[quote="Pascual2005"]Colombia es Pasion, aprendamos de neustros paises:

[url]http://www.youtube.com/watch?v=lXMZQ8gY3WU[/url][/quote]
Ese video lo quitaron de internet ahora lo encunetran aca:
http://www.youtube.com/watch?v=pXtvGsYS0UM&search=colombia

En ese caso...
[url]http://www.youtube.com/watch?v=ZBDHdJuOoJk&search=mexico%20en%20la%20piel[/url]
Jajaja, Todos los países latinos son preciosos. De eso no cabe duda

Perú tambien tien lo suyo

http://www.youtube.com/watch?v=Co3Zy0LLWuI&search=peru

[quote="carlosbr"]Aja....
no lo habia visto antes..
muy buena ... Marcos ;)

Aprovecho para extender mis saludos a todos los peruanos de este forum ;)
estamos de fiestas el 28 de Julio..
muchos de seguro viajaran este fin de semana ..
yo hoy estoy saliendo de viaje a Cañete ;)
nos vemos....

Ah.. claro para los peruanos... [color=red][size=134]Felices Fiestas Patrias!!![/size][/color]

Carlos Bravo ;)
Lima - PERU[/quote]
Lo mismo que tu Carlos, saludar a toods los peruanos por un año mas de fiestas patrias


If f(x)=4^x then f(x+1)-f(x) equals what?

I was just wondering, but what does the star(*) stand for???

                                                                          sorry,
                                                                                I'm new


inequalities
calculus
derivative
function
search
inequalities unsolved

Some time ago I found in mathlinks following prb. But I have no sln. Anayone can help me?

Let x, y, z > 0. Show that
$ \sum_{cyclic} \frac{\sqrt{xy+4yz+4zx}}{x+y} \ge \frac{9}{2}$

Christian Reiher's square root inequality!! Does anybody have a solution?  :?

[b]PS.[/b] I see where it was posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=3769 . But no proof was given.

[b]PPS.[/b] Christian says that this inequality implies Iran 1996.

  darij

Hm, I think we must kill the square roots first.I have some inequality for this job.For example:
(Assume that $ x \geq y\geq z $).
[tex] i) \sqrt{(zx+4yz+4yx)(zy+4xz+4xy)} \geq 2z(x+y)+4xy+2\frac{xyz}{x+y} [/tex]
[tex] ii) \sqrt{(xz+4xy+4zy)(xy+4xz+4yz)} \geq 2xy+3xz+4yz [/tex].
[tex] iii)\sqrt{(yz+4xy+4zx)(xy+4xz+4yz)} \geq 2xy+3yz+4xz [/tex]
What a pity, the inequality I received is not(not a symmetric) what I hope.But it may be right.Can someone use computer with heavy computation  :( to check if this one is right( expand it and take $ x=z+a;y=z+b $):
 [tex] \sum(\frac{xy}{x+y}+\frac{4z}{x+y}) +2\frac{16xyz+(6xy+7z^2)(x+y)+6z(x^2+y^2)}{(x+y)(y+z)(x+z)} \geq \frac{81}{4} [/tex]   ;)
However,this is still not a very good idea.Do you have any ideas or a smart way to kill the square roots?
This is really a tough inequality,at least in my opinion,it costs me near two days but without any results. :(

Why this mistary around which inequality is more powerful? It is absolutely clear that this problem implies Iran 1996. I have no idea to solve this one. Darij, do you have Christian's proof? Can you post it?

Or at least some hints.I prefer it than a full solution,because it is an exercise for me, and I shouldn't give up when the time has not end yet. :)

[i]To cuong: Thật khó nói bất đẳng thức nào là mạnh hơn lắm.Mình chứng minh được rằng nếu có một đi?u kiện nhất định của $ x,y,z $ thì:
       $ A \geq \frac{1}{4}B $
trong đó $ A $ là bất đẳng thức ta đang cần chứng minh,$ B $ là bất đẳng thức Iran 1996(ở dạng $ A \geq 0,B \geq 0$ )  :) [/i]

[quote="harazi"]It is absolutely clear that this problem implies Iran 1996.[/quote]

Why? That's anything but clear to me...

[quote="harazi"]Darij, do you have Christian's proof?[/quote]

Alas, I don't have it. Christian said that it is ugly. Anyway, that should not really say a lot since he said the same about the inequality itself (and urged me to keep it secret, what I did until I saw it already posted on MathLinks in this thread).

Have fun with this inequality. As for me, I have given it up for some time, but maybe I'll later return to it...

  darij

[quote="harazi"]Thanks a lot, I have more interesting things to do than thousands of useless computations.  :D[/quote]
I think at least we must have a solution first, although it might be ugly. Finding a nice solution is later job.
I even can't do it by brute force way, so it must hide some smart ideas.

[quote="cuong"]i spent nearly three days solving it with null results :( ;
when using iran1996 ineq ,this maybe stronger :D[/quote]

where can I find that inequality???

[quote="Namdung"]
The first step and quite interesting exercise is to prove inequality in special case y = z. [/quote]
After that you can apply Vasc's (n-1)EV principle, though, to justify its application, one has to check that, for any positive A, the [b]third[/b] derivative of the function $\frac{\sqrt{A+3z(1-z)}}{1-z}$ doesn't change sign on $(0,1)$, which is a doable, but not very pleasant exercise. Anyway, some proof is thus found. Now it is time to search for a "nice " proof, though I'm not sure it exists.

[quote="Namdung"]So, this problem still unsolved. Let try again, boys!!![/quote] [quote="Namdung"]The equality holds when x = y = z or x = 0, y = z (x = y, z = 0; x = z, y = 0). That why this inequality is so hard. We will fail immediately if we use classical inequality like Cauchy or AM-GM. The first step and quite interesting exercise is to prove inequality in special case y = z. Try it![/quote]I found an elementary solution which reduces the inequality to
$2(x^2+y^2+z^2)-yz+2x(y+z) \geq 3\sqrt{y^2z^2+4x^2(y^2+z^2)}$.
Can somebody give a nice solution to this last ineq?

i see that your friend is a veeery strong ineq solver.did he show you the solution for the one on "i cannot solve it" topic :blush: ?

He didn't show the solution of the topic. Therefore, I cannot solve it!  :D

Congratulations to your friend, Parkdoosung, for his magnificent solution. :lol:
I found out that my elementary solution is wrong, because I reversed a sign.  :oops:

To Sung-yoon : 제 친구인데 누군지 잘 모르실거에요..

A similar but shorter solution is the following.
Squaring and setting $A=bc+4a(b+c)$, $B=ca+4b(c+a)$, $C=ab+4c(a+b)$, the inequality becomes
$\sum\frac{A}{(b+c)^2}+2\sum\frac{\sqrt{BC}}{(c+a)(a+b)} \geq \frac{81}{4}$.
Soong-Yoou Kim Inequality
$\sqrt{BC} \geq \frac{2a(b^2+c^2)+4bc(b+c)+6abc}{b+c}$
give us
$2\sum\frac{\sqrt{BC}}{(c+a)(a+b)} \geq \frac{12\sum{bc(b+c)}+36abc}{(a+b)(b+c)(c+a)}$.
On the other hand, by Iran's Inequality we have
$\sum\frac{A}{(b+c)^2} = \sum\frac{ab+bc+ca}{(b+c)^2}+3\sum\frac{a}{b+c}\geq \frac{9}{4}+3\sum\frac{a}{b+c}$.
Then, it suffices to show that
$3\sum\frac{a}{b+c}+\frac{12\sum{bc(b+c)}+36abc}{(a+b)(b+c)(c+a)} \geq 18$,
which is equivalent to
$\sum a^3+3abc\geq \sum bc(b+c)$.
This is just Schur's Inequality.

[quote="Vasc"]...Soong-Yoou Kim Inequality
$\sqrt{BC} \geq \frac{2a(b^2+c^2)+4bc(b+c)+6abc}{b+c}$...[/quote]
Vasc, I don't understand this. What's it? :blush:

[quote="Parkdoosung"]To Sung-yoon : 제 친구인데 누군지 잘 모르실거에요..[/quote]

Is he KOREAN or Not?

I'm very curius

[quote="Namdung"]Some time ago I found in mathlinks following prb. But I have no sln. Anayone can help me?

Let x, y, z > 0. Show that
$ \sum_{cyclic} \frac {\sqrt {xy \plus{} 4yz \plus{} 4zx}}{x \plus{} y} \ge \frac {9}{2}$[/quote]

There exists a proof by Holder for this nice inequality. ;) :)

[quote="Namdung"]Let $ x, y, z > 0.$ Show that

$ \sum{\frac {\sqrt {yz + 4zx + 4xy}}{y + z}} \ge \frac {9}{2}.$[/quote][quote="can_hang2007"]There exists a proof by Holder for this nice inequality. [/quote]Using Hölder inequality,

$ \left(\sum{\frac {\sqrt {yz + 4zx + 4xy}}{y + z}}\right)^2\sum{(y + z)^2(yz + 4zx + 4xy)^2\left(x^2 + 2yz + zx + xy\right)^3}$

$ \geq\left[\sum{(yz + 4zx + 4xy)\left(x^2 + 2yz + zx + xy\right)\right]^3,}$

we only need to prove

$ \left[\sum{(yz + 4zx + 4xy)\left(x^2 + 2yz + zx + xy\right)\right]^3}$

$ \geq\frac {81}{4}\sum{(y + z)^2(yz + 4zx + 4xy)^2\left(x^2 + 2yz + zx + xy\right)^3}.$

But $ 4\left[\sum{(yz + 4zx + 4xy)\left(x^2 + 2yz + zx + xy\right)\right]^3}$

$ - 81\sum{(y + z)^2(yz + 4zx + 4xy)^2\left(x^2 + 2yz + zx + xy\right)^3}$

$ = \sum{x^2(y - z)^2\Big\{(y + z - x)^2\Big[256x^5(y + z) + 2x^4\left(95y^2 + 1892yz + 95z^2\right) + 4x^3(y + z)\left(219y^2 + 2362yz + 219z^2\right)}$

$ + 11x^2\left(86y^4 + 505y^2z^2 + 86z^4\right) + 8799xy^2z^2(y + z)$

$ + 2yz\left(512y^4 + 4391y^3z + 9561y^2z^2 + 4391yz^3 + 512z^4\right)\Big]$

${ + 108x^5yz(23x + y + z)\Big\}}\geq0.$

See also : http://www.mathlinks.ro/viewtopic.php?t=59771

[quote=Sung-yoon Kim]Doosung, whose solution is this?[/quote]

Ji Chen


First number is smallest possible, and second biggest possible:

$ 457\minus{}98\equal{}\boxed{359}$


linear algebra
matrix
algebra
polynomial

Let $ A \equal{} \left( \begin{array}{cc} \minus{} \omega ^ 2 & \minus{} \omega \\
1 & 0 \end{array} \right)$ with $ \omega ^ 3 \equal{} 1,\ \omega \neq 1$.

Let denote $ \sum_{n \equal{} 1}^{2009} A^n \equal{} \left( \begin{array}{cc} a & b \\
c & d \end{array} \right)$, find the value of $ a \plus{} b \plus{} c \plus{} d$.

The characteristic polynomial of $ A$ is $ x^2\plus{}\omega^2x\plus{}\omega\equal{}\frac{x^3\minus{}1}{x\minus{}\omega^2}.$ Note in particular that $ A^3\equal{}I$

$ \sum_{n\equal{}1}^{2009}A^n\equal{}\sum_{n\equal{}0}^{2009}A^n\minus{}I\equal{}\sum_{k\equal{}0}^{669}A^{3k}(I\plus{}A\plus{}A^2)\minus{}I\equal{}670(I\plus{}A\plus{}A^2)\minus{}I.$

$ I\plus{}A\plus{}A^2\equal{}\begin{pmatrix}1\minus{}\omega^2&1\minus{}\omega\\1\minus{}\omega^2&1\minus{}\omega\end{pmatrix}$

so the sum of the entries in $ I\plus{}A\plus{}A^2$ is $ 2(1\minus{}\omega\minus{}\omega^2)\equal{}4.$

That makes the answer to the problem $ 670\cdot 4\minus{}2\equal{}2678.$

[quote="Kent Merryfield"]
so the sum of the entries in $ I \plus{} A \plus{} A^2$ is $ 2(1 \minus{} \omega \minus{} \omega^2) \equal{} 4.$[/quote]

Kent Merryfield, we made a same mistake. :(

The sum of the entries in $ I\plus{}A\plus{}A^2\equal{}2(2\minus{}\omega\minus{}\omega^2)\equal{}6.$

That means the answer to the problem is $ 670\cdot 6\minus{}2\equal{}4018.$


We are selecting from set $ \{1,2,3,4,...n\}$ ten elements $ a_{1},a_{2},...,a_{10}$ . If  $ a_{1}\plus{}...\plus{}a_{5}\equal{}a_{6}\plus{}...\plus{}a_{10}$ then find maximum value of $ n$.

:wink:  There is no maximal value of $ n$ since if you may select these integers from $ \{1, \cdots, n \}$ then you may select them from $ \{1, \cdots , n\plus{}1 \}$....
In another hand, the minimal value is clearly $ n\equal{}11$ :
- You must have $ n \geq 10$.
- if $ n\equal{}10$, then if $ S$ denotes the common value of the sums, then $ 1\plus{}2\plus{} \cdots \plus{} 10 \equal{} 2S$, which id impossible since the LHS is odd.
- for $ n\equal{}11$, you may select $ 1,4,6,10,11$ and $ 3,5,7,8,9$.

Pierre.


vector
trigonometry
calculus
calculus computations

Let $ \bold{r}(t) \equal{} t \cos t  \bold{i} \plus{} t \sin t \bold{j} \plus{} \bold{k}$ and $ \bold{v}(t) \equal{} (\minus{}t \sin t \plus{} \cos t) \bold{i} \plus{} (t \cos t \plus{} \sin t) \bold{j}$.

Are there any values of $ t$ for which $ \bold{r}(t)$ is parallel to $ \bold{v}(t)$?

So we want to find a $ \lambda \in \mathbb{R}$ such that $ \bold{r}(t) \equal{} \lambda \bold{v}(t)$.

But I don't think there are because of the extra term in $ \bold{v}(t)$. Is this correct?

You are almost right. What remains is to exclude the possibility $ \bold v\equal{}0$ ;)


calculus derivative function geometry MIT college Harvard

Hi,

I was wondering what is the latest age you can be accepted into the RSI program? I am currently 15 and 9 months.

Will i be able to apply for this program next year (with my age in consideration)?

I have a passion for mathematics (as would most people on this forum would have). I am Australian. In school i have accelerated by 2.5 years in mathematics and i am currently ranked first in my school (for mathematics). I have only just been introduced into Olympiad activities and as such, have not been to Math Camp. But i intend to make this years December math camp (its like the MOP). I also maybe starting university mathematics via distance education next year [if not i will be trying to make IMO ;) ]. 2 years ago i had done some 'research.' It was more investigations, for example, i investigated the "a"th derivative of some functions, conjectured a formula for the "a"th derivative, and then proved that its true.

So by next year, i will be well up to 2nd year University mathematics.

Would i be a suitable candidate for this program next year?

-Cheers

[quote="t0rajir0u"][quote="kurt.math"]I was wondering what is the latest age you can be accepted into the RSI program? I am currently 15 and 9 months.

Will i be able to apply for this program next year (with my age in consideration)?[/quote]
The only requirement is that you are not in your last year of high school.  Typically students at RSI have just completed their junior year and are around 17.

[quote="kurt.math"]Would i be a suitable candidate for this program next year?[/quote]
Try your luck; why not?  I believe that RSI accepts two Australian students every year, so competition may be somewhat tough.  It is more important that you come willing to learn than that you come already knowing a lot.[/quote]
It seems like he will know quite alot though. However, making Australian MOP won't be on your app unless you guys decide really early

Not to intimidate either of you, but one of the Australian students in my year knew more [url=http://en.wikipedia.org/wiki/Algebraic_geometry]algebraic geometry[/url] at the beginning of the summer than I do now.  :P

[quote="mathemonster"]
However, making Australian MOP won't be on your app unless you guys decide really early[/quote]

They decide in October each year, so if i get in this year, i would be able to put it down :)

I think if you are an international student you should have a different admission way, maybe an organization in your country choose the participants. You should contact RSI.

Well for international students here (Saudi Arabia) they do a great competition and filtering for only 11 graders and from it the best 4 (out of a lot) they send them to RSI then John Hopkins in the second place and so on. I believe it's seriously much harder for international students.

[quote="the future"]I believe they accept something like a 50/30 (domestic/international) split or maybe a 55/25 split for about 80 overall (perhaps its more next year). [/quote]
That's usually about right, but you have to keep in mind that countries choose on an individual basis to participate in the program and each country has a [b]fixed[/b] quota of students they can send every year.  As hasan4444 mentioned, 4 students are sent from Saudi Arabia, for example.

[quote="the future"]I advise you to apply as a junior and not anytime earlier.[/quote]
You can only apply as a junior (unless you're graduating early, in which case I don't know what you would do): "The Research Science Institute is open to students who have completed their third year of high school or the
equivalent, by the summer of 2009, and plan to matriculate to a college or university in 2010."

This year's split was 41/29 (domestic/international), not including the five from Singapore who couldn't attend.

For me I seriously want to participate in RSI next year I just hope I could make it  :wink: I know someone in my school who really enjoyed RSI and it was an amazing experience for him I believe t0rajir0u knows him.
And by the way for anyone knows does MIT and Harvard or other good universities in America really gives admission or scholarships if your performance in RSI was great?

The top five projects and five presentations at RSI get recognized, so if you were to get one of those, you could indicate that to universities. Beyond that (ten alone were ITS finalists this year), they won't have any way of know if your performance was "great" unless you enter ISEF, ITS, Siemens, etc. with your project from RSI. Unfortunately, that doesn't come in time for college apps, but you can still get merit scholarships.

But I believe that RSI helps a lot in the admission decision even though I'm still not sure  :wink:

I get your point thanks maybe I get confused because RSI participant are already good  :)

[quote="hillarryous"]RSI doesn't help in the admission decision as much as you might think. As many as 1/5 of each year's participants get rejected from MIT, so it's not an "auto-admit" factor. There is such a huge correlation between RSI and top universities simply because RSI participants are already so achieved academically. Colleges don't necessarily look at RSI as an achievement in itself.[/quote]
although it almost is auto-admit b/c 1/5 is nothing compared to the 90 something percent who normally don't get in

"mathemonster" what "hillarryous" means that they get admitted not because of RSI because they are the best so it's normal to find most of them admitted.

After completing your RSI project, it's also much easier to polish it so it can be submitted to Intel, Siemens, or ISEF, and of course if you do well in those competitions that certainly counts for something.  Generally we're fairly well-represented in all of these competitions.

[quote="mathemonster"][quote="the future"][quote="kurt.math"]Whats the closing date for the application?[/quote]

I applied this year from the US (did not get in), and the application process was due for me sometime in mid-January.  Also, I believe that the results came back sometime in early March.  However, its been a couple of months, so my dates could be messed up.  Nevertheless, generally, to apply, you will need to write a couple of essays describing what you like to do in science/math and outside of science/math.  You also write about a couple of questions that you are interested in, your future goals, and your awards in science/math.

I believe they accept something like a 50/30 (domestic/international) split or maybe a 55/25 split for about 80 overall (perhaps its more next year).  Not to scare you, but I would say the chances of making RSI is smaller than chances of making any U.S. college for a domestic applicant (for international, I do not know because I am not sure about the number of international applicants and if they have a different format or not)

I advise you to apply as a junior and not anytime earlier.[/quote]
you applied as a sophmore?[/quote]

No I applied as a junior and did not get in.  I wish they waited 1 more month (decisions came out extremely early--earlier than other years, before the Olympiads this year) until decisions came out, but I guess its cool.  Im doing research anyways at a univeristy this summer.

[quote="hillarryous"]This year's split was 41/29 (domestic/international), not including the five from Singapore who couldn't attend.[/quote]

Wow only 41/29?  Hmm that seems really hard.  Two people from my state supposedly got in, and I did not think they would ever take two from our state.  As a result I thought more than 41 got in from the U.S.  Anyways, congrats and best of luck at RSI, hillarryous.


geometry articles percent calculus ARML Harvard college

Hi,

I'm thinking of *maybe* going Penn State (This is different school than University of Pennsylvania, which is one of the Ivy League schools) and I wonder how good the school is. I heard some people around my area who went there and said that it's a good school but I want to hear from more intelligent people. :D

If I go, I'll probably major in biology and minor in chem or math. Anyway, any ideas or things you heard about Penn State?

Penn State is definitely a reasonably good school. It should be on the application list of most Pennsylvania residents. I have been to Penn State campus; the facilities look good, and State College looks like a good living environment.

I am no pro but a friend of mine from school goes there for engineering, I think it is the top 20 in that. I think it is an averagely good school meaning that is probably on the better side of the non top (ivy schools) so I am sure you can do well there. I just think that, its probably one of the better schools in Pennsylvania. Maybe you want to come to Drexel? This is where I go, we can meet. LOL

It's one of the AMRL sites, so I was there last year and I absolutely hated it.  The buildings all look the same and the campus was just unattractive in general.  The dorm rooms are small and they have this weird waffle-shaped ceiling.  One person on my team got ill from eating the cafeteria food.

I went to Penn State last weekend to watch soccer game (GO SOCCER!!  :lol: ) between UCLA and Penn State. Unfortunately, Penn State lost but it was a good game. Anyway, the school seemed like a really nice school.

I know 5 guys who graduated last year and is at Penn State. They all say they like the school, and I heard it's a good school too. I just wanted second opinions.

[quote="hello"]I am no pro but a friend of mine from school goes there for engineering, I think it is the top 20 in that. I think it is an averagely good school meaning that is probably on the better side of the non top (ivy schools) so I am sure you can do well there. I just think that, its probably one of the better schools in Pennsylvania. Maybe you want to come to Drexel? This is where I go, we can meet.LOL[/quote]

Haha..  :D Is the Drexel one that is in Philadelphia? My friend (who is at IUP now.. he also graduated last year)'s cousin graduated Drexel and heard it's pretty good school for science and math. Or engineering. I don't remember the exact words I heard.

I have another question. Anybody knows anything about medical departments in Penn State? Like I'm pretty sure to go on medical fields, and I want to know where Penn State stands on that area. I'll probably have to go to grad school after the college though.

Keep responding! I appreciate this very much since I'll be "hopefully" taking SAT next year as junior (and aim for perfect on math --> I should seriously, and decent grade in vocabulary and writing).

Drexel is most famous for its engineering programs. Its current overall ranking in undergraduate is 109th which might not be high but I can tell you that the University's President gets the 6th largest salary among all American University Presidents, he had a Wall Street Journal article anbout him in February.

This is completely unrelated to Penn State, but UPenn has a very high acceptance rate, like 25% or higher.

I don't call acceptance rates "very high" unless they are 75 percent or higher. Moreover, the pool of applicants to Penn (the University of Pennsylvania, a privately operated school that is part of the Ivy League) is a self-selected group of students who largely have fairly competitive admissions statistics. It's not IMPOSSIBLE to get into Penn, of course, but it's prudent to apply somewhere else too if applying to Penn.

I have only two reasons going against UPenn..

1) It will cost pretty lot of money since it's part of Ivy League school.
2) Since the school's tough, I may or may not be able to keep my GPA really high (I'll get A's in any college I apply but I want to have really high A's like 98-100) and well, college isn't that important. To me, grad school is more important since I'll have to study a lot for doctor.  :roll:

Reader’s Digest named PSU’s Schreyer Honors College among America’s 100 Best (along with Disneyland):  “Offers an Ivy League-style education minus the sticker shock …”
http://www.rd.com/content/openContent.do?contentId=14809
http://www.scholars.psu.edu/

PSU Medical Program is a 6/7-year BS-MD program for exceptional candidates. It was a 5-year program in my good old days, and produced lots of Doogie Howsers then.
http://www.science.psu.edu/premed/6yrpremedprogram.html

For PA residents, applying to PSU (as target or backup)  is a no-brainer. For out-of-state applicants, Schreyer, the Med Program  and PSU’s Dickinson Law program make more financial sense than the rest of the PSU curriculums, since otherwise PSU is not necessarily the best bang for the buck if an applicant has an equally excellent choice in his/her own state with low resident tuition.

[quote="Silverfalcon"]I went to Penn State last weekend to watch soccer game (GO SOCCER!!  :lol: ) between UCLA and Penn State...[/quote]You went to the wrong PSU-UCLA soccer match! The PSU woman soccer team beat UCLA in a super thriller, and is currently 2nd nationally among 300+ NCAA Division 1 teams. http://www.gopsusports.com/pressreleases/pressrelease.cfm?anncid=8491

Wow, that program sounds great! I just told my mom about it and she agrees that it sounds like a good program.

Just for facts, here are classes I'm taking this year:
[hide="Click"]
[color=blue]
Calculus I
World History
Chemistry
CP English 10 (My school does not have AP English.. We have one that's lower, which is called CP - College Prep)
Computer Tech 1
Spanish II
Health, Gym (only one class per week)

If my ranking is same as last year, I'm 1st in my class (our class is pretty smart but so far, my lowest grade is 97. Hopefully, this can be continued on and on).

I don't think I did big deal on the extracurriculum activity but:

Boy's Soccer JV from 9th to 10th and currently play both JV and Varsity (Go SOCCER! :D )
Tennis

Oh, and I'm in Gifted Program.

My school is in rural, and well, I try my best to excel as the people in larger cities like Philadelphia and New York City (AoPS made me to realize how many smarter people are out there).[/color] [/hide]

I know this is pretty tough question to answer but is there any chance I can get into the special medical program? I'm pretty sure making myself into Penn State won't be a huge big deal. I know I can do that. But my more concern is into the special medical program, and according to what I heard, getting into medical grad school after colllege requires you to take a exam (which is NOT easy). So, this complete program set that gives B.S. degree and M.D. degree sounds good.

I just want to know if the picking of students go by larger school first then smaller schools. I don't think anyone in my school went as that program but I'm pretty sure if I keep up my work, I can get there.

ASAP repsonse will be great. Thanks for this GREAT info!

P.S. I didn't think it was bad game though. Penn State dominated the first half but on the second half, I think they got troubles and eventually lost a goal.

[quote="Silverfalcon"]... is there any chance I can get into the special medical program? ...[/quote]

Many members (especially the admired moderator of this college forum) have given great advice here about college application and I doubt that I can add anything meaningful. If you keep your grades high for 2.5 more years and not take your 2nd half of senior year off entirely, and have solid test scores, of course you have a good shot. It’s probably been said – if you are truly interested in a particular program, find a way to start your connection already (with the admissions office and the program faculty). Displaying a strong intention to enroll, if accepted, should be a plus in your application. I don't believe PSU (and probably most colleges) care if you are from a large or small high school as long as you can prove your worth.

From all that I can gather, getting into a medical school after graduating from college is definitely not easy. So getting accepted to, then be able to stay in, a BS-MD program does sound appealing for those who are certain about pursuing a medical degree.

I am going to sign off the right way this time, and hopefully I can still proudly do so after this Saturday’s football game vs. Northwestern. And yes, in 11 days PSU football will start the great turn-around by beating the very team that brought it down in 1999! Nothing (else) personal against the great state of Minnesota, Tokenadult  :D!

WE ARE, PENN STATE!

Furious, was this at ARML?

Oh man, I'm lucky our team got a hotel. I hear it's like 100 degrees in those dorms. I've been there a couple of times. It looks really friendly and clean. I might just go there because it's in state and also because my undergrad. should be affordable. I mean, I want U-Penn, but the price is rediculous.

Whitehorse, I know 2 ppl that went to Upenn this year. I know for certain that one of them had to pay only a couple of grand. I think if u received some financial aid plus the scholarship stuff, it doesnt cost a lot.

[quote="furious"]well, it's in the middle of nowhere and i've had the pleasure of sleeping in one of the non-air conditioned dorms. :mad: it has a nice campus though.[/quote]

Isn't Clemson out in the middle of nowhere!?

Anyway, I imagine most northern universities don't have AC.  Think about it.  College begins around August/September and end before the summer.  Since you're going to be up north, your weather will be either bearably warm, cool, cold, or freezing.

I've been to Penn State for ARML twice now.  We've stayed in the same type of dorms both times so I think they might only have that one type.  Out of the dorms that I've seen/stayed at (i.e. College of Charleston, USC, Penn State, and Brevard) the ones at Penn State seem to be a little sub-par.  If you're picky, this might be a problem.  If you can sleep anywhere and eat anything, like me :D , you won't have a problem.  Other than the dorms and campus of Penn State, I really don't know anything about them.

However, I suggest you don't lock yourself into Penn State right now, Silverfalcon.  Keep looking at schools.  I wasn't even considering UMich last year, but now it's one of my top choices.

Also, focus on school and math this year.  Worry about college over the summer and your senior year.

I'm glad you're really considering Michigan Joe. Have you applied yet?

I've still got those darn essays to write :(  I was hoping to do them this weekend but I seem to have run out of time.  But I should have those done this week and I think that's all I have left.

[quote="shinwoo"]he he

I live in Pennsylvania and I guess I have the most knowledge about Penn State than any of you guys

Silverfalcon, I think you shouldn't. You are extremely smart and applying for it would be just waste of money..... Penn State Honors class is for smart people with full scholarships..... , but Penn State only requires 1200 on SAT I to be accepted... Silverfalcon, I know you are like one of the top highschool kid in the nation. I would rather apply for University of Pennsylvania than go for Penn State...It's very obivious that you'd make it.. Beside, Penn State is usually for people with Economical Issues..( now I hope it helps..) <- most of my school teachers graduated Penn State...I'm pretty sure, middle school teacher isn't what you are hoping to be for your career...[/quote]

True. Applying to regular Penn State wouldn't be a problem. I wouldn't worry about that much. I mean, the most awesome way would be go to UPenn and go to Harvard Medical School or medical school in Boston. That will be a good plan but as we all know, that's a lot of $\$$.

Of course, UPenn is definitely better than Penn State for sure because it's one of Ivy League schools. But you have to know that even I'll get good grades on UPenn for studying hard, I don't think I'll want to have "too pressure" on my college life. Not saying that I want to be idle but to me, my studying comes to maximum if I get some rest now and then. I don't think UPenn will satisfy me that. I don't worry about Penn State for its dorm or food because I can sleep almost anywhere and eat everything (exception of a small allergy I have).

Also, UPenn will cost pretty a lot money. It may not be a lot compared to Harvard, Yale, Princeton, or MIT but I still want to consider the money seriously.

Acceptance is other issue. While Penn State grants me a good chance of acceptance, I'm not sure how much I'll get on the UPenn. I'm not saying I don't have a chance (I'm pretty sure I do) and probably in senior year, I'll shoot an application there.

I'm not saying that I'll go to Penn State for 100% so don't worry much. I just think I have pretty good chance of going there.

[quote="joml88"]However, I suggest you don't lock yourself into Penn State right now, Silverfalcon. Keep looking at schools. I wasn't even considering UMich last year, but now it's one of my top choices.

Also, focus on school and math this year. Worry about college over the summer and your senior year.[/quote]

Of course, my main goal this year is to be on PA ARML and AIME and you know the things come after that.  :D

I heard UMich is pretty good school, especially on math department. Is that what you're going to major on? (not surprising since you're one of moderators in Pre-Olympiad forums now).

The schools I really want to go are ones in Boston. I think most schools in there are quite decent. If I was going for law, I heard Georgetown (in D.C.) isn't too bad neither.

Honestly, if you make USAMO your junior year, that will help you out with colleges so much (that along National Merit).  If you're really passionate about math, as you are, go for USAMO.  Even if you don't make it, you'll most likely end up in a better position than where you started.

I'm not sure what I want to major in.  Math is definitely something I'm considering.  But I have so many other interests that I could go and major in just about anything.  That is also why U Michigan caught my attention.  The LSA (literature, science, and arts) department there requires you to take classes in a variety of subjects.  Also, since their student body is enormous, they have all kinds of classes.  Also, I like them because there is an opportunity for me to get some merit based financial aid.

Don't shut the door on schools like UPenn because you don't think you can make it or they cost too much, not this early at least.  Good luck with the college process, silverfalcon.


Does anyone know if there are plans to release AoPS books in eletronic formats? (e-book, kindle, ...) I'll be heading off to Exeter in a couple days, and since I like having all my math stuff with me during the summer so that I can work on it, it's going to mean a massive amount of shipping fees over the next couple years.

No plans currently.  We probably will some day, but want to see the technology mature a bit more first.

It is too easy for them to get "shared", presently.  :P


probability
real analysis
real analysis unsolved

Given a set $ X_1$, a probability space $ (X_2, \mathcal B_2, m_2)$ and a map $ T : X_1 \to X_2$, prove that there exists a $ \sigma$-algebra $ \mathcal B_1$ of subsets of $ X_1$ and a measure $ m_1$ on $ \mathcal B_1$ such that $ T : (X_1, \mathcal B_1, m_1) \to (X_2, \mathcal B_2, m_2)$ is a measure-preserving transformation (that is, $ m_1(T^{\minus{}1}(B)) \equal{} m_2(B)$ for all $ B \in \mathcal B_2$).

The solution is as follows: We put $ \mathcal B_1 \equal{} T^{\minus{}1}\mathcal B_2$, that is, $ \mathcal B_1 \equal{} \{ T^{\minus{}1} B : B \in \mathcal B_2\}$. Then for each $ B \in \mathcal B_1$, we let $ m_1(B) \equal{} m_2(TB)$. Clearly $ (X_1, \mathcal B_1, m_1)$ is a measurable space and $ T : (X_1, \mathcal B_1, m_1) \to (X_2, \mathcal B_2, m_2)$ is a measure-preserving transformation.

My question is are these $ \sigma$-algebra and measure uniquely defined?

[quote="transseries"]You will need $ T$ surjective or something.[/quote]

Thanks, let me think about one counterexample :)


function
LaTeX
limit
real analysis
real analysis unsolved

Hello, I'm new to the forums. I don't know exactly which latex bbcode to use or how to render the latex images.

Problem:
Let $ A$ be a subset of $ R^n$. The function $ F: R^n \rightarrow R$ defined by

$ f(x) \equal{} \left\{ {\begin{array}{*{20}c} {1} & {x \in A} \\
{0} & {x \notin A} \\
\end{array}} \right.$

is called the characteristic function of $ A.$ Show that $ \lim_{x \to x*} f(x)$ exists if and only if $ x*$ is either an interior or exterior point of the set $ A.$
=====================
Since $ A$ is a subset of $ R^n,$ then let $ x*$ be a point in $ R^n.$

If $ x*$ is an interior point then there is an open ball about $ x*$ that is contained in $ A.$

My question is that if $ x*$ is an interior point, then the $ \lim_{x \to x*} \equal{} 1$ since $ x*$ is contained in $ A$ from $ x \in A?$

Similarly, if $ x*$ is an exterior point(provided that there is an open ball about $ x*$ that is contained in $ R^n \minus{} A$ then  $ \lim_{x \to x*} \equal{} 0$ since $ x \not\in A?$

Thank you very much for your help.

[quote="paperwings"]Hello, I'm new to the forums. I don't know exactly which latex bbcode to use or how to render the latex images.[/quote]
You put the $ \text{\LaTeX}$ between dollar signs. I've edited the dollar signs into your post - click on the edit button to see what I've done.

I didn't change anything except change one \notin to \not\in. There are a few things I'd have done differently: \begin{cases} instead of \begin{array}, \mathbb{R}^n instead of R^n, x^* instead of x*.

I haven't looked at the substance of your post yet.

Thank you. I realized it when I saw the dollar signs, I've edited the piecewise function a little.

On the substance of your post: Yes, those are both correct statements, and $ f$ is continuous on both the interior and exterior of $ A.$ $ f$ would be discontinuous on the boundary of $ A.$ In fact, the boundary of $ A$ is precisely the set of discontinuities of $ f.$

key observations -- the interior and exterior of $ A$ are open.


calculus
derivative
calculus computations

Take the gradient, set them to zero, find the Hessian.

1. Start by finding the first order partial derivatives.

[hide="Answer"]$ \frac {\partial z}{\partial x} = \frac {y - \frac {3}{4}x^2y - y^3}{\sqrt {1 - \frac {x^2}{4} - y^2}}$,

$ \frac {\partial z}{\partial y} = \frac {x - 3xy^2 - \frac {x^3}{4}}{\sqrt {1 - \frac {x^2}{4} - y^2}}$[/hide]

2. Set and solve the system $  \{ \begin{array}{c} \frac {\partial z}{\partial x} = 0 \\
\,\, \\
\frac {\partial z}{\partial y} = 0 \end{array}$ in order to find the critical points.

[hide="Answer"]$  \{ \begin{array}{c} \frac {\partial z}{\partial x} = 0 \\
\,\, \\
\frac {\partial z}{\partial y} = 0 \end{array} = >  \{ \begin{array}{c} y - \frac {3}{4}x^2y - y^3 = 0 \\
\,\, \\
x - 3xy^2 - \frac {x^3}{4} = 0 \end{array} = >  \{ \begin{array}{c} y (1 - \frac {3}{4}x^2 - y^2 ) = 0 \\
\,\, \\
x (1 - 3y^2 - \frac {x^2}{4}) = 0 \end{array}$

and we get the following seven critical points: $ P_1 = (0,0)$, $ P_2 = (0,1)$, $ P_3 = (0, - 1)$, $ P_4 = (1, \frac {1}{2})$, $ P_5 = (1, - \frac {1}{2})$, $ P_6 = ( - 1, \frac {1}{2})$, $ P_7 = ( - 1, - \frac {1}{2}).$[/hide]

3. Find the second order partial derivatives $ \frac {\partial ^2 z}{\partial x^2}$, $ \frac {\partial ^2 z}{\partial x \partial y}$, and $ \frac {\partial ^2 z}{\partial y^2}$, and compute their values on each of the seven critical points.

4. Compute the Hessian determinant for each point, $ H(P_i) =  | \begin{array}{cc} \frac {\partial ^2 z}{\partial x^2} (P_i) &  \frac {\partial ^2 z}{\partial x \partial y} (P_i) \\
\,\, & \,\, \\
\frac {\partial ^2 z}{\partial x \partial y} (P_i) & \frac {\partial ^2 z}{\partial y^2} (P_i) \end{array} |$, and decide the character of each point according to the following: $ P_i$ is an extremum iff $ H(P_i) > 0$, and is a minimum if $ \frac {\partial ^2 z}{\partial x^2} (P_i) > 0$ and a maximum if $ \frac {\partial ^2 z}{\partial x^2} (P_i) < 0$.


logarithms
ratio
analytic geometry
geometry
power of a point
radical axis
geometry unsolved

part i. In the Poincare disk model, show that the locus of points with given Poincare distance $d$ from the origin is represented by a Euclidean circle of radius$<d$.

part ii. Use a suitable inversion to show that the locus of points with given Poincare distance $d$ from any given point $P$ is represented by a Euclidean circle.

part iii. Let $C$ be a Euclidean circle with center $(0,s)$ and diameter $s$. It represents a Poincare circle as above. Find its Poincare radius and Poincare center. Is the Poincare radius bigger or smaller than $s$?

[b](i)[/b] Let $\mathcal D$ be the disk with center O and radius R. Let A, B be 2 points inside $\mathcal D$ and let the hyperbolic line AB (the circular arc through A, B perpendicular to $\mathcal D$) meet $\mathcal D$ at P, Q, so that P, A, B, Q follow on the hyperbolic line in this order. The hyperbolic distance of A, B is $d_{AB}= R \ln \left(\frac{BP}{AP}\cdot \frac{AQ}{BQ}\right).$ If $B \equiv O$ is identical with the disk center, than the hyperbolic line OA is the diameter of $\mathcal D$ through A, $\frac{d_{OA}}{R}= \ln \frac{1+\frac{OA}{R}}{1-\frac{OA}{R}}$ or $y = \ln \frac{1+x}{1-x},$ where $y = f(x) = \frac{d_{OA}}{R},$ ${x = \frac{OA}{R}}.$ Assuming $d_{OA}= d_{OA'},$ we have

$\ln \frac{1+\frac{OA}{R}}{1-\frac{OA}{R}}= \ln \frac{1+\frac{OA'}{R}}{1-\frac{OA'}{R}},\ \ \ \frac{1+\frac{OA}{R}}{1-\frac{OA}{R}}= \frac{1+\frac{OA'}{R}}{1-\frac{OA'}{R}}$

$(R+OA)(R-OA') = (R+OA') (R-OA),\ \ \ 2R(OA-OA') = 0,\ \ \ OA' = OA$

Since $f(0) = 0$ and $f'(x) = \frac{1-x}{1+x}\cdot \frac{2}{(1-x)^{2}}= \frac{2}{1-x^{2}}\ge 2,$ $\frac{d_{OA}}{R}= y > x = \frac{OA}{R}$ for $x \in (0, 1).$

[b](ii)[/b] Let $\mathcal P$ be the locus of points with hyperbolic distance d from P and $A,\ B \in \mathcal P.$ Let SPAT, UPBV be circular arcs perpendicular to $\mathcal D,$ i.e., the hyperbolic lines PA, PB. The point P can be transformed to the center O of the disk $\mathcal D$ by a hyperbolic isometry. A hyperbolic isometry is a transformation, which leaves the disk $\mathcal D$ in place and preserves hyperbolic distances, i.e., cross ratios of the points SPAT, UPBV. Thus inversion in any circle perpendicular to $\mathcal D$ is a hyperbolic isometry. To find the inversion carrying P to O, just draw a normal to OP at P intersecting $\mathcal D$ at E, F. Perpendicular to OE at E meets OP at the inversion center I. Inversion in (I) with radius IE leaves $\mathcal D$ in place, carries P into O, and the locus $\mathcal P$ into an euclidean circle $\mathcal P'$ centered at O. Hence $\mathcal P$ is also an euclidean circle.

[b](iii)[/b] Let C be center of the circle $\mathcal C.$ Let R be the intersection of the radical axis of  the circles $\mathcal C,\ \mathcal D$ with their center line OC. Since $\mathcal C$ is inside of $\mathcal D,$ R is outside of $\mathcal D.$ Let (R) be a circle centered at R and perpendicular to $\mathcal D.$ This circle is also perpendicular to $\mathcal C.$ Let (R) meet the center line $OC \equiv OR$ at I, J outside and inside of $\mathcal D.$ Let (I) be a circle centered at I and perpendicular to $\mathcal D.$ This circle is not perpendicular to $\mathcal C.$ Inversion in (I) carries the circle $\mathcal D$ perpendicular to the inversion circle to itself, the center line OC passing through the inversion center I also to itself, and the circle (R) passing through the inversion center I to a line $r \perp OC \equiv OR.$ Since (R) is perpendicular to both $\mathcal D$ and OC, so is its inversion r. Hence r is a normal to OC at O. Likewise, since $\mathcal C$ is perpendicular to both (R) and OC, its inversion $\mathcal C'$ is perpendicular to both OC and r. Therefore, $\mathcal C'$ is centered at O, concentric with $\mathcal D.$ The hyperbolic center of $\mathcal C$ is the inversion of O in (I). I am not sure about placement of the circle $\mathcal C,$ I think s should be its radius and not diameter and $\mathcal C$ should pass through the center O of $\mathcal D.$ Then the euclidean radius of $\mathcal C'$ is greater than s and its hyperbolic radius greater than euclidean by [b](i)[/b].

thanks for the reply yetti..

for iii, do you have a more algebraic way of proving it, using the actual coordinates? thanks

If $s$ is really diameter of the circle $\mathcal C$ centered at (0, s), as the posted problem states, it does not pass through the center O of the disk $\mathcal D$ and the algebraic computation may get quite ugly. If $s$ is radius of $\mathcal C,$ as I think, this circles passes through O and the computation may be easier. Would you tell me what $s$ really is, diameter or radius of $\mathcal C$ ?

actually, sorry.. i apologize for my mistake..
i forgot to put in the conditions::

$s$ is actually the radius (not the diameter) and also, $s<1/2$.

OK, this calculation is for $s$ the diameter of $\mathcal C,$ I did not notice your last post. Then $s < \frac{2}{3}R$, otherwise $\mathcal C$ would intersect $\mathcal D$ and it would no longer be a hyperbolic circle. Let OC meet $\mathcal D$ at P, Q and let $\mathcal C$ meet OC at A, B, such that P, O, A, C, B, Q follow on the line OC in this order. Hyperbolic diameter of $\mathcal C$ is

$s_{h}= R \ln \left(\frac{BP}{AP}\cdot \frac{AQ}{BQ}\right) = R \ln \left(\frac{R+\frac{3}{2}s}{R+\frac{1}{2}s}\cdot \frac{R-\frac{1}{2}s}{R-\frac{3}{2}s}\right) = R \ln \frac{R^{2}+Rs-\frac{1}{4}s^{2}}{R^{2}-Rs-\frac{1}{4}s^{2}}$

and hyperbolic radius half of this, $r_{h}= \frac{s_{h}}{2}.$ The question is, which of $s,\ r_{h}$ is greater. Denote $x = \frac{s}{R}$ and $y = \frac{r_{h}}{R}.$ Then

$y(x) = \frac{1}{2}\ln \frac{1+x-\frac{1}{4}x^{2}}{1-x-\frac{1}{4}x^{2}},\ \ 0 < x < \frac{2}{3}$

Clearly, $y(0) = 0$ and

$y'(x) = \frac{1}{2}\cdot \frac{1-x-\frac{1}{4}x^{2}}{1+x-\frac{1}{4}x^{2}}\cdot \frac{2+\frac{1}{2}x^{2}}{(1-x-\frac{1}{4}x^{2})^{2}}= \frac{1+\frac{1}{4}x^{4}}{1-\frac{3}{2}x^{2}+\frac{1}{16}x^{4}}$

The denominator has 2 positive roots $x^{2}_{12}= 12 \pm 8 \sqrt{2}.$ As long as $0 < x^{2}< 12-8 \sqrt{2},$ the denominator is greater than zero and we have

$y'(x) > 1\ \Longleftrightarrow 1+\frac{x^{2}}{4}> 1-\frac{3x^{2}}{2}+\frac{x^{4}}{16}\Longleftrightarrow 28 > x^{2}$

The 1st condition is stronger, hence $y'(x) > 1$ when $x^{2}< 12-8 \sqrt{2}.$ But $x < \frac{2}{3},\ x^{2}< \frac{4}{9}$ and this is smaller that the smaller positive root. Therefore, $y'(x) > 1$ for $x < \frac{2}{3},$ which means $r_{h}> s.$
_____________________________________________________________________

Inversion in the circle (I) carries the circle (R) passing through the inversion center I into the normal $r \perp OC$ at O. Thus the other intersection J of (R) with OC is the inversion of O in (I), the hyperbolic center of the circle $\mathcal C.$ For the distance OR of the radical axis of $\mathcal C,\ \mathcal D$ from O, we have $R^{2}-OR^{2}= \frac{s^{2}}{4}-(OR-s)^{2},$ yielding $OR = \frac{4R^{2}+3s^{2}}{8s}.$ Radius of $(R) \perp \mathcal D$ is then $r = \sqrt{OR^{2}-R^{2}}= \frac{\sqrt{16R^{4}-40R^{2}s^{2}+9s^{4}}}{8s}$ and distance OJ is $OJ = OR-r = \frac{4R^{2}+3s^{2}-\sqrt{16R^{4}-40R^{2}s^{2}+9s^{4}}}{8s}.$

And this calculation is for $s$ the radius of $\mathcal C.$  Then $s < \frac{1}{2}R$, otherwise $\mathcal C$ would intersect $\mathcal D$ and it would no longer be a hyperbolic circle. Let OC meet $\mathcal D$ at P, Q. $\mathcal C$ meets OC at O and at another point B, such that P, O, B, Q follow on the line OC in this order. Hyperbolic diameter of $\mathcal C$ is

$t_{h}= R \ln \left(\frac{BP}{OP}\cdot \frac{OQ}{BQ}\right) = R \ln \left(\frac{R+2s}{R}\cdot \frac{R}{R-2s}\right) = R \ln \frac{R+2s}{R-2s}$

and hyperbolic radius half of this, $s_{h}= \frac{t_{h}}{2}.$ The question is, which of $s,\ s_{h}$ is greater. Denote $x = \frac{s}{R}$ and $y = \frac{s_{h}}{R}.$ Then

$y(x) = \frac{1}{2}\ln \frac{1+2x}{1-2x},\ \ 0 < x < \frac{1}{2}$

Clearly, $y(0) = 0$ and

$y'(x) = \frac{1}{2}\cdot \frac{1-2x}{1+2x}\cdot \frac{4}{(1-2x)^{2}}= \frac{2}{1-4x^{2}}> 2 > 1$

for $0 < x < \frac{1}{2},$ which means $s_{h}> s.$
_____________________________________________________________________

Inversion in the circle (I) carries the circle (R) passing through the inversion center I into the normal $r \perp OC$ at O. Thus the other intersection J of (R) with OC is the inversion of O in (I), the hyperbolic center of the circle $\mathcal C.$ For the distance OR of the radical axis of $\mathcal C,\ \mathcal D$ from O, we have $R^{2}-OR^{2}= s^{2}-(OR-s)^{2},$ yielding $OR = \frac{R^{2}}{2s}.$ Radius of $(R) \perp \mathcal D$ is then $r = \sqrt{OR^{2}-R^{2}}= \frac{R\sqrt{R^{2}-4s^{2}}}{2s}$ and distance OJ is equal to $OJ = OR-r = \frac{R^{2}-R\sqrt{R^{2}-4s^{2}}}{2s}.$

thank you very much for your reply, yetti.


I’ve been looking at [url=http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf]Olympiad Inequalities[/url] by Thomas J. Mildorf. The solution given to Example #2 appears rather messy – so I was wondering why Mildorf didn’t consider using Cauchy–Schwarz and AM–GM.

Cauchy–Schwarz would have given $ a\plus{}b\plus{}c\leqslant\sqrt{3(a^2\plus{}b^2\plus{}c^2)}$ and then by AM–GM. $ a^2\plus{}b^2\plus{}c^2\geqslant3$ as $ abc\equal{}1.$

I thought I’d mention this as the author himself says in the article that the easiest method should always be tried first when solving an inequality. Certainly Cauchy–Schwarz / AM–GM looks easier than the proposed solution. :)

It can also solved easily with only AM-GM this way: $ a^2 \plus{} 1 \plus{} b^2 \plus{} 1 \plus{} c^2 \plus{} 1\ge (a \plus{} b \plus{} c) \plus{} (a \plus{} b \plus{} c) \ge (a \plus{} b \plus{} c) \plus{} 3\sqrt [3]{abc} \equal{} (a \plus{} b \plus{} c) \plus{} 3$

But you also have to think of when you're solving a problem and you see a solution you don't think immediately how it can be done more easily.

[quote="ValentineA"]I’ve been looking at [url=http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf]Olympiad Inequalities[/url] by Thomas J. Mildorf. The solution given to Example #2 appears rather messy – so I was wondering why Mildorf didn’t consider using Cauchy–Schwarz and AM–GM.

Cauchy–Schwarz would have given $ a \plus{} b \plus{} c\leqslant\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}$ and then by AM–GM. $ a^2 \plus{} b^2 \plus{} c^2\geqslant3$ as $ abc \equal{} 1.$

I thought I’d mention this as the author himself says in the article that the easiest method should always be tried first when solving an inequality. Certainly Cauchy–Schwarz / AM–GM looks easier than the proposed solution. :)[/quote] He used Muirhead because by the same way you could prove the generalization: $ \sum_{i \equal{} 1}^n a_{i}^m \ge \sum_{i \equal{} 1}^n a_{i}^k$ when $ \forall a_{i}\ge0$ and $ m\ge k\ge 0$


Express the sum in terms of .

The answer is [hide][/hide] according to the place I got it from, but does anyone know how to derive that? I tried working with it a bit, but couldn't get any simple form.

[quote="Elemennop"]Ah, I see now.

Wow, I don't think I would've ever seen that on a ciphering problem.[/quote]

When faced with a problem with [b]binomial[/b] coefficients a common technique is to try and find a way to use the [b]binomial[/b] theorem.  Of course, it doesn't work every time but it is a very important tool to have, knowing how to apply the binomial theorem to these types of problems that is.


$ ABC$ is a triangle and $ D \in (BC),E \in (CA), F \in (AB).$ If the quadrilaterals $ BCEF,CAFD,ABDF$ are inscribed then prove that $ ABC$ is acute-angled and $ DEF$ is orthic.

It's similar to a south african olympiad problem. Note that the sides of ABC are external bisectors of DEF...


Let $ n\equal{}\dfrac{a^{2p}\minus{}1}{a^2\minus{}1}$ where $ a>1$ is an integer and $ p$ is an odd prime such that $ p\nmid a(a^2\minus{}1)$.  Show that $ n$ is a pseudoprime to base $ a$.

Nevermind I found a nice proof.  For those interested, see
http://primes.utm.edu/notes/proofs/a_pseudoprimes.html


Hi, I'm currently learning for my math exam on monday, and I'm learning about inverse functions. Now, my course says that to find the inverse function, you just have to change variables, and put y in function of x. Example:
\[ f = \left\{(x,y) : y = x^2 + 2\right\} \]
Than \[ f^{-1} = \left\{(x,y) : x = y^2 + 2\right\} \]
So \[ f^{-1} = \left\{(x,y) : y = \pm \sqrt{x-2} \right\} \]

That's all very nice, but how do we find the inverse of a more complicated function, for example \[ f(x)=\frac{x^2}{5x^3+2x+6} \]
Find $f^{-1}(x)$

Can somebody please help me with this? Is there a general way to solve these kind of problems?

First of all, you have to check if your function is bijective, else, it hasn't any inverse ;)

Yes of course, but then you can find multiple parts of your inverse, like $x^2$, it has $\sqrt{x}$ and $-\sqrt{x}$. You see what I mean? The inverse of a function is not always a function, so you can split it up in several pieces. I don't know if I made myself clear.

I thought I saw something a while ago on Mathworld, some tool like an integrator, it was software to find the inverse of a function. Does anyone have this software?

I looked it up, it is called Mathematica.

You can see the part on inverse functions [url=http://documents.wolfram.com/mathematica/functions/InverseFunction]here[/url]

Yes, I know Mathematica, but, for your exam, will you have the right to use it? :p

Yeah, I got what you meant with $f(x) = x^2$.

In fact, the method is always the same (as the one mentioned in your first post), but, it can easily get a bit messy :)

You can always write a piecewise function as a single expression using the Heaviside function, H(x) = 0 if x < 0 and 1 if x >= 0. Of course, this is really only useful if you want to do a Laplace transformation, which is not something you guys are likely to need for a while yet.

Thanks.
So you are telling me it is actually impossible for me to calculate the inverse of such functions without advanced or higher mathematics?
Pity... Anyway thank you, I'm almost sure my teacher is not going to ask the inverse of functions more difficult than $y = 3x+2$, so I'm safe there, even without Mathematica, mathmanman.  :)

Yes, that's right :)

If you want to practice, I just found this cool site : http://archives.math.utk.edu/visual.calculus/0/inverse.6/

Indeed nice site, that's were Google brought me before I asked for help here. But unfortunately, they do not discuss more complicated rational functions whose inverse is in fact no function...  :)

[quote="Jan"]That's all very nice, but how do we find the inverse of a more complicated function, for example \[ f(x)=\frac{x^2}{5x^3+2x+6}  \]
Find $f^{-1}(x)$[/quote]

Well, finding $f^{-1}\left(x\right)$ would mean that, for a given t, you want to find an x such that $f\left(x\right)=t$, i. e. such that $\frac{x^2}{5x^3+2x+6}=t$. This rewrites as $x^2=t\left(5x^3+2x+6\right)$, and this is a cubic equation for x. In order to solve it, you need the Cardano formula (or some equivalent fact), and the solution will be a messy long term with cube roots and square roots (and sometimes with complex numbers). So the explicite formula for the inverse function will be a messy expression with roots. Surely, you won't need something like this in your exam.

Generally: computing inverse functions = solving equations.

  Darij


Evaluate $ \displaystyle\sum_{k\equal{}1}^{101}\cos k$.

Whoo, tough problem. I'll try and match it with a riemann sum.

We have $ \int_{0}^{a} \cos(x) \equal{} \lim_{n \rightarrow \infty} \frac{a}{n} \sum_{k\equal{}1}^{n} \cos\left(\frac{ak}{n}\right)$

In our problem, we have $ a\equal{}1$ and $ n\equal{}101$

I'm not sure how you would attain the exact answer, but it would be $ \approx \sin(1) \minus{} \sin(0) \equal{} \sin(1)$

i think there is a formula for evaluating sum of cos's in which the angle is in ap.
I think it is this: $ cos\alpha \plus{} cos(\alpha \plus{} d) \plus{} .... \plus{} cos(\alpha \plus{} (n\minus{}1) \cdot d) \equal{} \frac {cos(\alpha \plus{} (n\minus{}1) \cdot \frac {d}{2}) \cdot sin(n \cdot \frac{d}{2})}{sin \frac {d}{2}}$

[hide="hint"]
It can be simplified to the form $ \frac{1}{2}\left(\displaystyle\frac{f(101.5)}{g(0.5)}\pm 1\right)$, where $ f,g$ represent a trig function.

[/hide]

[quote="Prime factorization"]Whoo, tough problem. I'll try and match it with a riemann sum.

We have $ \int_{0}^{a} \cos(x) \equal{} \lim_{n \rightarrow \infty} \frac {a}{n} \sum_{k \equal{} 1}^{n} \cos\left(\frac {ak}{n}\right)$

In our problem, we have $ a \equal{} 1$ and $ n \equal{} 101$

I'm not sure how you would attain the exact answer, but it would be $ \approx \sin(1) \minus{} \sin(0) \equal{} \sin(1)$[/quote]

This is not correct for several reasons, the biggest of which is that the problem is (presumably) in degrees, not radians.  The second problem is that $ a \equal{} 101$.

[hide="Another hint"] $ \cos x \equal{} \frac{e^{ix} \plus{} e^{\minus{}ix}}{2}$.  This allows you to easily derive the formula madness mentions.  :) [/hide]


inequalities
quadratics
inequalities proposed

For all $x,y> 0$ we have $\displaystyle \frac 1 {xy} \geq (\frac  2 {x+y})^2$ with equality if and only if $x = y$.
Thus, the lhs of the desired inequality is $\displaystyle \sum_{i<j} \frac 1 {a_ia_j} \geq  4 \sum_{i<j} (\frac 1 {a_i+a_j} )^2 \geq 4( \sum_{i<j} \frac 1 {a_i+a_j} )^2$,
the last inequality following from the inequality between quadratic and arithmetic means.
An equality occurs if and only if $a_1 = \cdots = a_n$.

Pierre.


[quote="Phạm Th� nh Quang"]Let $ abc\leq 1$.Prove that

$ \sum\frac{1}{a^{3}(b\plus{}c)}\leq 1$[/quote]
It's wrong. Try $ a \equal{} b \equal{} c \equal{} 1.$  :wink:

In IMO $ abc\equal{}1$
I think it has same method.
$ a\equal{}k\frac{x}{y}$
$ b\equal{}k\frac{y}{z}$
$ c\equal{}k\frac{z}{x}$
$ 1\geq k$

When $ abc\ge 1$ our problem also true . :lol:

[quote="quangpbc"]Do you mean : prove that :

$ \sum_{\text{a,b,c}}\frac{1}{a^{3}(b\plus{}c)}\ge\frac{3}{2}$.

It is a problem from IMO 1995. Prove by Cauchy-Schwarz  :wink:[/quote]

With $ x \equal{}\frac{1}{a}$ It becomes VERY EASY to prove  :lol:


It's a good idea, but it would be very hard to set up and run.

I agree with what 12markkram34 said. It [b]is[/b] a good idea, and I never thought of it, but it'd be really hard to do.

NOTE: i will not do it

how is it hard?  i mean all u need is like a microphone and software that records.  then all u do i upload it, right?

Well yes and no. If it's going to be done with different people from a bunch of different computers (that barely know each other) it might be difficult, even just agreeing on simple ideas like topics, background music, etc.

Not saying a podcast isn't a good idea, but it might need a bit more thought.

[quote="qwertythecucumber"]OK, what happened to my Geometry Question post? It seems to have disappeared... :o[/quote]
Go to search and then do a search on your username.

It hasn't disappeared...I've been posting on it.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208925

[quote="rrusczyk"][quote="matt919"]:ddr: a podcast by y'all would be awsome. everyone would listen to it! :idea:[/quote]

We're working on it.  We might have a couple up by the end of the summer.[/quote]
Yay! let us know!!

This should probably get locked now. This is just developing into spam.

Haha! True, it should be locked now!  :rotfl:

P.S. Hey fishy, who are you calling spammer, eh?  :P


A set $M$ has seven elements. Let $A_{1}, A_{2}, ...A_{7}$ be proper subsets of $M$ such that
(i) each pair of elements of $M$ belongs to exactly one of the subsets
(ii) each subset has at least 3 elements
Prove that every two subsets have exactly one common element.

[hide="Solution"]First, we notice that every two subsets have at most 1 common element. Then if there exists two subsets have no common elements,WLOG, call it $A_{1}$ and $A_{2}$, we consider two cases:
If one of them,$A_{1}$ has 3 elements namely $a_{1},a_{2},a_{3}$ . Then each of them must belongs to at least two other subsets (applying i), no pairs belong to a subset except $A_{1}$ and there are only 6 subsets but $A_{1}$. Then one of them must belong to $A_{2}$. Contradiction.
If both of them have 2 elements. Let $A_{1}=\{a_{1},a_{2}\}$. And each of these two elements also belongs to at least 3 others elements so one of them must belongs to $A_{2}$. Contradiction!
$\Rightarrow\mathbb{QED}$[/hide]


function
modular arithmetic
algebra unsolved
algebra

let $ f$ is a function
$ f: N\equal{}\equal{}>N$
$ n\equal{}\equal{}>f(n)$
such that
$ (\forall n \in N) f\circ f(n)\equal{}n\plus{}3$

1. prove that $ (\forall n \in N)f(n\plus{}3) \equal{} f(n) \plus{}3$
2. prove that $ (\forall k \in N) (\forall n \in N) f(n\plus{}3)\equal{}f(n\minus{}3k)\plus{}3k\plus{}3$
[hide]for fun :D
if $ f(0)\equal{}2006$ find $ f(2007)$
if $ f(1)\equal{}2$ find $ f(2008)$[/hide]

Actually, no such function exists.
[hide]
If $ f$ were a function with $ f(f(n))\equal{}n\plus{}3$, then we get
\[ f(n\plus{}3)\equal{}f([f(f(n))])\equal{}f(f([f(n)]))\equal{}f(n)\plus{}3\]
In particular, $ m\equiv n\pmod{3}$ implies $ f(m)\equiv f(n)\pmod{3}$. So $ f$ maps congruence classes modulo 3 to congruence classes modulo 3. And certainly every congruence class modulo 3 is in the range of $ f$ (since 4,5,6 are: $ f(f(1))\equal{}4$ etc.).

We also show that there is a positive integer $ a\leq 3$ such that $ f(a)\equiv a\pmod{3}$. This is certainly true if $ f(1)\equiv 1\pmod{3}$. If $ f(1)\equiv 2\pmod{3}$, then $ f(2)\equiv f(f(1))\equal{}4\equiv 1\pmod{3}$. So $ f(3)\equiv 0\pmod{3}$ (otherwise $ f(n)$ would never be congruent to 0 modulo 3). Similarly, if $ f(1)\equiv 0\pmod{3}$, then $ f(2)\equiv 2\pmod{3}$.

Now $ \frac{f(a)\minus{}a}{3}$ is a nonnegative integer, so
\[ a\plus{}3\equal{}f(f(a))\equal{}f\left( a\plus{}3\cdot \frac{f(a)\minus{}a}{3}\right) \equal{}f(a)\plus{}3\cdot \frac{f(a)\minus{}a}{3}\equal{}2f(a)\minus{}a\]
\[ \Rightarrow 2(f(a)\minus{}a)\equal{}3\]
a contradiction. So $ f$ cannot exist.

I think you can show that if $ f$ is a function from $ \mathbb{N}\rightarrow \mathbb{N}$ and $ c$ is an integer such that $ f(f(n))\equal{}n\plus{}c$ for every $ n\in \mathbb{N}$, then $ c$ is even, using a similar method.
[/hide]

[quote="greatestmaths"]let $ f$ is a function
$ f: N \equal{} \equal{} > N$
$ n \equal{} \equal{} > f(n)$
such that
$ (\forall n \in N) f\circ f(n) \equal{} n \plus{} 3$

1. prove that $ (\forall n \in N)f(n \plus{} 3) \equal{} f(n) \plus{} 3$
2. prove that $ (\forall k \in N) (\forall n \in N) f(n \plus{} 3) \equal{} f(n \minus{} 3k) \plus{} 3k \plus{} 3$
[hide]for fun :D
if $ f(0) \equal{} 2006$ find $ f(2007)$
if $ f(1) \equal{} 2$ find $ f(2008)$[/hide][/quote]
Yes this function not exist.
You can use $ a_n\equal{}f_{n}(x)$ to prove it not exist .


integration
function
calculus
real analysis
real analysis solved

I have no idea whether or not I've posted this before: calculate the primitive  \int (2+cos(x))^1/2 dx. My teacher showed it to me in class and said he couldn't solve it. I don't know if that means it's real;ly tough or he was too lazy for the ugly calculations  :D Can some1 help me?

it seems that there's no elementary primitive functions to this integral.
this integral has connections with elliptic integral...

Yup, that's what I think. I plugged it inot Mathematica and I got something connected to the elliptic integral, just as you said. I don't know where my teacher could have found sth like this. He even gave it as a homework for crying out loud! :D  :D That's absurd since we don't study the elliptic integral in high school...


If a,b,c >0 prove that
$ bc\left(a^3+a^2+1\right)+ac\left(b^3+b^2+1\right)+ab\left(c^3+c^2+1\right)\geq abc\left(6+ab+bc+ca\right) $

By rearrangement/Muirhead/AMGM/Whatever, we have \[a^3bc + ab^3c + abc^3 \geq a^2b^2c + a^2bc^2 + ab^2c^2\] and by AMGM we have \[bc(a^2 + 1) + ca(b^2 + 1) + ab(c^2 + 1) \geq 2abc + 2abc + 2abc = 6abc.\]

Adding these inequalities easily gives the result.

I have another solution in my mind. If noone give this I will post it soon

Is this the method you mean? :)
dividing both sides by abc, we get:
$\left(a^2+a+\frac1a\right)+\left(b^2+b+1/b\right)+\left(c^2+c+1/c\right)\geq \left(6+ab+bc+ca\right) $, which follows trivially from $a^2+b^2+c^2 \geq ab+ca+bc$ and $x+1/x \geq 2$.

In that case, I will have to disappoint you a bit, because it is EXACTLY the same solution as Arne's. Maybe denoted a bit cleaner, but the idea behind and the operations performed are simply the same. ;)


geometry
geometric transformation
algebra
polynomial
function
Putnam
linear algebra

Hi everybody

I'm taking my fist course in algebra and I have a very subtile question. I just realized that in Dutch the translation of 'indeterminate' and 'variable' is the same. This confused me completely. I tried to clear up everything, but since in our Dutch textbooks the difference is not exposed, I would really appreciate it if someone confirms my ideas about this.

 [b]Variabe[/b] A variable is an unknown which is [i]a priori[/i] known to be solvable. (So we know that it is solvable!). Like for example:

$2+3x=a+bx$, where $a,b\in\mathbb{Q}$.

Then we can solve for $x$ and find $x=(a-2)/(3-b)$

[b]Indeterminate[/b] If $x$ is unknown and not known to be solvable [i]a priori[/i], then we have a whole other scenario. Again the example

$2+3x=a+bx$, where $a,\,b\in\mathbb{Q}$.

yields now that: $2=a$ and $3=b$.

This was not yet the specific context I found this subtile difference. I was doing stuff in polynomials rings. So let $R [X]$ be a polynomial ring in the indeterminate $X$. Every element of this ring must be thought of a sequence with finitely many non-zero entries. It's wrong to think the elements are functions $p: x \mapsto p(x)$. (in this case $x$ is a variable). Now after this formal definition, we can consider an [i] a priori known[/i] element $u$ that is in a larger ring together with $R$. Now $u$ is something concrete, like $\sqrt{2}$ or $\pi$. According to the fundamentel homomorphism property of $R[X]$ we can find a unique homomorphism that maps $x$ to $u$ and $R$ on itself. Because of this we have the induced isomorphism $R[u] \cong R[x]/I$ where $I$ is an ideal such that $R \cap I = 0$. Now again the subtile difference between 'indeterminates' and 'variables' comes in play. If for example $u=\sqrt{2}$ then we know that $u$ is algebraic over $R=Q$. And [i]we know [/i] that the polynomial is solvable in the extension field. In this case we can call $x$ a variable (in the extension field). Now if we take $u=\pi$ for example and consider $Q[X]$, then mapping $Q[x] \to Q[\pi]$ is injective. And the symbol $\pi$ is nothing more than a replacement -because $\pi$ is transcedental over $Q$- for the symbol $X$ and hence is also called an indeterminate.

Can someone correct me :( Or am I correct on these things (a small reply in that case is very welcome, since I'm still a bit uncertain right now)

Thanks in advance

It sometimes helps to switch the focus from "what an object is" to "how it is used". For example, I'd have a trouble explaining what I mean by "a complex variable", but I know what "a function of a complex variable" is.

In your case you may want to concentrate on the distinction between [i]polynomials[/i] and [i]polynomial functions[/i]. For instance, $x^{2}$ and $x^{3}$ are distinct polynomials over $\mathbb Z_{2}$, even though they define the same polynomial function $\mathbb Z_{2}\to\mathbb Z_{2}$. If you keep this distinction in mind, you probably will not have to worry about "indeterminates" vs "variables". (Note that there is a natural homomorphism from the ring of polynomials onto the ring of polynomial functions.)

I looked up this topic in Lang's Algebra.
[quote]We want to give a meaning to an expression such as $a_{0}+a_{1}X+\dots a_{n}X^{n}$, where $a\in A$ and $X$ is a "variable". There are several devices for doing so, and we pick one of them.[/quote]
It seems that Lang does not really care what $X$ is, or what a polynomial is: it only matters what you can do with them.
[quote]
Let $A$ be a subring of a commutative ring $B$. Let $x\in B$. If $f\in A[X]$ is a polynomial, we may then define the associated polynomial function $f_{B}\colon B\to B$ by letting $f_{B}(x)=f(x)=a_{0}+\dots a_{n}x^{n}$... The association $f\mapsto f(x)$ is a ring homomorphism of $A[X]$ into $B$. This homomorphism is called the [b]evaluation homomorphism[/b]... We now see that the subring $A[x]$ generated by $x$ over $A$ is the ring of all polynomial values $f(x)$, for $f\in A[X]$. If the evaluation map $f\mapsto f(x)$ gives an isomorphism of $A[X]$ with $A[x]$, then we say that $x$ is [b]transcendental[/b] over $A$, or that $x$ is a [b]variable[/b] over $A$. In particular, $X$ is a variable over $A$.[/quote]

Putnam 2002 B6:

Let p be a prime number. Prove that the determinant of the matrix
\[\begin{pmatrix}x & y & z \\ x^{p}& y^{p}& z^{p}\\ x^{p^{2}}& y^{p^{2}}& z^{p^{2}}\end{pmatrix}\]
is congruent modulo $p$ to a product of polynomials of the form
$ax+by+cz,$ where $a,b,c$ are integers. (We say two integer
polynomials are congruent modulo $p$ if corresponding coefficients are congruent modulo $p.$)

In working this problem, one must always keep the distinction mlok mentions between the polynomial and the polynomial function.

In particular, $x^{p}-x$ is not the zero polynomial even though (by Fermat's Theorem), it evaluates to zero for every $x.$ The fact that it evaluates to zero simply means that we know how to factor it as a product of factors of the form $x-a$ where $a$ ranges over all of the possibilities.


induction
combinatorics unsolved
combinatorics

Hi guys first of all i want to thanks in advance the people than try to help me!! I have this problem related to the binomial theroem.
Prove that forall positive integer $ k \leq n$ the equality holds.

$ \sum_{i\equal{}0}^{k}$ $ \left( \begin{array}{c}n\\k\end{array}\right)$ $ \cdot$ $ (\minus{}1)^{k}$ = $ \left( \begin{array}{c}n\minus{}1\\k\end{array}\right)$ $ \cdot$ $ (\minus{}1)^{k}$

The first things that i m not able to understand is, if it is correct the i under the summed ?
anyway i  write exactly the problem form the book.

Thanks a lot

Best Regards.

All the $ k$ is the summation except the upper $ k$ on the sigma have to be replaced by $ i$.
Then, induction on $ k$ will do the job...

Pierre.

Can you find a combinatorial proof using the [url=http://en.wikipedia.org/wiki/Inclusion-exclusion_principle]principle of inclusion-exclusion[/url]?


According to Christians, those who accept Christ will go to heaven.

What about the people before Jesus existed? Obviously, they will not all go to hell; they were never given the opportunity to accept Jesus. Logically, you could deduce that some, if not all, will go to heaven.

Well, if they went to heaven without having to believe in Jesus, then I can go to heaven without believing in him too, correct?

Moreover, what is the importance of Jesus, then? If God can decide who goes to heaven and hell without him, why is he needed?

By Bible, there was God before Christ. It's not Christ just by himself whom you should believe, but the Holy Three (did I write it well, or is it other expression in English?)

OMG what a question... Read Bible, seriously, if you want to know more... Seriously.

[quote="Bugi"]the Holy Three (did I write it well, or is it other expression in English?) [/quote]  "Trinity," not "Three."

Almost all of them will go to hell for their sins. Life was not easy under Bible release 1.0

[url=http://www.faithalone.org/journal/2003i/bryant.pdf]This[/url] pretends to be a brief summary of the "official" Christian view on the subject. I surmise that not everyone will agree but, lacking time and transreality vehicles, our only means of figuring such things out are interpreting passages from old texts and psychological speculations. So, unless somebody wishes to share a recent revelation he or she had about this question, this thread doesn't promise to be very exciting.

Well honestly nobody really knows the bible is not really clear about that. But again I might be wrong. Relavtion is a good book to read if you want to know about heaven and hell. I think that the prophets like Noah and Moses are in heaven because relavation talks about Old testment saints going to heaven.

I'm not completely sure about this, either, but I do have some ideas.
Jesus only replaced the old covenant that God had made with Israel... It's not that everybody back then went to hell 'cuz there wasn't Jesus to believe in. For those who live now, it says in Hebrews
    "He (referring to Jesus) became sin (all of the sins of the world were laid on Him when He was on the cross), who knew no sin, that we might become the righteousness of God."
    That might be kind of paraphrased, but that just means that through faith in Jesus we are made righteous before God. However, in the Old Testament, it says that
    "Abraham believed God, and it was credited to him as righteousness."
    Obviously, Abraham went to heaven. Anyways, that was one thing I thought of. Hope it makes things clearer!

Thanks JBL (stupid me I knew it somewhere in my head)

I think it was enough for the people before Jesus to believe in God, as there was no Jesus  :wink: Otherwise even the best people would go to hell...

But my real opinion:
Only God knows the truth. Wait 'till you die.

I take Dante's [i]Inferno[/i] to be the seminal text on this issue.

Basically, all the cool people (Socrates, Plato, whoever Dante liked) are in Purgatory and everyone else is in the inner rings of hell. Scholars circa Dante's time took much contention with this issue, as they believed that the wise pagans probably belonged much closer to the frozen lake.

Look, as I said, we can't know it for sure. We can guess, make theories. But we cannot be sure.

[quote=theel]According to Christians, those who accept Christ will go to heaven.

[b]What about the people before Jesus existed? Obviously, they will not all go to hell; they were never given the opportunity to accept Jesus. Logically, you could deduce that some, if not all, will go to heaven.
[/b]
Well, if they went to heaven without having to believe in Jesus, then I can go to heaven without believing in him too, correct?
[b]
Moreover, what is the importance of Jesus, then? If God can decide who goes to heaven and hell without him, why is he needed?[/b][/quote]

[b]What about the people before Jesus existed? Obviously, they will not all go to hell; they were never given the opportunity to accept Jesus. Logically, you could deduce that some, if not all, will go to heaven.
Answer:
[/b]

They will go to hell. Sadly, they will.

They didn't believe, so they had too.

[b]
Moreover, what is the importance of Jesus, then? If God can decide who goes to heaven and hell without him, why is he needed?
Answer:[/b]
You have to believe in him to go to heaven. If you don't, you go to hell. It's a hell or heaven sort of thing.

[quote=bm2004]ouchie revive. There has been longer revives, though. Anyway, Do you realize that the religion of Christianity was written by... "abnormal" people? God and Christ may not even exist. That's my philosophy. I'm an atheist. Just because you don't believe in someone not showing signs of their existence doesn't mean you're a horrible person and deserve to go to hell. If an omniscient person thinks you deserve eternal torture just because you don't believe in his existence, while you had a very selfless life, that's the god's hole in rationality, not yours.[/quote]

not true. Read the bible! As john 3:18 states,

"Whoever believes in him is not condemned, but whoever does not believe stands condemned already because they have not believed in the name of God."

[quote=bm2004]I inquire how you know that line. Did you study the Bible in your free time? Also, what does that line prove for or against my argument?[/quote]

I do study the bible all the time.

This:

"Whoever believes in him is not condemned, but whoever does not believe stands condemned already because they have not believed in the name of God."

Answers This:

Just because you don't believe in someone not showing signs of their existence doesn't mean you're a horrible person and deserve to go to hell. If an omniscient person thinks you deserve eternal torture just because you don't believe in his existence, while you had a very selfless life, that's the god's hole in rationality, not yours.

I mean, everybody goes to heaven. Unless you either don't believe in heaven, or you think of yourself as not deserving to go there. Think about it like this: Christians believe that you have to believe in Jesus to go to heaven. Socrates believed in his going to heaven, and everybody else deserved eternal torture. Everybody believes they will go to heaven, except those of the aforementioned cases. And you can't decide where other people go, honestly, because where they are when they're dead depends on them and their religion, not you and yours.

EDIT: This was written right after the above post. I had no chance to read it, but I will leave this post as a record of where the conversation would have gone. Feel free to answer whichever you choose.

Not everybody goes to heaven. You have to believe, the bible states it. read it.

since not everybody believes, you should tell others so they could be saved too.

[quote=hello120471][quote=equinox145111]If I don't believe in Jesus, am I going to hell?

(cynical)[/quote]

yes you are, sadly  :noo: . So you better believe  :yup: ![/quote]

As I told you, this is a stupid lapse in rationality on God's side. I'm an atheist, and if this is true, God can be truly an idiot.

Or, depending on which religion you believe in... I guess everybody of a different religion would think of Christians as dominating their afterlife. Muslims, Hindus, or Jews, none of them really do this or at least to this extent.

Interesting to think about: If you don't believe in Vishnu, you will be sentenced to horrible torture after you die. Would you believe? Try thinking about it from their perspective.


function
algebra
polynomial
quadratics
Rational Root Theorem

Let $ f(x) \equal{} x^3 \minus{} 2x^2 \minus{} 5x \plus{} 6$ and $ g(x) \equal{} 2x \minus{}7$. If $ R$ is the largest root of $ f(x) \equal{} 0$, where $ x$ is any real number, then determine the polynomial function $ h(x)$ of degree $ 3$, with integer coefficients having no common factors, which has roots: $ g(R)$, $ g(f(R))$, and $ g^{\minus{}1}(R)$.

Hello
$ f(x) \equal{} (x \minus{} 1)(x \minus{} 3)(x \plus{} 2)$
$ R \equal{} 3$
$ g(R) \equal{} \minus{} 1$,$ g(0) \equal{} \minus{} 7,g^{ \minus{} 1}(3) \equal{} 5$
thus
$ h(x) \equal{} x^3 \plus{} 3x^2 \minus{} 33x \minus{}35$
Thank u

$ h(x)\equal{}x^3\plus{}3x^2\minus{}33x\minus{}35$ if I'm not mistaken

[hide="Explanation"]It is easy to see that $ f(1)\equal{}1\minus{}2\minus{}5\plus{}6\equal{}0$, so $ x\equal{}1$ is a root of $ f(x)$.
(Alternatively, rational root theorem combined with synthetic division can be used to find $ x\equal{}3$ and $ x\equal{}\minus{}2$ as roots).
$ \frac{f(x)}{(x\minus{}1)}\equal{}x^2\minus{}x\minus{}6$ by synthetic division. The quadratic is easily factored into $ (x\minus{}3)(x\plus{}2)$, so $ f(x)\equal{}(x\minus{}3)(x\plus{}2)(x\minus{}1)$.
The largest root is clearly $ R\equal{}3$.
$ g(3)\equal{}\minus{}1$, $ g(f(R))\equal{}g(0)\equal{}\minus{}7$.
If $ 2y\minus{}7\equal{}x$, $ y\equal{}g^{\minus{}1}(x)\equal{}\frac{x\plus{}7}{2}$
$ g^{\minus{}1}(R)\equal{}5$.

Now $ \boxed{h(x)\equal{}(x\minus{}5)(x\plus{}7)(x\plus{}1)}$, which can be expanded if desired.[/hide]


I would like to know how to number the questions (and the sub-questions) in this form:

1.  [text]
_a. [text]
_b. [text]

and so on where '_' refers to a space.

The alignment (roughly like the above sketch) is of importance.

Please enlighten me on the suitable commands.

Thank you.  :)

You need to use the enumerate environment - see [[LaTeX:Layout#Lists]].
[code]\renewcommand{\labelenumii}{\alph{enumii}.}  % changes secondary numbering from (a) to a.
\setlength{\leftmarginii}{1mm} % change indentation of secondary list
\begin{enumerate}
\item one
	\begin{enumerate}
	\item two
	\item three
	\end{enumerate}
\item four
\end{enumerate}[/code]
If you want to further customise the way the lists look it's best to use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=enumitem]enumitem package[/url].

Thanks a lot. That was exactly what I wanted :)


A $ 200\%$ increase is the same as a $ 50\%$ increase followed by what other percent increase?

Assume the original number is x. A 200% increase would make the number 3x. A 50% increase would make it 1.5x. An increase from 1.5x to 3x is $ \frac{3x-1.5x}{1.5x} \cdot 100=100%
$


pigeonhole principle
LaTeX
number theory proposed
number theory

Let $c$ be fixed natural number. Sequence $(a_n)$ is defined by:

$a_1=1$, $a_{n+1}=d(a_n)+c$ for $n=1,2,...$.

where $d(m)$ is number of divisors of $m$. Prove that there exist $k$ natural such that sequence $a_k,a_{k+1},...$ is periodic.

Let us first prove that for all $m$, $d(m) \leq \frac{m}{2} + 1$. Indeed none of the integers between $[m/2]+1$ and $m-1$ can divide $m$, so $m-d(m) \geq m-1-[m/2] \geq m/2-1$, which gives the desired result.

Now take $n$ the smallest integer, if it exists, such that $a_n\geq 2c+2$. We must have $n>1$. We thus get $d(a_{n-1})\geq c+2$, so $a_{n-1}\geq 2(d(a_{n-1})-1)\geq 2c+2$, which contradicts the minimality of $n$. Therefore, for all $n$, $a_n \leq 2c+1$.

Let us consider then the set $\{a_j, j=1,2,...,2c+2 \}$. We must have by pigeonhole principle that $a_i=a_j$ for some $i$,$j$ in $\mathbb{N}$. This yields easily the conclusion.

[i]moderator: corrected $\LaTeX$-error[/i]


A circle has a radius of three units. It is divided into four congruent sectors. A circle is inscribed in one of the sectors. What is the radius of the inscribed circle?
Solutions please

[hide="Hint"]
Try drawing a square with verticies at the centers of the two circles.
[/hide]

[hide="i was thinking of that"]
Draw the square and call the side $r$, since it is the radius of the small circle.
Then diagonal is $r\sqrt 2$
If you extend that diagonal, you draw a radius of the quarter circle.
So $3=r+r\sqrt 2$
$3=r(1+\sqrt 2)$
$r=\frac{3}{1+\sqrt{2}}$
Then simplify to get
$\frac{3(1-\sqrt{2}}{-1}$
$\frac{3-3(\sqrt{2})}{-1}$
$3\sqrt{2}-3$
[/hide]

[hide]The distance from the center of the inscribed circle to the center of the larger circle is $r\sqrt{2}$, where r is the radius of the inscribed circle.

Distance between Centers+r=radius of larger circle

$r\sqrt{2}+r=3$
$r=\frac{3}{\sqrt{2}+1}$

multiplying both the numerator and denominator by $\sqrt{2}-1$, we get

$r=3\sqrt{2}-3$[/hide]


2x2 matrix (to the 18th power):

[    cos (pi/12)     -sin(pi/12)]  ^18
[   sin (pi/12)       cos (pi/12)]

A) 0 1  B)0 -1   C) 1 0   D) 0 -1
   -1 0    1 0        0 -1      0  0

This is to be done without a calculator. Thanks in advance!

Can you give a geometric description of this mapping?
\[ \begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}\cos\theta&\minus{}\sin\theta\\\sin\theta&\cos\theta\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}\]


algebra
function
domain
inequalities unsolved
inequalities

[quote="2Tam"]Let $ABC$ be a triangle with $\large A \leq B \leq C$. Prove that :
$\large 8.sin\frac{A}{2}.cos^{3}\frac{A}{2}-10.(sin\frac{A}{2}+1).cos\frac{A}{2}+6\sqrt{3}\geq 0$[/quote]
 :blush: this solution is not pretty
Solution:
Let$sin\frac{A}{2}=x, cos\frac{A}{2}=\sqrt{1-x^{2}}$,
the ineq in the question changes to$\sqrt{1-x^{2}}(4x^{3}+x+5)\leq 3\sqrt{3}$
$f(x)=(1-x^{2})((4x^{3}+x+5)^{2}$ and we need to prove that $f_{max}(x)=27$ on a domain of$0<x<0.5$
Then$f'(x)=(4x^{3}+x+5)(1-2x)[(2x-1)^{2}+8x^{3}+x]$
So, when$x=\frac{1}{2}$, we have$f(\frac{1}{2})=27$

[quote="2Tam"]Let $ABC$ be a triangle with $\large A \leq B \leq C$. Prove that :
$\large 8.sin\frac{A}{2}.cos^{3}\frac{A}{2}-10.(sin\frac{A}{2}+1).cos\frac{A}{2}+6\sqrt{3}\geq 0$[/quote]
sorry but where is the rull of $\large A \leq B \leq C$ and that they are triangles angel .

[quote="2Tam"]Thanks for your help, andy_tok![/quote]
 :P You're welcome

[quote="arkhammedos"]sorry but where is the rull of $\large A \leq B \leq C$and that they are triangles angel .[/quote]
I guess you mean "rule" and "angle" ^_^
I'm not sure what u meant by "rules". sorry :oops: maybe sth like"$\angle{A}\leq 60^\circ$"?


A circle with radius 7 is inscribed in an equilateral triangle. The circle has 3 points tangent to the triangle. 3 more smaller circles are inscribed in the triangle. They are tangent to 2 sides of the triangle and the large circle. 3 more smaller squares are inscribed in the triangle. They are tangent to 2 sides of the triangle and the smaller circle. The process is repeated an infinite number of times. What is the sum of the circumferences of the circles?

[hide]The diameter of the incircle of an equilateral triangle is equal to two-thirds of its height. The diamter of the first smaller inscribed circle is then

$\frac 13 \cdot \frac 23 \cdot 21 = \frac {14}3$.

The diamter of the second smaller inscribed circle is

$\frac 13 \cdot \frac 23 \cdot 7 = \frac {14}9$

because all the circles are inscribed in similar equilateral triangles that have $\frac 13$ the height of the previuos equilateral triangle. We have an infinite geometric series with a common ratio of $\frac 13$ and a first term of $\frac {14}3$.

$\frac {\frac {14}{3}}{\frac 23} = 7$

We multiply this by three to find the sum of the diameters of the circles in all three corners of the equilateral triangle. The diamter of the first inscribed circle is $14 \pi$

The final answer is then $14 \pi + 21 \pi = 35 \pi$[/hide]

This is aa interesting problem. It involves infinite geometric series. I agree with everything tarquin said except for $\displaystyle \frac {{14}{3}}{\frac 23} = 7$. i think its probably a typo  :D

He has $\frac {\frac {14}{3}}{\frac 23} = 7$ which is true.  He doesn't have $143$ on top, he has $\frac {14}3$

That's only because I quickly edited it after realizing my mistake :)

You dog :)

Good solution.  I did not know diameter of incircle is $\frac 23$ height.  It makes sense though, because diameter is collinear with any median, and the center of the circle is the (point where medians meet), which separates all the medians into $\frac 23$.


Prove that iff $a,b,c$ are $3$ lengths of a triangle and $a+b+c=3$ then
\[ \frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}} \ge \frac{9}{ab+bc+ca} \]

put $x=a+b-c , y=b+c-a , z=c+a-b$ then $x+y+z=3$
we have to show that $\sum\frac{1}{\sqrt{x}}\geq \frac{36}{9+xy+yz+zx}$
or that
$(\sum \sqrt{a})(9+ \sum a)\geq 36 \sqrt[4]{abc}$ where $a=xy , b=yz , c=zx$but
$\sum \sqrt{a}\geq 3\sqrt[6]{abc}$ and $9+\sum a \geq 12\sqrt[12]{abc}$

Nice solution, Aquady, But I don't like expending.  :blush:

I feel that Albanian Eagle has somthing wrong when substution.

I think that   Albanian Eagle's proof is true and nice. :)


logarithms
calculus
derivative
function
real analysis
real analysis unsolved

For $ 1\leq n\in\mathbb{N}$ let $ u_{n}(x) \equal{} \frac {ln(1 \plus{} x^{n})}{n}$.  For which  $ x\in \mathbb{R}$ does $ f(x) \equal{} \sum_{n\geq 1}u_{n}(x)$  converge ?  For which $ x\in\mathbb{R}$  is $ f$ is differentiable?

-------------------------------------------

I couldnot use the convergent tests to solve for this problem? Are there any trick here? Thanks

The first "trick" is to break into cases before trying anything else.  Obvious points at which to consider separating the analysis are $ x \equal{} 1$, $ x \equal{} 0$ and $ x \equal{} \minus{} 1$.  That should be enough to help you answer the first question, at least.

For $ x\ge 1,$ we have $ \ln(1\plus{}x^n)\ge\ln 2$ and the (positive) series diverges by comparison with $ (\ln 2)\sum\frac1n.$

For $ x\le\minus{}1,$ we have $ \ln(1\plus{}x^n)$ undefined as a real number whenever $ n$ is odd, so the series cannot converge.

For $ 0\le x<1,$ we have $ \ln(1\plus{}x^n)\le x^n,$ so the series converges absolutely by comparison to the convergent power series $ \sum\frac{x^n}{n}.$

Now suppose $ \minus{}1<x\le 0.$ For $ x\le y\le 0$ we have that $ \frac yx\ln(1\plus{}x)\le\ln(1\plus{}y)\le 0$ and $ |\ln(1\minus{}y)|\le|\ln(1\plus{}y)|.$ Hence,

$ |\ln(1\plus{}x^n))\le \frac{|x|^n}{|x|}\ln(1\plus{}x)\equal{}C_x|x|^n.$

Once again, we have $ \sum\left|\frac{\ln(1\plus{}x^n)}n\right|\le C_x\sum\frac{|x|^n}n,$ which converges.

The series converges absolutely for $ \minus{}1<x<1.$ The convergence is uniform on any compact subset of that interval but not uniform on the whole interval.

A very similar argument (which I won't give in detail) shows that the series of derivatives of these terms converges uniformly on any compact subset of $ (\minus{}1,1).$ Hence the sum is differentiable on $ (\minus{}1,1).$

We can represent $ f$ by the double sum
\[ f(x)\equal{}\sum_{n\equal{}1}^{\infty}\sum_{k\equal{}1}^{\infty}\frac{(\minus{}1)^kx^{kn}}{kn}\]
although I haven't yet come up with any clever ways of using that expression.

What kind of number theoretic function of $ m$ is $ \sum_{k\mid m}(\minus{}1)^k\ ?$

Because I see that $ f(x)\equal{}\sum_{m\equal{}1}^{\infty}\left(\sum_{k\mid m}(\minus{}1)^k\right)\frac{x^m}m.$

The [url=http://www.research.att.com/~njas/sequences]OEIS[/url] entry is [url=http://www.research.att.com/~njas/sequences/A048272]A048272[/url].  (Actually that's for the negative of what you wanted, but close enough.)


An ant starts at one vertex of a unit cube and walks to the opposite vertex along the surface of the cube.  What is the minimum distance the ant can walk?

[hide="Hint"]NETS[/hide]

Without looking at the hint
[hide]Draw the net. It makes a right triangle with legs of 2 and 1 and a hypotenuse of $\sqrt 5$[/hide]

Could someone explain?  (With a picture?)

[quote="Treething"]You don't really need a net if you're decent at visualizing things. The shortest distance between two points is a straight line, so connect the two vertices with a straight of a line as possible - one edge plus half an edge, times two.

sqrt(1 + 1/4)
sqrt(5/4)
sqrt(5)/2

times two is sqrt(5).

It might seem longer, but it saves a ton of time.[/quote]

why not just do $\sqrt{1+4}=\sqrt{5}$...?


Hi i'm new to this forum, I just hope I am posting this in the right section..  :blush:

In grade 8 math I have a test tomorrow, and i'm stuck on how to solve questions like these.. if anyone could help it would be great.

Example:

Draw the net. Then, estimate and calculate the surface area of each polyhedron.

1.  [img]http://img296.imageshack.us/img296/915/mathproblem1351co.jpg[/img]

Break the 3-D figure into 2-D figures.
First, we look at the triangle closest to us.
We need the base and height to find the area. The base is 7 and the height is 6
Plugging into the formula
$A=\frac 12 bh$, we get
$A=\frac 12 (7)(6)$
Simplifying gives $A=21$, so the area is $21$
There is an exact same triangle on the other side so the total area of those two triangles is $42$
Now, looking at one of the rectangles,
it has a base of $15$ and a height of $7$. The area formula is
$A=bh$
Plugging in gives
$A=105$
There are 3 of such rectangles so the total are of the rectangles is $315$
Adding together with the triangles, the total surface area is $315+42=\boxed{357}$
As for estimating it, you can just guess "the triangle looks about 20 and rectangles 100 so $20+20+100+100+100=360$
For the net part, "unfold" the polyhedron and draw it.


AMC
AMC 12
AMC 10
AIME
USA(J)MO
USAMO
AMC 10 B

Which exam did you take and what was your score?

[quote="Elemennop"]AMC12 - 108.5

So many mistakes. I really need to focus more and not rush through a problem (especially problems such as #23).[/quote]

I thought you were a sophomore.  Taking the AMC10 gives you an advantage in making the USAMO.  Takers of the AMC10 only need to score the AIME floor to make the USAMO -- the lowest AIME score of any of the index qualifiers.

Yes, but I felt I would get more practice out of the AMC12, and do my best in my junior/senior year.

The only thing is I'm unsure of whether I should take the AMC10B or the AMC12B; I would've gotten 125+ if it weren't for careless mistakes, and several problems that I skipped without even seeing how easy it would be (#23).

Well, if you take the 10 B then you still give yourself that shot at qualifying through the AIME floor.

So with my current AMC12 score, I would be unable to get into the USAMO regardless of anything short of a near-perfect AIME score?

I'd like to ask some question about several problems:

You went over problem #24 in the Math Jam today, and I understood how there's 27 'possible' answers; however, I got lost when you canceled out five of them. Why exactly does 1 as a root of Q(X)-1, 2 as a root of Q(x)-2, and 3 as a root of Q(x)-3 not be applicable roots? I also completely didn't understand how you subtracted the last two incorrect roots as well.

Edit> I figured out this morning what all that information meant. It's a very elegant problem.

Also, can you explain #16, or at least give a hint of what to do? All attempts to find r/s resulted in me recieving an identity of the equation, which doesn't really help (and also wastes time, unfortunately).

[quote="Elemennop"]So with my current AMC12 score, I would be unable to get into the USAMO regardless of anything short of a near-perfect AIME score?[/quote]

An 11 would get you in most likely.

I got 112.5 on the AMC12.

One reason we're taking the 12 over the 10 is that our ARML team qualification is ranked by AMC12 scores (along with AIME and an ARML-style qualification round), so an AMC10 score makes it hard to evaluate. I guess now that I've gotten my 12 score, I can try the 10, but I don't want Kevin and Zhenya qualifying over me for ARML. :( (They both got 122, I think.)

About what would I need on AMC10 + AIME to qualify for USAMO?

It is impossible to tell what kind of scores you need without first knowing the results.

Taking the AMC 10, it's usually a good bet that you need a 7 or 8 on the AIME to qualify for the USAMO (I think?).

Here's what the AMC site says:

[quote]# The first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
# The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.
# The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
[/quote]

http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html

The USAMO index last year was 210, so aiming for more than that should be a reasonable goal in terms of trying to get into the USAMO.

I'm going to take the AMC12; I've really been practicing organizing my work (to help combat all those errors I make), and have been going through AoPS Volume 2 that we finally got at our school.

Note that last year's AIME floor was 7.  It's usually 6-8.


topology
group theory
abstract algebra
advanced fields
advanced fields unsolved

1)let the group $G$ be complete and have the property that for every $\epsilon>0$,every element $a\in G$ has a representation $a=a_{1}a_{2}\dots a_{n}$ where $d(a_{k},e)<\epsilon$.is the group connected in sense of hausdourf?(that is to say it cannot be represented as a sum of two disjoint closed set)
2)if answer to 1) is yes is is it arcwise connected?
(1) is still open but for locally compact answer is known :)

oh my gush i dont know why i post here . :blush: I do apologize moderator :(

But where to move it to? The problem with putting it in the Algebra section is that this question also has topology in it. OK, Advanced Topics it is.

I think I can prove (1) for locally compact groups without any assumptions of metrizability. More precisely:

Let $G$ be a locally compact group with the property that for every neighborhood $V$ of the identity $1$, the set of finite products of elements in $V$ is $G$. Then $G$ is connected.

Call the property in the statement $(P)$. Notice that if a group has property $(P)$, then so do all of its quotient groups. This observation allows us to factor through the connected component of the identity (which is a closed, connected, normal subgroup of $G$). Since the quotient is totally disconnected, we may as well assume that $G$ was totally disconnected to begin with. The aim is now to prove:

Let $G$ be a locally compact totally disconnected group with property $(P)$. Then $G$ is trivial.

Assume the contrary. We can then find $U\ne G$, an open neighborhood of $1$ whose closure $\overline U$ is compact. Because in a compact totally disconnected space the clopen sets form a basis, we can find a compact open neighborhood $W$ of $1$ which is contained in $U$. If we manage to find an open neighborhood $D$ of $1$ such that $WD=\{wd\ |\ w\in W,d\in D\}\subset W$, then we’re done: we can assume WLOG that $D\subset W$, and then the set of finite products of elements in $D$ cannot be the whole $G$, contradicting $(P)$.

Assume we cannot find $D$ as above. Then the system of compact sets $(WU_\alpha\setminus W)_\alpha$ where $U_\alpha$ ranges through the compact neighborhoods of $1$ has the finite intersection property, as can easily be seen, so it has non-empty intersection. Let $p$ be a point belonging to all $WU_\alpha\setminus W$. $p$ lies in the set $X\setminus W$ and, at the same time, in the closure of $W$, which is $W$. We have our contradiction.


Two eight-sided dice each have faces numbered 1 through 8. When the dice
are rolled, each face has an equal probability of appearing on the top.
What is the probability that the product of the two top numbers
is greater than their sum? Express your answer as a common fraction.

If $ xy>x\plus{}y$, then $ (x\minus{}1)(y\minus{}1)>1\equiv(x\minus{}1)(y\minus{}1)\geq2$. Thus, the only time that this does not hold is when $ 1\leq x,y\leq2$, so there are $ 4$ bad cases.

Finally, our probability is $ 1\minus{}\frac4{8^2}\equal{}\boxed{\frac{15}{16}}$.

1 is the multiplicative identity, so 1*x doesnt increase while 1+x does.
Remove the eight multiplications 1 * x  and the 7  x* 1. total 15
2 * 2 = 2 + 4 so remove that too.

16 don't work so 64-16 do.  so 48/64 = 3/4

Math154, for your way we must discount

11
12
13
14
15
16
17
18
21
31
41
51
61
71
81
22

OR 16 possibilities.

Yeah, that's right. I am good at making careless mistakes.

I solved this by making an 8 by 8 table and checking each sum and product.  I like the smart algebraic solution and wish I'd thought of it. 


How many integers are common solutions to these three inequalities?

\begin{tabular}{cccc}
(1) & $ -3y$ & $ \geq$ & $ y+7$ &
(2) & $ -2y$ & $ \leq$ & 12 &
(3) & $ -4y$ & $ \geq$ & $ 2y+17$
\end{tabular}

[quote="GameBot"]How many integers are common solutions to these three inequalities?

\begin{tabular}{cccc}
(1) & - 3y & \geq & y + 7 &
(2) & - 2y & \leq & 12 &
(3) & - 4y & \geq & 2y + 17
\end{tabular}[/quote]

Look at the second equation.  All solutions for that are $ y > -6$.

For the first equation, all solutions are $ y \le -2$

For the third equation, all solutions are $ y \le -3$

So our $ \boxed{\text{four}}$ integral solutions are $ \{-3,-4,-5,-6\}$.

is it just me or are all the equations red? also i think you ment for the second equation greater than or equal to.