Sonia Gandhi, a lady who is Italian, didn't attempt to learn Hindi until six years ago, is still "uncomfortable" with it, and, is ITALIAN and only related to Indian politics by virtue of her former husband is set to become the Prime Minister of India! I just can't understand why so many real Indian senior politicians, and basically, the entire Indian race would allow the country to return to being led by a Westerner, independence or not. Geez, I just....this makes me so angry. I shudder when I think about what will happen to the country under her leadership.

[quote]Sonia Gandhi, a lady who is Italian, didn't attempt to learn Hindi until six years ago, is still "uncomfortable" with it, and, is ITALIAN and only related to Indian politics by virtue of her former husband is set to become the Prime Minister of India![/quote]

The fact that she is Italian doesn't mean she can't be a good politician. Okay, her Hindi-speaking is very limited... Thats OK, Bush can't speak good english and plenty of people voted for him! Seriously though, I understand your complaint as to the fact that she is not politically experienced, but so what if shes Italian? Maybe one day I'm going to run for political office in the US, and I wasn't born here. I don't think that its fair to criticize someone just on the basis of nationality.

[quote]I just can't understand why so many real Indian senior politicians, and basically, the entire Indian race would allow the country to return to being led by a Westerner, independence or not.[/quote]

Well, in a democratic election, the senior Indian politicians have really very little power. As for the entire Indian race, probably a good amount of voters were not very informed about the candidates... this is something that's very difficult to change. And again, so what if she's foreign? I mean, it doesn't mean she won't do a good job. Plus, she's not subjugating all of India, and she's not a supreme ruler. So this isn't a return to colonial India.

[quote]Geez, I just....this makes me so angry. I shudder when I think about what will happen to the country under her leadership.[/quote]

Is that due to the fact that she is Italian or to the fact that she is politically inexperienced?

tetrahedr0n wrote:Quote:Geez, I just....this makes me so angry. I shudder when I think about what will happen to the country under her leadership.

Is that due to the fact that she is Italian or to the fact that she is politically inexperienced?

I have to admit this struck me, too.  Nothing else you said in your post seems like a good basis for this particular sentiment.  Personally, I think getting the government out of the hands of a Hindu nationalist party is probably a good thing, although I hope the change of government doesn't mess up the India-Pakistan peace process [hide]... as if something else won't.[/hide]

[quote="tetrahedr0n"]Thats OK, Bush can't speak good english and plenty of people voted for him![/quote]

And he's done wonderfully!

India is definitely not a country of immigrants, like America. About her experience, she may have been a legislator for a while, but I doubt it's because of her plans for the people. Her connections and her name were probably the deciding factor that helped her gain that experience. Just take a look at all of those actors and actresses that have gained seats in the Lok Sabha. What do you think got them elected? Returning to Sonia Gandhi, at her website, SoniaGandhi.org, it states that control of the Congress Party was offered to her on May 21, 1991, after her husband was killed by a suicide bomber. Why? Earlier, the biography said that she was disgusted with politics. So someone who has no experience in it, and even hates it, should be offered the top job over more seasoned others? In another article on that site, it states that she only truly got involved with the Congress Party in 1998. Is eight years experience enough?

[quote]And he's done wonderfully![/quote]

You and I can argue that later.

[quote]India is definitely not a country of immigrants, like America.[/quote]

What do you mean by this? Sure, there isn't a significant amount of immigration, but there are immigrants. And the fact that there isn't a big immigrant base doesn't mean that people who aren't natives cannot be voted into office. Leadership positions aren't accorded by racial/nationalilty majority.

[quote]About her experience, she may have been a legislator for a while, but I doubt it's because of her plans for the people. Her connections and her name were probably the deciding factor that helped her gain that experience.[/quote]

How many politicians have really gotten places just by doing things for the people? A large amount of Indian politicians have gotten to their respective positions by connections or even corruption. And one gains experience in office, regardless of how one got there.

[quote]Returning to Sonia Gandhi, at her website, SoniaGandhi.org, it states that control of the Congress Party was offered to her on May 21, 1991, after her husband was killed by a suicide bomber. Why? Earlier, the biography said that she was disgusted with politics. So someone who has no experience in it, and even hates it, should be offered the top job over more seasoned others?[/quote]

It frequently happens that after the loss of a major political leader, his spouse can take his place. It happened to a senator in the US, whose name evades me. The results were not bad. Besides, politics don't just go along the lines of experience or time in office. Otherwise, Strom Thurmond would have been US president a while ago.

[quote]In another article on that site, it states that she only truly got involved with the Congress Party in 1998. Is eight years experience enough?[/quote]

For the right person, 8 years is plenty. I mean, whats to say that after 8 years she can't do the job plenty well?

[quote="tetrahedr0n"][quote]
How many politicians have really gotten places just by doing things for the people? A large amount of Indian politicians have gotten to their respective positions by connections or even corruption. And one gains experience in office, regardless of how one got there.[/quote][/quote]

yes, i agree with you on this one. but then again, look at the state of affairs in india. how are things ever going to change for the hundreds of millions of desperately poor if politician after politician are only elected for their name instead of for their deeds? many, many elected officials there are corrupt, and only run for the prestige of the office.

[quote="polymorphic"]in the US at least, i thought you had to be born in the US to run for president.[/quote]

The [url=http://www.law.cornell.edu/constitution/constitution.articleii.html#section1]exact Constitutional language[/url] is a "natural born citizen," that is, someone whose citizenship derives from facts of birth rather than from naturalization. I am sure that my third son, who was born in [url=http://www.taipeitimes.com/News/edit/archives/2004/05/22/2003156522]Taiwan[/url] but has no claim to any national citizenship but citizenship in the United States (by virtue of having two American parents), would be an eligible candidate for the presidency once he reaches the age of thirty-five and accumulates fourteen years of residence in the United States. It is interesting that Sonia Gandhi thus achieves an office analogous in India to an office she could never achieve in the United States.

There are plenty of members of the United States Senate and at least some state governors from time to time who are first-generation immigrants, so the citizenship from birth qualification is narrowly applied in the United States.

The meaning of "Natural Born Citizen" is vague. No one really knows what it means. So far there's never been a challenge to it, but it's just a matter of time until someone does and then all you-know-what is going to break loose.

[quote="BHorseMath"]hmm.  TA I was born in Germany and my parents looked this up just to see and they talked to several people here and it stands that I can not run for president[/quote]

Who told you that? Did any of the people who told you that have legal training, or cite a court precedent to that effect? I'm pretty sure that this is an entirely unlitigated issue, but I have no doubt how the litigation would turn out if a candidate came forward claiming eligibility to run even though born overseas.

Yes TA, several of the people we talked to did in fact have lega training.

One of the few things I learned in my government class in my senior year of high school was that you don't actually have to be born in the US to become president and that BHorseMath could in theory become president.

hmm
interesting.  I have been told countless times that if I wished to become president that I would have to fight to be allowed.

[quote="BHorseMath"]I have been told countless times that if I wished to become president that I would have to fight to be allowed.[/quote]

My contention is that you would win that fight. As a lawyer who used to work in an appellate court, I feel pretty confident that you could present a winning case if indeed you meet the other Constitutional requirements of age above 35 and fourteen years of residence in the United States, and you can demonstrate that you have no claim to any nationality but United States nationality from the circumstances of your birth. I think that would be enough, especially because the [url=http://www.law.cornell.edu/constitution/constitution.amendmentxiv.html]fourteenth amendment[/url], which made clear the citizenship of black people born in the United States, uses a different phrasing that specifically refers to location of birth.

Hey, since you're older than my third son, how about you be the test case (in which case I would be glad to be your lawyer  :)  ) and then we can see what happens?

This is of course mostly unrelated, but it is worth pointing out that our first few presidents were most certainly not born in the United States of America.  Van Buren was the first president born after 1776.

[quote="JBL"]This is of course mostly unrelated, but it is worth pointing out that our first few presidents were most certainly not born in the United States of America.[/quote]

Aw, JBL, I thought you would check the citation to the Constitution I provided. The Constitutional language is "No person except a natural born citizen, or a citizen of the United States, at the time of the adoption of this Constitution, shall be eligible to the office of President," and that makes clear how the earlier-born Patriots were treated, doesn't it? British subjects who joined in the revolution were American citizens once they established a new country.

(After edit:) Doesn't everyone read the Constitution in childhood? It's in every copy of the World Almanac, and I had read it through at least twice by the time I was finished with junior high. Actually, I already know the answer to my rhetorical question, as I had law school classmates tell me when we graduated that they had never read the Constitution even once  :cry:

[quote="tokenadult"]as I had law school classmates tell me when we graduated that they had never read the Constitution even once  :cry:[/quote]

Would love to know how many members of Congress can honestly say the same, sigh.

I actually remembered that such language existed, although I didn't go read the relevant passages.  We read the constitution in my government class, but I was already reasonably familiar with it via my parents, the newspaper, and having done debate-type-activities.

I actually have several classmates who were born outside of the country (under different circumstances) and my school's Quiz Bowl team spent about half a practice meeting one time discussing who could run for president and who couldn't. It was actually pretty funny...

I'm not sure exactly what sources were used, but apparently, my friend who was born to American citizens outside of the country could run for president (once she reaches 35, that is) but my other friend (who wasn't born to American citizens) can't. According to those sources, anyway. But then again, I'm no law professor.

[quote="tetrahedr0n"]You and I can argue that later.[/quote]

It was sarcasm, don't worry.

Oh yeah, the 9-11 of the Indian economy occured yesterday when Sonia Gandhi's Prime Minister-ship became official.

Ah yes, updates!

A case in the Supreme Court has been requested by two lawyers from Lucknow, stating that Sonia Gandhi becoming Prime Minister was a threat to India's independence and security, because of her foreign origins. Also, they claim that her citizenship, which she received in 1983, is void, since she didn't use her real name on her citizenship application, and also because at that time one could not become a citizen by marrying an Indian, which she claimed as her right to citizenship. Her husband changed the Indian Citizenship Act to allow that, but that addition became law in 1987!

Also, Vajpayee's Labor Minister started an Independent Indian Army demonstration in the capital.

[quote="BHorseMath"]hmm.  TA I was born in Germany and my parents looked this up just to see and they talked to several people here and it stands that I can not run for president even though I am an American Citizen and never have been any other nationality.  I was even born of a military base over seas, but according to the constitution that doesn't matter I can not run for president (not that I have any intention of doing so ever) as it stands now.[/quote]

um u don't have to be president. We had two awesome jewish (german, and czech) secretary of states (which i think should be third in power in the executive branch). Kissinger and Albright. and Albright is female and democrat which makes her twice as awesome as kissiger

[quote="Scrambled"]Sonia Ghandi is no longer running to be the prime minister.

On a side note, who the heck is zhu yuan chen?[/quote]

Uh that is a catchy thing so u would click on the link to johnkerry.com which u probably did. HEHE! Thanks to Osrisis who taught me how to url links.


inequalities
geometry
circumcircle
geometry proposed

Given triangle $ABC$ centroid $G, R_{a},R_{b},R_{c},R$ are circumradius of triangle $GBC,GCA,GAB,ABC$ resp prove that
\[R_{a}+R_{b}+R_{c}\ge 3R\].

[quote="N.T.TUAN"]This is equivalent to $\sum\frac{m_{a}m_{b}}{ab}\geq \frac{9}{4}$ and it is posted before![/quote]
That's right it is the thing that anybody can see and show but do you have solution for that inequality, can you show the link or can you post solution?

[color=purple] The similar problem for inceter denote element similarly ( $R'_{a}$ for triangle IBC) we have $R'_{a}+R'_{b}+R'_{c}\le 3R$  :D and from two problem we had$R_{a}+R_{b}+R_{c}\ge R'_{a}+R'_{b}+R'_{c}$ :lol:
 [/color]

[quote="N.T.TUAN"]I can not find it now. But by Ptolemy's inequality we can solve following problem
For all points $P$ in plane of a triangle $ABC$ we have $a\cdot PB\cdot PC+b\cdot PC\cdot PA+c\cdot PA\cdot PB\geq abc$.
This problem is from China National Olympiad.  :wink:[/quote]
Really by Ptolemy can you show clearly  :D


[quote="0123456789"]Can someone tell me if the intermediate Algebra Book is more focused on competition math or school math? In addition, can someone tell me the approximate difficulty of the text...perhaps someone can give me a few sample question to look at?

Thanks in advance :)[/quote]

It focuses on both.  We believe the problem solving like that you see in contests should be the way to teach much of what you think of as 'school mathematics'.

There are tons of problems from various contests in the book.  You can take a look at some excerpts [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=300]here[/url].  You'll see plenty of both 'school math' and 'contest math'.


How long did it take everyone here to finish AOPS I? I'm currently still working on it; I started about a month ago, and I'm about half way through. Is that about right? I only spend about 30 minutes a day. Am I taking too long?

There's no need to rush. You should spend as long on it as it takes you to feel confident using all the techniques presented. What I like to do when I go through problem-solving books is to go through it really fast at first (spend 5 hours a day and go through it in 2 or 3 days for example) and then go back through it and pick up the finer details that I missed the first time around. I don't know if that works for everyone, but it works for me.

AOPS I took me 2 or 3 weeks to finish. My method of studying was just to go through the thing and do every problem. I was spending between two and four hours a day during that period of time. Also I didn't read it until I'd already been in competition for three years and was familiar with many of the concepts.

You're not taking too long... two months to finish AoPS is definitely not bad, especially on 30 minutes a day. The most important part is to make sure you understand every concept and know how/when to apply them.

eh...she's right...if it makes you feel any better, I've been on book I for the last...*counts*...10 months...on and off of course (mostly off) but still...as long as you understand everything (and dont forget 93.72% of it like i do) ur doing fine

Yah. your doing it at a good pace.
I was careful not to do too much at once, since I might forget some important stuff. I took like 5-6-7 months...

im still working on it :lol:

dont worry, just like, finish it sometime, and remember whats there. also try to refresh on the previous chapters. something i should do also

It would help if I'm not stuck on some chapter in book 2...
I keep getting interrupted... maybe I should just put a sign on my door when I do AOPS.

yeah, im stuck at a chapter also. not that its hard, but boring. i should skip it, and come back. that way, i wouldve gotten more done. oh well, ill trot through it somtime

eh...im taking a 7-month break from AoPS 2...

lol, hmmmm...i suggest you dont, but meh

what grade are you in??? but it doesn't matter since MC and AMC and that kinda stuff are all next year... a 2 week break would be nice

lol...i mean...i know i should continue AoPS...but i just dont have the time right now...

by 7-month break i dont mean i havent looked at it once...i just havent worked a lot in it...


logarithms
Euler
number theory
prime numbers
number theory unsolved

prove that $\sum \frac{1}{p}$ is unbounded.
$p$ is a prime.
I think it is a nice and challenging problem.Anybody here want to solve it?

This is really well known, I think. First show that $\prod \left(\frac 1{1-\frac 1p}\right)$ is unbounded (it's $1+\frac 12+\ldots$), and then take the logarithm from that. You get $\sum\ln\left(1-\frac 1p\right)=-\infty$. Now compare $-\frac 1p$ to $\ln\left(1-\frac 1p\right)$.

[quote="grobber"] Now compare $-\frac 1p$ to $\ln\left(1-\frac 1p\right)$.[/quote]
very very nice .but can you  explain for this in detail.
and I think it can be proved that $\sum \frac{1}{p}$   is  equal to $lnlnN$.
can you kindly prove it in detail.

Suppose that:
\[
\sum\nolimits_p {\frac{1}{p}}  = A \prec  + \infty
\]
But:
\[
\begin{array}{l}
 e^x  = 1 + x + \frac{{x^2 }}{{2!}} + ......... \\
 e^x  \ge 1 + x \\
 \end{array}
\]
Now let
\[
n \ge 2
\]
be any positive integer. Then
\[
e^A  = e^{1/2 + 1/3 + 1/5 + ... + 1/q}  = \prod\limits_{p \le n} {e^{1/p} }  \ge \prod\limits_{p \le n} {\left( {1 + \frac{1}{p}} \right)}
\]
\[
 = \left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right)...........\left( {1 + \frac{1}{q}} \right) \ge \sum\limits_{k \le n} {\frac{1}{k}}
\]
We then conclude that for any
\[
n \ge 2
\]
then

\[
\sum\limits_{k \le n} {\frac{1}{k} \le e^A  \prec }  + \infty
\]
A contraddiction because this series diverge.

[quote="Nené"]

$\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right)...........\left( {1 + \frac{1}{q}} \right) \ge \sum\limits_{k \le n} {\frac{1}{k}}$
[/quote]

Are you sure about this? You only get the square-free numbers which are less than $n$, not all numbers $\le n$. What I had in mind is very similar, except that you use the entire $\prod_p\left(1+\frac 1p+\frac 1{p^2}+\ldots\right)$ instead of just $\prod_p\left(1+\frac 1p\right)$. I mean, in your LHS you only have prime numbers, right?

Anyway, when I said "compare $-\frac 1p$ to $\ln\left(1-\frac 1p\right)$" I meant using something like $-\frac 1p<\left(1-\frac 1p\right)\cdot \ln\left(1-\frac 1p\right)$. From here and from the fact that $\sum\ln\left(1-\frac 1p\right)=-\infty$, we easily get $-\sum\frac 1p=-\infty$, which is what we want.

OK Grobber. With

\[
\sum\limits_{K \le n} {\frac{1}{k}}
\]

I mean the sum of the reciprocals of all square free integers less than or equal to n(including 1). This series diverge

Yes, it does, but in this case we have to use the extra result that that sum diverges :). Why do that when we can avoid it?

$grobber$,you are right.And I think there is a small mistake in $Nené$'s solution.can you try to correct it?

First, note that \[(1+ \frac{1}{p})(1 + \frac{1}{p^2} + \frac{1}{p^4} + \ldots + \frac{1}{p^{2k}}) = 1 + \frac{1}{p} + \frac{1}{p^2} + \ldots + \frac{1}{p^{2k+1}}.\] Let $p_i$ denote the $i$-th prime & let $S_k$ be the set of all positive integers whose prime factors are $\leq p_k$. It follows tt \[(1+ \frac{1}{p_1})(1+ \frac{1}{p_2}) \dots (1+ \frac{1}{p_k}) \sum_{n \in S_k} \frac{1}{n^2} = \sum_{n \in S_k} \frac{1}{n}.\] Since $\sum_{n \in S_k} \frac{1}{n^2}$ converges to $\frac {\pi^2}{6}$ and $\sum_{n \in S_k} \frac{1}{n}$ diverges as $k \to \infty$, we thus have $(1+ \frac{1}{p_1})(1+ \frac{1}{p_2}) \dots (1+ \frac{1}{p_k}) \to \infty$ as $k \to \infty$.
Now we know tt $\forall x > 0, e^x > 1 + x$. Taking $x = \frac{1}{p_i}$, we have \[e^{(\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \ldots +\frac{1}{p_k})} > (1+ \frac{1}{p_1})(1+ \frac{1}{p_2}) \dots (1+ \frac{1}{p_k}).\] Thus letting $k$ tend towards $\infty$, we see that $\sum_{p \text{ prime}} \frac{1}{p}$ diverges and we are done.

[quote="dblues"]$S_k$ be the set of all positive integers whose prime factors are $\leq p_k$. It follows tt \[(1+ \frac{1}{p_1})(1+ \frac{1}{p_2}) \dots (1+ \frac{1}{p_k}) \sum_{n \in S_k} \frac{1}{n^2} = \sum_{n \in S_k} \frac{1}{n}.\]$ [/quote]
explain it better,please.thank you inadvance

Well... $\prod_{i = 1}^{k} (1 + \frac{1}{{p_i}^2} + \frac{1}{{p_i}^4} + \ldots)$ when expanded out, is a sum of terms of the form $\frac{1}{r^2}$ where $r \in S_k$. This is because for each $r^2$, its canonical representation consists of primes $p_j$ such that $1 \leq j \leq k$ (by definition, as $r \in S_k$), and these primes $p_j$ have even exponents in the canonical representation of $r^2$, so each such $\frac {1}{r^2}$ term corresponds to each term in the expansion of $\prod_{i = 1}^{k} (1 + \frac{1}{{p_i}^2} + \frac{1}{{p_i}^4} + \ldots)$. Thus, we have \[\prod_{i = 1}^{k} (1 + \frac{1}{{p_i}^2} + \frac{1}{{p_i}^4} + \ldots) = \sum_{n \in S_k} \frac{1}{n^2}.\] Similarly, we can easily see that $\prod_{i = 1}^k (1 +  \frac{1}{p_i} + \frac{1}{{p_i}^2} + \frac{1}{{p_i}^3} + \ldots) = \sum_{n \in S_k} \frac{1}{n}.$ (This is called Euler Product I think...)

Thus, we use $(1+ \frac{1}{p})(1 + \frac{1}{p^2} + \frac{1}{p^4} + \ldots + \frac{1}{p^{2k}}) = 1 + \frac{1}{p} + \frac{1}{p^2} + \ldots + \frac{1}{p^{2k+1}}$ by letting $k$ tends towards $\infty$, and by ranging $p$ from $p_1$ to $p_k$ and considering its product, we get $(1+ \frac{1}{p_1})(1+ \frac{1}{p_2}) \dots (1+ \frac{1}{p_k}) \sum_{n \in S_k} \frac{1}{n^2} = \sum_{n \in S_k} \frac{1}{n}.$


Study the simple and uniform convergence for:

$f_{n}: [-1,1] \to R$, $\displaystyle f_{n}(x) = \frac{x}{1+n^{2}x^{2}}$ where $n \in N^{*}$ and $x \in [-1,1]$.

The usage I'm used would use the word "pointwise" instead of "simple."

If I were going to use this as a homework problem in the course in which we introduce the notion of uniform convergence, I would also have the sequence

$g_n(x)=\frac{nx}{1+n^2x^2}$ for $x\in[-1,1].$

Obviously it converges [i]simply[/i] toward $0$ and furthermore $|f_n(x,n)| \leq \frac{1}{2\,n}$

And $g_n(1/n) = 1/2$ is a major problem for uniform convergence  :lol:

Pointwise convergence; simple convergence. Is that another one of our American usage/European usage questions, or does it split along some other axis? (tµtµ doesn't give a location in his profile.)

[quote="tµtµ"]And g_n(1/n) = 1/2 is a major problem for uniform convergence[/quote]
Indeed it is! Although $g_n$ also tends simply/pointwise/pick your terminology to zero.

At this point we can simply way that the usage does not converge uniformly  ;)


Welcome to the first AoPS Newbie Game!

The once quiet town of Placid Hills has been recently struck by a terror. The rumors of a mafia infestation have been confirmed, and 9 citizens gather in a last ditch effort to save the town. But two among them work for the other side...

Alive:
2. AceOfDiamonds
3. math_explorer (replacing 15!!!)
5. lotsofmath
6. percy314
7. Royalflower
8. isabella2296 (IC)
9. pacman2812 (IC)

Dead:
1. [color=green]monkeygirl13 (replacing 2lnx), Townie, Lynched Day 1[/color]
4. [color=green]randomguy64, Townie, killed Night 1[/color]

Current Game Status: Day 2!

Previous Days:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319081&start=280]Day 1 Ends on Post 297/298[/url]

The [url=http://mafiascum.net/wiki/index.php?title=F11]F11[/url] setup is the current game format used in Newbie games.  To prevent the development of game-breaking strategies, these Newbie Games are Semi-Open, wherein the actual game setup is randomly chosen from one of four possible setups as detailed below:

[list][*]1 Mafia Goon, 1 Mafia Roleblocker, 1 Sane Cop, 1 Sane Doctor, 5 Townies
[*]1 Mafia Goon, 1 Mafia Roleblocker, 7 Townies
[*]2 Mafia Goons, 1 Sane Cop, 6 Townies
[*]2 Mafia Goons, 1 Sane Doctor, 6 Townies[/list][color=red][b]If either of the first two setups are in play, and the situation occurs where the Goon is lynched first, the Roleblocker will still be able to submit both night kills and roleblocks (roleblocking is a function of the Roleblocker; night kills are a function of the scum team as a whole).[/b][/color]

These are the Role PMs that have potentially been sent to the players:

[b][u]Mafia-Aligned players:[/u][/b]

[quote="Mod"]
[b]Mafia Goon:[/b]

You are a member of the Mafia along with your partner, [b]name[/b]. Since your plot to quietly overtake the town has failed, you will attempt take it by force by eliminating the others until you outnumber the populace.
[list][*]During the day, try to blend in with the normal Townies, and attempt to get someone lynched.
[*]During the NIGHT cycle ONLY, you may talk with your fellow Mafia (via Quicktopic) and choose another player in the town to kill that night.
[*]You may communicate privately with your fellow Mafia up until the game thread is opened.  Once the game begins, all outside communication must cease until the night cycle.
[*]You win when the number of Mafia equals or exceeds the number of remaining Town players, whether you survive to the end or not.[/list][/quote]

[quote="Mod"]
[b]Mafia Roleblocker:[/b]

You are a member of the Mafia along with your partner, [b]name[/b]. Since your plot to quietly overtake the town has failed, you will attempt take it by force by eliminating the others until you outnumber the populace.
[list][*]During the day, try to blend in with the normal Townies, and attempt to get someone lynched.
[*]As a Roleblocker, you have the ability to distract any player throughout the night, effectively negating their night ability, if they have one.
[*]Send the moderator a PM with your choice of player that you want to block.  If they have a Night action, it will have no effect.  You will not be informed as to whether your roleblock worked or not.
[*]During the NIGHT cycle ONLY, you may talk with your fellow Mafia (via Quicktopic) and choose another player in the town to kill that night.
[*]You may communicate privately with your fellow Mafia up until the game thread is opened.  Once the game begins, all outside communication must cease until the night cycle.
[*]You win when the number of Mafia equals or exceeds the number of remaining Town players, whether you survive to the end or not.[/list][/quote]

[b][u]Town-Aligned players:[/u][/b]

[quote="Mod"]
[b]Townie:[/b]

You are a normal townsperson, trying to make a living and survive this insanity.
[list][*]You have no abilities at Night other than getting a good night's rest.
[*]Although you do not have any special abilities, your voice and vote are powerful weapons in their own right.  Use them to your best advantage!
[*]At no time may you privately communicate with any other player.
[*]You win when all the Mafia players are gone, whether you survive to the end or not.[/list][/quote]

[quote="Mod"]
[b]Cop:[/b]

You are the local law enforcement, and can tell the good guys from the bad.
[list][*]Each NIGHT, you may send the moderator a PM with your choice of player that you want to investigate.
[*]The Moderator will inform you as to that person’s guilt or innocence.
[*]At no time may you privately communicate with any other player.
[*]In a Newbie game, you will always get a correct result (no sanity issues).
[*]You win when all the Mafia players are gone, whether you survive to the end or not.[/list][/quote]

[quote="Mod"]
[b]Doctor:[/b]

You are the local physician, and can protect people from harm.
[list][*]Each NIGHT, you may send the moderator a PM with your choice of player that you want to protect.
[*]If someone attempts to kill that player during the Night, they will be spared.
[*]You may not protect yourself.
[*]At no time may you privately communicate with any other player.
[*]In a Newbie game, your protection target will always live through the night (no quacks).
[*]You win when all the Mafia players are gone, whether you survive to the end or not.[/list][/quote]

And finally:

[size=150][b][u]The Rules (please read!)[/u][/b][/size]:[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*]Please be attentive and [b]unvote[/b], if necessary, before casting a new vote.  This is not [i]required[/i], but I’d appreciate it.
[*]Executions will require a simple majority of votes.  Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may [b]vote: no lynch[/b] – majority votes of this kind are necessary to end the day without a death.
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
[*]Once your death scene has been posted, you’re dead.  Stop typing.
[*]Don’t edit/delete previously submitted posts
[*]Don’t quote any PMs from me.
[*]If you have a night choice to make, it is due by the posted deadline.  I will not wait for you – if you do not submit a choice to me, tough!  No choice will be made.
[*]To make sure you read the rules, please make the first character of your first post in this thread the same as the second character in your username.
[*]Communication outside of the thread, unless explicitly stated in your role PM, is prohibited. This is a modkillable and bannable offense. Talking after death, and discussing games that are ongoing, even if you are dead, are also prohibited, and bannable/modkillable offenses.
[*]Editing or deleting posts once they have been entered is also bannable/modkillable, even if no one has posted after you. If you need to make an edit, post again. Like, seriously.
[*][b]If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread.[/b]  If for a 72-hour period, you make no posts without first informing me of your absence, you will be sent a prod through PM. You will have 48 hours to respond to the prod in-thread, or be replaced. I will give a maximum of two prods; the third time you will be replaced.
[*]Remember, it’s a game.  It’s supposed to played.  It’s supposed to be [i]fun[/i].  Do your part.[/list]

I confirm.

[b]vote: AceOfDiamonds[/b] for posting a message that consists only of "confirm" and an empty bold tag.

No need for such cautions Ace :wink:
Confirmation Stage is often very fun, but after all no votes there count (well... that depends on the mod).
Now please start this, Brute :)

Let the games begin!

[u][b]First Votecount[/b][/u]
[b]AceOfDiamonds (1)[/b]: [i]randomguy64[/i]
[b]Not Voting (8)[/b]: [i]2lnx, AceOfDiamonds, 15!!!, lotsofmath, percy314, Royalflower, isabella2296, pacman2812[/i]
[i]With 9 alive, it takes 5 to lynch[/i].

By the way, I think 6 (or 7) of you should read the rules.  :wink:

The mafia will probably swoop in and kill before anyone can unvote and kill the mafia.

[quote="Brut3Forc3"]He's saying to vote for the person who makes someone L-1, not the person who gets L-1'd.[/quote]
ik ik
No townie would want to put someone at L-1 if we follow this system :P

Well, that's not necessarily true. If the town comes to some consensus before anyone makes it vote, then it's okay...

Who is going to create the next newbie mafia game?

When I posted "... LyLo is very serious too, so I'll withhold my vote just in case. "
I was also thinking "percy's so scummy, he must be town!" but I thought that would look suspicious, was bad logic anyway, and could possibly get me lynched. Either way we would have lost, I guess.

Also, I did some incredibly stupid (game-wise and ettiquette-wise) things in previous games.

[quote="percy314"][quote="percy314"]Ok. I'm a cop. Investigated Yoshi and got that he was mafia, thats why I was the first one to vote him. I didnt have any legit reasons for voting him, so it looked like I just voted him for no reason. I didnt know that I could give wrong results though. Last night I investigated lotsofmath, and got townie. What does that mean?[/quote]

Man, if there were actually insane cops in this game, town would've won. My plan failed.[/quote]
maybe not, since I would've counterclaimed. People would believe me more, since I had a breadcrumb hidden in one of my first couple posts, and the fact that you didn't investigate yourself(as cops in multi-cop games should) would be helpful to me

[quote="AceOfDiamonds"][quote="ernie"]It seems to be that people forgot that this was an [u]open[/u] set-up.[/quote]
you mean semi-open?[/quote]

Yep.  But you can't have insane cops either way. (cough percy314)

[quote="math_explorer"]What breadcrumb? What's a breadcrumb? I thought it was one of those "Home >> Other Topics >> Games >> Mafia!" things.

Anyway, I'm too lazy today to look up any of these things.[/quote]

Breadcrumb is, AFAIK, in mafia terms, when you secretly roleclaim so you can refer back to it earlier to "prove" your role.

For example, you might be breadcrumbing by making the first letters of each sentence I, a, m, t, h, e, c, o, and p, which would say I am cop.

you don't have to breadcrumb a roleclaim.
it can be anything that you might need, like someone's innocence/guilt you don't want to announce right away, or a night action that will happen(vig=psychic!), or something else someone might do.
breadcrumb roleclaims aren't always true :P but I won't reveal the post where I breadcrumbed, in case I use the same breadcrumbing technique in other games.

I think I found your breadcrumb post, by the way :P

Which is somewhat ironic. I won't say why for the sake of keeping AoD's technique secret :P

lastpost (this is used to preserve order of locked topic)


Hi to all!!!

This is my first post here in mathlinks.ro hope answer the following: (this is the Philippine’s Regional Screening Test for the National’s Mathematical Olympiad.)

2. If 2A99561 is equal to the product when 3 x (523 + A) is multiplied by itself, find the digit A.

3. The perimeter of a square inscribed in a circle is p. What is the area of the square that circumscribes the circle?

4. The sum of the first ten terms of an arithmetic sequence is 160. The sum of the next ten terms of the sequence is 340. What is the first term of the sequence?

5. It is given that triangle CAB is congruent to triangle EFD. If AC = x+y+z, AB = z+6, BC = x+8z, EF = 3, DF = 2y-z, and DE = y+2, find x^2+y^2+z^2.

6. Container A contained a mixture that is 40% acid, while container B contained a mixture that is 60% acid. A chemist took some amount from each container, and mixed them. To produce 100 liters of mixture that is 17% acid, she needed to pour 70 liters of pure water to the mixture she got from containers A and B. How many liters did she take from container A?

Again, PLEASE don't triple post! It's like spamming.

Mods plz delete this topic from here and Intermediate Topics.


Princeton college Gauss MIT calculus probability topology

Does anyone have any suggestions for self-studying differential equations? I have a textbook by Boyce & DiPrima, but saw some negative reviews of it on Amazon.

I would suggest finding something else to self-study.  Diff-eq does not at all strike me as a subject ammenable to being learned on ones own.

I don't think that's true. Differential Eq's is pretty ammenable to being learned on one's own. You just need a good book. I personally think that Boynce and DiPrima is pretty good. It's used in a lot of college courses - including the ODE classes at Princeton, Cal Tech, and Duke.

The books [u]Differential Equations[/u] by Blanchard, Devany, and Hall and [u]Differential Equations[/u] by Polking, Boggess, and Arnold are pretty good also. A more straightforward look at Diff. Eq.'s is [u]Differential Equations and the Applications[/u] by Martin Braun. All of these books mainly focus on methods of solving Diff. Eq.'s.

More advanced books that go into the mathematical theory of ODEs and the analysis of their solutions are [u]Ordinary Differential Equations[/u] by Birkhoff and Rota, and [u]Differential Equations, Dynamical Systems, and an Introduction to Chaos[/u] by Hirsch, Smale, and Devaney. Both of these are very good for that.

There are LOTS of Diff. Eq. books out there though. It all depend on what depth you want to go to, and what you want to learn it for. There are books that are geared toward certain specialities - like engineers or physicists or economists; and then there are ones for more general use. I personally think Boynce and DiPrima is pretty good for a general introduction to Diff. Eq.'s though.

For a large list of Diff. Eq. books, you can look [url=http://www.ericweisstein.com/encyclopedias/books/OrdinaryDifferentialEquations.html]here[/url].

Wow, thanks Gauss!

To elaborate, in case anyone else has ideas - I'm researching options for someone who wants to test out of diff. eq. at MIT or Caltech a year from now. He prefers the theoretical subjects like abstract algebra and toplogy to the practical subjects like calc and ODE, so would prefer to get diff. eq. out of the way w/o spending a lot of time in a class.

The book used in the normal ODE class (18.03) at MIT is [u]Elementary Differential Equiations with Boundary Value Problems[/u] by Edwards and Penny.

And the book used in their Honors ODE class (18.034) is [u]Differential Equations, A Modeling Perspective[/u] by R. Borrelli and C. Coleman.

You can find the homepages to these courses, along with their syllibi here: http://math.mit.edu/classes/concise-classes.html

I think Cal Tech still uses Boyce and DiPrima for both their Practical ODE class and their analytic ODE class (Math 2b).

Homepages for these courses can be found here: http://www.math.caltech.edu/courses/2term03.html

Is this course a requirement for graduation or something? I mean, many fine matematicians just never took DifEq... Is it necessary?

Some school require Diff. Eq's for a math degree and some don't. I never had to take a Diff. Eq's class.

Some schools incorporate it into their Freshman-Sophomore Calculus sequences (Cal Tech); and some schools include the basic elements of Diff. Eq's in their Linear Algebra and Complex Analysis classes (Princeton), in addition to having more advanced courses focused solely on ODE's.

It's really not as boring a subject as its sometimes made out to be. Like Linear Algebra and Probability and Statistics, it can either be taught rather dryly, or with a level of theory and depth. It all depends on where you take it, and how the teacher wants to teach it.

[quote="mathnovice"]What is the mathematics sequence after calculus?

Calculus I
Calculus II
Multivariable Calculus
Linear Algebra
Differential Equations

what next?[/quote]

There typically is no set sequence after those. Though some schools (such as Princeton) encourage math majors to take an Introductory Analysis class before taking upper division courses.

After those though, you're usually prepared to take any of the core courses for a math degree (Abstract Algebra, Real and Complex Analysis, and Topology), and most of the other introductory courses in the math department.

I usually recommend for students to take Abstract Algebra and Topology early in their math career - as they introduce a lot of themes and terminology that reappear in other classes. But it all depends on your interests, and who's teaching what-when. Some people who are interested in specific things might take introductory Graph Theory, Set Theory, Number Theory, PDEs, Discrete Math, or whatever else early on so that they can take more advanced classes later.

The Harvard math sequence for math majors is Precalculus, Calculus, linear algebra and multivariable calculus (in that order, but in what is meant to be a full-year course), and then whatever you want, so long as it includes an analysis class, an algebra/number theory class, and a topology/geometry class.  Next year, I'll probably take Algebra I and II, Complex Analysis, and something topology-related ... maybe Differential Geometry.

Just on a side note JBL,  what year in college are you?  That is how long have you been in college.  Are you majoring in math?  I would assume that you are since you have such a strong background in math.  Also,  what is Algebra I and II like?  They caught my attention because they have the same names as high school classes and I know they must be tons more advanced than the high school classes.

Hehe.  There's a thread around the site somewhere about Algebra ... I think Other Problem Solving Topics, with someone asking for what book to buy.  I just finished my freshman year.  I am a math major.  Algebra is, well, the study of the structure of numbers, I guess I would say.  I'm sure someone can explain it better than that.
[quote="Mathworld"]One use of the word "algebra" is the abstract study of number systems and operations within them, including such advanced topics as groups, Rings, invariant theory, and cohomology. This is the meaning mathematicians associate with the word "algebra." When there is the possibility of confusion, this field of mathematics is often referred to as abstract algebra. The word "algebra" can also refer to the "school algebra" generally taught in American middle and high schools. This includes the solution of polynomial equations in one or more variables, and basic properties of functions and graphs. Mathematicians call this subject "arithmetic," reserving the word "algebra" for the more advanced aspects of the subject.[/quote][quote="Mathworld"]Abstract algebra is the set of advanced topics of algebra that deal with abstract algebraic structures rather than the usual number systems. The most important of these structures are groups, rings, and fields. Important branches of abstract algebra are commutative algebra, representation theory, and homological algebra. Linear algebra, elementary number theory, and discrete mathematics are sometimes considered branches of abstract algebra. Ash (1998) includes the following areas in his definition of abstract algebra: logic and foundations, counting, elementary number theory, informal set theory, linear algebra, and the theory of linear operators.[/quote]


we are to show a=(1/2) closed loop integral over [r x dl]

I suppose this can be done formally from the alternative form of Stokes' theorem that can be obtained by replacing the vector field in curl theorem by VxC where C is a constant vector

The identity is :

surface int [(da x grad) x V]=closed loop integral over [dl x V]

The RHS matches.But how to show that LHS leads to the required value?

[quote="kolahalb"]we are to show a= (1/2) closed loop integral over [r x dl]

I suppose this can be done formally from the alternative form of Stokes' theorem that can be obtained by replacing the vector field in curl theorem by VxC where C is a constant vector

The identity is :

surface int [(da x grad) x V]=closed loop integral over [dl x V]

The RHS matches.But how to show that LHS leads to the required value?[/quote]

$\vec{a}= \frac{1}{2}\oint \vec{r}\times dl$

$\int_{S}[(d\vec{a}\times \nabla) \times V] = \oint dl \times V$

Is that what you mean?

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php#operators]You can find the LaTeX here.[/url]


Well this is gonna be my first mathcounts year, and I was wondering what score you need on average to get to the next section for school,chapter, and state. BY this I mean combined target and sprint. If you can't give a very good estimate, plz just give a range of 10

It depends on what state you're in.
Most of the time, you have to get in the 30~46 range.

Anyways, I think what you should do, ihatepie, is to go to the forum corresponding to where you live and ask there.  You'll get much more accurate results

[quote="gauss1181"]You probably need a score of around, i'd say, more than 32 or so.[/quote]
NNNNNNOOOOOOOOOOOO!!!!
I have 31's.... :( *sigh*

it's hopeful for me :)

I get an average of 36s on state

Yeah, try asking ur mc coach or teammates.  they probably know best, cuz the scores needed to move in like CA is way higher than say, WY.

KY is OK  I guess....... Evan Miller is pretty good....

For example. in my state (NJ), a 32 will probably get you in 10th (with some luck). A 34 got eigth last year. For top four, though, at least 40.

how many ppl from each "Chapter" go to state, and how many from each state go to nats.?

It's different for each chapter;but only 4 people get to go to Nats from anywhere.

I think for chapter the top 20 percent no more than 10 individuals + the top 3 school teams get to go to state.

The chapter where I live is a 40 cutoff.

I had a 36 in sixth grade.

=[

6th grade i got 28 but that was cuz i was stupid...not really. i got 3rd, but our chapter is hard...somewhat.....not really...

right now on practice chapter i get 37-40 mostly states 35-40 i guess


function
trigonometry
calculus
calculus computations

Can someone show me how to solve this equation in  terms of sinus and cosinus?

y``+y/9=0

This should probably be somewhere else, but if you have done differential equations before, this is a standard 2nd degree linear DE with $ b \equal{} 0$. Use the standard characteristic equation route. Bear in mind the roots will be complex.

The characteristic equation of $ y'' \plus{} \frac {1}{9} y \equal{} 0$ is $ \lambda ^2 \plus{} \frac {1}{9} \equal{} 0 \implies \lambda \equal{} \pm \frac {1}{3} i$

Thus, any solution will be a linear combination of the functions $ e^{\frac {1}{3} ix}$ and $ e^{ \minus{} \frac {1}{3} ix}$, which are linearly independent.

Using Euler's Formula, we have that:
$ e^{\frac {x}{3} i} \equal{} \cos \frac {x}{3} \plus{} i \sin \frac {x}{3}$
$ e^{ \minus{} \frac {x}{3} i} \equal{} \cos \frac {x}{3} \minus{} i \sin \frac {x}{3}$

Adding the two equations, we see that $ 2 \cos \frac {x}{3}$ is a solution, and thus $ \cos \frac {x}{3}$ is a solution.
Subtracting the two equations, we see that $ 2i \sin \frac {x}{3}$ is a solution, and thus $ \sin \frac {x}{3}$ is a solution.

We check to make sure we've preserved linear independence, and this is indeed the case, so linear combinations of $ e^{\frac {1}{3} ix}$ and $ e^{ \minus{} \frac {1}{3} ix}$ can also be expressed as linear combinations of $ \cos \frac {x}{3}$ and $ \sin \frac {x}{3}$.

Thus, the solution space of this differential equation is the set of functions of the form $ \boxed{y \equal{} c_1 \cos \frac {x}{3} \plus{} c_2 \sin \frac {x}{3}}$

[quote="Cartman"]Thank you very much. What about this problem?

y``+y/9=cos (at)[/quote]

$ y'' \plus{} \frac{1}{9}y \equal{} \cos at$

$ * a \not \equal{} \pm \frac{1}{3}$ : $ y \equal{} C_{_1} \cos \frac{t}{3} \plus{} C_{_2} \sin \frac{t}{3} \plus{} \frac{9}{1 \minus{} 9a^2} \cos at$

$ * a \equal{} \pm \frac{1}{3}$ : $ y \equal{} C_{_1} \cos \frac{t}{3} \plus{} C_{_2} \sin \frac{t}{3} \plus{} \frac{t}{2a} \sin at$


:maybe:  Suppose that ABC is an isosceles triangle. :D

AF intersects BC at point F.
CD intersects AB at point D.
CD intersects AF at point E.
D is a point which lies on AB.
And DF intersects BC at point F.
GIVEN:
AB=AC
$\angle(DBF)=50^\circ$
$\angle(DAE)=20^\circ$
$\angle(ECF)=30^\circ$

 :?: What is $\angle(DFE)=?$ :?:

[hide]The first triangle has what we are given, the second has what can be found:

[img]http://www.initdotd.com/~nikita/Pictures/Angles.gif[/img][/hide]

Uh, you can only find the sum of $m\angle{EDF}$ and $m\angle{EFD}$ (it's 80). From there, there is no way to determine how much of it is in each angle. Unless I'm not seeing something that I should.....

[quote="13375P34K43V312"][quote="i_like_pie"]You're right, there's no way to determine the exact angle. It can be anything from $1^{\circ}$ to $80^{\circ}.$ Sadi, are you sure you didn't forget anything?[/quote]

It can be anything from 0.000............001 to 79.999............999.[/quote]
Yes, but everything else so far is integral.


superior algebra
superior algebra unsolved

In $ S_{4}$, I have counted $ \binom{3}{2} \equal{} 6$ $ 2$-cycles as well as $ \binom{3}{1} \equal{} 3$ other elements of order $ 2$ (in the form (ab)(cd) where a,b,c,d are different).  This gives $ 9$ elements of order $ 2$.

For elements of order $ 3$, I looked at $ 3$-cycles of the form (1ab) and found $ \binom{3}{2} \equal{} 6$ such $ 3$-cycles.

For elements of order $ 4$, I looked at $ 4$-cycles of the form (12ab) and found $ \binom{3}{2} \equal{} 6$ such $ 4$-cycles.

As for elements of higher order, since there are only four letters in $ S_{4}$, we can't write disjoint products of anything greater than two $ 2$-cycles.  So, adding in the order of the identity ($ 1$), I have found $ 1 \plus{} 9 \plus{} 6 \plus{} 6 \equal{} 22$ elements.  Where did I miss the other elements of $ S_{4}$?

Thank you.

You lost two 3-cycles by assuming that all of them contain $ 1$.
You also did something wrong on the 4-cycles but still got the correct number.

So for the $ 4$-cycles I should have done (1abc) and the answer would then be $ 3!$, right?  Thanks again.


Show how to dissect a square into at most five pieces in such a way that the pieces can be reassembled to form three squares of (pairwise) distinct areas.

The 5 pieces of the 7 x 7 square rearrange to give the 6 x 6 square shown and the yellow 2 x 2 and the orange 3 x 3 squares.


integration
function
real analysis
real analysis unsolved

Hello everyone!
I have a problem:
Assume $ (\Omega, \Sigma, \mu)$ is a measure space and $ f\in L^p, \hspace{1mm} 1\leq p < \infty$. Then prove that for every $ \varepsilon > 0$ there exists $ g\in L^{\infty}$, $ g\in L^{p}$ such that $ \int_{\Omega}|f \minus{} g|^p d\mu < \varepsilon$.

I have almost solved this problem, but with very bad way. Can anyone help me.

Let $ g_n = \text{sgn}(f)\cdot\min(|f|,n).$

We have that $ \|g_n\|_p\le \|f\|_p$ and $ \|g_n\|_{\infty}\le n,$ so $ g_n\in L^P\cap L^{\infty{.}}$

Note also that $ |f - g_n|\le |f|,$ or more to the point, $ |f - g_n|^p\le |f|^p,$ which is integrable. And $ f - g_n\to 0$ almost everywhere. So by the Dominated Convergence Theorem,
\[ \int_{\Omega}|f - g_n|^p\to 0\text{ as }n\to\infty.\]
Note that the sequence $ g_n$ is (probably) unbounded in $ L^{\infty}.$ If you could approximate $ f$ by an $ L^{\infty}$ function and maintain any control over the $ L^{\infty}$ norm of your approximating function, then $ f$ would be in $ L^{\infty}$ in the first place.


Try to prove that they concur at the circumcenter.

btw, one of the quickest approach would be stating triangles ABC and DEF are orthologic, but this is probably too strong to kill this problem ;)

It's obvious that $AP$ and $AD$ are isogonal conjugate lines in $\Delta ABC$. The same holds for $BQ$ and $BE$, and for $CR$ and $CF$. Since $AD$, $BE$ and $CF$ are concurrent at the orthocentre $H$, $AP$, $BQ$ and $CR$ must be concurrent at the isogonal conjugate of $H$, which is the circumcentre $O$.

The lines $AP,\ BQ,\ CR$ are concurrent in the circumcenter $O$ of the triangle $ABC$ because the sides $EF,\ FD,\ DE$ of the orthic triangle $DEF$ are parallel with the tangents $AA,\ BB,\ CC$ respectively to the circumcircle of the triangle $ABC\ \ : \ \ AA\parallel EF\ \ \wedge\ \ AO\perp AA\Longrightarrow AO\perp EF$ a.s.o. ... and STOP.

[quote="warut_suk"]Let $ABC$ be an acute-angled triangle, and let $D$, $E$, $F$ be the feet of its altitudes from the vertices $A$, $B$, $C$, respectively. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $A$, $B$, $C$ to $EF$, $FD$, $DE$, respectively. Prove that the lines $AP$, $BQ$, $CR$ are concurrent.[/quote]

A more basic proof:

We will prove that the lines AP, BQ, CR are concurrent at the circumcenter O of triangle ABC.

In fact, these lines AP, BQ, CR are the perpendiculars from the points A, B, C to the lines EF, FD, DE. Hence, we have to show that the perpendiculars from the points A, B, C to the lines EF, FD, DE are concurrent at the point O; equivalently, we have to prove that $AO\perp EF$, $BO\perp FD$, $CO\perp DE$.

The point O is the circumcenter of triangle ABC, thus the center of a circle through the points A, B, C. Hence, if we use directed angles modulo 180°, the central angle theorem yields < OBC = 90° - < CAB. Equivalently, < (BO; BC) = 90° - < CAB. On the other hand, since < CFA = 90° and < CDA = 90°, the points F and D lie on the circle with diameter CA, and consequently, < CDF = < CAF. In other words, < (BC; FD) = < CAB. Thus,

  < (BO; FD) = < (BO; BC) + < (BC; FD) = (90° - < CAB) + < CAB = 90°,

so that $BO\perp FD$. Similarly, $AO\perp EF$ and $CO\perp DE$. This completes the proof.

I gave a different proof of $AO\perp EF$, $BO\perp FD$, $CO\perp DE$ in my note [i]Isogonal conjugation with respect to a triangle[/i], which can be found at http://www.cip.ifi.lmu.de/~grinberg/geometry2.html and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18472 (scroll to Theorem 16 there).

  darij

I think it's easy with Ceva in its trigonometric form. :D


integration
LaTeX
calculus
real analysis
real analysis theorems

Hello,
I want to show that the following hold:
If $f\in R[a,b]$ and g is bounded on $[a, b]$ and $f(x) = g(x)$ $\forall x\in [a,b]\setminus A$, where A is a finite subset of $[a, b]$.
Then $g\in R[a,b]$ and $\int^{b}_{a}{f(x)}= \int^{b}_{a}{g(x)}.$

I know how to start but I want to be clear about the whole structure of the proof because I am going to present the proof to my classmates. So I will appreciate any help.

$\LaTeX$ tip: the set symbol you wanted was \setminus, not /. I've already edited your post to make that change.

Are you using upper and lower sums in your integration and integrability proofs? I'll assume you are.

The set $A$ has $n$ elements. Let $M_{1}$ be an upper bound for both $|f(x)|$ on the whole interval and then let $M=\max(M_{1},\max(\{|g(x)|: x\in A\})).$

Given an $\epsilon,$ find a "good enough" partition for the integral of $f.$ Refine this partition by including points to make intervals about each point in $A,$ with each of these new intervals being only $\frac{\epsilon}{nM}$ in width. Argue that changing from $f$ to $g$ can only affect the upper and lower sums on those short intervals, and that the cumulative effect is no bigger than $\epsilon.$ Finish from there, using the integrability of $f.$

You'd probably want to go into more detail, but that's intended to be an overview.

Thank you so very much Professor Kent Merryfield.
I will fill in the missing details.

My teacher suggested that it is enough to just show that $\int^{b}_{a}{f(x)}= \int^{-b}_{-a}{f(-x)}.$ I can see how this implies the desire result because one can take a partition so that f and g differs only at one point in each subinterval.


geometry perimeter inradius calculus algebra polynomial inequalities

Find the triangle with given angle A and given inradius r with the smallest perimeter.

Here's a solution: (in spoiler...)

[hide]2r(cot(x/2) + 2cot( :pi: /4 - x/4)), where x = measure of angle A and r = inradius.

If we draw two tangents of a circle O that intersect at point A, and call the point of tangencies D and E, call the measure of angle A x and the radius of O r. Thus, the length of the tangents is r*cot(x/2). We then extend AD to some point B, with the measure of angle B = y, and then draw the tangent from B to circle O, tangent at point F. Then, of course, we extend both BF and AE until they come to an intersection, which we call C. Tada, here is our triangle.

Now the perimeter of the triangle is thus 2r(cot(x/2) + cot(y/2) + cot(90-x/2-y/2)). However, using simple calculus, we can verify that cot(y/2) + cot(90-x/2-y/2) is minimal when y/2 = 90 - x/2 - y/2. In other words, our triangle is isoceles.

[/hide]

I hate the phrase "using simple calculus."  Anyhow, can you do it without calculus?

lol.

JBL, I love calculus. Its godly.

Anyway, the way you can do it without calculus is that you simply reason the graph of area vs. length of DB will be pseudo-parabolic, and the isoceles triangle should logically be the vertex of that parabola.

Yeah, but what if it was a 4th degree polynomial with a relative max at isoceles and relative mins on each side of zero?  I feel like there should be a triangle inequality + similar triangles solution or something like that.

Hmm. Thats an interesting point. I would like to see a proof with similar triangles.

And yes, I suppose I did assume that parabolic idea with the knowledge that the derivative of tangent increases from 0 to  :pi: /2. Haha, I'm so used to using calculus for everything.

I should brush up on my geometry theorems though...


Cassette tapes regularly priced at $ \$9.60$ are on sale for $ \$7.68$. What is the percent of discount?

The amount of discount is one minus the sale price over the original price, so the percent here is $ 1\minus{}\frac{\$7.68}{\$9.60}\equal{}0.2\equal{}\fbox{20\%}$.


let p be a prime number of form 4k+1 such that (2^{p}-2) divisble p^2
prove that: there exists a prime number q,divisor of (2^{p} -1) such that
2^p>(6p)^p

Maybe there is a typo in the statement, there is no q in the conclusion.


Okay, I have one more proof I'm stuck on:

Prove for any rational root of a polynomial with integer coefficients,

$a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$, ($a_{n}\not=0$)

if written in lowest terms as $p/q$, that the numerator $p$ is a factor of $a_{o}$ and the denominator $q$ is a factor of $a_{n}$.

According to the book, the reason for doing this is it permits us to obtain all rational real roots and hence demonstrate the irrationality of any other real roots.

Thanks. The book I'm using has problems that are much more advanced (to me at least) than the material presented.

[hide]

It makes sense once you know: for all roots are contained by the set of factors of $a_{0}$ divided by $a_{n}$

The proofs follows very quickly from this fact

[/hide]

[hide][hide]If you divide the polynomial by a(n) you get a polynomial that is still has the same roots as the original.

We can then factor this into (x-a)(x-b)(x-c)(x-d)...where a,b,c,d...are the roots of the polynomial.  If a root is not a rational number, that root times its conjugate is a rational number.

Since this factored product yields a last term that is equal to a(n)/a(0) all the rational roots have a numerator and a denominator that are factors of the last term (but can have the opposite sign like -3 of x^2+7x+12)

Edit: and if there is a root that has a numerator or a denominator that is not a factor of the last term's, even if there is another root to balance it out (as 1/x is to x) so that the roots still multiply to an integer term, no other root will have the same denominator so the second term (sum of the roots) will not be an integer, a contradiction.
[/hide][/hide]

I need help on proofs so I tried to write one

so, if x = Pn/Qn for all n such than P/Q are rational roots, then (QnX - Pn) is a factor of the polynomial. Therefore, when you multiply everything out....(Q1X - P1)(Q2X-P2).....(QnX-Pn) you see than all Pn are factors of the constant term of the polynimial while all Qn are the factors of the leading coefficient.


geometry
3D geometry
tetrahedron
circumcircle

How do we know that the height of a tetrahedron intersects the centroid of the base?

I mean like the one in this problem.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=743933#p743933

Why is it the centroid-I thought it was the circumcenter in that question it doesnt matter because the centroid is the circumcenter-but in general isnt it the circumcenter?

Yeah, but why is that?  I want to visualize why it is.

It's only the circumcenter if the three edges joined to the top vertex have equal lengths.  In general, it isn't any particular point: for any triangle and any point on the plane there exist (infinitely many) tetrahedra with that triangle as a base and the top vertex directly above the chosen point.

To see why it's the case, draw three right triangles: they have one leg (the height) equal, so if they have their hypotenuses equal then the other legs must be equal as well.  In that case, the distances from the three base vertices to the projection of the top vertex are equal $\Longrightarrow$ circumcenter.

In this case, just imagine a regular tetrahedron, and look straight down on it.  Everything is symmetric, so it damn well better be directly above the center.  (For equilateral triangles, of course, "the center" is unambiguous.)

If this doesn't help, you need to go draw a bunch of pictures and acquire a bunch of toothpicks to play around with.


conics
ellipse
geometry
analytic geometry
graphing lines
slope
rotation

Let $ P,\ Q$ be the points of intersection of the ellipse $ x^2\plus{}\frac{y^2}{4}\equal{}1$ and the line $ y\equal{}x\plus{}k$.
Find the maximum area of $ \triangle{OPQ}$ where $ O$ is the origin.

First, let's solve this problem: what is the maximum area of the triangle obtained by intersecting a line with a circle of radius $ 1?$ If we orient the line to be vertical, we can make this the maximum of $ xy$ subject to the constraint $ x^2\plus{}y^2\equal{}1.$ It's fairly routine to solve that to find that the maximum area of $ \frac12$ occurs when $ x\equal{}y\equal{}\frac1{\sqrt{2}}.$ If we used any other slope or orientation for the line, we'd still obtain the same maximum area, as we could always rotate the picture without changing any areas.

Now, let's look at our problem. The transformation that sends $ (x,y)$ to $ (x,u)$ where $ u\equal{}\frac y2$ sends lines to lines (albeit with a change in slope), sends triangles to triangles, sends the ellipse we're concerned with to the circle $ x^2\plus{}u^2\equal{}1,$ and changes all areas by a factor of $ \frac12.$

Thus a line which gives us the triangle of maximum area would, after that transformation, give us the maximum area within the circle. From above, we know that that maximum area is $ \frac12.$ Undoing the transformation, we see that the solution to the problem is a triangle of area $ 1.$

Do anyone usuallly solve the problem in that way in USA? In Japan few students slove the problem by transforming the ellipse to a circle.

Solving for x, we get: $ x \equal{} \frac { \minus{} k \pm 2\sqrt {5 \minus{} k^2}}{5}$. Plugging this back into the equation $ y \equal{} x \plus{} k$ to get y, we then have:
$ P \equal{} (x_{P}, y_{P}) \equal{} \left(\frac { \minus{} k \plus{} 2\sqrt {5 \minus{} k^2}}{5}, \frac {4k \plus{} 2\sqrt {5 \minus{} k^2}}{5}\right)$,
$ Q \equal{} (x_{Q}, y_{Q}) \equal{} \left(\frac { \minus{} k \minus{} 2\sqrt {5 \minus{} k^2}}{5}, \frac {4k \minus{} 2\sqrt {5 \minus{} k^2}}{5}\right)$, and $ O \equal{} (0, 0)$.
Then the area of triangle $ OPQ \equal{} S(k) \equal{} \frac {1}{2}\left|x_{P}y_{P} \minus{} x_{Q}y_{Q}\right| \equal{} \frac {2}{5}\times k \sqrt {5 \minus{} k^2} \leq \frac {2}{5}\times \left(\frac {k^2 \plus{} 5 \minus{} k^2}{2}\right) \equal{} 1$.
Therefore, the maximum is 1.

That's a typical silution. I add to say that $ k$ should be $ - \sqrt {5} < k < \sqrt {5}$.

Here is my solution:

We can set the point of intersection as $ (\cos \theta ,\ 2\sin \theta )\ (0\leq \theta < 2\pi)$, so the points are also on the line $ y = x + k$, we have $ k = 2\sin \theta - \cos \theta$. The condition for which the triangluar equation has two solutions is $ - \sqrt {5} < k < \sqrt {5}$. Thus let $ P(\cos \alpha ,\ 2\sin\alpha),\ Q(\cos \beta ,\ 2\sin \beta)\ (0\leq \alpha < \beta < 2\pi)$, $ \triangle{OPQ} = \frac {1}{2}|\cos \alpha \cdot 2\sin \beta - 2\sin \alpha a \cdot \cos \beta | = |\sin (\alpha - \beta)|\leq 1$, the equality holds , say, when $ \alpha - \beta = \pm \frac {\pi}{2}$, yielding the desired area is 1.


Indeed, both inequalities are quite easy...  :)  :P

[quote="Pain rinnegan"]1)Let $ a,b,c > 0$ such that $ \frac {a^4}{b^2} \plus{} \frac {b^4}{c^2} \plus{} \frac {c^4}{a^2} \equal{} 3$ . Show that :
\[ 3\ge a^2 \plus{} b^2 \plus{} c^2\ge a^3 \plus{} b^3 \plus{} c^3\]
[/quote]
$ \bullet$ By Cauchy, we have
\[ \left( \sum \frac {a^4}{b^2} \right) \left( \sum b^2\right) \ge \left( \sum a^2\right)^2.\]
Cancelling $ \sum a^2$ and using the initial condition, the above inequality is equivalent to $ 3\ge \sum a^2$, which is what we wanted to prove.
In order to prove $ \sum a^2 \ge \sum a^3$, homogenise the inequality:
\[ \left( \sum a^2 \right)^2 \left( \sum \frac {a^4}{b^2} \right) \ge 3 \left( \sum a^3 \right)^2.\]
By Cauchy, we have
\[ 3\sum a^2 \ge \left( \sum a \right)^2 \textrm{ and } \left( \sum a^2 \right) \left( \sum \frac {a^4}{b^2} \right) \ge \left(\sum \frac {a^3}{b} \right)^2.\]
Hence it's enough to prove
\[ \left( \sum a \right) \left( \sum \frac {a^3}{b} \right) \ge 3\left( \sum a^3 \right).\]
But this is obvious using Cauchy, Muirhead, etc.
[quote="Pain rinnegan"]2)Let $ a,b,c > 0$ such that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \equal{} \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2}$ . Prove that :
\[ a^4 \plus{} b^4 \plus{} c^4\ge 4(a^2 \plus{} b^2 \plus{} c^2)\]
[/quote]
$ \bullet$  By Hölder,
\[ \left( \sum \frac {a^3}{b^2} \right)^2 \left( \sum b^4\right) \ge \left( \sum a^2 \right)^3.\]
After using the initial condition this inequality is equivalent to $ \sum a^4 \ge 4\sum a^2$, which is what we wanted to prove.

(All sums are taken cyclic)


quadratics
modular arithmetic
combinatorics unsolved
combinatorics

Suppose $A\subseteq \{0,1,\dots,29\}$. It satisfies that for any integer $k$ and any two members $a,b\in A$($a,b$ is allowed to be same), $a+b+30k$ is always not the product of two consecutive integers. Please find $A$ with largest possible cardinality.

OK I HOPE THIS WORKS  :)

[b]Problem.[/b] Suppose $A\subseteq \{0,1,\dots,29\}$ and that for any integer $k$ and any two members $a,b\in A$ (not necessarily distinct), $a+b+30k$ is always not the product of two consecutive integers. Find the maximum number of elements of $A$.

[b]Solution.[/b] Let us call a nonnegative integer $s$ [i]bad[/i] iff there exists an integer $k$ such that $s+30k$ is the product of two consecutive integers.  Thus, $s$ is bad iff $s+30k = n(n+1)$ has a solution with $n$ an integer.  In other words, regarding the equation as a quadratic in $n$, we have $s$ is bad iff the discriminant $\Delta = 1+4(s+30k)$ is a perfect square.  We need only to consider $0 \leq s \leq 58$ for this problem (why?).

Note that there exists an integer $k$ such that $\Delta = 1+4s+120k$ is a perfect square iff there exists an integer $x$ such that $x^2 \equiv 1+4s \pmod 120$.  Clearly, $x$ must be odd so we have

$(x+30)^2 \equiv x^2+60x+900 \equiv x^2+60+60 \equiv x^2 \pmod {120}$

Thus, after a (somewhat tedious!) check of $x^2 \pmod {120}$ for $x = 1,3,\dots 29$, we conclude that the only possible values of $x^2 \pmod {120}$ are $1,9,25,49,81,105$.  Thus, $1+4s$ must be congruent to these values, so we obtain all bad values: $s = 0,30,2,32,6,36,12,42,20,50,26,56$.

Now, in particular, $0,15,1,16,3,18,6,21,10,25,13,28 \not\in A$. (why?)  Thus, the only possible values of elements in $A$ are the following:

$2,4,5,7,8,9,11,12,14,17,19,20,22,23,24,26,27,29$ and checking shows that the following pairs are the only even ones that can't be in $A$. (Note that the bad values of $s$ are all even, so we divide this up into even and odd).

$(2,4), (2,24), (4,8), (4,22), (4,26), (8,12), (8,22), (8,24), (12,24), (12, 20), (12,24), (14,22),(20, 22), (24,26)$

and the best we can do is $2,14,20,26 \in A$ (these are the only ones which appear twice above, everything else appears four times... The proof is pretty boring and left as an exercise to the reader :) Well, I hope it's right... I haven't checked it)

Similarly, we deal with the odd case as follows.

$(5,7), (5,27), (7,19), (7,23), (7, 29), (9, 11), (9, 17), (9, 23), (9, 27), (11, 19), (17, 19), (19, 23), (23, 27), (27, 29)$

and we can get by with $5,11,17,29 \in A$  (same comments as above)

Thus, the maximum number of elements in $A$ is $4 + 4 = 8$.  (I hope  :D )

hmm.. I think I've found a 10 number solution, I'll provide the details later.

EDIT: Ok, here's my solution:

First notice that \[(n+15)(n+16) = n^{2}+31n+240 = n^{2}+n = n(n+1) \quad (\textrm{mod }30).\] So a product of two consecutive numbers must be $0, 2, 6, 12, 20, 26 \textrm{ mod }30$. (check all possibilities of n)

Now the condition $a+b+30k$ is never a product of two consecutive integers is equal to the condition that $a+b=0, 2, 6, 12, 20, 26 \textrm{ mod }30$. Then notice that $a$ cannot be $0, 1, 3, 6, 10, 13 \textrm{ mod }15$. We deduce that \[A \subseteq \{2, 4, 5, 7, 8, 9, 11, 12, 14, 17, 19, 20, 22, 23, 24, 26, 27, 29\}\] Since products of two consecutive numbers are always even, it suffices to check the cases that $a$ and $b$ are both even or both odd.

If $a$ and $b$ are both even, notice that there are 9 even numbers, and that \[2+4=6; 8+12=20; 14+22=36; 24+26=50,\] so there are fewer or equal to 5 even elements in $A$.

If $a$ and $b$ are both odd, notice that there are 9 odd numbers, and that \[5+7=12; 9+11=20; 17+19=36; 23+27=50,\] so there are fewer or equal to 5 odd elements in $A$.

So there are fewer or equal to 10 elements in $A$. Now it is easy to check that \[A =\{2,5,8,11,14,17,20,23,26,29\}\] satisfies the conditions. (checking mod 6, or even mod 3, might be easier, now I look at this...)

So the maximum value of $|A|$ is 10.

Here is the solution.$n(n+1)$ can be congruent to $0,2,6,12,20,26$ , so if $a+b\equiv 0,2,6,12,20,26 \pmod {30}$ then there exists $k$ such that $a+b+30k=n(n+1)$ .Proof; Let  $a+b=30m+t$ with $t=(0,2,6,12,20,26)$ . Now equation looks like this $a+b+30k=30m+t+30k=n(n+1)$
$30m+t+30k=n(n+1)$   We take $n\ge 10^{m+300} $(take it large so that $r$ is bigger than m) and $n(n+1)\equiv t \pmod {30}$ ,so  $n(n+1)=30r+t$ and $r> m$  ;  $30m+t+30k=n(n+1)=30r+t$ Now just take  $k=r-m$ and proof is finished.Now we can see that $0,1,3,6,10,13,15,16,18,20,21,25,28 \not\in A$
(because if some $x\in A$ then $2x$ is congruent to $0,2,6,12,20,26$) Next pair rest of them like this $(2,4),(5,7),(9,11),(8,12),(14,22),(19,23),(24,26),(27,29)$ .  From any pair we can take at most $1$ (if we take two then their sum is  $0,2,6,12,20,26$) .So from pairs we take at most 8 numbers and we are left with two that are not in any pair and that are not disregarded ($17$ and $20$ ). From this follows that $|A| \le 8+2=10$ and equality is all that are congruent to $2$ $mod 3$ (sum of any two can not be $n(n+1)$ because their sum is $\equiv 1 \pmod {3}$ but  $n(n+1)\equiv 0 or 2  \pmod {3}$ ) and we are done .


function
algebra
polynomial
limit
real analysis
real analysis solved

Let $f:\mathbb{Q}\to\mathbb{Q}$ be a monotonous one-to-one onto (bijective) mapping.

a) Prove that there exists an unique continuous function $F:\mathbb{R}\to\mathbb{R}$ such that $F(x)=f(x)$ for all $x\in\mathbb{Q}$;

b) Give an example of a non-one-to-one (non-injective) polynomial function $G:\mathbb{R}\to\mathbb{R}$, such that $G(\mathbb{Q}) \subset \mathbb{Q}$ and its restriction $G |_{\mathbb{Q}}$ is one-to-one.

Nice problem! The subjects in 11-th grade are really nice! The second question is an old (easy) nice AMM problem.

For the second part (that's the interesting part, right? :)) we may take $x(x^2-5)$.

I agree with grobber that b) part is more interesting,
but for 1) part how do you prove existence of such function ?

I don't know I think we can prove it using that $Q$ is countable and
construct it's graphically step by step. Or how ?

To prove that it is unique, consider inversly.
It means there exits interval $[a,b]$ where $a,b$ are from $R\Q$
such $F_1(x) > F_2(x)$ on it.
Then using that $Q$ is dense in $R$, exists there $c$ in $Q$ and in$[a,b]$.
So $F_1(c) > F_2(c)$.

If $f:\mathbb Q\to \mathbb Q$ is increasing and bijective, for each $x\in \mathbb R\setminus\mathbb Q$ let $F(x)$ be the supremum of $\{f(y)|y<x\}$. Since the $f(y)$ in this set are $<f(z)$ for some $z>x$, the set is bounded from above, so the supremum exists as a real number. Similarly, the infimum of $\{f(z)|z>x\}$ also exists. To see that $F$ is continuous is the harder part.

First, note that the sup and inf above are equal. If not, let $q$ be a rational number $\sup < q<\inf$. By bijectivity, there exists $r\in \mathbb Q$ such that $f(r)=q$ and by construction $r\geq x$ and $r\leq x$ so $r=x\notin \mathbb Q$ contradiction.

If $x<y$ then $\{f(a)|a<x\}\subset \{f(a)|a<y\}$ so $F(x)< F(y)$ so $F$ is increasing.

Now, let $x$ be an irrational number. For every increasing sequence $x_n$ converging to $x$ we have $F(x_n)$ is an increasing sequence bounded from above by $F(x)$. But there exists a sequence $y_n<x_n<y_{n+1}<x_{n+1}<\cdots<x$ of rational numbers. So $f(y_n)<F(x_n)<f(y_{n+1})<F(x_{n+1})<\cdots<F(x)$. Since by definition $F(x)=\lim f(y_n)$ it must be that $F(x_n)$ is also converging to $F(x)$ so $F$ is continuous.


number theory
greatest common divisor
number theory unsolved

A sequence $ \{a_i\}$ of integers is given. It is known that $ \gcd(a_i,a_j)=a_\gcd(i,j)$. Prove that

$ \prod_{i=k}^{n+k-1}a_i$ is divisible by $ \prod_{i=1}^{n} a_i$.

For $ k \equal{} 1$, the statement is obvious. Let it is right for $ k \equal{} k_0 \minus{} 1$.

$ \gcd(a_{k_0 \minus{} 1},a_{n \plus{} k_0 \minus{} 1}) \equal{} a_{\gcd(k_0 \minus{} 1,n \plus{} k_0 \minus{} 1)} \equal{} a_{\gcd(k_0 \minus{} 1,n)}$, but $ \gcd(k_0 \minus{} 1,n)\le n$, consequently there exists $ a_j \equal{} \gcd(a_{k_0 \minus{} 1},a_{n \plus{} k_0 \minus{} 1})$ for some $ j\le n$. Let $ a_{k_0 \minus{} 1} \equal{} a_j.t$ and $ a_{n \plus{} k_0 \minus{} 1} \equal{} a_j.s$, where $ \gcd(t,s) \equal{} 1$.

$ \Rightarrow \prod_{i \equal{} 1}^{n} a_i \mid \prod_{i \equal{} k_0 \minus{} 1}^{n \plus{} k_0 \minus{} 2} a_i \equal{} \prod_{i \equal{} 1}^{n} a_i.A.t$, $ A\in \mathbb{Z}$.
but $ \prod_{i \equal{} k_0}^{n \plus{} k_0 \minus{} 1} a_i \equal{} a_{n \plus{} k_0 \minus{} 1}\prod_{i \equal{} k_0}^{n \plus{} k_0 \minus{} 2} a_i \equal{} a_j.r\frac {\prod_{i \equal{} 1}^{n} a_i.A.t}{a_j.t} \equal{} \prod_{i \equal{} 1}^{n} a_i.A.r$.
Therefore $ \prod_{i \equal{} 1}^{n} a_i \mid \prod_{i \equal{} k_0}^{n \plus{} k_0 \minus{} 1} a_i$ and the statement follows by induction.

PS If $ a_i\equal{}\textrm{const}$, then A won't be an integer, but in all other cases $ A\in \mathbb{Z}$.

[quote="BG Yoda"]
PS If $ a_i \equal{} \textrm{const}$, then A won't be an integer, but in all other cases $ A\in \mathbb{Z}$.[/quote]
Why?

BTW, a very beautiful problem. What is the source? And are there any well-known sequences that have the given property, except for the Fibonacci numbers and $ a_i\equal{}\lambda i$?

I don't really know the source. Probably, $ c\cdot F_n$? Seriously, there has to be some.

Any sequence of the form $ c(a^n \minus{} b^n)$ has this property, including the Fibonacci numbers and, say, the sequence $ a_n \equal{} 2^n \minus{} 1$ as well-known special cases.  (It is only necessary that the terms of the sequence be integers, not that $ c, a, b$ themselves be integers.)


Bueno ...
 para quienes se estan iniciando en el mundo de Latex....

pueden ir a la seccion de MathLinks [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]Latex Help[/url]
dedicada a este fin
donde encontraran las nociones generales de la edicion de texto en latex, los comandos mas utiles
y mucha referencia al respecto!!! ........

Sin embargo, estoy colgando una guia (en español )que muchos deben haber revisado ...
(quienes ya han utilizado latex)
es una de las referencias mas rapidas de como elaborar textos en latex desde cero, de manera comoda y practica

Espero les sirva..

Carlos Bravo :)
Lima -PERU

Aqui añado un pdf .........

adicional, donde encontraran un listado adicional de los comandos
que puedan necesitar en la edicion de texto en latex..

esta bastante completo...

Carlos Bravo :)
Lima -PERU


[quote="joseph14i"]If an ellipse is not centered at the origin of an x-y coordinate system, but again has its major axis along the x-axis, it may be specified by the equation

where (h,k) is the center.

why????[/quote]

it just has to be [b]parallel [/b]to the x-axis.   do you get the case when it is centered at the origin?


geometry
3D geometry
sphere
tetrahedron
trigonometry
circumcircle
geometry proposed

In $xyz $ space, four points $A,B,C,D$ satisfy the following conditions.

\[AB=1,AC=2,AD=3,\ \angle BAC=\angle CAD=\frac{\pi}{3},\angle DAB=\frac{\pi}{2}\]
Let $E$ be a point which equidistant from these four points. Find the length of the line segment $AE$.

The point $E$ is the center of a sphere circumscribed around the tetrahedron $ABCD$. The sides of the tetrahedron are

$d_{AB} = AB = 1$
$d_{AC} = AC = 2$
$d_{AD} = AD = 3$

$d_{BC} = BC = \sqrt{AB^2 + AC^2 - 2AB \cdot AC \cos{\frac \pi 3}} = \sqrt 3$

$d_{BD} = BD = \sqrt{AB^2 + AD^2} = \sqrt{10}$

$d_{CD} = CD = \sqrt{AC^2 + AD^2 - 2 AC \cdot AD \cos{\frac \pi 3}} = \sqrt 7$

and the circumradius is

$d_{AE} = AE = R$
$d_{BE} = BE = R$
$d_{CE} = CE = R$
$d_{DE} = DE = R$

The 4-dimensional volume $V_4$ of a 5-vertex object $ABCDE$ in 4-dimensional space is given by the [url=http://mathworld.wolfram.com/Cayley-MengerDeterminant.html]Cayley-Menger determinant[/url]:

$\displaystyle V_4^2 = -\frac{1}{2^4 \cdot (4!)^2}\ \left| \begin{array}{cccccc} 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{AB}^2 & d_{AC}^2 & d_{AD}^2 & d_{AE}^2 \\ 1 & d_{AB}^2 & 0 & d_{BC}^2 & d_{BD}^2 & d_{BE}^2 \\ 1 & d_{AC}^2 & d_{BC}^2 & 0 & d_{CD}^2 & d_{CE}^2 \\ 1 & d_{AD}^2 & d_{BD}^2 & d_{CD}^2 & 0 & d_{DE}^2 \\ 1 & d_{AE}^2 & d_{BE}^2 & d_{CE}^2 & d_{DE}^2 & 0 \\ \end{array}\right|$

Since the 4-dimensional volume of the tetrahedron $ABCD$ with its circumcenter $E$ is zero, we get the following equation for the circumradius $R$:

$\displaystyle \left| \begin{array}{cccccc} 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 4 & 9 & R^2 \\ 1 & 1 & 0 & 3 & 10 & R^2 \\ 1 & 4 & 3 & 0 & 7 & R^2 \\ 1 & 9 & 10 & 7 & 0 & R^2 \\ 1 & R^2 & R^2 & R^2 & R^2 & 0 \\ \end{array}\right| = 0$

Calculating the determinant by adding/subtracting rows to get at most 2 non-zero elements in a column and then expressing the determinant as a sum of 2 minors, etc., the result is

$R^2 = \frac 5 2$

$R = \frac{\sqrt{10}}{2} \doteq 1.58114$

Thank you for your reply, yetty.You are right.
Your solution is difficult for me.

kunny

It is the same argument as using the Cayley-Menger determinant for 3-D volume $V_3$ of a tetrahedron $ABCD$ and then saying that 3-D volume $V_3$ of a triangle $\triangle ABC$ with its circumcenter $D$ is zero. This gives the following equation for the circumradius $R$:

$\displaystyle \displaystyle \left| \begin{array}{ccccc} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & c^2 & b^2 & R^2 \\ 1 & c^2 & 0 & a^2 & R^2 \\ 1 & b^2 & a^2 & 0 & R^2 \\ 1 & R^2 & R^2 & R^2 & 0 \\ \end{array}\right| = 0$

$(2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4 - b^4 - c^4) R^2 = a^2b^2c^2$

where $AB = c, AC = b, BC = a$. Factoring the left side leads to:

$[4a^2b^2 - (a^4 + b^4 + c^4 + 2a^2b^2 - 2b^2c^2 - 2c^2a^2)] R^2 = a^2b^2c^2$

$[(2ab)^2 - (a^2 + b^2 - c^2)^2] R^2 = a^2b^2c^2$

$(2ab + a^2 + b^2 - c^2)(2ab - a^2 - b^2 + c^2) R^2 = a^2b^2c^2$

$[(a + b)^2 - c^2][c^2 - (a - b)^2] R^2 = a^2b^2c^2$

$(a + b + c)(a + b - c)(c + a - b)(c - a + b) R^2 = a^2b^2c^2$

$16 s(s - c)(s - b)(s - a) R^2 = a^2b^2c^2$

$\triangle^2 = s(s - c)(s - b)(s - a) = \left( \frac{abc}{4R} \right)^2$

where $s, \triangle$ are the triangle semiperimeter and area.

[quote="yetti"]

$\triangle^2 = s(s - c)(s - b)(s - a) = \left( \frac{abc}{4R} \right)^2$

where $s, \triangle$ are the triangle semiperimeter and area.[/quote]

Yes! This is Heron's formula. ;)


Ha... I have dial-up, and so I got about 10 seconds of music at a time followed by about the same amount of "buffering" :roll:

[quote="drunner2007"]http://profile.myspace.com/index.cfm?fuseaction=user.viewprofile&friendID=70505041

click on "dream chaser"
i think its great![/quote]

wow that's cool :lol:
how do you download it?

[quote="junggi"]
how do you download it?[/quote]

3

yeah i want to download it to. i might need to use a P2P service like ares and search for it.

if you're having trouble with the buffering, try reloading the page.

I found another song that I think is great

http://profile.myspace.com/index.cfm?fuseaction=user.viewprofile&friendID=72213713


geometry
symmetry
circumcircle
geometry unsolved

Let triangle ABC and the line $(l)$. Let $A',B',C'$ are the symmetry point of $A,B,C$ through the line $(l)$ . Let $(l_{1}) , (l_{2}) , (l_{3})$ are the line through $A',B',C'$ and parallel $BC,CA,AB$ respectively.
Prove that $(l_{1}), (l_{2}), (l_{3})$ are concurrent.

We denote the intersection points of the lines through A', B', C' and parallel to BC, AC, AB respectively, as follows:

     - As M, the intersection point of the lines through A', B'.
     - As M', the intersection point of the lines through  B', C'.
     - As M'', the intersection point of the lines through  A', C'.

     It is enough to prove that the points M, M', M'', are coincided.

     It is clear that the circumcircles (o), (o'), of the triangles ABC, A'B'C' respectively, are congruent and also symmetrical each other, with respect to the line (l).

     Because of MA' // BC and MB' // AC we have that <A'MB' = <ACB, (1).

     From the triangles ABC, A'B'C', we have that <ACB = <A'C'B' (2).

     From (1), (2), =>  <A'MB' = <A'C'B', (3).

     From (3), we conclude that the point M, lies on the circumcircle (o') of A'B'C'.

     By the same way we can prove that the points M', M'', lie also on the circle (o').

     Hence, we conclude that M ≡ M' ≡ M'' and the proof is completed.

     Kostas Vittas.


[quote="Ravi B"]Is there a way to view all the topics from the previous 24 hours?  I tried to use the Search button, setting the "Search previous" option to "1 day".  But the Search feature seems to require a nontrivial keyword or author string.  Until about a year ago, I could just put * in either of those fields, but now the query seems to require at least 3 non-wildcard characters.

Thanks.[/quote]Yeah there were some security issues with allowing a search after * to be performed. Very soon there is going to be an update to the site and this particular issue is being dealt with.


ARML
geometry
3D geometry
tetrahedron
incenter
Duke
college

I got the idea from Archie's email.

What should our next ARML t-shirt look like?

I personally like the hypercube idea and the pink is certainly noticeable...

Dude pink worked last time I say we stick with what works. I'm pretty open about keeping or changing the hypercube, though it would be good to have this shirt not be last year's shirt with a 6 changed to a 7. So if we keep the hypercube, make it a different hypercube!

Yeah! Go pink!!!

how bout a hypersphere eh? that would be 1337

wait quick question, so how exactly do you qualify for arml again?

Hypersphere, you know, the cool two-sided fountain-looking thing.  :D

You qualify for ARML thru AMCs, State math contest, ARML practices, other contests (eg Duke Math Meet)

Yes, I know what it is. A hypersphere projected into 2D looks exactly like a circle...

.............

a 3d representation-ish thing flattened into 2d maybe

(as I described...)

I hope I'm right, I got that from books

[quote="not_trig"]a 3d representation-ish thing flattened into 2d maybe[/quote]

i.e., a circle. Check your books again...

[quote="jb05"]Perhaps you have a suggestion for how to draw a hypersphere?

What about a tetrahedron with its insphere and exspheres? Just a spur of the moment thought, but it may be interesting. Even the two-dimensional version looks pretty cool. And if you give me a formula for the incenter and excenters, I can draw it.[/quote]

Hmm, that doesn't sound too bad.  A picture would be nice (or how to generate a picture with Mathematica or something similar would also be nice).

What about a figure like one of Escher's?  Of course it would have to be simplified, but maybe something similar...

[img]http://home.uchicago.edu/~janie/escher/EscherBoxbrown2.jpg[/img]

Here's what I was talking about... If you actually want it, I can *try* to add the extensions of the planes, but the machine doesn't appear to want to do that. Also, I can change colors, camera position, and lighting if you want :).

Wow, I like this even more than I thought I would.

interesting...

if a hypersphere->2d = circle, then a hypercube->2d=square... (of course, squares are different from circles, but still.... >.<)

No! A hypersphere is smooth, a square has vertices. Big difference!

It is certainly not true that sphere->2D is a circle, thus cube->2D is a square.

A square actually is one possible projection of a hypercube into 2D.

ok you win :maybe:

your tetrahedron is interesting, just not very symmetric...

Actually, that was purposeful. If you would like me to make it more symmetric, give me the vertices of the tetrahedron and the centers/radii of the spheres. I'm not really up for trying to do that again. If you want to do it, find the equations of the four planes (extended faces), let the center of the sphere be (a,b,c) and the radius r, and let all of the distances from (a,b,c) to the planes be r, and then solve for all six spheres, and find the extraneous one.

So who's up for a black shirt with that on it?

Hmm, picture looks nice, but how difficult would it be to put that on a t-shirt?


$ \text box{rank}(AB)=\text box{rank}(BA)$ ????

If $ A$ or $ B$ is invertible then $ AB$ and $ BA$ are similar.
Else find $ A,B$ of dimension 2 s.t. $ AB\equal{}0$ ($ Im(B)\subset{ker(A)}$) and $ BA\not\equal{}0$.


inequalities
LaTeX
calculus
inequalities unsolved

prove the following inequality for a,b,c>0

 $\sum_{\text{cyclic}}\frac{a}{\sqrt{4a^{2}+ab+4b^{2}}}\leq 1$?

P.S Edited.
Thanks. N.T.Tuan

is it $\sum_{\text{cyclic}}\frac{a}{\sqrt{4a^{2}+ab+4b^{2}}}\leq 1$?

learn Latex here http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php  :wink:

If we take $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$, inequality is equivalent to \[\sum\frac{1}{\sqrt{4+x+4x^{2}}}\leq 1\] with $xyz=1$, which can be done uzing Lagrange's multipliers.

Well, I'm sorry Goblin but I am not familliar with lagrange multipliers. I have not studied calculus yet.

Is there another simple solution?

Problem proposed by Zhaobin in Mathematics Reflection.

[quote="ywl"]Well, I'm sorry Goblin but I am not familliar with lagrange multipliers. I have not studied calculus yet.

[/quote]

[b]lagrange multipliers[/b]: http://en.wikipedia.org/wiki/Lagrange_multipliers


induction
number theory proposed
number theory

hi,
consider the sequencens $ (a_{n})_{n \geq 1}$ and $ (b_{n})_{n \geq 1}$ s.t.
\[ a_{n} \equal{} \frac{2a_{n\minus{}1}}{1\plus{}2a_{n\minus{}1}^{2}}, a_{1} \equal{} 1\]
and
\[ b_{n} \equal{} \frac{1}{1\minus{}2a_{n\minus{}1}^{2}}.\]
Prove that $ a_{n}$ contains only of irreductible fractions and $ b_{n}$ contains only of squares of integers.

Define two sequences $ p_n,q_n$ via $ p_1\equal{}2,q_1\equal{}3$ and $ p_{n\plus{}1}\equal{}2p_nq_n$ and $ q_{n\plus{}1}\equal{}q_n^2\plus{}2p_n^2$.

[b]Claim:[/b] $ \forall n: q_n^2\minus{}2p_n^2\equal{}1$

[i]Proof[/i]: We perform induction on $ n$ and can verify the case $ n\equal{}1$. Making the induction hypothesis for $ n$ we conclude $ q_{n\plus{}1}^2\minus{}2p_{n\plus{}1}^2\equal{}(q_n^2\plus{}2p_n^2)^2\minus{}8p_n^2q_n^2\equal{}(q_n^2\minus{}2p_n^2)^2\equal{}1$.

Since $ q_{n\plus{}1}\equiv q_n\bmod{2p_n}$ and $ q_{n\plus{}1}\equiv 2p_n^2\bmod{q_n}$ we conclude that $ p_n$ and $ q_n$ are coprime for all $ n$. Since $ a_2\equal{}\frac{p_1}{q_1}$ and $ \frac{p_{n\plus{}1}}{q_{n\plus{}1}}\equal{}\frac{2\frac{p_{n}}{q_{n}}}{1\plus{}2\left(\frac{p_{n}}{q_{n}}\right)^2}$ we  can identify $ a_{n\plus{}1}\equal{}\frac{p_{n}}{q_{n}}$ and $ b_{n\plus{}2}\equal{}\frac{q_n^2}{q_n^2\minus{}2p_n^2}\equal{}q_n^2$. Hence $ \forall n>1: \,a_n\in\mathbb{Q}\setminus\mathbb{Z}$ and for all $ n>2$ the term $ b_n$ is a perfect square (note, that $ b_2\equal{}\minus{}1$ is not a square)


1.Radium dissassociates at an average rate of $ 2.24 * 10^{13}$ $ \alpha$ particles per minute.these particles then capture two electrons from air and get converted to Helium atoms.After 420 days the collected helium gas occupied $ 0.5 mL$ when measured at $ 27^{0}C$ and $ 750 mm$ of $ Hg$.From these data calculate the Avogadro's constant

2.find the volume occupied by 2 mole $ H$ atoms at NTP.

1. Strategy:

1) From the given thermodynamic data for Helium gas, compute the number of corresponding moles, assuming ideal gas behaviour.
2) Compute the total number of Helium atoms from the given rate of dissociation of Radium, and from the number of minutes that exist in 420 days.
3) Finally use  number of particles = Avogadro's constante x number of moles.

2. Where is the difficulty?

1.i already had proceeded that way but i don't get the answer

2.can we apply the gas laws directly?

1. I also don't get the exact value of Avogadro's constant, but the reasoning is the more important.

2. I don't see other way, although hydrogen atoms in the gaseous state at NTP seems strange.

yes my reasoning was same as urs
not so surprisingly for the second answer is $ 22.4 ~l$ but doesn't that seem to go against gas laws

Perhaps the explanation it that, at NTP, 2 moles of hydrogen atoms correspond to 1 mole of hydrogen molecules. That seems a problem stated in a strange way.

but suppose 2 moles of $ H$ did exist as a gas then what would have been the answer.
like i can have $ 6$ moles of $ O_{2}$ or $ 4$ moles of $ O_{3}$ from $ 12$ moles of $ H$ so what's the reasoning

but i sincerely beleive the question is wrongly stated

Then the 2 moles of H would behave like two moles of Helium, and the answer would be doubled.

Try These!

[b]1st Part[/b]

One method of calculating Avogadro's number is by measuring radioactive decay. Radium-226 decays to Lead-210 via several intermediates. Only $ \alpha$ and $ \beta$ decays are involved.

a) How many of each type of decay occur in this chain?

The experiment involves leaving a readium-226 sample for several months and measuring the volume of helium gas generated by the $ \alpha$-particles. The first step is to measure the rate of decay of radium.

b) It is found that when a $ 6.3\times 10^{\minus{}5}g$ sample of radium-226 is placed in a vacuum and exposed to a screen of area $ 1.0mm^{2}$ placed $ 1.5m$ away from the sample, an average of $ 46$ $ \alpha$-particles hit the screen in $ 10.0$ minutes. During this short period of time, assume that only one step has taken place in the decay chain. Calculate the total number of decayse per gram per second of radium-226.

c) Why must this be done in a vacuum?

Ernest Rutherford conducted an experiment and showed that a sample of $ 192mg$ of radium-266 left for $ 83$ days produced $ 6.83mm^{3}$ of helium gas, measured at $ 0^{\circ}C$ and $ 1$atm.

d) Calculate the number of moles of helium produced.

e) Calculate the number of helium atoms produced.

f) Hence calculate Avogadro's number.


Who is doing the UCSB Contest and when does the second round start?

I'll see if I can remember to ask Prof. Ryavec tomorrow when the second round is. Good job on the first round by the way.

How did I do on the first round, Sir Simon the Great?  I am unaware of the result, and probably will be for several more days.

William J. Nygard IV

Hm. So I'm Sir Simon the Great now? Prof. Ryavec told me not to tell people their scores. I figured they would have arrived by now since I had them a week ago. I'll tell you that you got a lot of questions right and not very many wrong. Anyway, he's always telling me how good you are at math. You should be flattered I suppose.

Thank you for the compliment.  Your knowledge and kindness spans over all hyphens and the occasional apostrophe or ampersand.

Sir Simon, are you responsible for that monster qualifying exam? :D

Anybody took that?

[quote="jelyman"]Is there a site that has the problems?[/quote]

Don't think so. Also each student must turn in the test so copy of the problems is not available either.

really? u mean the university of santa babara right? our teacher let us keep the contest


Express the simplest relation between a,b and c, not all equal,if a^2-bc=b^2-ca=c^2-ab.

$ a^{2}-bc=b^{2}-ca\Longleftrightarrow a^{2}-b^{2}+c(a-b)=0$
$ \Longleftrightarrow (a-b)(a+b+c)=0$, similarly $ b^{2}-ca=c^{2}-ab\Longleftrightarrow (b-c)(b+c+a)=0$.
Since $ a,\ b,\ c$ are distinct numbers, we have $ a+b+c=0$.

Related problem:

Solve the following system for $ (x,y,z)$ in terms of $ t$:
$ \begin{cases}x^{2}+tyz=y^{2}+tzx=z^{2}+txy\\ \end{cases}$

[You need to break it into cases for $ t$.]


geometry
3D geometry
sphere
vector
calculus
derivative
real analysis

Let  $f$ and $g$ be funtions from $\mathbb{R}^3$ to $\mathbb R$. Suppose $f$ is differentiable and $%Error. "grad" is a bad command.
f = g(\mathbf{x})\mathbf{x}$, where $\mathbf{x}$ is vector. Show that $f$ is constant on spheres centered at origin.

Some hypotheses are missing or its obviously false.

Run your mouse over that formula and you'll see what's wrong: liyi wrote "\grad f = g(\mathbf{x})\mathbf{x}", so he meant $\nabla f({\bf x})=g({\bf x}){\bf x}$. In other words, the gradient of $f$ maps each vector to a collinear vector.

I think it's pretty much the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=perpendicular&t=26689]this one[/url], except that this time we're working in $\mathbb R^4$ instead of $\mathbb R^3$.

I know that the derivatives of $f$ in all directions which are perpendicular to the sphere is 0. How to prove rigorously that $f$ constant on the sphere?


vector
trigonometry
analytic geometry
geometry
3D geometry
sphere
integration

If :    $ \vec{M}\equal{}\hat{\phi }\frac{r}{R}sin\theta M_{o}$
How much are these:
$ J_{b}\equal{}\vec{\bigtriangledown }\times \vec{M}$
$ K_{b}\equal{}\vec{M}\times \hat{n}$

I find that the first is:
$ J_{b}\equal{}\frac{2M_{o}}{R}(cos\theta \hat{r}\minus{}sin\theta \hat{\theta })$
but i don't know how to work the K.

Thank you a lot.

If we only had the slightest idea of what all those letters with small hats over them mean... :roll:

They're meant to be unit vectors in the radial and angular directions.

Thanks JRav, that's exactly what i mean.

For the   $ K_{b}$ i found:
$ K_{b}\equal{}\frac{r}{R}sin\theta M_{o}\hat{\theta }$

Do you agree?
Thanks a lot...

I don't exactly know what these things are but it seems like $ K_{b}$ is nothing more than a cross product.  I also don't know how you're orienting your axes.

We have $ \vec{M} \equal{} \frac {M_0}{R}r\sin\theta\hat{\phi} \equal{} M_\phi$, so $ M_\theta \equal{} M_r \equal{} 0$
Now it's just computing the [url=http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates]curl in spherical coordinates[/url]. If your symbols mean what I think they mean, and hopefully I don't miscalculate, then I get
$ \nabla\times\vec{M} \equal{} \frac {1}{r\sin\theta}\frac {\partial}{\partial \theta}(M\sin\theta)\hat{r} \minus{} \frac {1}{r}\frac {\partial}{\partial r}(rM)\hat\theta$

$ \equal{} \frac {1}{r\sin\theta}\frac {r}{R}M_02\sin\theta\cos\theta \hat{r} \minus{} \frac {1}{r}\frac {M_0\sin\theta}{R}2r\hat\theta$

$ \equal{} \frac {2M_0}{R}\cos\theta\hat{r} \minus{} \frac {2M_0}{R}\sin\theta\hat\theta$

$ \equal{} \frac {2M_0}{R}(\cos\theta\hat{r} \minus{} \sin\theta\hat\theta)$

BlueVelvet thanks a lot.
We agree to this solution and you didn't do mistake with the symbols.

Now, my real problem is with these current bounds $ K_b, J_b$ which is the [b]A(r)[/b] at each place, inside and outside the sphere???

$ \vec{A}(\vec{r})\equal{}\frac{\mu_0}{4\pi} \int_{\mathcal{V}} \frac{\vec{J}_b}{|\vec{r}\minus{}\vec{r}|}dV'\plus{}\frac{\mu_0}{4\pi} \oint_{\mathcal{S}} \frac{\vec{K}_b}{|\vec{r}\minus{}\vec{r}|}dA'$

[quote="BlueVelvet"]I think it's best if this topic is moved to the physics section. Could you give some information on the situation? All we have is some formulas; I don't know what you are trying to calculate.[/quote]

You are right, this is from a physics section but I think a mathematician is also able to help.
I wrote the formula of A(r).
So we try to calculate A(r) which is:
    $ \vec{A}(\vec{r})\equal{}\frac{\mu_0}{4\pi} \int_{\mathcal{V}} \frac{\vec{J}_b}{|\vec{r}\minus{}\vec{r}|}dV'\plus{}\frac{\mu_0}{4\pi} \oint_{\mathcal{S}} \frac{\vec{K}_b}{|\vec{r}\minus{}\vec{r}|}dA'$

wih:
   $ J_{b}\equal{}\frac{2M_{o}}{R}(cos\theta \hat{r}\minus{}sin\theta \hat{\theta })$
and $ K_{b}\equal{}\frac{r}{R}sin\theta M_{o}\hat{\theta }$
 and V is the Volume of the sphere and S is the Surface of the sphere.


A teacher asks for a group of volunteers from a class of 6 students to participate in a class project. Assuming that at least one student volunteers, how many combinations of volunteers are possible?

This is $ 2^6\minus{}1\equal{}\boxed{63}$.

To clarify, there are $2^n$ ways to field a group of $n$ people. This includes a group of 0 people, which does not satisfy the criteria of the problem. $n=6$ in this case, so we use AIME's math, $ 2^6-1=\boxed{63} $