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My tributes to the man who did not bow to the US verdicts till the end.
He had his faults, but he is far better than tha people who hanged him.

The Iraqis and the Americans will pay for this. The fact that papa Bush now sits comfortably somewhere in his son's kingdom is especially stinging. :mad:

Killing people is not nice.

Interpret the above as you wish...

look who's talking!
who told me once 'i have an instant urge to stab you'?

[quote="beta"]Yeah Saddam Hussein's defiance to the last minute is worthy of respect.[/quote]

If non-defiance would have changed his situation in any aspect, this would make sense.

[quote="bubka"]He had his faults, but he is far better than tha people who hanged him.[/quote]

Please do specify the criteria for "good" you are using. Unless the post is meant just as another provocation to start a flamewar/spam topic.

  darij

[quote="beta"]Yeah Saddam Hussein's defiance to the last minute is worthy of respect.[/quote]
Nothing about that man is worthy of respect. Have you seen the atrocities he and his son committed?

That said, killing people is still [i]not nice.[/i]

EDIT: And what's up with Bubka subtitling this 'the martyr'? Honestly, this is a new low, even for Bubka.

He's no martyr.  He committed genocide; he was a brutal tyrant and a dictator and in no way worthy of anyone's respect.

I think the Iraq war was a bad idea from the start, and I oppose the death penalty (I think life imprisonment would be a better sentance), but I find it hard to get upset over his death, or feel any sympathy for him.

I have this question, how many of the people in this forum had heard about Sadam's crimes [i]before[/i] the begginning of hostilities that finally led to Iraq's war and Sadam's capture?
The question is relevant to me because I hear all these opinions about Sadam's being a brutal tyrant and all which I'm sure he was but were you people aware [i]before[/i] the western world campaign against Sadam started.

That trial was a fraud, pretty conveniently orchestrated from the US and of course taking place in Iraq where law allows these quick executions. Had the trial taken place in the US, there was no way you could convict Sadam and some days later execute him. That, besides the fact that the conviction came right on time for the elections, curious, pity it didn't do the republicans any good.

And the Iraq war is for oil. The idea of US coming to save the world of the opression of the tyrant is pretty but c'mon!. The accusations made to Iraq prior to the war had nothing to do with the killing of many shii'te or kurds and were false and I don't buy they didn't know they were. And that's is despicable.

As for Sadam's hanging, I truly hope someday we won't have DP anymore. As other people have said I'm just against DP, whoever it is, I don't see the point. And it does not fall in the way of making our society better.

Also it is no joke to add the martyr thing in the title of this thread. That is also an insult to many people who suffered Sadam's regime. No matter [b]bubka[/b] what your political views on the subject are, which btw you can expose rightfully in the thread, I can't figure out the point of putting it in the title so it is probably only to make people angry about it. That is a very childish behaviour

Ursula

[quote="deimos"]we already had a discusion about the death penalty.
sure there were some other ways than killing him...[/quote]
Yes but the Iraqi law did allow executions and there have been executions of ordinary criminals since the fall of the regime.  So it would make no sense to pardon him and not those others.

[quote="ursula"]I have this question, how many of the people in this forum had heard about Sadam's crimes [i]before[/i] the begginning of hostilities that finally led to Iraq's war and Sadam's capture?
The question is relevant to me because I hear all these opinions about Sadam's being a brutal tyrant and all which I'm sure he was but were you people aware [i]before[/i] the western world campaign against Sadam started.[/quote]
Well I had, but only after the first Gulf War erupted.

[quote]That trial was a fraud, pretty conveniently orchestrated from the US and of course taking place in Iraq where law allows these quick executions. Had the trial taken place in the US, there was no way you could convict Sadam and some days later execute him. [/quote]
There was a proposal to let the trial take place in Sweden too, because they knew if he would have been sentenced to death, Sweden would have never given him back.

[quote]That, besides the fact that the conviction came right on time for the elections, curious, pity it didn't do the republicans any good.[/quote]
That was silly.

[quote]And the Iraq war is for oil. The idea of US coming to save the world of the opression of the tyrant is pretty but c'mon!. The accusations made to Iraq prior to the war had nothing to do with the killing of many shii'te or kurds and were false and I don't buy they didn't know they were. And that's is despicable.[/quote]
It's amazing how many people have forgotten that back in 1988,the Bush administration tried to blame Iran for the Halabja incident... :|

[quote]Also it is no joke to add the martyr thing in the title of this thread. That is also an insult to many people who suffered Sadam's regime. No matter [b]bubka[/b] what your political views on the subject are, which btw you can expose rightfully in the thread, I can't figure out the point of putting it in the title so it is probably only to make people angry about it. That is a very childish behaviour[/quote]
It's the same kind of behavior as Chavez's, hanging out with other people who don't like the USA (he hangs with Lukashenko, the Iranian president, Castro,..)
Bubka's ideas tend to frighten me.  He seems to defend communism at all costs, including democracy and human rights.  And anti-USA sentiment, at all costs.

Okay here is a simple question:
between saddam and papa bush, who deserves more to be hanged? (On second thoughts, I'll add drawn and quartered.)

I'll come to the 'martyr' part later.

There are many ways for a man to be dead.

Just think about it...

It doesn't mean just hanged or executed.

He could be killed in many other ways, not violently or with any weapons....

make him ashamed..

[quote="bubka"]Okay here is a simple question:
between saddam and papa bush, who deserves more to be hanged? (On second thoughts, I'll add drawn and quartered.)
[/quote]

I'd still have to say Saddam- Bush hasn't committed any sins that Saddam hasn't committed many times over.

However I oppose the death penalty on general principle, and I would have to agree with bubka on the martyrdom thing- martyrdom is in the eyes of the beholder, and to be honest, Saddam's defiance, and his call for nonviolence at the end of his life will probably bring him to the status of martyr in the eyes of many.

In other words, the government probably shouldn't have given him the death penalty- life in prison would have been far more successful at erasing his influence.

[quote="bubka"]oh dear. poor bush. mother's icicle.

anyway what about donald ramsfeld who ordered the torture in abu ghraib?[/quote]

Can you find evidence for that statement?

[quote="fredbel6"][quote="Go Around the Tree"][quote="kstan013"]In a land using the Code of Hammurabi, Saddam was treated, er, justly.
@bubka What if I said when that the tsunami hit southern India, I was very happy although some Americans died. And the more frightening thing about 9/11 was the TERROR. Living in the US is not as happy jolly as you might think. :wink:[/quote]

I quite agree. TERRORISTS ARE TERRIBLE.

SADDAM WAS A TERRORIST. SADDAM=TERRIBLE

Get the picture?

I don't see why they hanged him so soon though.  :maybe:

were you happy, guys, that the WTC was knocked down?

You support Saddam?

WHYYY????

SADDAM IS A TERRORIST. ALTHOUGH BUSH IS A BAD PRESIDENT IN MY OPINION, SADDAM IS STILL MORE EVIL .[/quote]
Does Saddam have anything to do with the attacks on the World Trade Center?

[quote]NUMBER OF PEOPLE KILLED ON THE 9/11 ATTACKS

2973

Now...

Darfur conflict, Sudan(ongoing): ~400 000(mostly civilians)
Second Sudanese Civil War, Sudan (1983-2005): ~2 000 000 deaths
Tutsis-Hutus conflict, Rwanda (three months of 1994): ~1 000 000 deaths (that's 10 000 per day)
Iraqi conflict, Iraq(ongoing): 3000+ US soldiers, the number of Iraqi civilian casualties differ greatly, from 30 000 to 600 000 deaths
Chechenyan War, Chechenya(1994-1996): 30 000 ~ 100 000 deaths many more thousands in the second conflict.
Gulf War, Kuwait: 100 000 Iraqi deaths
Kosovo War: ~12 000 deaths, NATO bombing in 1999, 2000 civilian deaths
Croatian Independence War: ~15 000 deaths
Eritrea Ethiopia war (1998-2000): 70 000~170 000 deaths
Algerian Civil War, (1992-2001): 100 000 ~ 200 000 deaths
Sierra Leone Civil War(1991-2001): ~75 000 deaths [/quote]

You left out the Second Congo War(1998-2003) : more than 3 500 000 civilian casualties  :ninja:

It's time we simply accept that people have double standards.  Posters here are doing their best to express grief for the people who died in Manhattan but seem to forget that action like those are a consequence of US policy and MORE IMPORTANTLY : will lead to much more of the same suffering among others (the Iraqi people first)
People get bombed everyday, I suggest you watch Fahrenheit 9/11 if you wanna see real wounded peopple, not the clean version.
And now this has led to retaliations in Spain and London...[/quote]

Bush really didn't have a GOOD reason. He just associated Terrorists with WTC with Saddam.

Even though Saddam didn't cause WTC and wasn't involved, Bush uses the tone that makes us believe he is in with the same group and that is used as the reason why we are in Iraq.

I agree that the media takes what they think will sell the most papers and blows it out of perspective.  Yet, despite this, many Americans are becoming aware of other problems, such as the Darfur conflicts.  Recently, that has become a big thing here, and I have even seen a few disaster relief fund commercials.

Also, on a separate note, even as a conservative Republican, I do not support the war in Iraq, nor do I agree with many of Bush's policies. What a lot of foreign people don't get is that a strong majority of Americans do not support the war in Iraq.  America is around 50% Republican (Bush's party) and 50% Democratic.  However, Bush's approval rating dipped below 40% at the end of last month, and less than 35% of Americans approve of the war in Iraq.

[quote="hsiljak"][quote="ursula"][quote="fredbel6"][

You left out the Second Congo War(1998-2003) : more than 3 500 000 civilian casualties  :ninja:
[/quote]

You are quite right. I firstly wrote the conflicts I remembered and then went through Internet looking for the death tolls I didn't knew about. But I knew that you were going to be there if I missed any :)

Ursula[/quote]

You missed the toll in Bosnia too (150.000-200.000)...  :!:[/quote]

OK thanks, :) I'm editing the post, so it is more complete.

Ursula

THE AMERICAN WAY OF LIFE
according to bubka

Everyone goes to the beach every day,
nobody ever has to work,
we all sit in recumbent lawn chairs all day sipping pina coladas out of coconut shells,
oil is flowing out of our ears and we burn it for fun,
we worship Bush and never ever disagree with him,
we kill Iraqis in our free time

Bubka, we donated millions of dollars in relief efforts to help the tsunami victims and what did we get for 9/11? Not a dime.

I realize that the Iraqi way of life was bad, don't get me wrong, but when I said [quote="kstan013"]Living in the US is not as happy jolly as you might think.  :wink: [/quote]I was referring to how you think Americans live.

[quote="kstan013"]Bubka, we donated millions of dollars in relief efforts to help the tsunami victims and what did we get for 9/11? Not a dime.
[/quote]

I'm sorry my friend but that argument is ridiculous.
1) A lot of developed countries and not developed gave help to the tsunami victims one way or another. I don't know about the actual numbers but certainly the US weren't the only good souls in the world. I'm not sure about this one but it tends to happen that the help the US gives is absurd compared with what other countries do.

2) There are not a lot of countries who can presume of being able to give something to the US that they can not proveide themselves. If furthermore such countries observe that after 9/11 you rather spend the money in multiple wars - and that's a lot of money -, money that could be spent in reconstruction or in helping the families of the victims, or the families of the firemen now sick, or en fin preventing Katrina, in such a case, you question the point of helping the US.

3) I remember you that a long time ago there was this UN agreement about donating 0.7 percent of the GIP of all developed countries to the developing ones. The greatest GIP of the world, that is, the US, set an example by donating about 0.15 percent some years. I don't know how the thing is now, but your behavior towards the developing countries is full of stains.

Ursula

I don't understand why you all find a necessity to make juxtapositions with other items that have no implication on the absolute value of a deed.

Bush vs. Saddam
Hurricane deaths vs. trade center

Simply put Bubka, let's take things for face value.  1 death due to evil is wrong.  That's it.  There is no need to compare it to a natural disaster or what Americans did or so on.  Looking at everything from an individualistic standpoint ellicits a neutral position on the subject that is unbiased by the influence of needless comparisons.  This is the most logical way of handling it.

I am no judge, but if you consider yourself a good person you will admit that evil is evil at face value.

[quote="Potato Theory"]
I am no judge, but if you consider yourself a good person you will admit that evil is evil at face value.[/quote]

[b]Yay for Ad Hominem arguments![/b]

Though I am still not sure why this thread hasn't undergone serious deletion by mods.

At any rate, it would be kinda cool for people, instead of accusing everyone else of making stuff up, to give us credible sources that back their arguments (by credible I mean actually credible, if you don't know what I mean, well, at the very least don't cite random news articles or something that are all going to be obviously biased).

[quote="JSteinhardt"]
Though I am still not sure why this thread hasn't undergone serious deletion by mods.
[/quote]

Well, it is a serious discussion, although totally off topic thanks to our friend bubka (and me too for responding to him, I guess).

[quote="JSteinhardt"]
[b]Yay for Ad Hominem arguments![/b]
[/quote]

The best kind!   :wink:

[quote="ursula"][quote="kstan013"]Bubka, we donated millions of dollars in relief efforts to help the tsunami victims and what did we get for 9/11? Not a dime.
[/quote]

I'm sorry my friend but that argument is ridiculous.
1) A lot of developed countries and not developed gave help to the tsunami victims one way or another. I don't know about the actual numbers but certainly the US weren't the only good souls in the world. I'm not sure about this one but it tends to happen that the help the US gives is absurd compared with what other countries do.[/quote]

[url=http://en.wikipedia.org/wiki/Humanitarian_response_to_the_2004_Indian_Ocean_earthquake#List_of_Donors]link[/url]

[quote]2) There are not a lot of countries who can presume of being able to give something to the US that they can not proveide themselves. If furthermore such countries observe that after 9/11 you rather spend the money in multiple wars - and that's a lot of money -, money that could be spent in reconstruction or in helping the families of the victims, or the families of the firemen now sick, or en fin preventing Katrina, in such a case, you question the point of helping the US.[/quote]
True, after 9/11 we heard the usual "stand as one" talk... but what does it really mean?  Lots of firemen are now sick and unable to provide for themselves.  In the mean time 3000 US soldiers have died in Iraq in a war that has cost more than 350 $10^{9}$ US dollars.

But what you say about helping the US is not entirely correct.  Even Afghanistan sent some money after Katrina even though that was probably symbolic.  But Belgium sent some soldiers to Louisiana to help them coordinate the relief efforts... even though we were quite surprised that a superpower able to wage several costly wars at the same time was not able to help its own citizens in its backyard... The Dutch, masters when it comes to manipulating water, have also offered help reconstructing the levees.

EDIT: post removed...

One thing I would like to mention is that those of you who want to discuss "hot" issues like this should try to do so in a less abrasive manner. While I hate to say it, I have learned one thing from my learning argumentation in english. What you have to say is less important than how you say it. This is why I suggest [note I claim no authority] that people, like Bubka, consider rogerian argumentation. That way, you don't look like a troll who just wants to start a flame war [not only this topic, but the topic about the cost of WOOT].

[quote="fredbel6"]But what you say about helping the US is not entirely correct.  Even Afghanistan sent some money after Katrina even though that was probably symbolic.  But Belgium sent some soldiers to Louisiana to help them coordinate the relief efforts... even though we were quite surprised that a superpower able to wage several costly wars at the same time was not able to help its own citizens in its backyard... The Dutch, masters when it comes to manipulating water, have also offered help reconstructing the levees.[/quote]

[quote="fredbel6"]But what you say about helping the US is not entirely correct.  Even Afghanistan sent some money after Katrina even though that was probably symbolic.  But Belgium sent some soldiers to Louisiana to help them coordinate the relief efforts... even though we were quite surprised that a superpower able to wage several costly wars at the same time was not able to help its own citizens in its backyard... The Dutch, masters when it comes to manipulating water, have also offered help reconstructing the levees.[/quote]

I didn't actually meant the US didn't receive any help for Katrina :). Actually even [url=http://www.cnn.com/2005/WEATHER/09/03/katrina.castro/]Cuba[/url] proposed to send help to the US. But I see it can be misunderstood.
One point more in my argument could be that compared to the tsunami, 9/11 is a minor catastrophe in terms of human lives lost aswell as of economical damage.

Also [url=http://www.unmillenniumproject.org/involved/action07.htm]here you have[/url] the link about the 0.7 percent the rich countries agreed to donate to the developing ones. There you have data for year 2005. You see that the US is along with Portugal the developed country who gives [i]the least[/i], you can also see that is one of the few who have NOT set a timetable for doing it. Draw your own conclusions

Ursula


Here's another change to the ARML rules for 2009:

[quote]In order to keep the proportions of points as they were prior to the addition of the fifth individual round, the scoring structure will be altered as follows.

    * Individual Rounds: 1 pt each (total of 10 points possible per person, 150 points possible per team)

    * Team Round: 5 pts each (total of 50 possible points per team)

    * Power Round: 50 points possible points per team

    * Relay Rounds: correct answer submitted in 3 minutes 3 pts, correct answer submitted in 6 minutes 2 pts. (15 points possible per team for Relay 1, 15 points possible per team for Relay 2)[/quote]

So instead of 120/40/40/40, it's going to be 150/50/50/30.  It doesn't actually keep the proportions of points the same, but I guess they only meant that to apply to individual, team, and power.  Comments?

I guess I like it.  Relays always seem kind of unpredictable, and it's too easy to blow a big lead with a bad relay.

By the way, NYSML has made the same change, except relays are worth 50 points (5 for 3 minutes, 2 for 6 minutes).

I was under the impression that in order to both preserve proportions for the events and make the relay scoring gentler on B teams, the values were going to change to 5 pts / 3 minutes and 3 pts / 6 minutes.

The scoring Silas mentions indeed makes the 6 minutes less penalizing than before, since it is now worth more than half the previous scoring. I know that relays seem to elicit the most contentious opinions from ARML students and coaches, so I'm curious how reducing the overall impact of the relays will be met.

I guess it may depend somewhat on how seriously you take the competition. I see ARML as more "for fun" than anything else at this level, so I don't mind the random element that the relays seem to impart. From a coaching standpoint, the possibility of a massive swing in the standings heading into the final event is a good motivator for my students... no matter how bad a day they might have had up to that point, they can still jump several spots if they buckle down and do a good job.

I think of all the events, the relays is the one that is hardest to "carry," with a couple good people, that is, you need many smart people to score well on relays, whereas for team and power, a couple strong people (2 in the case of my ARML team last year) can lead the team to 30+ scores.

Yeah, I would have thought 5/3 also.  I would also support that scoring.  I would object to 3/1 or 5/2, because I've always thought 6-minute answers should be worth at least half as much as 3-minute answers.

Going to 5/3 relays would also give a way to make a rough comparison between new-format and old-format scores.

An excellent change! I am completely in favor of it.

I emailed the head writer and I think indeed that there was a miscommunication regarding the relay scoring rules. Hopefully this will be clarified soon  :lol:

I heard from Dr. Merryfield that the relay score won't be changing at all... so not 40 to 30, but just 40 still.

Brut3Forc3, my source was this thread, which I seem to have misread. Believe what you see here.

A lot of ARML officials have been emailing each other voicing their opinions on the relay scoring change. It is sounding like the 3/2 scoring rubric sent out earlier was the result of some miscommunication, and that the intended scoring was 5/3. Hopefully that will be made official soon  :)

Apparently things have been taken care of. It will in deed be 3 points for a 6 minute answer, 5 points for a 3 minute answer, and 23 points for a 36-second answer.

[quote="generating"]Apparently things have been taken care of. It will in deed be 3 points for a 6 minute answer, 5 points for a 3 minute answer, [b]and 23 points for a 36-second answer.[/b][/quote]

Huh? I hope you were kidding.

Which is why I was going to wait until tomorrow to ask him about the first part of that sentence.


hi, I'm new and I thought I would post a problem... at least to see what happens :D

Here it is

Laura is 3 times as old as Sara was when Laura was as old as Sara is now. In 2 years Laura will be twice as old as Sara was 2 years ago. How old are they now?

Well, what happens is people like me respond with a answer and a solution.
[hide]We can deduct that Laura has to be a multiple of six, because her age is a multiple of 3 and has to be able to be divisible by two because an odd +2 is always odd.
So plug in 6 and see if it works, which it doesn't
Then try 12 and see if it works, and it does, so
[b]12[/b] is the answer[/hide]
EDIT- I mean plug in 6 and 12 for the age of Laura.

me too! :)
[hide]i found that for every 3 years Laura goes up, Sara goes up by two. so, i estimated and trial and error two or three times, and i got Laura = 18 and Sara = 12.[/hide]

[hide="an albegraic solution :)"]
Call Laura's age L
Sara's age S

For the second statement, it is easy to decode.
$L+2=2(S-2)$
$L+2=2S-4$
$2S-L=6$
For the first,
If the difference in ages is $d$,
then when Laura was Sara's age, Sara was $S-d=S-(L-S)=2S-L$
Thus $L=3(2S-L)$
$L=6S-3L$
$4L-6S=0$
$2L-3S=0$

Now we have to solve.
Change the first equation to
$-2L+4L=12$
$2L-3S=0$

Add
$L=12$

Substitute back in to get $S=8$[/hide]


for one of my usamts files i have 8 syntax errors but i cant find anyof them and the doc is only like 2 paragraphs long wut do i do i no i cant post it for people to look at because of the honor code
also is there a way to get a spell check in latex?

Usually, it's one error that causes several errors. For Example (not breaking honor code)...

blablabla
\ begin*{eqnarray}

blabla...

\end*{eqnarray}

will create about 100 errors while

blablabla
\ begin*{eqnarray}
blabla...
\end*{eqnarray}

will create none.

Usually, when you have a string or two of LaTeX, it's hard to determine where the error is. So I recommend deleting like the second row and see if the problem persists. If it does, then delete more until you have just the code that shows up right and try to fix from there. I used to look at entire string but what happens is sometimes, you actually make correct code to be wrong, thinking it was the wrong code and therefore, make the problem even more difficult.

are you saying that u dont put a spare line after \begin{document}
and before \end{document} because i did
ill try removing the lines

i just found a section of the forum for latex so ill stop posting latex stuff here


Solve in $ \mathbb{P}$ the equation $ 2^{x\plus{}1}\plus{}y^2\equal{}z^2$.

(z+y)(z-y)=2^(x+1). As z+y > z-y, z+y is a multiple of z-y

All primes greater than 3 can be expressed as 6k+1or 6k-1. So, either z+y is a multiple of 3 or
z-y is a multiple of 3. So, not possible for y>3.
For y=3,
z+3 is a multiple of z-3. So, z<7. The only possible prime value for z=5
For y=2,
z<5. So, no possible primes for this case.

(2,3,5) is the only solution in P.

i have a question
why you said
$ x\plus{}y>x\minus{}y\Longrightarrow x\minus{}y\mid x\plus{}y$
i didnt understand  :blush:

[quote="pelao_malo"]i have a question
why you said
$ x \plus{} y > x \minus{} y\Longrightarrow x \minus{} y\mid x \plus{} y$
i didnt understand  :blush:[/quote]
because x+y and x-y are both powers of 2, which means x+y is a greater power of 2 than x-y


calculus
integration
limit
real analysis
function
calculus computations

What condition is necessary so that:

$ \lim_{x\to\infty}\left(\int_{a}^{b}f(x,y) \ dy\right) \equal{} \int_{a}^{b}\lim_{x\to\infty}f(x,y) \ dy$

?

I don't want to get into asking about necessary conditions; I suspect there really isn't anything nice you can say. In other words, sometimes you "shouldn't" be able to interchange the limit and integral, but it works anyway.

There are several reasonable sufficient conditions.

1. If $ f(x,y)\to g(y)$ [b]uniformly[/b] as $ x\to\infty,$ then this is justified. That is, $ \forall\,\epsilon>0\,\exists\,N\equal{}N(\epsilon)$ such that if $ x>N,$ then $ |f(x,y)\minus{}g(y)|<\epsilon.$

2. If there exists $ h(y)$ defined almost everywhere on $ [a,b]$ such that $ |f(x,y)|\le h(y)$ and $ \int_a^bh(y)\,dy<\infty,$ then this is justified. Of course, this is simply the Lebesgue Dominated Convergence Theorem, with some minor stuff about turning limits on arbitrary sequences into limits at infinity.

If the functions in question are non-negative, the interval is finite, the limit exists a.e. or in measure, and the integral of the limit function is finite, then eventual equiintegrability (or whatever the right term in English is) is necessary and sufficient. The property is just that for every $ \varepsilon>0$ there exists $ \delta>0$ such that whenever the measure of a set $ E\subset[a,b]$ is less then $ \delta$, one has $ \int_E f(x,y)\,dy<\varepsilon$ for all sufficiently large $ x$. Another way to state the same is to say that there exists a function $ \Psi: [0,\plus{}\infty)\to[0,\plus{}\infty)$ such that $ \lim_{t\to\plus{}\infty}\frac{\Psi(t)}{t}\equal{}\plus{}\infty$ and $ \limsup_{x\to\infty}\int_a^b \Psi(f(x,y))\,dy<\plus{}\infty$. For sign changing functions it is as close to a necessary and sufficient condition as you can possibly wish but, of course, as Kent said there is nothing that is absolutely necessary.


[quote="STARS"]Find all pairs of positive integers $ (a, b)$ such that
\[ ab \equal{} gcd(a, b) \plus{} lcm(a, b).
\]
[/quote]

$ (a;b)\equal{}k$----->$ a\equal{}nk;b\equal{}mk$-----$ (m,n)\equal{}1$---->$ [a;b]\equal{}nmk$----->$ ab \equal{} gcd(a, b) \plus{} lcm(a, b)\equal{}(a;b)\plus{}[a;b]$----->

$ nmk^{2}\equal{}k\plus{}mnk$-----> $ mnk\equal{}1\plus{}mn$----->$ k\equal{}\frac{1}{mn}\plus{}1$---->$ mn\equal{}1$-----> $ m\equal{}n\equal{}1$ and $ k\equal{}2$----> $ a\equal{}b\equal{}2$

It is well known that $ ab \equal{} \gcd(a,b)lcm(a,b)$, so

$ \gcd(a,b)lcm(a,b) \equal{} \gcd(a,b) \plus{} lcm(a,b)$
$ (\gcd(a,b) \minus{} 1)(lcm(a,b) \minus{} 1) \equal{} 1$

$ \gcd(a,b) \equal{} lcm(a,b) \equal{} 2$.

$ (a,b) \equal{} (2,2)$


Four people think of different positive integers. They add them and their sum is 10. What is the greatest integer?

Lets say biggest integer is more then 5. Then other 3 integer have to sum up to equal or less then $ 5$.
Smallest sum of different positive integer is $ 1 \plus{} 2 \plus{} 3 \equal{} 6$ Therefore biggest integere has to be $ 4$ or less.
We can easily see that integers are $ 1,2,3,$ and $ 4$ Therefore $ 4$ is biggest integer


Determine all function $f: \mathbb N \to \mathbb N$ such that
\[f(m+n)f(m-n) = f(m^{2}),\]
for all $m,n \in  \mathbb N$.


inequalities
induction
inequalities proposed

Prove that for all positive integers $0<a_1<a_2<\cdots <a_n$ the following inequality holds: \[ (a_1+a_2+\cdots + a_n)^2 \leq a_1^3+a_2^3 + \cdots + a_n^3 . \]
[i]Viorel Vajaitu[/i]

I will prove it by induction. It is true for $n=1$, so suppose it is true for $n$. We have:
\[ \sum\limits_{i=1}^n a_i ^3 \geq \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^n a_ia_j  \]
Adding $a_{n+1}^3$ to both sides we have to prove that:
\[ \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^{n} a_ia_j +a_{n+1}^3 \geq \sum\limits_{i=1}^{n+1} a_i ^2 + 2\sum\limits_{i<j}^{n+1} a_ia_j  \]
equivalently:
\[ a_{n+1}^2+2a_{n+1}\sum\limits_{i=1}^n a_i \leq a_{n+1}^3  \]
\[ a_{n+1}+2\sum\limits_{i=1}^n a_i \leq a_{n+1}^2  \]
I will prove this by induction as well - suppose it is true for $n+1$, adding to both sides $a_{n+1}+a_{n+2}$ we are left to prove:
\[ a_{n+1}^2 +a_{n+1}+a_{n+2} \leq a_{n+2}^2  \]
\[ a_{n+1}(a_{n+1}+1)\leq a_{n+2}(a_{n+2}-1)  \] which is true due to condition $a_{n+1}<a_{n+2}$
Hence the result

[quote="Valentin Vornicu"]Prove that for all positive integers $0<a_1<a_2<\cdots <a_n$ the following inequality holds: $(a_1+a_2+\cdots + a_n)^2 \leq a_1^3+a_2^3 + \cdots + a_n^3 $
[i]Viorel Vajaitu[/i][/quote]

Even simpler:

$\displaystyle \left(\sum_{i=1}^n{a_i^3}\right)-\left(\sum_{i=1}^n{a_i}\right)^2=$ $\displaystyle \left(\sum_{i=1}^n{a_i^3-a_i^2}\right)-2\left(\sum_{i=2}^n{\left(a_i \sum_{j=1}^{i-1}{a_j}\right)}\right)$ $\displaystyle \ge \left(\sum_{i=1}^n{a_i^3-a_i^2}\right)-2\left(\sum_{i=2}^n{\left(a_i \sum_{j=1}^{i-1}{(a_i-j)}\right)}\right)$ $=\displaystyle \sum_{i=1}^n{a_i(a_i-i)(a_i-i+1)}$, that is a sum of non negative integers. []

An easy one!

We can use induction and the facts that:

1) $a_n-a_i>=n-i$
2)$a_n>=n$

Hmmm strange... the identity $\sum_{i=1}^n{i^3}=\left(\sum_{j=1}^n{j}\right)^2$ kills this problem without thought.  A curious choice of a TST problem...


Can we color some $1\times1$ squares in $1377\times1377$ square that each $1\times1$ square has odd colored neighbors.(Each square has at most 8 neighbors).

I solved this problem today.
This is very beautiful problem I suggest you solve it.

I think I solved it. The answer is no.
My solution is hard to understand without a sheet of paper.
I prove that the number of coloured bordering squares must be odd, the even so reaching a contradiction.
Please tell me if my answer is correct.
I would like to see your solution

I did'nt solved it in this way but I like to see your solution.
I send my solution later.

Hi!
I don't know about iura, but when I last asked him (10 May), he said his solution was wrong. Anyway, I would like to see you solution, Omid.
I have an idea, similar to another problem.
Make a graph with each square and put some numbers in each square and some numbers in each edge of the graph. The numbers must be picked in such a way, so that if one square is colored black, the parity doesn't change. In the initial situation, if it would be possible, then we would heve 1 parity, in the final, when all are colored, the sum must be another parity.
This is like another IRAN problem ,1998, Titu Andreescu Math Olymp Chalenges

I tell you ahint.
Choose all cells (i,j) that both i and j are both odd.....

We prove the problem with contradiction. Suppose there exists such coloring. Consider set $S$ of cells of tables $(i,j)$ that both $i,j$ are odd. For each cell in $S$ consider the number of its black neighbors and let sum of these numbers to be $A$. Because $|S|=\frac{1378^{2}}4=689^{2}$, thus $A$ is odd. Also each black cell is counted as the number of its neighbors that are in $S$. Because for each cell this number is even, so $A$ is even. Contradiction proves everything. :)

[quote]Also each black cell is counted as the number of its neighbors that are in . Because for each cell this number is even[/quote]

I don't think this is true...

That's true and is obvious. Also you see each cell has even neigbors in $S$. (Read the problem carefully and note that each cell has 8 neighbors (We consider vertex neighbors.) )

could you write the problem one more time??

A black cell at (1, 1) only has one neighbor in S...

Of course $(1,1)\in S$ and each cell is not neighbor of itself, and it does not have any other neighbors.


algebra
polynomial
logarithms
algebra unsolved

For $x\neq 1$, $x^n-1=(x-1)\cdot \frac{x^n-1}{x-1}=(x-1)(x^{n-1}+x^{n-2}+\cdots\cdots +x+1)$

Since $x_1,x_2,\cdots x_{n-1}$ are the roots of the equation of $x^{n-1}+x^{n-2}+\cdots x+1=0$, we have

$x^{n-1}+x^{n-2}+\cdots x+1=(x-1)(x^{n-1}+x^{n-2}+\cdots\cdots +x+1)$. Taking logarithm of the both sides,

we have $\log |x^{n-1}+x^{n-2}+\cdots\cdots +x+1|=\log |x-x_1|+\log |x-x_2|+\cdots +\log |x-x_n|$,

Differentiating of both sides with respct to $x$, for $n=2,3,\cdots$, we get

$\frac{(n-1)x^{n-2}+(n-2)x^{n-3}+\cdots +2x+1}}{x^{n-1}+x^{n-2}+\cdots x+1}$

$=\frac{1}{x-x_1}+\frac{1}{x-x_2}+\cdots +\frac{1}{x-x_{n-1}}$

Plugging $x=1$ into the both sides of this equation, we obtain $\frac{1}{1-x_1}+\frac{1}{1-x_2}+\cdots +\frac{1}{1-x_{n-1}}=\frac{\frac{(n-1)n}{2}}{n}=\frac{n-1}{2}$. Q.E.D.

I think I'm missing a huge shortcut, now that I read the title.

But it's really very simple.

We want to show that

$\sum_i \prod_{j\not=i} (1-x_j)=\frac {n-1}{2}\prod_k (1-x_k)$

Using Viete's formulae, it is very easy to show that

$\prod_k (1-x_k)=1-\sum_k x_k+\sum_{k<l} x_kx_l=1-(-1)+(-1)(-1)-\ldots=n$

Similarly, by a direct counting argument, we get

$\sum_i \prod_{j\not=1} (1-x_j)=(n-1)-(n-2)\sum_i x_i+(n-3)\sum_{i<j}x_ix_j=(n-1)+(n-2)+\ldots+1$

which easily evaluates to

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

as desired.

I think we can prove by the equation:
  f'(x)/f(x)= 1/(x-x1)+...+1/(x-xn) with x1,..xn are roots of polynomial f(x)

Another method is to formulate the eqn. whose roots are $ x_{i} -1 $. Now , using this new eqn. we can easily evaluate the result we want

Let $\omega$ be a root of $x^n - 1 = 0$ such that $\omega \neq 1$.  Then $\overline{\omega} = 1/\omega$ is also such a root.  Note that
\[\frac{1}{1 - \omega} + \frac{1}{1 - \overline{\omega}} = \frac{1}{1 - \omega} + \frac{1}{1 - 1/\omega} = \frac{1}{1 - \omega} + \frac{\omega}{\omega - 1} = 1.\]
It's pretty easy from there.

Say: $r_i = \frac{1}{1-x_i}$ which gives us $\displaystyle\sum_{i=1}^{n-1} r_i$ our desired sum. We know that $f$ with roots $x_1, x_2, \ldots, x_{n-1}$ is of the form:
$f(x) = \frac{x^n - 1}{x - 1} = x^{n-1} + x^{n-2} + \cdots + x^2 + x + 1$.

$r_i = \frac{1}{1-x_i} \Leftrightarrow x_i = \frac{r_i - 1}{r_i}$

$\implies 0 = f(x_i) = f\left(\frac{r - 1}{r}\right)$

$= \left(\frac{r - 1}{r}\right)^{n-1} + \left(\frac{r - 1}{r}\right)^{n-2} + \cdots + \left(\frac{r - 1}{r}\right)^2 + \left(\frac{r - 1}{r}\right) + 1$

$\implies 0 \cdot r^{n-1} = (r-1)^{n-1} + r(r-1)^{n-2} + \cdots + r^{n-3}(r-1)^2 + r^{n-2}(r-1) + r^{n-1}$.

Since each of the $n$ terms contributes a power of $r^{n-1}$ we know that the coefficient of $r^{n-1}$ is $n$. The coeffecient of $r^{n-2}$ is easily found by expanding to get:

$Cr^{n-2} = \displaystyle\binom{n-1}{1}(-1)^1r^{n-2} + r\displaystyle\binom{n-2}{1}(-1)^1r^{n-3} + \cdots + r^{n-3}\displaystyle\binom{2}{1}(-1)^1r + r^{n-2}\displaystyle\binom{1}{1}(-1)^1 + 0r^{n-2}$

$= r^{n-2}\cdot(-1)\cdot\displaystyle\sum_{i=1}^{n-1}\displaystyle\binom{i}{1} = r^{n-2}\cdot(-1)\cdot\displaystyle\sum_{i=1}^{n-1}i = r^{n-2}\cdot(-1)\cdot\frac{(n-1)n}{2}$

So, we know $r_1, r_2, \ldots, r_{n-1}$ are roots of the polynomial:

$0 = g(r) = nr^{n-1} - \frac{(n-1)n}{2}r^{n-2} + \cdots = 0$.

So, by Viete's formulas, we get:

$\frac{1}{1-x_1} + \frac{1}{1-x_2} + \frac{1}{1-x_3} + \cdots + \frac{1}{1-x_{n-1}}$

$= \displaystyle\sum_{i=1}^{n-1}\frac{1}{1-x_i} = \displaystyle\sum_{i=1}^{n-1} r_i = -\frac{-\frac{(n-1)n}{2}}{n} = \frac{n-1}{2}$
QED


another one for you all

While attempting to solve a quadratic eqn, Christobel interchanged the coefficient of X^2 with a constant term and solved it accurately.  one of the roots she got was 2 and the other was the root of the original eqn.  What is the sum of the squares of the roots of the original eqn?

i wasnt sure if you needed to use the roots of the eqn i.e using alpha, beta, gamma etc , but im thinking that i dont have enough info to use those formulas, or try to work it out somehow using the quadratic eqn formulae.

thanks

[hide]
WLOG, assume the coefficient of the $x^2$ term is $1$ (it can be scaled to anything).

So $x^2+ax+b$ is the original quadratic. Let $r_1$ and $r_2$ be the roots.

Then we have $bx^2+ax+1$ as the second polynomial. Let $r_1$ and $2$ be the roots.

So $bx^2+ax+1 = b(x-r_1)(x-2)$.

From Viete's Formulas, we get:

$r_1+r_2 = -a$
$r_1r_2 = b$
$b(r_1+2) = -a$
$2r_1b = 1$

See if you can work it from there...[/hide]

thanks for the help, the solution is right there but i just cant get it.  Im not sure if im looking for a real number solution like its 7 or something or if its a theoretical answer like its -a^2b or something.

I tried to solve eqns for r1 and r2 and a and b and all that but i cant get it to work out so i only have 2 unknowns in 2 eqns to solve simultaneously.  I always seem to have 3 unknown terms in an eqn.  I'm sure the key is b(r1+2)=1 and i need to do something with that.  but i need a start.

Cheers

Go to the below URL for more information about quadratics and Viete's Formulas.

http://mathworld.wolfram.com/QuadraticEquation.html

ok, so all i have been doing this whole time is trying to solve those eqns, i'm afraid though all i have been doing is finding ways to prove the vietes formulas as given to me by paladin8.  I'm no closer to solving this thing.  I pretty much understand the viete's formula thing but no lightbulbs are going off.

Help please

[hide]
$r_1+r_2 = -a$ (1)
$r_1r_2 = b$ (2)
$b(r_1+2) = -a$ (3)
$2r_1b = 1$ (4)

So we have from (1) and (3), $r_1+r_2 = b(r_1+2)$.
But from (2), we have $r_2 = \frac{b}{r_1}$, so it becomes $r_1+\frac{b}{r_1} = b(r_1+2)$
And from (4), we have $r_1 = \frac{1}{2b}$, so then $\frac{1}{2b}+2b^2 = b(\frac{1}{2b}+2)$.

Multiply through by $2b$ and rearrange to get $4b^3-4b^2-b+1 = (b-1)(2b+1)(2b-1)$

So $b = 1, \pm \frac{1}{2}$. However, if $b=1$, we get the same polynomial in both cases, so we don't really care about that case.

Thus $b = \pm \frac{1}{2}$.
From (4), $r_1 = \pm 1$.
From (2), $r_2 = \frac{1}{2}$ in both cases.

Thus $r_1^2+r_2^2 = (\pm 1)^2+\left(\frac{1}{2}\right)^2 = 1+\frac{1}{4} = \frac{5}{4}$.[/hide]


trigonometry AMC AIME MATHCOUNTS USA(J)MO USAMO search

Hi,

I'll be taking my first AMC-12 this year (I decided not to take AMC-10 since I'm better off with trig and complex numbers) and I'm very nervous. My usual score is around 104-110 but I want to make it around 120 so it's more safer that I'll be qualifying to AIME.

Any tips on raising the score?

I've read AoPS and did some practice AMC-12's, AIMEs, and questions of the same caliber. My AMC-12 score this year was either 98 or 101.5, I took it again over the summer (with no memory of the answers hehe) and got around 130-140.

Of course preparing for the AIME will automatically get you AMC-12 points :D

I'd actually have to say time management is quite important, at least it is for me.

On mock's and on the real one this year, I ran a little short of time.

Oh yea, if it wasn't obvious, little mistakes kill :D

Solve 20 and leave 5 blank. That shouldn't be too hard, and you can still afford a couple of dumb mistakes.

that is not always as easy as it sounds for some of us, darktreb. Of course for you it would be fairly easy given your ability ;)

practice past aime problems.

it will help with the AMCs more than doing past AMC problems.

What topics would you say are important in doing well on the AMC? (Oh wise experienced AMC-er)

What topics are most important for the AIME?

Generally, AoPS Vol 1 for AMC

AoPS Vol 2 for AIME

I guess a hint I'd have to share is to [b]CHECK YOUR WORK[/b].  If there's one thing I've learned, mistakes kill.  Last year my score dropped from a 132.5 to a 114.5, almost a 20-point drop, because of not checking my work.

[quote="darktreb"]Well practice until you can get 20 and leave 5 blank. If you're spending a lot of time on this site, and spending a lot of time practicing, that really shouldn't be THAT hard. Especially since I've seen Silverfalcon writing problems and stuff so he's obviously getting the hang of things.

To be perfectly honest, 20 right and 5 blank is a good plateau, but a true goal should be 25/0. What's the point if you don't set high goals? I don't think I've ever left a problem blank on any AMC.[/quote]

That's a good idea. Maybe I should try to solve more problems because when I took practice AMC-12's, I usually don't try or put answer on most of the later problems in case I get wrong (and not qualify). Maybe I should set the goal high (not 25/0 but I'll wish for it. Why not? :rotfl: ). I'll probably take AMC-12 Problem Series too so that should get me ready..

[quote="darktreb"][quote]I guess a hint I'd have to share is to CHECK YOUR WORK.[/quote]

Heh. Easier said than applied.

Yeah I was thinking maybe I'd write up some short thing about random knowledge I've gathered over the years regarding tips for taking MathCounts, AMC, AIME, and USAMO. Remind me in a couple of weeks after I get settled in college and we'll see if there's anything worth putting on paper.[/quote]

Hm, iono. Maybe something about achieveing the fine balance between checking too much and having confidence in your answer?

I agree with darktreb in that 20 answered and 5 blank shouldn't be that hard because you can get 5 whole answers wrong!

For Silverfalcon its not going to matter whether he gets 5 wrong or 0 wrong because once he qualifies only the floor on the AIME matters for USAMO

So, this method allows you to make 5 mistakes.

And in response to the question about #17-25, go in planning to not even look at the last 5 questions. You should be able to solve the first 16 and still have a bit of time left for a few more questions. Then if one of the problems 17-20 looks intimidating or you know you can't do it, take a glance at the last 5 problems and search for one that looks easy.
Sometimes one of the last 5 happens to be easy/ier than one of 17-20.

I guess the strategy depends on whether you can take the B test or not.

I'd say that on the A test go all out, do and MARK all you did.

Fall back on the B test for security

You should be able to do at least 2 of the final 5 and at least half of the final 10, so if the question is not too formidable, just try it.

[quote="Silverfalcon"]Sorry to double post but I have another question.

Is there any tip to solve the problems in say, #17-25? Those are usually the ones that I can't solve or come close to solve them when the time's done :( .

Is there any test that has similar level problems? I'll be very grateful if there's anyway I can get them because if I find them, I'll do them EVERYDAY till I solve all.  :) (and I don't joke by that.)[/quote]

Heh, stop looking for so many tips and tricks dude. Just get good at solving problems. The "technique" aspect of things (fine balance between checking and confidence or whatever) is extremely MINOR compared to just plain old problem solving skill. It's like 95% skill, 5% techniques or tricks. The 5% often gets focused on but unless you want to be MathCounts National Champion or get 150 or 15 on AMC/AIME, it really doesn't come into play that much.

Heh (darktreb read my mind :rotfl: )

Okay, thanks for the great replies. I think I'll start to concentrate more on problem solving and see where it leads me.

Thanks again!


prove that l_a+l_b+m_c <=p.sqrt(3),where l_a,l_b are the lengts of the bisectors of the angles A,B of the triangle ABC ,and m_c is the lents of the median of th side AB,p is the semiperimter

Hey this is very hard inequality!!
Here is a proof of my friend, he is in Viet Nam :D
we change notion we will prove $m_{a}+l_{b}+l_{c}\le \frac{\sqrt{3}}{2}(a+b+c)$
Proof:
Put $x=s-a,y=s-b,z=s-c,s=\frac{a+b+c}{2}\Rightarrow x+y=c,y+z=b,z+x=a$
we have
$m_{a}=\frac{1}{2}\sqrt{2b^{2}+2c^{2}-a^{2}}=\frac{1}{2}\sqrt{2(x+z)^{2}+2(x+y)^{2}-(y+z)^{2}}=\frac{1}{2}\sqrt{4x^{2}+4x(y+z)+y^{2}-2yz+z^{2}}=\sqrt{(x+\frac{y+z}{2})^{2}-yz}=\frac{1}{\sqrt{3}}\sqrt{3(x+\frac{y+z}{2}-\sqrt{yz})(x+\frac{y+z}{2}+\sqrt{yz})}\le\frac{1}{\sqrt{3}}\frac{3(x+\frac{y+z}{2}-\sqrt{yz})+(x+\frac{y+z}{2}+\sqrt{yz})}{2}=\frac{2x+y+z-\sqrt{yz}}{\sqrt{3}}$
we known
$l_{b}=\frac{\sqrt{ac}}{a+c}\sqrt{s(s-b)}\le\sqrt{s(s-b)}=\sqrt{y(x+y+z)}$
similar$l_{c}\le\sqrt{z(x+y+z)}$
thus
${m_{a}+l_{b}+l_{c}\le\frac{2x+y+z-\sqrt{yz}}{\sqrt{3}}+\sqrt{x+y+z}(\sqrt{y}+\sqrt{z}) =\frac{2x+y+z-\sqrt{yz}}{\sqrt{3}}+\frac{2}{\sqrt{3}}(\sqrt{x+y+z})\frac{\sqrt{3}}(\sqrt{y}+\sqrt{z})}{2}\le\frac{1}{\sqrt{3}}(2x+y+z-\sqrt{yz}+x+y+z+\frac{3}{4}(\sqrt{y}+\sqrt{z})^{2}) =\frac{1}{\sqrt{3}}(3(x+y+z)-(\frac{\sqrt{y}-\sqrt{z}}{2})^{2}\le\sqrt{3}(x+y+z)=\frac{\sqrt{3}}{2}(a+b+c)$
we are done!
 a corollary
$m_{a}+l_{b}+h_{c}\le\frac{\sqrt{3}}{2})(a+b+c)$ :D


Hi, it's Tom Lovering. If you are one of the people from Bath and you came here because Geoff recommended it, and you decided the most natural place to come to talk to British mathmos might be the British section, you have probably discovered that you are wrong.

However, an excellent website that does exist (although I don't think the discussion of IMO camps, apart from Birmingham, is allowed generally) is NRICH, administered by a load of Cambridgey people. The site is [url=http://nrich.maths.org]here[/url], and "Ask Nrich" seems to be the best section for UK-based mathematical discussion (the rest of the site features largely fairly trivial problems and open investigative mathematical tasks, although they are sometimes interesting, but probably too easy for someone beyond BMO1 level).

Enjoy :D

N.B. This doesn't mean I don't recommend MathLinks. It is probably my favourite maths site on the web, but NRICH is handy especially to network with UK people and talk about STEP, Cambridge, etc.

I see you mention 'Geoff recommends it'. I take it from that that Geoff is Leader again this year- though having read through the IMO squad email archives, he said that he would leave as Leader after Slovenia. Has he decided to stay on?

(Note that I in no way want Geoff to have finished as leader: I think that he is a very good leader, in fact)

He certainly appeared to be 'leader' at the Bath camp, although geographically this is a logical thing for him to do. I got the impression he was continuing as team leader.


Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?

It will take $ 3$ times less for him to go the second half, so it takes $ 6\plus{}\frac63\equal{}\boxed8$ minutes total.


limit
logarithms
calculus
calculus computations

[quote="nemesis"]Could anyone explain how to determine the fallowing limit:
$\lim_{ x->{\infty}} {(x {(1+ \frac{1}{x})}^{x}-ex) }$
Please give me a general technic because I have lots of problems of this type and I can not solve them by using fundamental limits or by applying l'Hospital rule ( I get $\infty - \infty$)[/quote]

I assume by ex you mean exp(x). You have:

x*(1+1/x)^x-exp(x) =
x*exp(x*ln(1+1/x))-exp(x) =
x*exp(x*(1/x+o(1/x)))-exp(x) =
x*exp(1+o(1))-exp(x), which goes to +oo as x->+oo.

If by ex you meant exp(1)*x then you need to develop further:

x*exp(x*(1/x-1/(2x^2)+o(1/x^2)))-e*x =
x*exp(1-1/(2x)+o(1/x))-e*x =
e*x*(-1/(2x)+o(1/x))-e*x,

and now the limit is -3e/2.

P.S: I didn't check the numerical values above so be careful, but the idea is here: using taylor developments.

As $y\to0,$ $\ln(1+y)=y-\frac{y^2}2+\frac{y^3}3+\cdots.$

So $\ln\left[\left(1+\frac1x\right)^x\right]=x\ln\left(1+\frac1x\right)$

$=x\left(\frac1x-\frac1{2x^2}+O(x^{-3})\right)=1-\frac1{2x}+O(x^{-2})$

Now, $e^{1+u}=e\cdot e^u=e\left(1+u+O(u^2)\right)$

So $\exp\left\{\ln\left[\left(1+\frac1x\right)^x\right]\right\}=\dots$

(To be completed later, although you're welcome to finish it yourself.)

[quote="nemesis"]no, it is x*(1+1/x)^x-ex ,[/quote]
Not so literal minded! Can't you see that I'm working on parts of the problem, to be assembled in the end? Have a little imagination.

[quote="nemesis"]By the way, what is it $O(u^2)$ [/quote]
See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31517]this topic.[/url]

julien_santini had the right idea and came close, but some of the details slipped.

Continuing from #3,

$\left(1+\frac1x\right)^x= \exp\left\{\ln\left[\left(1+\frac1x\right)^x\right]\right\}= \exp\left(1-\frac1{2x}+O(x^{-2})\right)$

$=e\cdot\left(1-\frac1{2x}+O(x^{-2})\right)$

So, $x\left(1+\frac1x\right)^x=ex-\frac{e}2+O(x^{-1})$

and $x\left(1+\frac1x\right)^x-ex =-\frac{e}2+O(x^{-1})$

The limit as $x\to\infty$ is $-\frac{e}2.$

If the original problem had been $\lim_{x\to\infty}x^2\left(1+\frac1x\right)^x-ex^2+\frac{ex}2$

then we would have had to extend our analysis by one more term, which would be quite a bit more work.


A,B,C are 3vectors,
Prove that $|A+B+C|+|A|+|B|+|C|>|A+B|+|A+C|+|B+C|$


Ok here is the problem: There is a number that ends on 2 (the last digit is 2 and the number of digits is unknown)
and if you place the last digit (2) in the begging of the number then the number doubles. Whats the number?

I hope it makes sence, if it doesn't then let me know and I'll give you an example.

When you place the 2 at the beginning, do you take it away from the last digit?
so it's like __________2 * 2 = 2__________? or 2________2?

Like if you had 12 it would become 21, does that asnwer your question?

I think you will be thinking for quite a few minutes, lol unless you just randomly pick a number ending on 2 and it turns out to be right, lol.

:rotfl: Can't belive it, tell me your method please, master... lol

Ok, so since the second number has to start with 2.. and it's the first number multiplied by 2.. I just did long division. a number that starts with 2 divided by 2.. gets 1, and then so the second digit has to be 1.. and then the third 0, cause that comes out to 0 on top.. etc until you get the last digit to be 2.

You know that the tens digits has to be 4.  Then the hundreds is 8.  So on, until you get a 1, which becomes a two when the number is doubled.

Clever's my middle name.  :lol:
(no really it is.. just in a different language)

and yeah, what k81o7 said was good.. I didn't think of that. well maybe i did but i didn't realize..

[quote="ambierona"]Ok got it.

105263157894736842
x2 =
210526315789473684[/quote]

Interesting.  If you take the original number move the 1 from the front  to the back and place a decimal before the 0 and put  a repetend sign over it (.052631578947368421 repeating) you get 1/19 in exact form.

You did do it the same way...you just went in the other direction.  You went left to right and divided by 2 instead of multiplying by 2.  So it's basically the same approach.

That's no accident Shogia. It's actually part of a more general method for solving problems like this.

Let N be a repeating decimal number whose decimal expansion looks like $\displaystyle N = .\overline{a_1a_2a_3 \ldots a_n2}$, and such that $\displaystyle 2N = .2\overline{a_1a_2a_3 \ldots a_n2}$.

In other words, its a repeating decimal number such that multiplying the number by 2 moves the digit 2 from the last of the repeating part to the front of the repeating part.

But notice that $2N = .2 + \frac{N}{10}$. So solving for N we get that $N = \frac{2}{19}$. And sure enough if you look at the repeating part of the decimal expansion of 2/19 (using a computer because its too big to fit in a calculator) it is 105263157894736842.

Bonus Problem: See if you can apply this method to find the smallest positive integer such that when you move its last digit to the front of the number it equals the original number multiplied by 3. There was also a USAMTS problem similar to this on one of the rounds this year.

I'm actually quite aware of that.  The only reason i noticed the 1/19 thing was because of the reading I have done on using Vedic mathematics (if anyone is familiar with it) to compute recurring decimals.

It is a form of mathematics that is a bit different from the standard method.  It makes some problems much easier to solve and the even the methods for multiplication and division are different.  I would probably never even have heard of it if my friend did not have a book on it.  I will try to find a site with it if you like.

The bit on recurring decimals allows you to calculate things like 1/19, 1/29, 1/23... in less than a minute with plenty of time to spare.  One cool bit that I learned from it was that  if you have 1/x you can determine what the last digit, or last digit in the repetend, is.  If x ends in a 1 then it will 1/x will "end" in a 9, 2-->5, 3-->3, 5-->2, 7-->7 of course you can figure out the others since they arent prime.


inequalities
algebra
polynomial
Computer solutions
4-variable inequality
Symmetric inequality

yep. it has been posted before....

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112435#p112435
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=190407#p190407

[quote="bodom"]shobber wasn't this problem already posted?since i can't find the link right now i will tell the main ideas of two solutions:one posted here(you can also find it in old and new inequalities since this is vasc's problem posted in "gazeta matematica")and one that i found.
one elegant idea would be to use the fact that $1/(a+1)^2 +1/(b+1)^2\ge 1/(ab+1)$
however if you are not inspired enough to see that you may also try to kill it with sturm and a little "strategy" since from sturm you get that $1/(a+1)^2 +1/(b+1)^2$is decreasing when $a-b$ decreases(when assuming that a>b)if $ab\ge1$.so you always pick two numbers such that their product is bigger than one.if anybody wants to know more about this solution i will post it[/quote]

Can you post your solution bodom?

ok.i will do that in the evening because now i'm a little busy

Let a,b,c> 0 abc=1
Using the previous lemma $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac{1}{1+ab}=\frac{c}{c+1}$
We only need to prove that : $\frac{1}{(1+c)^2} + \frac{c}{c+1} \geq\frac{3}{4}$
which's equivalente  after expanding to $( c-1)^2\geq0$

[quote="leminscate"]These bashes are really ugly! :([/quote]

No, they're not. :lol:

I think from the first we must use CS inequality, then Muirhead's inequality.

The following inequalities are also true.
Let $a,b,c >0$ and $abc=1$. Prove that: \[\frac{1}{(a+1)^2} + \frac{1}{(b+1)^2} + \frac{1}{(c+1)^2} + \frac{1}{{a + b + c +1}} \geqslant 1.\]
Let $a,b,c,d >0$ and $abcd=1$. Prove that:
\[ \frac{1}{(3a-1)^2}+\frac{1}{(3b-1)^2}+\frac{1}{(3c-1)^2}+\frac{1}{(3d-1)^2} \geq 1. \]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16027]here[/url]

we can prove:
$ f(a,b,c,d)\geq f(\sqrt{ab},\sqrt{ab},c,d)\geq 0 $ with $ f(a,b,c,d)=\sum (\frac{1}{(1+a)^{2}}-\frac{1}{4}) $

[quote="sqing"]The following inequality is also true.
Let $a,b,c,d >0$ and $abcd=1$. Prove that:
\[ \frac{1}{(3a-1)^2}+\frac{1}{(3b-1)^2}+\frac{1}{(3c-1)^2}+\frac{1}{(3d-1)^2} \geq 1. \]
[/quote]
It's Jensen! :D

generalization to 6 variables:
let $x_1,x_2,\dots ,x_6$ be 6 positive numbers such that $x_1x_2\dots x_6=1$ and $\prod_{1\le i<j\le 6}(x_ix_j-1)\ge 0$. then
$$\sum_{i=1}^{6}\frac{1}{(x_i+1)^2}\ge \frac{3}{2}$$

[quote=sqing][quote="sqing"]The following inequality is also true.
Let $a,b,c >0$ and $abc=1$. Prove that:
\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\][/quote]
Let $a=\frac{y}{x},b=\frac{z}{y},c= \frac{x}{z},$ we have (《Mathematical communication》(China Wuhan) N0.7、8(2014) Q183 ):
Let $x,y,z$  be positive real numbers . Prove that
\[ \frac{x^2}{(x+y)^2}+\frac{y^2}{(y+z)^2}+\frac{z^2}{(z+x)^2}\geq\frac{3}{4}.\][/quote]

Just is my LBQ105.
BQ

[quote=sqing]The following inequality is also true.
Let $a,b,c >0$ and $abc=1$. Prove that:
\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\][/quote]

Also substituting $ a=xy/z^2$ and likewise does the trick.

[quote=War-Hammer]For prove the problem , we will prove that :

\[ \frac {1}{(1+a)^2} + \frac {1}{(1+b)^2} \ge \frac {1}{1+ab} \]
\[ \frac {1}{(1+c)^2} + \frac {1}{(1+d)^2} \ge \frac {1}{1+cd} \]

Proof :

\[ \frac {1}{(1+a)^2} + \frac {1}{(1+b)^2} \ge \frac {1}{1+ab} \]
\[ \Longleftrightarrow (1+ab)((1+a)^2+(1+b)^2) \ge (1+a)^2(1+b)^2 \]
\[ \Longleftrightarrow a^3b+b^3a-a^2b^2-2ab+1 \ge 0 \]
\[ \Longleftrightarrow  1+ab(a^2+b^2-ab-2) \ge 0 \]

Use AM-GM on $LHS$ we have

\[ \text {LHS} \ge 1+ab(2ab-ab-2) = 1+ab(ab-2)\1+a^2b^2-2ab \ge 2ab - 2ab = 0 \]

Similarly we can prove second one. Now use $abcd=1$. So we have :

\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \ge \frac {1}{ab+1} + \frac {1}{cd+1} \]
\[ = \frac {cd+ab+2}{cd+ab+abcd+1} = \frac {cd+ab+2}{cd+ab+2} = 1 \][/quote]

thank you for Lemma

[quote=shobber]Let $a,b,c,d >0$ and $abcd=1$. Prove that:
\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1 \][/quote]
[url=https://artofproblemsolving.com/community/c6h1392755p7773884]here[/url] [url=https://artofproblemsolving.com/community/c6h1387835p7725948]here[/url] [url=http://www.360doc.com/document/13/0712/11/10693281_299362805.shtml]here:[/url]

Let $a,b,c >0$ and $abc=1$. Prove that
$$\dfrac{1}{(3a^2+1)^2}+\dfrac{1}{(3b^2+1)^2}+\dfrac{1}{(3c^2+1)^2}+\dfrac{1}{(a+b+c+1)^2}\geq\dfrac{1}{4}$$
[quote=shobber]Let $a,b,c,d >0$ and $abcd=1$. Prove that:
\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1 \][/quote]
[url=https://om.mimuw.edu.pl/static/app_main/camps/oboz2018.pdf]Poland 2018, MO Science Camp[/url]
[url=https://artofproblemsolving.com/community/c6h383409p2126599]h[/url]
 [url=https://artofproblemsolving.com/community/c6h2200804p16594994]Romanian magazine Gazeta Matematica, 1999, No. 11[/url]

[quote=sqing]The following inequality is also true.
Let $a,b,c >0$ and $abc=1$. Prove that:
\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\][/quote]
Let $a,b,c\in(0,2)$ and $abc=(1-a)(2-b)(3-c).$ Prove that$$\dfrac{a^2}{3}+\dfrac{b^2}{12}+\dfrac{c^2}{27}\geq\frac{3}{4}.$$

[quote=arqady][quote="shobber"]Let $a,b,c,d >0$ and $abcd=1$. Prove that:
\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1  \][/quote]

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq\min\{1,\frac{2}{1+abcd}\}$
for all non-negative $a,$ $b,$ $c$ and $d.$
It is stronger. ;)[/quote]

Let us not forget this math problem

This problem is a special case of this inequality:
Let $abcd=1$ as above. Then
$\frac{1}{(1+a)^k}+\frac{1}{(1+b)^k}+\frac{1}{(1+c)^k}+\frac{1}{(1+d)^k} \geq 2^{2-k}$.

[quote=sqing]Thanks.
Let $x_1,x_2,\cdots,x_n>0$ and $x_1x_2\cdots x_n=1$. Prove that:
\[ \frac{1}{(1+x_1)^2}+\frac{1}{(1+x_2)^2}+\cdots+\frac{1}{(1+x_n)^2} \geq \frac{n}{4}. \][/quote]

This is wrong for $n\geq 5$. Try $x_1=...=x_{n-1}\to +\infty$.


$ (\frac{a}{b})^ 3 \plus{} (\frac{c}{d})^ 3 \equal{} 6$

$ a, b, c, d$ are positive integers. a couple of them may or may not be equal.

find $ a, b, c, d$.

I think it can be proved with infinite descent, we have $ m^3\plus{}n^3\equal{}6p^3$, by mod 6 either m and n are both multiples of 3, thus p is too and it divides out?

dgreenb801, $ m$ and $ n$ do not need to be multiples of $ 3$.  All residue in mod 6 are cubic residue.

But $ (ad)^3 \plus{} (bc)^3 \equal{} 6(bd)^3$ is a good start...

$ (ad)^3 \equal{} 6(bd)^3 \minus{} (bc)^3$, so $ b|ad$.  Similarly, $ d|bc$.  Let $ bk\equal{}ad$, $ dj\equal{}bc$, then $ bkj\equal{}adj\equal{}abc \implies kj\equal{}ac$

I'm a bit stuck here


geometry
3D geometry
MATHCOUNTS
floor function
AMC
AIME

The increasing sequence $2,3,5,6,7,10,11,\ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence.

[hide]Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than $500$. This happens to be $23^2=529$. Notice that there are $23$ squares and $8$ cubes less than or equal to $529$, but $1$ and $2^6$ are both squares and cubes. Thus, there are $529-23-8+2=500$ numbers in our sequence less than $529$. Magically, we want the 500th term, so our answer is the smallest non-square and non-cube less than $529$, which is $\boxed{528}$.[/hide]

This was similar to a mathcounts question.

[hide]500 is the what number term? After taking the 22 perfect squares and 7 perfect cubes away, and then adding the 2 perfect 6th powers that we eliminated twice, we get the 473rd. To get to the 500th term, we add 27 to 500 to get 527. However, we also have to skip over 512 in the process, so the actual 500th term is 528.   [/hide]

[hide="Answer"]We have $23$ squares up to $575$, and we will not need to go beyond that. In that same space, we have $8$ cubes. However, we also have two sixth-powers which overlap: $1^2=1^3=1^6=1$ and $8^2=4^3=2^6=64$, so we subtract those. Therefore, we have $29$ terms which are not used. However, this sequence has the $29^{th}$ of those terms as $529$, so we actually want the number before, which is $\boxed{528}$.[/hide]

[quote="4everwise"][quote="236factorial"]then adding the 2 perfect 4th powers that we eliminated twice[/quote]You mean 6th powers, right?  :)[/quote]

Oops, only had 6 hours of sleep the previous night  :blush:

[quote="dgreenb801"]We seek the number n such that
$ n \minus{} \lfloor \sqrt {n} \rfloor \minus{} \lfloor \sqrt [3] {n} \rfloor \plus{} \lfloor \sqrt [6] {n} \rfloor \equal{} 500$, in which case $ n \equal{} 528$, which we find by trial and error.[/quote]

This is an AIME question.. I don't think it would be a good idea nor is it a good use of time..

It's not nearly as bad as it looks, you see that there will be 22-23 squares less than it, 8 cubes less than it, and 2 6th powers less after only trying out one number, there's not that much to check.

[hide=Solution]We want the number satisfying $n - \lfloor \sqrt{n} \rfloor - \lfloor \sqrt [3]{n} \rfloor + \lfloor \sqrt [6]{n} \rfloor = 500$. Testing $500$ shows it's too small, so we try $529$, it being a perfect square. This works, but it can't be in the sequence, so the answer is one below, or $\boxed{528}$.[/hide]

[quote=xingzhe]this post is from over 11 years ago, i do not see why you are posting in a post that was from 11 years ago[/quote]

I believe its an AIME problem, so it is allowed

$n - \lfloor \sqrt{n} \rfloor - \lfloor \sqrt [3]{n} \rfloor + \lfloor \sqrt [6]{n} \rfloor = 500$, testing $500$ is too small, more testing gives us $n=\boxed{528}.$


Hi All,

I have a question about a question that was given out in our multivariable class.

"Find the maximum value of 4xy(10-x-y) over positive numbers x and y."

We are supposed to do it 2 ways, one with calculus, and the other without.

I was able to do it with calculus pretty easily by finding critical pts. by setting the partials to zero, and found f(5/2,5)=625 to be the maximum value of the function for positive x and y.

However, I am stuck on how to approach this problem without calculus. My guess is that I have to somehow apply AM-GM to the expression, but b/c I am trying to find the maximum, I need the arithmetic side to become constant..

Also, since x and y are positive, out of the three terms in the expression right now, the second and third terms will both be negative, so is AM-GM even an option here???  :maybe:

Actually, if $ f(x,y) \equal{} 4xy(10 \minus{} x \minus{} y)$, then $ f(5/2,5) \equal{} 125$, and that is not the maximum value since $ f(10/3,10/3) \equal{} 4000/27$

[hide="AM-GM hint"]
Apply AM-GM to $ x$, $ y$, and $ 10 \minus{} x \minus{} y$.[/hide]

[hide="solution with AM-GM"]
Since $ x$ and $ y$ are positive, $ (10 \minus{} x \minus{} y)$ must also be positive in order for the function to reach a maximum.

By AM-GM, $ \frac {x \plus{} y \plus{} (10 \minus{} x \minus{} y)}{3}\ge\sqrt [3]{xy(10 \minus{} x \minus{} y)}$
notice how the inequality is set up so that the x and the y cancel out on the left side.

Some final manipulations:

$ \frac {10}{3}\ge\sqrt [3]{xy(10 \minus{} x \minus{} y)}$

$ \frac {1000}{27}\ge xy(10 \minus{} x \minus{} y)$

$ \frac {4000}{27}\ge 4xy(10 \minus{} x \minus{} y)$

as desired.

This inequality reaches equality when $ x \equal{} y \equal{} 10\minus{}x\minus{}y$ or $ x \equal{} y \equal{} 10/3$.[/hide]

Clearly 10 - x - y must be positive or else we'll get a negative answer, which is less than a positive one.

By AM-GM,
\[ \frac{x \plus{} y \plus{} (10 \minus{} x \minus{} y)}{3} \ge (xy(10\minus{}x\minus{}y))^\frac{1}{3}\]
\[ \therefore 4 \left(\frac{10}{3} \right)^3 \ge 4xy(10\minus{}x\minus{}y)\]

Symmetrize. Let $ z\equal{}10\minus{}x\minus{}y$; now you want to maximize $ 4xyz$ under the constraint that $ x\plus{}y\plus{}z\equal{}10$, and at most one of the three is negative. (If two are negative, the product can be made arbitrarily large)
We can constrain further to $ x,y,z\ge 0$, since having one negative obviously won't generate the maximum.

You might also want to see how Lagrange multipliers work for the symmetrized version.

O shoot, I was looking at your solutions and wondering why it didn't match mine when I realized that I had typed the original function wrong.

it should be 4xy^2(10-x-y), which with calculus, should give you a mx of 625 at (5/2,5).

I was able to verify it with AM-GM by splitting it up into 4 terms, 2x, y, y, and 2(20-x-y), and I also reached an answer of 625.


geometry
minimization
triangle inequality
Triangle
optimization
IMO
IMO 1973

A soldier needs to check if there are any mines in the interior or on the sides of an equilateral triangle $ABC.$ His detector can detect a mine at a maximum distance equal to half the height of the triangle. The soldier leaves from one of the vertices of the triangle. Which is the minimum distance that he needs to traverse so that at the end of it he is sure that he completed successfully his mission?

Let $ABC$ the triangle, with side $a$. Let $h=a\frac{\sqrt{3}}{4}$ the half altitude

The soldier starts moving from $A$.

Let $D,E,F$ the midpoints of $AB,BC,CA$ respectively.

If he moves on the segment $DE$ (after $A\to D$), then he can reach the most points of the area but he will lose the vertice $C$.

The problem is of course at the vertices.

We construct the triangle $DEF$ and let $K,L$ the midpoints of $DE,EF$.

From $K$, he can reach the point $B$ since $BK=h$.
Also, from $L$ he can reach the point $C$.

If he could start from $K$, then the better way would be $K\to L\to M$ where $M$ is the midpoint of $DF$.
Actually, each of these points covers a large percentage of the triangle, for example the circle $(K,h)$ includes the triangle $BDE,DEF$ and much more area.

By the way, if could "jump" from $K$ to $L$ and then to $M$, the area would be covered perfectly.  :rotfl:

I think that, since he must start from $A$, the most convinience way is to get first to the point $K$ and then go to the point $L$

The vertical distance from $A$ to $K$ is $2h-\frac{h}{2}=\frac{3h}{2}$

The horizontal distance is $\frac{a}{8}$

$AK^2 = (\frac{3h}{2})^2+(\frac{a}{8})^2 =$

$(\frac{3a\sqrt{3}}{8})^2 + (\frac{a}{8})^2 =$

$= a^2 \frac{9 \cdot 3 +1}{8^2}=$

$= a^2 \frac{7}{16}\Rightarrow$

$AK = a\frac{\sqrt{7}}{4}$ and $KL = \frac{a}{4}$

$AK+KL = a\frac{\sqrt{7}+1}{4}$

I don't know if this is the minimum or if this is the better way to face the problem. At least I tried  :(

Maybe it should be easier using calculus?

This solution is wrong!
Check http://www.mat.itu.edu.tr/gungor/IMO/www.kalva.demon.co.uk/imo/isoln/isoln734.html
The answer is  $ a\frac {2\sqrt {7} \minus{} \sqrt {3}}{4}$ which is less then $ a\frac {\sqrt {7} \plus{} 1}{4}$

Geometric Problems on MAXIMA and MINIMA (by Author "Titu Andreescu")

"Let h be the length of the altitude of the given equilateral triangle ABC. Assume that the soldier's path starts at the point A. Consider the circles k1 and k2 with centers B and C, respectively, both with radius h/2. In order to check the points B and C, the soldier's path must have common points with both k1 and k2. Assume that the total length of the path is t and it has a common point M with k2 first and then a common point N with k1. Denote by D the common point of k2 and the altitude through C in triangle ABC and by L the line through D parallel to AB. Adding the constant h/2 to t and using the triangle inequality, one gets t+(h/2)≥AM+MN+NB=AM+MP+PN+NB≥AP+PB, where P is the intersection point MN and L. On the other hand, Heron's problem shows that AP+PB≥AD+DB, where equality occurs precisely when P=D. This implies t+(h/2)≥AD+DB,
i.e. , t≥AD+DE, where E is the point of intersection of DB and k1."

⇒shortest path length=a×{2×(√7/4) - (√3/4)}, where a is the length of equilateral triangle ABC.

Let d be the distance from any of the vetices to the mid point of the triangle's altitude. Let h be the altitude of the equilateral triangle. In order to do his task, the soldier must travel the distance d plus the diference (d - h/2). Hence:

 displacement = d + d - h/2 ~ 0.8899 units of length,

      as 77ant suggested

Let h be the altitude of triangle ABC and t be the radius of the detector, $r=\frac{h}{2}$

Let the soldier start from A, so the critical points to detect are B and C --> the paths to be taken are $A\to D$ and $D\to E$, where D is a point with the distance $r=\frac{h}{2}$ from B and E is a point with the distance $r=\frac{h}{2}$ from C.

The length of the paths is $s=\|A\to D\|+\|D\to E\|$

Let's imagine the soldier is already on D (although we don't know where). To take the shortest path from D to E, the soldier must walk in the direction to C
--> E is on the line DC. So $\|D\to E\|=\|D\to C\| - \|E\to C\|=\|D\to C\| - \frac{h}{2}$

--> $s=\|A\to D\|+\|D\to C\|-r$

s is minimal, if and only if $\|A\to D\|+\|D\to C\|$ is minimal. $\|A\to D\|+\|D\to C\|$ is minimal, if and only if  $\|A\to D\|=\|D\to C\|$.
--> D must be on the line from B to F, where F is the midpoint of line AC.

Because BF is an altitude of triangle ABC, so D must be the midpoint of line BF.

We can calculate s as a function of h, but it's not the answer from the question. The correct answer is:
[i]The soldier must walk from A to D and then to E. D is the midpoint of line BF (the altitude of the triangle ABC from B). E is on line DC with $EC=\frac{h}{2}$[/i]


geometry
geometric transformation
reflection
Support
ratio

[quote="PenguinIntegral"]http://en.wikinews.org/wiki/North_Korea_claims_it_has_conducted_a_nuclear_test

So, I'm wondering if China or Russia are going to do anything about it, or if everyone's just going to sit back and go "ho-hum". The US is sure way too unpopular and streached out to do anything. Early results look like a relatively low-magnitude bomb, and I don't think the North Koreans are planning on doing much with it at the moment. If they were to say, smuggle a nuke into Tokyo and set it off, they'd be able to take out a few thousand South Koreans before we bombed them back to the stone age. CNN projects that a war will have about 100,000 casualties.[/quote]
ok fine be racist and make a "koreans suck/are violent/are stupid" topic :mad:

"ho hum" is what they'll do because "ho hum" is pretty much all the UN and any of its councils do... in the short term, anyway.

but don't worry, time is what we have lots of. i'm calling the nuclear test a bluff, or at least heavily exaggerated in terms of its success - SK has to brag and stand tall, cuz when you're small, that's what gets you attention.

at any rate, it's irrational to think that nuclear attack is imminent in light of this development, since SK is neither irrational or suicidal.  the biggest threat i can think of right now is the sale of nuclear technology to fanaticist organizations - which is still very far-fetched, since A) it'll have to be the sale of nuclear weapons, since having the technology won't let just any small group to build nuclear devices like they do cell-phone-rigged-bombs, and B) SK is not suicidal, and wont do anything that will have the slightest chance of getting blamed on them, causing massive retaliation - remember, a war w/ 100,000 causualties may be unwanted from our side, but it'd be downright devastating on theirs.

and that concludes my first post here since like April :D

Why should we do anything but "ho hum" ? They have the right to make nukes, it's not our job to tell them what they can and can't make. We're not the rulers of them.

[quote="L_Li"]they have the right to make nukes as much as your neighbor down the street has the right to buy a pistol and a target with your picture on it.

p.s.: that profile pic is the weirdest game of Go i've ever seen... is that even a real endgame?[/quote]

Yes, my neighbor definitely has that right to bear arms as long as he wasn't just released from prison for having illegal arms or something.

Also, your example is bad, my neighbor is under the jursidiction of the US and its laws, however, we don't control North Korea, they are their own country with their own soverignty.

If he had a firefight then it's definitely illegal for him to be having a gun.

fine, he has ninja throwing stars, or a compound bow, or really pointy darts. or maybe he had a fistfight w/ your neighbor. sorry if i didn't have the perfect analogy. please try to see the bigger picture of the analogy. back on topic, now, please.

The thing is, it can be deduced that he has intent to kill which is [i]not allowed[/i]. However, whatever North Korea does IS allowed as it has FULL SOVERIGNTY except for any resolutions the UN may pass because it is a member.

[quote="PenguinIntegral"]If they were to say, smuggle a nuke into Tokyo and set it off, they'd be able to take out a few thousand South Koreans before we bombed them back to the stone age.[/quote]

Last I checked, Tokyo was in Japan. I think you're looking for Seoul :wink:.

[quote="PenguinIntegral"]http://en.wikinews.org/wiki/North_Korea_claims_it_has_conducted_a_nuclear_test

So, I'm wondering if China or Russia are going to do anything about it, or if everyone's just going to sit back and go "ho-hum". The US is sure way too unpopular and streached out to do anything. Early results look like a relatively low-magnitude bomb, and I don't think the North Koreans are planning on doing much with it at the moment. If they were to say, smuggle a nuke into Tokyo and set it off, they'd be able to take out a few thousand South Koreans before we bombed them back to the stone age. CNN projects that a war will have about 100,000 casualties.[/quote]

You speak in such an interesting fashion. I intend to reply the whole thing but just can't help quoting this one first "before we bombed them back to the stone age". I'm just surprised you missed the next line "Yeah baby, let's rock & bomb them!". That's quite a spirit in the first place you know? I mean, I still find it hard to understand the state of the mind in which one sits and states 'let's bomb these asians to the stone age". But that surely explains the US being unpopular.

So we have 'go ho-hum' as alternative number one, alternative number two, good ol' rock'n roll. I can perceive distaste for alternative number one, nevertheless, it is valuable maybe the fact that it doesn't implies war? Do you think it is a solution a 100 000 casualties belic 'incident'? I forget you are assuming they'll be giving the world a reason for it, the damn communists targeting Seoul or something. We've seen much already in this millenium and this might be an argument based ultimately on faith but the question I would ask inmediately would be, [i]what for?[/i]. In order to get their country levelled? Relations between Pyonyang and Seoul have seen a slight improvement before with the joint declaration and a reunification of Korea I can only imagine in the long run as good for both sides. So again, what makes you think NK will use its nuclear capability on actual targets? and are you sure they will be the first to strike in the case we're in the brink of war?

In the other hand, I think there must be criticism on this event, and it is not as simple as claiming sovereignity. Countries have indeed the right to develop a peaceful nuclear program but what is the intention of building a nuke? And this should be criticized and dealed with diplomatically. There are enough nukes in the world already, there's no need for more, be NK of the US who builds them. The nuclear bomb was a weapon in whose development historical motives played a significant role. Those motives are gone, in today's world there's no need for nukes, nor for war, for what matters. By withdrawing from the nuclear non proliferation treaty (which by the way was and maybe still is sistematically violated by western countries) and now by detonating a bomb NKorea is clearly stating that they ALSO consider the nuclear bombs a solution. This should be at least a concern for everyone who hopes for, in a few million years or so, a peaceful and balanced world.

Daniel

[quote="ben12345_5"]I'm quite worried about the situation... :maybe:

I fear that Japan will use this event as an excuse to build their own nuclear weapons, leading to a nuclear arms race between Japan and China  :o[/quote]

No, I don't think Japan will be doing any nuclear bomb development any time soon.

[list]$\bullet$Firstly, Japan's constitution prohibits use of its military other than in defense, and a nuclear bomb would not exactly be a defensive use of the military.

$\bullet$Secondly, I perceive Japan's culture to be fairly anti-war, at least among the younger generation. Japan is so used to being subservient to the US, their culture reflects this and the younger Japanese generation (I believe) as a whole would have little inclination to go to war.

$\bullet$Thirdly, because of this anti-war stance, the news that nuclear development was going on would result in loss of power by the ruling political party (which has been in power almost all of Japan's post-WWII history), and another political party would take control. And everyone knows that [i]politicians[/i] are assuredly spineless when it's their job on the line :D.[/list]

I will admit that I'm not Japanese and I haven't experienced Japanese culture firsthand.

[quote="L_Li"][quote="Ignite168"]
Yes, my neighbor definitely has that right to bear arms as long as he wasn't just released from prison for having illegal arms or something.[/quote]

No, but he never really got over that firefight he had w/ your friend and other neighbor a few years ago...

[quote="Ignite168"]Also, your example is bad, my neighbor is under the jursidiction of the US and its laws, however, we don't control North Korea, they are their own country with their own soverignty.[/quote]

your neighbor = NK
you = US

NK not under US jurisdiction, not violating international law (which debatably even exist)
neighbor not under your jurisdiction, not violating national law.

get the analogy right before you post again... or do i have to reduce myself to a humorless brick before we even start debating?[/quote]

I think Ignite is not allowed to keep the gun himself either. And if he has, he can't complain about his neighbour. So if the US agree to hang up their nukes, and Russia, China, India, Pakistan, Iran all such nuclear powers follow suit, sure we shall take them off Seoul as well. No problem. See?

[img]http://www.lokasenna.pe.kr/bb/denahi_spring.jpg[/img]

And also without the aid North Koreans would be starving and CCY I love how you just started saying that Ignite is a racist if you had not noticed Ignite said that the South Koreans were stupid after 2 of them started telling everyone that we should nuke Russia and China off the map because they exist and protest nuking North Korea as that would cause huge enviromental damage and probably kill millions of Chinese South Koreans and Russians from radiation. And the huge amounts of aid were proven to be a lot less then what you keep saying if you had not noticed.

But the thing is, you can't just say that after you see comments that were obviously not serious in their intent.  Look at the context of his first comment
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113976&start=60#648226
and tell me how that is justified.

Cutting off aid is not just to please Kim Jong-Il so he won't attack China, whatever the chances of that may be--it's to slow the flow of N. Korean refugees over the border.  China doesn't want to be responsible for those refugees overflowing into its borders.  But in the end, it's a problem on the N. Korean side--not developing their agriculture properly and/or not distributing this food properly to their people.

[quote="Ars"]That was because the 2 south koreans were saying to nuke North Korea off the map on the previous page. Yes I agree it was a bad thing to say but guess several posts later someone decides to blame the issue on China and Russia (which by the way even the person had no way to connect to this) and the koreans decided that they should be nuked also so I do not agree with what he said but what you are defending is worse[/quote]
dude, are you stupid?
1. treething NEVER says anything serious about these matters.
2. I was being sarcastic

yes but then Ignite is being accused of being a racist as he was also joking but some people did not understand it which makes it interesting as they see you guys joking but accuse Ignite for saying a joke. That is the problem

[quote="Ars"]That is the problem[/quote]
No.
The only problem is that there are about 6 posts in this 6-page-thread seriously concerning with the topic! The rest is a near-to-flame discussion about chinese-japanese conflicts, accusations of being racist, how stupid or not Koreans are, how bad USA are, and whatever offtopic else.

Am I the only one who has this impression¿

[quote="i_like_pie"][quote="ZetaX"]
The only problem is that there are about 6 posts in this 6-page-thread seriously concerning with the topic! The rest is a near-to-flame discussion about chinese-japanese conflicts, accusations of being racist, how stupid or not Koreans are, how bad USA are, and whatever offtopic else.

Am I the only one who has this impression¿[/quote]
You're not alone.

But you forgot to mention Treething suggesting USA nukes Russia... :mad:

[quote="Treething"]well just nuke the bastards, nuke russia and china too if they get pissed or something[/quote][/quote]
and you forgot that treething is never serious on these forums(ie its a dumb joke)

And also I tried to get this back towards the topic in my first post but that entirely failed.

i think that the n. koreans look like nazis
if i were indiana jones i'd be fighting them
really who is the US to tell them what to do?  :?:
NOTE: Funcia is not setting off nukes in north korea if you think this you are among the fake-moustached and obtused-glassesed people who are plotting against him.

[quote="funcia"]i think that the n. koreans look like nazis
[/quote]
I don't care if that's a joke
GO DIE

(wow this thread is so off topic)

My point was NOT that. But I am wrong about it in many ways but right in one way. And in that one way I was talking about the US too. That way is "patriotism". Which I don't really like and is how I say people look like nazis.
The US is acting like it owns the world. which i must point out is not really true.

Yeah I don't want to fight with junggi or anyone. If someone is a seer and can see what they'll do with the nukes-fine. Tell us. But we all are biased in some way. Last post Funcia will make in this thread. Bye!

So long, [b]funcia[/b]! And thanks for all the fish!!!

The title of the thread promises a lot. But the posts disappoint. We should really have more order on ML/AoPS.

On a Romanian forum (there the moderators are oppressive: you can actually get banned for repeated off topic and/or flaming - I guess it's a remnant from the "Communist Era"), I found a similar discussion. One of the users is a fervent conspiracy theorist... and I just can't get enough.

I have one question: Does anyone think that North Koreans are really THAT ignorant? Or you people just like to post (like me)?


MATHCOUNTS
rotation
analytic geometry
geometry
geometric transformation
trigonometry

When the point (4,1) is rotated 90 degrees counterclockwise about the point (1,0), what are the coordinates of the point at which it lands?

I find that because these are Mathcounts, all you need to do is graph it, because it's all lattice points in MC. xD

Anyway, a real solution:
OK think about it this way. Draw the lines $ x\equal{}1$ and $ y\equal{}1$, and the line containing both points (that were given). We form a right triangle here. We know that the two legs are of length 1 and 3. If we rotate the point (4,1), this is essentially rotating the triangle: the legs will stay the same. Because we are just rotating lines perpendicular to the axes, this is not too difficult to do.

OK. Imagine you had 2 coordinate planes on transparent sheets of paper. Align them perfectly one on top of the other. On the top plane, mark the point (4,1). Fix the bottom plane to the table so that it doesn't move. Now put a pin at the point (1,0). Now it is possible to spin the top plane around the point (1,0). Rotate it 90 degrees counterclockwise.

Before the rotation, the point (4,1) one unit higher and 3 units to the right of (1,0). With some careful thinking, you will realize that after the rotation, the point is 3 units higher and one unit to the left of (1,0). That is, directly over the point labeled (0,3) on the bottom, fixed plane.

In general, when you rotate point (a,b) 90 degrees counterclockwise about point (c,d), you get point ( c-b+d , d+a-c ). (or at least I think so... :maybe: )

Thanks math154 and leoxnlin. :)  Leoxlin, the way you explained it really helped me understand.

In a Cartesian plane, any $ (x,y)$ rotated counterclockwise through any angle $ \theta$ about the origin will result in $ (x',y')$...

$ x'\equal{}x\cos(\theta)\minus{}y\sin(\theta)$

$ y'\equal{}x\sin(\theta)\plus{}y\cos(\theta)$

To rotate a point $ (x_1,y_1)$ about $ (x_2,y_2)$ with an angle of $ \theta$...

$ x'\equal{}(x_1\minus{}x_2)\cos(\theta)\minus{}(y_1\minus{}y_2)\sin(\theta)\plus{}x_2$

$ y'\equal{}(x_1\minus{}x_2)\sin(\theta)\plus{}(y_1\minus{}y_2)\cos(\theta)\plus{}y_2$

So the answer is $ (0,3)$.

Sorry if I'm over complicating things.

Yeah, but this is MC forum and we're not supposed to use trig. :noo:  :o

Hey, it doesn't say anywhere in the MathCounts Rules that "Competitors are neither allowed to use Trigonometry to solve the problems nor think about Trigonometry while the contest is in session." In fact, I used Trig for a lot of my MathCounts problems.

Yeah but you're not expected to use trig...there's basically always a better way to do the things without trig.

no problem in mathcounts [b]requires[/b] the use of trigonometry, but you can and in some problems, it may be beneficial.

[quote="herefishyfishy1"]In a Cartesian plane, any $ (x,y)$ rotated counterclockwise through any angle $ \theta$ about the origin will result in $ (x',y')$...

$ x' \equal{} x\cos(\theta) \minus{} y\sin(\theta)$

$ y' \equal{} x\sin(\theta) \plus{} y\cos(\theta)$

To rotate a point $ (x_1,y_1)$ about $ (x_2,y_2)$ with an angle of $ \theta$...

$ x' \equal{} (x_1 \minus{} x_2)\cos(\theta) \minus{} (y_1 \minus{} y_2)\sin(\theta) \plus{} x_2$

$ y' \equal{} (x_1 \minus{} x_2)\sin(\theta) \plus{} (y_1 \minus{} y_2)\cos(\theta) \plus{} y_2$

So the answer is $ (0,3)$.

Sorry if I'm over complicating things.[/quote]
why are the top formula's true?

Consider a rotation of the axes by $ \theta$ degrees. Choose coordinates $ (x,y)$ such that they lie on one axis of the rotation. Basic trig will yield those formulas.

you can also convert to polar and back to show that.

remember that x=rcostheta and y=rsintheta to convert


geometry
3D geometry
sphere
icosahedron
ratio
rectangle
geometry proposed

12 points are arranged on the surface of a sphere with radius 1. Let "a" be the shortest distance between any two of the points. What is the maximum possible value of "a?"

the regular icosahedron inscribed in the sphere has 12 vertices and gives an arrangement with maximal $ a$: $ a$ is the length of an edge of the icosahedron and is easy to compute via straightforward geometry and trig.

The ratio betweem the radius of the sphere and the value "a" can be determined by observing a rectangle inside the icosahedron that connects opposite edges. This rectangle happens to be the golden rectangle (this can be shown by observing pentagons within the icosahedron). From there, the diameter of the sphere is the rectangles diagonal, and the value "a" is it's shorter edge length. "a" can now be solved for using the pythagorean theorem.


Hello,

A friend and I have the next problem: we collected the drafts of some lectures, those have been given by different people, and the author of each lecture typed it. Now, everyone defines different commands, and some of those would overwrite the ones of other people. Is there a way to say to the compiler in $\LaTeXe$ to take the macros on each document separately.

Let me rephrase, because probably it is not totally clear. Suppose that I typed lecture1, and I defined
[code]\newcommand{\R}{\mathbb{R}}
\newcommand{\r}{\rightarrow}
\newcommand{\rr}{\Rightarrow}[/code]
while John typed lecture2, and he defined
[code]\newcommand{\rr}{\mathbb{R}}
\newcommand{\ra}{\Rightarrow}[/code]
the command \rr is rewrited, but I don't want to have to open the file typed by John and "correct" all the \rr to \R, and change al the \ra to \rr. Is there a way to just say something similar to

[code]%
% Something in the preamble
%
\begin{document}

\include{lecture1}

\include{lecture2}

\end{document}[/code]

and that the compiler understands that it has to take the macros in lecture1 just for lecture1, and the ones in lecture2 just for lecture2?

It should be possible, because the journals usually receive their articles (I suppose) typed with the personal commands of each author defined on the preamble of the article, and they compile all of them together!

thanks :first:

[quote="DPatrick"]It seems to me that the simplest solution is to change all subsequent redefinitions of the same command to \renewcommand rather than \newcommand.  For instance, in your example above, all you would need to do is to change the \newcommand{\rr}{...} line in Lecture 2 to \renewcommand{\rr}{...}, and it should work fine.[/quote]Thank you! I will try that!


calculus
integration
number theory
relatively prime

Transform the equation $ x^{5}\plus{}25x^{4}\plus{}250x^{3}\plus{}1250x^{2}\plus{}3125x\plus{}3125\equal{}0$, so that its roots may be increased by $ 5\frac{1}{2}$. Find the sum of the numerical coefficients of the third degree term and the 1st degree term if the transformed equation has integral coefficients.
A) 70
B) 79
C) 81
D) 90
E) 89

$ (x\plus{}5)^5\equal{}0$

All roots are equal to -5

$ \minus{}5\plus{}5\frac{1}{2}\equal{}\frac{1}{2}$

$ (x\minus{}\frac{1}{2})^5\equal{}x^5\minus{}\frac{5}{2}x^4\plus{}\frac{5}{2}x^3\minus{}\frac{5}{4}x^2\plus{}\frac{5}{16}x\minus{}\frac{1}{32}$

I can't finish this problem, cause I just don't know what to do.

Just multiply it by 32 to get rid of the fractions:

$ 32(x \minus{} \frac12)^5 \equal{} 32x^5\minus{}80x^4\plus{}80x^3\minus{}40x^2\plus{}10x\minus{}1$

$ 10 \plus{} 80 \equal{} 90$.

So "intergral coefficients" means that all coefficients are relatively prime integers? :)

"integral" means that the coefficients are integers, not necessarily relatively prime. I picked the relatively prime solution because all the others would exceed the alternatives given (i.e. they would be higher than 90).


Let
\[ \mathsf{A} \equal{} \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \textup{ and } \mathsf{P} \equal{} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\]Do you expect $ \mathsf{PAP}^{\minus{}1}$ to be symmetric? Why?

$ P$ is not an orthogonal matrix, so at first sight there is no reason to expect $ B \equal{} PAP^{\minus{}1}$ to be symmetric. What you can say at a glance is that $ B$ is lower triangular.

So in general, for $ B \equal{} PAP^{\minus{}1}$ to be symmetric it is required that $ P$ is orthogonal?

Assume $ A\equal{}diag(\lambda_i)\in\mathcal{M}_n(\mathbb{R})$ where the $ \lambda_i$'s are pairwise distinct. Let $ P$ be an invertible matrix s.t. $ S\equal{}PAP^{\minus{}1}$ is symmetric. Then $ P\equal{}OD$ where $ O$ is an orthogonal matrix and $ D$ is a diagonal matrix with positive entries. Conversely these matrices $ P$ work.
Proof: Let $ (e_i)$ be the canonical basis of $ \mathbb{R}^n$ (it's an orthonormal basis). Then $ (P(e_i))$ is a basis of eigenvectors of $ S$. The eigenvalues of $ S$ are pairwise distinct. Hence $ (P(e_i))$ is an orthogonal basis. Thus the columns of $ P$ are pairwise orthogonal and we are done. The converse is easy.


Three ants 1,2,3 are placed at the three vertices of an equilateral triangle ABC of side length 'a' . They start moving towards each other with a velocity 'v' each. The three ants meet at the centroid of the triangle after a time of t= 2a/3v.

Find the Angular Velocity and the Angular Acceleration of ant 2 with respect to ant 1 at time t=0.

[quote="Anvay"]YEH TO HC VERMA KA SOLVED PROBLEM HAI[/quote]

Translates to : THIS IS HC VERMA'S SOLVED PROBLEM!

[quote="aadil"]anvay hindi mein baat muth karo.kuch logon ko hindi samajh mein nahin aaegi.[/quote]

Same goes for you!  :P

.

Look at the full question man!
I have asked about the relative angular acceleration, No where in H C Verma is it solved!
[b]Whatever is solved there, I have already supplied in the question![/b]


conics
ellipse
analytic geometry
geometry unsolved
geometry

recently i have came across this question which i dun think i can solve,
If the circle $ x^2\plus{}y^2\minus{}ax\equal{}0$ meet the ellipse $ \frac{x^2}{a^2}\plus{}\frac{y^2}{b^2}\equal{}1$ at three points, prove the the x coordinate of the centroid of the three points is $ a(a^2\plus{}b^2)/3(a^2\minus{}b^2)$

I am enclosing a pdf file with the solution of this problem. It is not so hard as is said in the ttle...


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Find out all X in M_2(R) so that we have X^2004-11X^2003+30X^2002=O_2.

X has eigenvalues 0,5 or 6.
So X is similar to a matrix A (X=TAT<sup>-1</sup> for some invertible X) from the set:
0, 5I, 6I, 6I+U, 5I+U, U

0 0
0 5

0 0
0 6

5 0
0 6

U is the matrix:
0 1
0 0

I have another solution, lets see if u find it! :D :D

the matrix has to have the proper values 0, 5 or 6, but 5 and 6 can appear at most one time.

so basically the char. polynomial of X can only be x^2, x(x-5), x(x-6) or (x-5)(x-6).

so the solutions of this eq. are exactly those matrices in M_2(R) which have the char. polynomial one of the 4 mentioned above.

they are fully (well almost :D) described by two elements from the first line.


[hide="solution"] Clearly $ \frac{1}{3^2}\plus{}\frac{1}{4^2}\plus{}...\plus{}\frac{1}{n^2}<\frac{1}{4}\plus{}\frac{1}{8}\plus{}\frac{1}{16}\plus{}...\frac{1}{2^n}<\frac{1}{4}\plus{}\frac{1}{8}\plus{}\frac{1}{16}\plus{}...$, where the last sum continues to infinity. This is true firstly because each of the corresponding terms in the first are smaller than the corresponding in the second and secondly because the third is an infinite series, not a finite.

The sum of the infinite series $ \frac{1}{4}\plus{}\frac{1}{8}\plus{}...$ is $ \frac{1}{2}$. Thus, $ \frac{1}{1^2}\plus{}\frac{1}{2^2}\plus{}\frac{1}{3^2}\plus{}...\plus{}\frac{1}{n^2}<\frac{1}{1^2}\plus{}\frac{1}{2^2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}\plus{}....\equal{}\frac{5}{4}\plus{}\frac{1}{2}\equal{}\frac{7}{4}<\frac{41}{24}$.[/hide]

[quote="xpmath"]Clearly $ \frac {1}{3^2} \plus{} \frac {1}{4^2} \plus{} ... \plus{} \frac {1}{n^2} < \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} ...\frac {1}{2^n} < \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} ...$, where the last sum continues to infinity. This is true firstly because each of the corresponding terms in the first are smaller than the corresponding in the second and secondly because the third is an infinite series, not a finite.

The sum of the infinite series $ \frac {1}{4} \plus{} \frac {1}{8} \plus{} ...$ is $ \frac {1}{2}$. Thus, $ \frac {1}{1^2} \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ... \plus{} \frac {1}{n^2} < \frac {1}{1^2} \plus{} \frac {1}{2^2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} .... \equal{} \frac {5}{4} \plus{} \frac {1}{2} \equal{} \frac {7}{4} < \frac {41}{24}$.[/hide][/quote]

No!  No!

$ \frac{7}{4} \equal{} \frac{42}{24} > \frac{41}{24}$

Whoops. Anyway, just use $ \frac{1}{9}\plus{}\frac{1}{27}\plus{}...$ in place and add that to $ \frac{1}{1^2}\plus{}\frac{1}{2^2}\plus{}\frac{1}{3^2}$ with the same argument.

the series $ \frac1{1^2}\plus{}\frac1{2^2}\plus{}\frac1{3^2}\plus{}...$ converges to $ \frac{\pi^{2}}{6}$

[quote="xpmath"]Whoops. Anyway, just use $ \frac {1}{9} \plus{} \frac {1}{27} \plus{} ...$ in place and add that to $ \frac {1}{1^2} \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2}$ with the same argument.[/quote]

So you believe that $ \frac{1}{n^2} < \frac{1}{3^n}$ for all $ n > 3$?

Ok disregard everything I said. I wasn't thinking at all  :wallbash: . Sorry about that.