principle is really useful. Also Functional Equations might be useful but rarely one appears in INMO.
Anyway has any 9th std from Delhi cleared the RMO? {I think you know the full list}.

Yes I do have the Full List !

The Top 30 Students all All From Class 11th and 12th , except one who is from class 10th.

However there are 2 students from class 9th who are not in the top 30 but have been invited to attend the camp so that they can perform better the next year.


geometry
circumcircle
inequalities
geometry unsolved

The circle O lies entirely inside the equilateral triangle ABC. If AM, BN and CP are tangents to O, (M, N and P lying on O), then one can construct a triangle having as sides the lengths AM, BN and CP.

Best regards,
sunken rock

The idea is to use [url=http://mathworld.wolfram.com/PursersTheorem.html]Purser's theorem[/url], which is none other than a degenerate case of [url=http://mathworld.wolfram.com/CaseysTheorem.html]Casey's theorem[/url], coming from Ptolemy's theorem. For instance, if circle $ (O)$ touches the circumcircle $\mathcal{K}$ of $\triangle ABC$ at the small arc $BC,$ we have $ AM \equal{} BN \plus{} CP.$ But if $(O)$ does not touch $\mathcal{K},$ then Ptolemy's inequality holds and Casey/Purser relation become inequality in exactly the same way. Therefore, $ BN \plus{} CP\ge AM$ $ \Longrightarrow$ $ AM,BN,CP$ are sides of a nondegenerate triangle.


logarithms
function
limit
algebra unsolved
algebra

Find all real numbers $ x$ for which
\[ (x\plus{}1)^x\plus{}(x\plus{}2)^x\equal{}(x\plus{}3)^x\]

hello, the only real solution is $ x\equal{}2$.
Sonnhard.

[b]abbreviated　proof of only solution[/b]

Let, $ f(x) \equal{} (x\plus{}1)^x \plus{} (x\plus{}2)^x \minus{} (x\plus{}3)^x$.
So, $ f'(x) \equal{} (x\plus{}1)^x \left(\frac{x}{x\plus{}1}\plus{}\ln (x\plus{}1)\right) \plus{} (x\plus{}2)^x (x\plus{}2)^x\left(\frac{x}{x\plus{}2}\plus{}\ln (x\plus{}2)\right) \minus{} (x\plus{}3)^x (x\plus{}3)^x\left(\frac{x}{x\plus{}3}\plus{}\ln (x\plus{}3)\right)$.

And, let $ g_a(x) \equal{} \frac{x}{x\plus{}a}\plus{}\ln (x\plus{}a) \ \ \ x\in(\minus{}a, \infty)$.
$ \frac{x}{x\plus{}a}$, $ \ln (x\plus{}a)$ is concave function, so $ g_a(x)$ is concave function too.
$ \lim_{x \to \minus{}a} g_a(x) \equal{} \minus{} \infty$ and $ g_a(0) \equal{} \ln a \ge 0$. So, $ \exists! c_a$ s.t. $ \minus{}a < c_a \le 0$.

So, $ \lim_{x \to \minus{}1} f'(x) \equal{} \minus{}\infty$, $ \lim_{x \to \infty} f'(x) \equal{} \minus{}\infty$. and $ f'(1)>0$.
(and $ f'(0)\equal{}\ln 2 \minus{}\ln 3<0$)
So, $ f'(x)\equal{}0 \iff 0<\exists p<1<\exists q \ \text{s.t.} \ f(p)\equal{}0, f(q)\equal{}0$.

Therefore, $ \lim_{x \to \minus{}1} f(x) \equal{} \infty$, $ \lim_{x \to \infty} f(x) \equal{} \minus{}\infty$ and $ f(p)>0$, $ f(q)>0$.
So $ f(x)\equal{}0 \iff \exists! C$ s.t. $ 1 < q < C \le \infty$.

[b]yuriy[/b], sort of pythagorean triangle, if x = 2, then 3^2 + 4^2 = 5^2.


a) Find x if the first term of an arithmetic progression is x - 1, the third term is x + 1, and the fifth term is 2x+3.

b) Find x if the first term is x - 1, the second term is 2x - 7, and the third term is 4x + 9.

c) Find x if the first term is x - 1, the second term is 2x - 7, and the fourth term is 4x + 9.

Hello, rrusczyk!

I used a basic fact of A.P. to solve these.

The common difference  is the difference of two consecutive terms: $ d \;= \;a_{n+1} - a_n $

[quote]a) Find x if the first term of an arithmetic progression is x - 1,
 the third term is x + 1, and the fifth term is 2x+3.[/quote]
We have: $a_1 = x-1,\;a_3 = x+1,\;a_5 = 2x+3$

Since $a_3 - a_1 = 2d$, we have: $(x + 1)-(x-1) = 2d\;\;\Rightarrow\;\;d = 1$

Since $a_5-a_3 = 2d$, we have: $(2x+3) -(x+1) = 2\;\;\Rightarrow\;\;x = 0$

[quote]b) Find x if the first term is x - 1, the second term is 2x - 7,
 and the third term is 4x + 9.[/quote]
We have: $a_1 = x-1,\;a_2=2x-7,\;a_3=4x+9$

Since $a_2 - a_1 = d$, we have: $(2x-7) - (x-1) = d\;\;\Rightarrow\;\;x - 6 = d$

Since $a_3-a_2 = d$, we have: $(4x+9)-(2x-7) = d\;\;\Rightarrow\;\;2x + 16 = d$

Hence: $2x + 16 = x - 6\;\;\Rightarrow\;\;x = -22$

[quote]c) Find x if the first term is x - 1, the second term is 2x - 7,
and the fourth term is 4x + 9.[/quote]
Unless there is a typo in the statement of this problem, there is no solution.

We have: $a_1 = x-1,\;a_2 = 2x-7,\;a_4 = 4x + 9$

Since $a_2-a_1 = d$, we have: $(2x-7)-(x-1) = d\;\;\Rightarrow\;\;x-6 = d$

Since $a_4-a_2 = 2d$, we have: $(4x+9)-(2x-7) = 2d\;\;\Rightarrow\;\;2x + 16 = 2d\;\;\Rightarrow\;\;x + 8 = d$

Hence: $x-6 = x+8\;\;\Rightarrow\;\;-6 = 8\;\;??$


Im trying to figure out how to do this question. This is an example in the book i have. Im not sure how they got the answer.

Here is the example from the book:

Find the Transition Matrix P from the basis B={t+1, 2t, t-1} to B'={4(t^2)-6t, 2(t^2)-2, 4t} for the space R$ _{3}$[t].
A little computataion shows that 4(t^2)-6t:(-3, 2, -3)$ _{B}$,   2(t^2)-2: (-1, 1, 1)$ _{B}$ and   4t:(2, 0, 2)$ _{B}$. Therefore

$ P \equal{} \left(\begin{array}{ccc}\minus{}3 & \minus{}1 & 2\\2 & 1 & 0\\\minus{}3 & 1 & 2\end{array}\right)$

I'd like to know how the numbers for the transition matrix were found?! Any help  is appreciated.

As currently written, $ B$ is a linearly dependent set, which doesn't span the space of degree $ \le 2$ polynomials. The question cannot be answered.

I dont understand what you are saying.. :(
I just dont see how they got 4(t^2)-6t to be(-3, 2, -3), 2(t^2)-2 to be (-1, 1, 1) and 4t to be(2, 0, 2)...?


I have 10 distinguishable socks in my drawer: 4 white, 4 brown, and 2 blue.  In how many ways can I choose a pair of socks, provided that I get two socks of the same color?

[hide="Solution"]2c2=2 ways to choose blue.

4c2=6 ways to choose white

4c2=6 ways to choose brown. Thus 6+6+2=14 ways.[/hide]

[hide="Actually"]$\binom{2}{2}=1$ so the answer is $6+6+1=13$ ways.[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Hide your solutions.[/url][/color]


function
integration
limit
real analysis
real analysis unsolved

1. Suppose that $f$ is in the uniform closure of the space of step functions on $\left[a,b\right]$.  Prove that if $f\left(x\right)\geqslant 0$ for all $x\in\left[a,b\right]$ then

\[\int_{a}^{b}f\left(x\right)\,dx\geqslant 0.\]
2. Prove that if $a<c<b$ then

\[\int_{a}^{b}f\left(x\right)\,dx=\int_{a}^{c}f\left(x\right)\,dx+\int_{c}^{b}f\left(x\right)\,dx\]
for $f$ in the uniform closure of the space of step functions on $\left[a,b\right]$.

Well-known theorem, in a fairly complete version: If $f_{n}$ is a sequence of Riemann-integrable functions on $[a,b]$ and $f_{n}\to f$ uniformly, then $f$ is Riemann-integrable and

$\lim\int_{a}^{b}f_{n}=\int_{a}^{b}f.$

This implies that all of the $f$ in this space are Riemann integrable. Since #1 and #2 are both standard theorems true for any Riemann integrable function, they are true in this case.


LaTeX
MATHCOUNTS
calculus
quadratics
algebra
quadratic formula

Ok there are 5 questions here and i would appreciate it if you guys could post your answer and preferrably your solutions to these problems as i have no answer key for it.

3. Take 111th root of all of them to get $ 1^8, 2^7, 3^6,4^5, 5^4 ,6^3, 7^2 ,8^1$ I think you can arrange them now.
4. I think I know, but I am not sure. First try to get as much 2 as you can, so 10 2's. Which is 2^10 and then the last one is 3. [hide]So $ 3072$[/hide]

Yeah, I stink at Latex

thanks shentang, for number two did anyone happen to get [(root2)-1]/2?

oh dude, I was experimenting with an idea pretty much like problem 2

number 1
[hide]
mary is 24 years old.  she is twice as old as ann was when mary was as old as ann is now. how old is ann?

so, 24 is twice as old as ann was...
and 24=2*12
then, 12 is as old as ann was when mary was as old as ann is now

lets say ann is age A right now.
24-A=X (so X is the difference between their ages, no matter what their ages are)

going back to the first part:

12 is ann's age when mary was age A.

A-12=X
24-A=X

it's pretty easy from there

2A=36
A=18[/hide]

For 4, you can do better.

I don't like these problems on tests like MathCounts, because to do them rigorously requires some bigger stuff/thinking.

for #2, I got $ 3\minus{}2\sqrt{2}$.  Be careful, the circle's diameter is not the distance from the point on the circle to the corner.

number 4

i'm pretty sure shentang is wrong.  i've done stuff like this before, and you want to get everything as close to 3 as possible.[hide] $ 3^7*2\equal{}4374$

the only other solution worth testing is $ 3^5*4^2$, which is only 3888. [/hide]

i'm trying to remember why, give me a minute

[hide="4 again"] if you take a smaller number, like 12, you can see that 3 works the best

$ 1^{12}\equal{}1$
$ 2^6\equal{}64$
$ 3^4\equal{}81$
$ 4^3\equal{}64$
$ 5*5*2\equal{}50$

Also, all twos will always be the same as all 4's, because $ 2*2\equal{}2\plus{}2\equal{}4$  It makes sense then that the peak is inbetween them.
[/hide]

"for #2, I got . Be careful, the circle's diameter is not the distance from the point on the circle to the corner."

Um Alan, I understand what i did wrong because now that i look at it carefully you could keep inscribing circles to an infinite point, but how did you derive your answer?
If it helps in the explaining i drew an artistic little diagram:

[hide] call the radius of a little circle r.  draw 2 radii so that the form a right angle with the sides of the square.  the radii, with the sides of the big square form a little square with diagonal r*root 2.  do the same to the circle on the other end of the diagonal of the larger square. [/hide]

if you need more than that ask

How cool is it that all but one response on this thread comes from the same school?

Well, it's cool for me to see anyway!!

[quote="allyrb"]and you want to get everything as close to 3 as possible[/quote]

Good. Now, think about why. It helps to consider what happens if we eliminate the restriction that everything has to be an integer (but still positive). Then, use AM-GM (there may be a little bit of calculus involved, I'm not completely sure and I don't feel like writing anything down right now).

Oh yeah, opps sorry. :blush: I forgot, I knew it was either 2 or 3.

can someone check my #2 solution, because i get something different from everyone else

[hide="my answer"]
(2-sqrt2)/2

sorry i haven't downloaded latex yet
[/hide]

[quote="allyrb"]can someone check my #2 solution, because i get something different from everyone else

[hide="my answer"]
$ \frac{2\minus{}\sqrt{2}}{2}$
[/hide]
sorry i haven't downloaded latex yet
[/quote]
Allyrb, you don't need to download latex, just hit the quote button and look at the text I typed bewteen the dollar signs to make your answer be in latex.

That's how I learned latex, I just hit quote on peoples messages and then looked at how they typed it :lol:

[quote="Yongyi781"][quote="allyrb"]
sorry i haven't downloaded latex yet
[/quote]

You don't need to download $ \mathrm{L\!\!^{{}_{\scriptstyle A}} \!\!\!\!\!\;\; T\!_{\displaystyle E} \! X}$ to use it. Just put the math code in between dollar signs. The server handles all the conversion.[/quote]

thanks everyone, very helpful to know

After spending some time this is what i came up with,
Problem 2: the length of the radius of one of the smaller circles and and the leftover length of the diagonal which is not part of the small circle is (R root 2), then the length of the diameter of the small circle is
 [(R root 2) + R]. We also know that the length of the whole diagonal of the square is rt(2), and we can also say that [2R(root 2) + 2R +1= sqrt(2)]. Knowing that the diameter is equal to 2R, we can simplify the equation to [D(root 2) + D +1=sqrt(2)]. Now group the equation such that when it is squared, we can eliminate the square roots. such a solution is this, D + 1= sqrt(2)-D(root 2). squaring both sides, we get D^2+2D+1= 2-4D+2D^2. Simplifying this we have, D^2-6D+1=0 using the quadratic formula, we get two answers and since the diameter must be smaller then the length of the square, the only answer that makes sense is [b]3-2root2.[/b]


A rectangle with perimeter $ 176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles?
[asy]defaultpen(linewidth(.8pt));
draw(origin--(0,3)--(4,3)--(4,0)--cycle);
draw((0,1)--(4,1));
draw((2,0)--midpoint((0,1)--(4,1)));
real r = 4/3;
draw((r,3)--foot((r,3),(0,1),(4,1)));
draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]$ \textbf{(A)}\ 35.2\qquad \textbf{(B)}\ 76\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 84\qquad \textbf{(E)}\ 86$

[hide="Click for solution"]
Let $ x$ and $ y$ denote the width and height of one of the five rectangles, with $ x<y$.  Then $ 5x\plus{}4y\equal{}176$ and $ 3x\equal{}2y$.  Solving yields $ x\equal{}16$ and $ Y\equal{}24$>  Thus, the perimeter is $ 2(16\plus{}24)\equal{}80$, or $ \boxed{\textbf{(C)}}$.
[/hide]

Another method to solve this problem (not the best to use on a test but still interesting) is to use the fact that each of the 5 rectangles has equal area. Label the sides x and y of each rectangle such that 3x is at the top and 2y is at the bottom. On the sides you should have x+y. (x+y) * 2y = 5xy. Solve for y in terms of x or x in terms of y. Then plug that in to 5x +4y = 176 (the perimeter equation essentially given from the question). Then solve and you get the same answer.


Let $ a;b;c$ be real numbers satisfying $ a\plus{}b\plus{}ab\equal{}3$
Prove that:

$ \frac{3a}{b\plus{}1}\plus{}\frac{3b}{a\plus{}1}\plus{}\frac{ab}{a\plus{}b} \leq a^2\plus{}b^2\plus{}\frac{3}{2}$

[quote="chien than"]Let $ a;b;c$ be real numbers satisfying $ a \plus{} b \plus{} ab \equal{} 3$
Prove that:

$ \frac {3a}{b \plus{} 1} \plus{} \frac {3b}{a \plus{} 1} \plus{} \frac {ab}{a \plus{} b} \leq a^2 \plus{} b^2 \plus{} \frac {3}{2}$[/quote]
$ a \plus{} b \plus{} ab \equal{} 3 \Rightarrow (a\plus{}1)(b\plus{}1)\equal{}4$.
And: $ 3\equal{}a\plus{}b\plus{}ab\le a\plus{}b\plus{}\frac{(a\plus{}b)^2}{4} \Rightarrow (a\plus{}b\minus{}2)(a\plus{}b\plus{}6) \ge 0 \Rightarrow a\plus{}b \ge 2$.
We have:
$ \frac {3a}{b \plus{} 1} \plus{} \frac {3b}{a \plus{} 1} \plus{} \frac {ab}{a \plus{} b} \leq a^2 \plus{} b^2 \plus{} \frac {3}{2}$
$ \Leftrightarrow \frac{3a(a\plus{}1)\plus{}3b(b\plus{}1)}{(a\plus{}1)(b\plus{}1)}\plus{}\frac{3\minus{}(a\plus{}b)}{a\plus{}b} \le a^2\plus{}b^2\plus{}\frac{3}{2}$
$ \Leftrightarrow \frac{a^2\plus{}b^2}{4}\plus{}\frac{5}{2} \ge \frac{3(a\plus{}b)}{4}\plus{}\frac{3}{a\plus{}b}$
By Cauchy-Schwarz inequality, we have:
$ \frac{a^2\plus{}b^2}{4} \ge \frac{(a\plus{}b)^2}{8}$
So we need to prove that:
$ \frac{(a\plus{}b)^2}{8}\plus{}\frac{5}{2} \ge \frac{3(a\plus{}b)}{4}\plus{}\frac{3}{a\plus{}b}$
Set $ a\plus{}b\equal{}x \Rightarrow x \ge 2$
The inequality becomes:
$ \frac{x^2}{8}\plus{}\frac{5}{2} \ge \frac{3x}{4}\plus{}\frac{3}{x}$
$ \Leftrightarrow x^3\minus{}6x^2\plus{}20x\minus{}24 \ge 0$
$ \Leftrightarrow (x\minus{}2)(x^2\minus{}4x\plus{}12) \ge 0$
It's true because $ x \ge 2$ and $ x^2\minus{}4x\plus{}12\equal{}(x\minus{}2)^2\plus{}8>0$

[quote="NguyenDungTN"]

$ (a \plus{} b \minus{} 2)(a \plus{} b \plus{} 6) \ge 0 \Rightarrow a \plus{} b \ge 2$.
[/quote]
$ a\plus{}b \geq 2$ or $ a\plus{}b \leq \minus{}6$

[quote="chien than"][quote="NguyenDungTN"]

$ (a \plus{} b \minus{} 2)(a \plus{} b \plus{} 6) \ge 0 \Rightarrow a \plus{} b \ge 2$.
[/quote]
$ a \plus{} b \geq 2$ or $ a \plus{} b \leq \minus{} 6$[/quote]
But $ a,b,c>0$ ?$ \Rightarrow a\plus{}b\geq 2$ (That right?)[/list]

[quote="chien than"][quote="NguyenDungTN"]

$ (a \plus{} b \minus{} 2)(a \plus{} b \plus{} 6) \ge 0 \Rightarrow a \plus{} b \ge 2$.
[/quote]
$ a \plus{} b \geq 2$ or $ a \plus{} b \leq \minus{} 6$[/quote]
If $ a\plus{}b<\minus{}6$ then $ \frac{3(a\plus{}b)}{4}\plus{}\frac{3}{a\plus{}b}<0<\frac{a^2\plus{}b^2}{4}\plus{}\frac{5}{2}$ :)
OK?


A car gets 12 miles per gallon uphill and 24 miles per gallon downhill. If the car goes to the top of Pike's Peak and back (48 miles uphill followed by 48 miles downhill), what is the car's gas mileage, in miles per gallon, for the entire trip?

The car took $ \frac{48}{12} \plus{} \frac{48}{24} \equal{} 4\plus{}2\equal{}6$ hours to get up and down the hills, because $ \text{time }\equal{}\frac{\text{distance}}{\text{rate}}$.  Because the car traveled a total of $ 48\plus{}48\equal{}96$ miles in $ 6$ hours, the overall rate is $ \frac{96}{6} \equal{}\boxed{16}$ miles per hour.

We could also use harmonic means here. The harmonic mean would be
$ \frac{2}{\frac{1}{24}\plus{}\frac{1}{12}}\equal{}\boxed{16}$.


geometry
3D geometry
perimeter
vector
cube
IMO
IMO 1962

Consider the cube $ABCDA'B'C'D'$ ($ABCD$ and $A'B'C'D'$ are the upper and lower bases, repsectively, and edges $AA', BB', CC', DD'$ are parallel). The point $X$ moves at a constant speed along the perimeter of the square $ABCD$ in the direction $ABCDA$, and the point $Y$ moves at the same rate along the perimiter of the square $B'C'CB$ in the direction $B'C'CBB'$. Points $X$ and $Y$ begin their motion at the same instant from the starting positions $A$ and $B'$, respectively. Determine and draw the locus of the midpionts of the segments $XY$.

I've got a proof using vectors, but for the life of me cannot figure out how to put it into LaTex.

Find the four midpoints of the four pairs of corresponding vertices of the squares. This gives C and three midpoints of faces of the cube.
In between, everything moves along straight lines, so connect the four points in the appropriate order to find a deltoid.


The odds in favour of a Pythag-Air-US airlines flight being on time are 3:1.
a) What is the probability that this airline's next eight flights will be on time?
b) What is the expected waiting time before a flight delay?

Looks like homework problems to me. (Might want to post these in middle school forums.)


trigonometry
algebra
polynomial
algebra unsolved

Solve this system in $ R$:

$ \begin {cases} 2(x^3\minus{}y^3)\minus{}x(x\plus{}1)(x\minus{}2)\equal{}1 \\ 2(y^3\minus{}z^3)\minus{}y(y\plus{}1)(y\minus{}2)\equal{}1 \\2(z^3\minus{}x^3)\minus{}z(z\plus{}1)(z\minus{}2)\equal{}1 \end {cases}$

hello, you must prove that we have $ x\equal{}y\equal{}z$ in this system.
Sonnhard.

$ x^3 \minus{} x^2 \minus{} 2x \minus{} 1 \equal{} 0$ doesn't factorise nicely.

A sketch gives one real solution which is a bit larger than $ 2$.

(My computer keeps giving the message "canot find server" and I have to shut down and start again, so I'm giving up  for this evening.)

$ y\equal{}(x^3\minus{}x(x\plus{}1)(x\minus{}2)/2\minus{}0.5)^{1/3}\equal{}f(x)$. $ f$ is an strictly increasing continuous function.
$ z\equal{}f(y),x\equal{}f(z)$.
1) Assume $ x\geq{y}\geq{z}$ ; then $ y\geq{z}\geq{x}$ and $ x\equal{}y\equal{}z$.
2) Assume $ x\geq{z}\geq{y}$ ; then $ y\geq{x}\geq{z}$ and $ x\equal{}y\equal{}z$.
and so on...
Conclusion: $ x\equal{}y\equal{}z$ where $ x^3\minus{}x^2\minus{}2x\plus{}1\equal{}0$ that is $ x\approx\minus{}1.25$ or $ \approx0.45$ or $ \approx1.80$.

In this kind of problem,we don't need f'(x) increasing or decreasing!
I have an example:
$ x\equal{}f(y)$
$ y\equal{}f(z)$
$ z\equal{}f(x)$
and $ f'(x)\equal{}\frac{x}{x\plus{}1}$

Hi Math pro,
that you say is absolutely false. If $ f$ is not increasing or decreasing ($ f'$ has nothing to do here) then you must solve $ x\equal{}f^{(3)}(x)$.

[hide=" The roots are "]
\[ 2\cos(\pi/7), 2\cos(3\pi/7), 2\cos(5\pi/7)\]

As a hint: $ \cos3a\ \plus{}\cos4a\ \equal{} 0$
for what values  of $ a$ ?

[/hide]

We have $ |f'(x)|<1$ for all x and f(x) is continous,then we use lagrange!

Hi Math pro,
for my $ f$ there exists $ x_0$ s.t. $ f'(x_0)\equal{}\infty$.

seasonal squirrel is right.
More generally $ cos(2\pi/n)$ is algebraic of degree $ 3$ iff $ n\equal{}7,14,9,18$.
For $ n\equal{}7$ the minimal polynomial is $ 8y^3\plus{}4y^2\minus{}4y\minus{}1$ that gives the roots:
$ cos(2\pi/7),cos(4\pi/7),cos(6\pi/7)$.
For $ n\equal{}14$ the minimal polynomial is $ 8y^3\minus{}4y^2\minus{}4y\plus{}1$ that gives the roots:
$ cos(\pi/7),cos(3\pi/7),cos(5\pi/7)$.

Bad button....
For $ n\equal{}9$ the minimal polynomial is $ 8y^3\minus{}6y\plus{}1$ that gives the roots:
$ cos(2\pi/9),cos(4\pi/9),cos(8\pi/9)$.
For $ n\equal{}18$ the minimal polynomial is $ 8y^3\minus{}6y\minus{}1$ that gives the roots:
$ cos(\pi/9),cos(5\pi/9),cos(7\pi/9)$.
Thus we obtained $ 12$ values for $ a$ s.t. $ cos(a)$ is algebraic of degree $ 3$.
Moreover expressions as $ \dfrac{ucos(a)\plus{}v}{ycos(a)\plus{}z}$, where $ u,v,y,z$ are integers s.t. $ uz\minus{}yv\not\equal{}0$, are also algebraic of degree $ 3$.

sorry loup blanc!but it 's my example to indicate we can also solve this kind of system with $ |f'(x)|$ is not $ 1$ for all x
here is my solution:
Cause $ |f'x)|$ continuous,so $ |f'(x)|>1$ for all a,or $ |f(x)|<1$ for all x
1.$ |f'(x)|<1$
then $ |x\minus{}y|\equal{}|f(y)\minus{}f(z)|\equal{}|f'(c)(y\minus{}z)|<|y\minus{}z|<|z\minus{}x|<|x\minus{}y|$
so $ x\equal{}y\equal{}z$
2.the same
->we have done!


inequalities
geometry
inequalities unsolved

Do you know inequalities which are solving with geometry? And do you know the famous inequalities' geometrical interpretation? For example aritmetic- geometric mean, cauchy etc.? I'm finding more original interpretations.
                                                                         Regards.

The classical story... Fix two variables, notice that the expression is convex in the third variable, ...

Or what also works is $(x - 1)(y - 1)(z - 1) < 0$, hence $x(1 - y) + y(1 - z) + z(1 - y) = x + y + z - (xy + yz + zx) < 1 - xyz < 1$... too easy, really.


real analysis
integration
real analysis unsolved

Prove that the set $A=[0,1]\cap\mathbb{Q}$ is not Jordan measurable, but it is Lebesgue measurable. ;)

what is jordan mesurable ? excuse my ignorance  :blush:

We say that $E \subset \mathbb{R}^n$ is Jordan measurable if $E$ is bounded and $\chi_E$ is Riemann integrable. In this case the Jordan measure of $E$ is defined to be $m_j(E) = \int \chi_E$.
The claim here is of course trivial. We know that $\mathbb{Q}$ is countable and thus $0 \le m^*(A) \le m^*(\mathbb{Q}) = 0$ $\Longrightarrow m^*(A) = 0$ which makes $A$ Lebesgue measurable. Here I denote $m^* = m_1^* =$  Lebesgue outer measure. Indeed $A$ is not Jordan measurable because
$\chi_{Q}$ is not Riemann integrable over any $[a,b]$, $a < b$.

E is Jordan measurable if it is bounded and its boundary has measure/content zero in the sense of Jordan. The boundary of your set is [0,1], which does not have content zero. In fact, its interior is non empty.


limit
integration
calculus
function
real analysis
real analysis unsolved

Ascertain $ \boxed {\ \lim_{n\to\infty }\int_{0}^{\infty}\frac {1}{x^{2}\plus{}1}\cdot\arctan \sqrt [n]x\ \mathrm {dx}\ }$

substitution $ x \equal{} \frac {1}{y}$ and use $ \arctan x \plus{} \arctan \frac {1}{x} \equal{} \frac {\pi}{2}$ you are done
answer will be $ \frac {\pi^{2}}{8}$

The limit equals $ \pi^{2}/8$.

We have that, by using the substitution $ t \equal{} 1/y$ in the second integral, that

$ I_{n} \equal{} \left(\int\limits_{0}^{1} \plus{} \int\limits_{1}^{\infty}\right)\frac {\arctan \sqrt [n]{t}}{1 \plus{} t^{2}}dt \equal{} \int\limits_{0}^{1}\frac {1}{1 \plus{} x^{2}}\left(\arctan \sqrt [n]{x} \plus{} \arctan\frac {1}{\sqrt [n]{x}}\right)dx$

$ \equal{} \frac {\pi}{2}\int\limits_{0}^{1}\frac {1}{1 \plus{} x^{2}}dx \equal{} \frac {\pi^{2}}{8}$,

since $ \arctan \frac {1}{\sqrt [n]{t}} \equal{} \frac {\pi}{2} \minus{} \arctan \sqrt [n]{t}$.

In conclusion we get that $ I_{n} \equal{} \frac {\pi^{2}}{8}$.

OK. Extremal has already posted the solution to this problem.

With less computation but more high-powered tools:

As $ n\to\infty,$ $ \arctan\sqrt [n]{x}\to\arctan(1) \equal{} \frac {\pi}4$ pointwise for each $ x\in(0,\infty).$ Furthermore, this sequence is bounded by $ \frac {\pi}2.$

Hence $ \frac {\arctan\sqrt [n]{x}}{1 \plus{} x^2}$ tends pointwise to $ \frac {\pi}4\cdot\frac1{1 \plus{} x^2},$ while being bounded by the integrable function $ \frac {\pi}2\cdot\frac1{1 \plus{} x^2}.$

The Lebesgue Dominated Convergence Theorem applies, and the limit is

$ \frac {\pi}4\int_0^{\infty}\frac {dx}{1 \plus{} x^2} \equal{} \frac {\pi}4\cdot\frac {\pi}2 \equal{} \frac {\pi^2}{8}.$


calculus
derivative
integration
calculus computations

Prove that
\[
\frac {d} {dx} \displaystyle \left ( \left ( \int^x_0 e^{-t^2}\,dt \right)^2 +\displaystyle \int^1_0 \frac {e^{-x^2(1+t^2)}} {1+t^2} \,dt \right )=0
\]

This is fairly straightforward. Using the chain rule and the Fundamental Theorem of Calculus, we see that the derivative of the first term is $2e^{-x^2}\int_0^x e^{-t^2}dt$. Differentiating under the integral sign in the second term, we get $-2x\int_0^1e^{-x^2(1+t^2)}\,dt$. Pulling the factor $e^{-x^2}$ outside and making the change of variable $t\to xt$, we see that the derivative of the second term is just the derivative of the first one with minus sign.

Now
\[
 \displaystyle \left ( \left ( \int^x_0 e^{-t^2}\,dt \right)^2 +\displaystyle \int^1_0 \frac {e^{-x^2(1+t^2)}} {1+t^2} \,dt \right )= k
\]
If $ x \to 0 $ the value of k is $ \frac {\pi} {4} $ then if $ x \to +\infty $:
\[
\displaystyle\int^{+\infty}_0 e^{-t^2}\,dt =\frac {\sqrt {\pi}} {2}
\]
Another beautiful way to find  a known integral!

please  can anyone  tell me in post #2,  how  in that last step  answer  comes out  as           :?: square root  (pi)/  2.

Using Leibnitz rule, I get:

$ 2\int_{0}^x e^{\minus{}x^2}e^{\minus{}t^2}dt \minus{} 0 \plus{}0\plus{}0 \minus{}\int_0^1 2xe^{\minus{}x^2(1\plus{}t^2)}dt$

[quote="b555"]but thats  not what i aksed  , i asked  how  did  he get  root (pi)/ 2 in last step in post #2[/quote]

Sorry! I have no idea! My post was a reply to the question! :oops:

Basic thing to know about Gaussian integrals. See [url=http://en.wikipedia.org/wiki/Gaussian_integral]here[/url].

ya  now  i gopt it thanks.  in fact  this is te first time i heard of this gaussian integral and wondering how i missd this all these days!!!!!!! :o