Hello, everyone. this is my first post here. i do not do physics, i am asking these questions on behalf of my cousin who needs it urgently and does not have access to a computer at the moment. it's quite a few problems. for the first, i tried using the suvat equation v^2 = u^2 + 2as to find the acceleration and then use F = ma to find the force, but the numbers don't look right to me. for the rest, i have no clue. any assistance is much appreciated

[color=red]Please be advised, i posted these questions on two other forums, so you may check to see if a question you're planning on answering has been addressed. so far, (4.9) has been answered, i will try to keep you up-to-date[/color]

http://www.mathhelpforum.com/math-help/urgent-homework-help/20141-jhevon-needs-help-physics.html

http://www.physicsforums.com/showthread.php?p=1458680&posted=1

4.9)

A 0.140 kilogram baseball traveling 35.0 m/s strikes the cather's mitt, which, in bringing the ball to rest, recoils backward 0.11 meters. What was the average force applied by the ball on the glove?

4.25)

One 3.20 kilogram bucket is hanging by a massless cord from another 3.20-kilogram bucket, also hanging by a massless cord. (a) is the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.60 meters per second squared by the upper cord, calculate the tension in each cord.

4.54)

A roller coaster reaches the top of the steepest hill with a speed of 6.00 kilometers per hour (1.67 meters per second). It then descends the hill, which is at an average angle of 45.0 degrees and is 45.0 meters long. What is its speed when it reaches the bottom? Assume the kinetic coefficient of friction is 0.18.

5.15)

How many revolutions per minute would a 15-meter-diameter Ferris wheel need to make for its passengers to feel "weightless" at the topmost point?

5.37)

A typical white dwarf star, which was once an average star like our sun but is now in the last stage of its evolution, is the size of our moon but has the mass of our sun. What is the surface gravity on this star?

5.48)

During an Apollo lunar landing mission, the command module continued to orbit the moon at an altitude of about 100 kilometers. How long did it take to go around the moon once?

The problems were addressed at [url]http://www.mathhelpforum.com/math-help/urgent-homework-help/20141-jhevon-needs-help-physics.html[/url] if any of you are interested.

5.37)

A typical white dwarf star, which was once an average star like our sun but is now in the last stage of its evolution, is the size of our moon but has the mass of our sun. What is the surface gravity on this star?

The radius of the moon is roughly 1737.5 kilometers (about .27272 times Earth's mean radius), and the sun's mass is 332,950 Earths. So, using the formula $ g\equal{}(m/r^{2})$ where $ g$ is the surface gravity, expressed as a multiple of Earth's, $ m$ is the mass, also a multiple of Earth's, and $ r$ is the radius of the object, I get:

$ g\equal{}(332,950/.27272^{2})$ or about $ 4,476,567$ times Earth's gravity. Multiply by $ 9.8 m/s^{2}$ to get around $ 43,870,351 m/s^{2}$ for the surface gravity of the white dwarf.


http://www.artofproblemsolving.com/Classes/IntermAlg/PostTest.pdf

Can somebody post their solutions to these problems? They're from the Post-Test of the AoPS Intermediate Algebra course. Thanks.

We would prefer that students not start threads on these problems in particular so that the problems will be completely fresh for any student who first looks at those tests to determine if they can answer all of the problems on their own.


Princeton college AMC USA(J)MO USAMO AIME calculus

I will be a freshman at Princeton this fall and am wondering what course I should register for. The placement guide says 215, but I am worried that I will be in that class with a bunch of USAMO guys. Does anyone have experience with the honors track at Princeton or have knowledge of the level of mathematical ability needed to succeed in it (as measured by AIME score or whatever other metric you think appropriate)?

Uh I never qualified for USAMO and I aced MAT215 freshman year. That probably is the right class to start out with if you're considering majoring in math (assuming you haven't studied analysis before). Is Gunning teaching it? He's a pretty good prof.

I'd say if you've done contest math and you've written proofs for contest math before, you'll be fine for 215. If you haven't ever written proofs before, then you will have to work a bit harder, but you aren't [i]expected[/i] to know how to write proofs before you take that class. There are usually a good number of people in the class who have little to no math contest experience. (For example, two of my roommates this year were in that situation when they took the class freshman fall. They both got good grades.)

ps I'm a math major at Princeton. If you have any other questions, I'll answer those too (as well as I can, at least).

I would describe my proof experience as minimal. I read all of the proofs in my Stewart Calculus text until I understand them, but I have very little occasion to write proofs of my own aside from the occasional extra credit problem. At April Hosting I talked to Oliver Conway (Professor Conway's son) and he said 215 was easy, so I figure that means if I work really hard I can get an A-. I've already taken linear algebra so 217 doesn't worry me too much. It's 218 that I'm pretty nervous about. Could you comment on the difficulty of this course? Also, were you inconvenienced in other subjects such as physics and economics by not taking multivariable until sophomore year?

I haven't taken 218 and I'm not going to. That class probably won't teach you very much that's useful for computational purposes in classes like physics or economics. I took a class at Texas A&M on multivariable calculus and another on diffyqs, which gave me the basics necessary for doing computations in physics. The physics textbooks they use teach you the multivar you need, so you can pick it up on the fly if necessary.

The difficulty of 218 varies a lot depending on who's teaching it. Also the topics covered vary a bit. You can skip this class by taking one of the "computational" multivar classes (201/203) or by taking one of the real analysis classes (314/332) (which would teach you the general theory of integration on $\mathbb{R}^{n}$).

If you work really hard you can get an A. The grading is not THAT harsh here. Also, Oliver Conway is a cool dude. He and I are in the same compsci class.

Ok thanks a lot for all of the info you have provided. One last question: this fall will I register for one year's worth of classes or one semester's worth?

How hard is it to persuade the advisors to let you take 5 classes in the first semester (two of them math)? I heard from a number of people that it's almost impossible...

It depends on your advisor. Some are more lenient than others. But actually, here are two things you should know:

1) Here, your advisor is just an advisor. You sign up for courses yourself. So you can just sign up for 5 classes even if your advisor tells you not to. Basically, the only thing that will happen is that if you end up doing poorly, the university won't have any sympathy for you. In freshman spring, I signed up for 6 classes and told my advisor I was going to pick one of MAT330 and MAT332, and I liked them both so I just stayed in all six classes. When I met him later he was kinda miffed, but I suffered no ill consequences.

2) If your writing seminar is fall semester, you might want to just take 4 classes anyway. I took 4 classes first semester, and that gave me the opportunity to get used to Princeton and to survive my writing sem. Ever since then I've been taking much more hardcore schedules. But it's okay to go easy one semester.

crap, this writing seminar seems like a really difficult class...
about how much time (hours) a week did your writing seminar take? hopefully this is not the lit/writing equivalent of math 55... :P

It's not horrible. I don't really remember how much time I spent on it. The problem isn't that it's a lot of work all the time -- that's what math classes are for. The thing that makes writing seminar hard is that occasionally you have a lot of work and occasionally you have none. You think, "oh I'll only have 4 essays over the whole semester, it's like nothing at all." But then on the weeks where you have your essays to write, you will have a hard time doing a good job on the essay and still getting all of your usual work done.

I spend a lot more time on my math classes now than I ever did on the writing seminar. It's just that high school classes (and even college classes at community colleges/large state universities) don't really prepare you for Ivy League type classes. You are expected to do a lot more on your own, the pace is much faster, and the work is much harder. It's good to be conservative your first semester so that you can get accustomed to what Princeton is like. Then you can adjust your workload in future semesters to match what you feel you can handle. I wouldn't be surprised if you are capable of doing well with 5 or 6 courses, but (in my opinion) it's not worth the risk to try that first semester. (You will also have an easier time convincing your advisor to let you do what you want after the first semester.)

Another reason to go easy on classes first semester is that there's stuff to do at Princeton other than classes and you want to be free to find out what some of that stuff is. The Math Club is very active (we often have 2 or more events in a week), so if you wanted to go to all of the Math Club events that's a nontrivial time commitment. If you want to join a music or acting or dancing performance group or a sports team, those have a fair amount of practice. I think it is good to be able to explore the extracurricular options during your first semester. (Of course, taking 5-6 classes won't stop you from doing that, but it would make it harder.)

i know that they can still make the class really hard, but what if you carefully went over the text for 215 during the summer? would it still be as hard? in other words, is it the material that's difficult or the problem sets?

good point about life... i didn't factor in the fact that i would probably not want to work all the time...

If you carefully go over baby Rudin during the summer, 215 will probably be easy for you. Those 21x classes usually aren't really hard unless Fefferman is teaching. When Conway teaches he tends to go very quickly and add in random stuff, but I don't think it's ridiculously hard. I'm not aware of any of the other profs who usually teach 21x classes having reputations of being hard in any way.

Math people probably won't find the first couple of math classes to be their hardest classes. I think the hardest first year classes for people who like math tend to be classes with a lot of lab work and classes with a lot of writing. My first semester grades were actually my worst grades (on average), even though it was the lighted course load in number of classes. I think this is because it was the only semester when I had just one math class. ;) Also, I wasn't prepared for how much effort I'd have to put into my classes outside of the lectures. But MAT 215 didn't cause me any trouble at all. In fact I loved it and because of that class I am pretty much for sure going into some part of analysis as my field.

(And again, I had no prior experience with analysis. I had proofwriting experience from math contests, but I never qualified for USAMO, so it's not like I dominated HS contests either.)

Any special bits of insight on Professor Gunning?

Prof Gunning is great. I had him for 217. He's also the undergraduate advisor.

Pros: He's funny. He goes at a good and consistent pace. He teaches exactly the material you need to know. He follows the book, but not blindly (i.e., the book is useful because it follows the lectures, but he's not just reading from the text). He speaks clearly.

Cons: Sometimes he will get mixed up with his notation in the middle of doing computations on the board. This can be confusing. Don't hesitate to speak up if something like this happens; it may be a bit embarrassing at first to "admit you don't know what's going on," but it's really better to stop that kind of problem as soon as possible.


OK, these are the rules for this rebirth of a word game created by abcadaea:

1) A word of at least seven letters is posted.
2) You must post words made with letters of the original word. Your score is the number of letters of the word. If you make an anagram of the complete word, your score is double!
3) If you think that there are no more anagrams, post FORFEIT, and post a new word. You receive a three point bonus.
4) If you post FORFEIT and someone else find another anagram, you receive a penalty of the number of letters in the word.
5) If you post a non-existent word, you receive a 3-point penalty, and the points you earned with the inexistant word are substracted, too.

First word: ONOMATOPOEIA (very easy!!!)

patio

1=2: 10
mz94: 10
alexhhmun: 5
dynamo729: 5

poet

mz94: 14
1=2: 10
alexhhmun: 5
dynamo729: 5

that was found by mz94.

mz94: 14
1=2: 10
alexhhmun: 5
dynamo729: 5

Oh sorry, I meant moon.

moan

mz94: 14
1=2: 10
alexhhmun: 5
dynamo729: 5
chenhsi: 4

he posted that in the first post, but ok I will think you mean moan.

potion

1=2: 16
mz94: 14
alexhhmun: 5
dynamo729: 5
chenhsi: 4

top

1=2: 16
mz94: 14
alexhhmun: 5
dynamo729: 5
chenhsi: 7

tempo

1=2: 21
mz94: 14
chenhsi: 7
alexhhmun: 5
dynamo729: 5

Point

1=2: 21
mz94: 14
chenhsi: 7
alexhhmun: 10
dynamo729: 5

meat

hmm... delicious meat

mz94: 86
alexhhmun: 73
chenhsi: 56
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8
pacman2812: 4

[quote="mz94"]meat

mz94: 35
1=2: 30
dynamo729: 15
alexhhmun: 14
chenhsi: 10[/quote]

Try again.

Name

mz94: 86
alexhhmun: 77
chenhsi: 56
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8
pacman2812: 4

Amen

after you pray you say amen

mz94: 90
alexhhmun: 77
chenhsi: 56
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

am

I am Bob

mz94: 90
alexhhmun: 77
chenhsi: 58
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

I

I am Bob
 :P

mz94: 91
alexhhmun: 77
chenhsi: 58
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

to

From here to there.

mz94: 91
alexhhmun: 77
chenhsi: 60
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

A

i saw a bug

mz94: 92
alexhhmun: 77
chenhsi: 60
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

I

I saw a bug :)

mz94: 92
alexhhmun: 77
chenhsi: 61
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

[quote="mz94"]I

I am Bob
 :P

[/quote]

try again :P

No

No, I saw a bug

mz94: 92
alexhhmun: 77
chenhsi: 62
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

on

i sat on the bug

mz94: 94
alexhhmun: 77
chenhsi: 62
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

moo

moo is the sound that a cow makes

mz94: 94
alexhhmun: 77
chenhsi: 65
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

can we post again if its been like 2 hours?

ape

you look like an ape

mz94: 94
alexhhmun: 77
chenhsi: 65
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8

pea

you look like an pea

mz94: 94
alexhhmun: 77
chenhsi: 68
1=2: 34
dynamo729: 15
mewto55555: 8
metafor: 8


3x^2 - 2y^2 = 1  where x,y are integer >1

  :?  :?  :?

dude, the question says 3x^2 - 2y^2 = 1, not 0.

I think there are infinitely many solutions, but still trying to find that magical generating algorithm... :D

it's Pell like equation just take $3x=t$:
$t^2-6y^2=3$
it has sloution:
$(3,1)$
the basic equation
$a^2-6b^2=1$
has solution $(5,2)$
so the solutions has form :
$t_0=3$  $y_0=1$
$t_{n+1}=5t_n+12y_n$
$y_{n+1}=2t_n+5y_n$
and
$x_n= \frac{t_n}{3}$

for example second solution is $(9,11)$

you forgot about $t=2y$ and also you still (cause that's not sufficient to find all pairs) have to check whether there aren't any other solutions left.


Hi.  for my geometry class, we are starting to get into circles and I often have to refer to arcs.  However, the closest I can get to an arc is $\widehat{arc}$.  is there anything better?


probability geometry rectangle inequalities

In the diagram, if $ x$ is chosen at random from the set $ \{1,2,3,4,5,6\}$, what is the probability that the area of the square is greater than $ 50\%$ of the area of the rectangle? Express your answer as a common fraction.
[asy]draw((0,0)--(10,0)--(10,6)--(0,6)--cycle);
draw((12,0)--(16,0)--(16,4)--(12,4)--cycle);
label("$x+1$",(0,3),W);
label("$x+3$",(5,0),S);
label("$x$",(14,0),S);
label("$x$",(16,2),E);[/asy]

We just plug in:
$ x\equal{}1$, $ 1^2<0.5\cdot (1\plus{}1)(1\plus{}3)$
$ x\equal{}2$, $ 2^2<0.5\cdot (2\plus{}1)(2\plus{}3)$
$ x\equal{}3$, $ 3^2<0.5\cdot (3\plus{}1)(3\plus{}3)$
$ x\equal{}4$, $ 4^2<0.5\cdot (4\plus{}1)(4\plus{}3)$
$ x\equal{}5$, $ 5^2>0.5\cdot (5\plus{}1)(5\plus{}3)$
$ x\equal{}6$, $ 6^2>0.5\cdot (6\plus{}1)(6\plus{}3)$

Two of the six work, so $ \frac{2}{6}\equal{}\boxed{\frac{1}{3}}$

If you wanted to, you could also solve the inequality $ x^2>0.5(x\plus{}1)(x\plus{}3)$, but since the numbers are small this isn't necessary for the problem


Find the value of $ m$ such that the complex roots of the equation $ x^3\minus{}x\plus{}m\equal{}0$ are all integer.

[quote="xxxxtt"]Find the value of $ m$ such that the complex roots of the equation $ x^3 \minus{} x \plus{} m \equal{} 0$ are all integer.[/quote]
$ m \equal{} 0$ trivial with roots $ \minus{} 1,0,1$

for $ m \neq 0$ let $ a,b,c$ be non zero integer roots of that equation then :
WLOG $ |a| \le |b| \le |c|$
$ \minus{} abc \equal{} m$
$ ab \plus{} bc \plus{} ca \equal{} \minus{} 1$
$ a \plus{} b \plus{} c \equal{} 0$

$ 0 \equal{} (a \plus{} b \plus{} c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca) \equal{} a^2 \plus{} b^2 \plus{} c^2 \minus{} 2$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
$ 1 \le a^2$ then $ 3 \le a^2 \plus{} b^2 \plus{} c^2$ contradiction.

Moreover you can generalize for $ x^3\minus{} k^2 x \plus{}m\equal{}0$ if $ 2k^2$ cannot be expressed as a sum of three positive square numbers.


So... anyone read The Catcher and the Rye??
Do you think Holden is a likeable character?
State your thoughts!

maybe we can get a debate going XD

I like him, he thinks the way everybody in the world thinks, yet he's nice as hell to everyone. Very likeable.  :idea:  :idea:

[quote]I like him, he thinks the way everybody in the world thinks, yet he's nice as - to everyone.[/quote]

lol.. I can tell you just finished reading that book....

I think Holden Caulfield is a cool guy.

Oh look, Aunt Sally from Huck Finn is here too.

lol nah i didnt finish reading it, im in the middle of it, reading it alot cuz i like it  :P

Holden is a cool guy definitely, but he is a little misguided in the way he looks at the world

But I hate when people try to psychoanalyze him, like hes a hypocrite and everything...

Anyway, it's a really good book

It's also the same book as the Odyssey. Holden Cauliflower is like Ulysses.

I just finished the book.
Holden is a tremendous character, he cares deeply of his sister and handles all situations well. He is a great guy to be with, very likeable. I dont understand, why do phsycokillers say this book cause them to become killers.... ?


Mrs. Jones has a back yard in the shape of a square that is $ 27$ feet on each side. After dividing each side with sides $ \frac{1}{3}$ the original sides, she wants to plant grass in the shaded areas show in the diagram. How many square feet of the backyard will remain without grass?

[asy]filldraw((0,0)--(6,0)--(6,6)--(0,6)--cycle,white);
filldraw((0,0)--(2,0)--(0,2)--cycle,black);
filldraw((4,0)--(6,0)--(6,2)--cycle,black);
filldraw((6,4)--(4,6)--(6,6)--cycle,black);
filldraw((0,4)--(2,6)--(0,6)--cycle,black);
filldraw((2,2)--(4,2)--(4,4)--(2,4)--cycle,black);[/asy]

[quote="GameBot"]Mrs. Jones has a back yard in the shape of a square that is $ 27$ feet on each side. After dividing each side with sides $ \frac {1}{3}$ the original sides, she wants to plant grass in the shaded areas show in the diagram. How many square feet of the backyard will remain without grass?

Each triangle is 1/2bh = 1/2(9)(9) = 40.5 X 4 = 162.

The square in the middle = 9 ft. per side for an area of 81. 162 + 81 = 243 sq. ft. of grassed area.

The whole yard is 27 * 27 = 729. The question asks for the sq. ft. of the backyard that will remain without grass so

729 - 243 = 486 grassless sq. ft.


geometry
3D geometry
algebra
polynomial
factorization
sum of cubes

I'm a bit interested to know how one might derive the sum of squares formula.

Also, it's kind of obvious for the sum of integers and the sum of cubes formulas, (just notice a pattern, write it down, prove it), but is there a "mathematical" derivation way? And how might this be applied to 4th powers and 5th powers and so on?

This has been posted a number of times recently; see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=222591]here[/url] for example.

First off, this topic probably belongs in Intermediate Topics.

The answer to the general case is highly complicated, but there are several general methods for finding $ \sum_{n\equal{}1}^k n^p$ for a given $ p$.

Method 1:
write $ (n\plus{}1)^{p\plus{}1} \minus{} n^{p\plus{}1} \equal{} Q(n)$, where $ Q$ is the binomial expansion of the LHS of the equation.  $ Q$ is then a polynomial with degree $ p$.  Now if we sum this equation up the LHS telescopes: $ \sum_{n\equal{}1}^k (n\plus{}1)^{p\plus{}1} \minus{} n^{p\plus{}1} \equal{} (k\plus{}1)^{p\plus{}1} \minus{} 1 \equal{} \sum_{n\equal{}1}^k Q(n)$.  If we let $ Q(n) \equal{} a_p n^p \plus{} a_{p\minus{}1} n^{p\minus{}1} \plus{} \cdots \plus{} a_1 n \plus{} 1$ we can rewrite the RHS: $ (k\plus{}1)^{p\plus{}1} \minus{} 1 \equal{} a_p\sum_{n\equal{}1}^k n^p \plus{} a_{p\minus{}1}\sum_{n\equal{}1}^k n^{p\minus{}1} \plus{} \cdots \plus{}a_1 \sum_{n\equal{}1}^k n \plus{} k$
So to find $ \sum_{n\equal{}1}^k n^p$ we need to find plug in all the formulas with smaller exponents.

So for $ \sum_{n\equal{}1}^k n^2$ we have $ (n\plus{}1)^3 \minus{} n^3 \equal{} 3n^2\plus{}3n\plus{}1$
Summing from $ n\equal{}1$ to $ n\equal{}k$ we obtain $ (k\plus{}1)^3 \minus{}1 \equal{} 3\sum_{n\equal{}1}^k n^2 \plus{} 3\sum_{n\equal{}1}^k n \plus{} k$
Now since $ \sum_{n\equal{}1}^k n \equal{} \frac{k(k\plus{}1)}{2}$ we have
$ k^3 \plus{} 3k^2 \plus{} 3k \equal{} 3\sum_{n\equal{}1}^k n^2 \plus{} \frac{3k(k\plus{}1)}{2} \plus{} k$
$ 2k^3 \plus{} 6k^2 \plus{} 4k \minus{} (3k^2 \plus{} 3k) \equal{} 6 \sum_{n\equal{}1}^k n^2$
$ 2k^3 \plus{} 3k^2 \plus{} k \equal{} 6\sum_{n\equal{}1}^k n^2$
$ \frac{k(k\plus{}1)(2k\plus{}1)}{6} \equal{} \sum_{n\equal{}1}^k n^2$

Method 2:
express $ x^p$ in terms of the newton polynomials, and then sum up this new representation using hockey stick theorem: [url]http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identities[/url].
Discussed here, in problem 1: [url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381[/url]

The general solution to this problem is Faulhaber's formula: [url]http://mathworld.wolfram.com/FaulhabersFormula.html[/url]


linear algebra
matrix
linear algebra unsolved

If A is a real skew symmetric matrix, show that B is orthogonal, where B = (A + I)(A - I)^-1 (inverse only on the second matrix).

I came as far as showing that BB^T = PQ^-1P^-1Q, where P is A + I, and Q is A - I. But I need to show some kind of commutativity to prove this. So close ...

$ P$ and $ Q$ trivially commute.  $ A$ generates a commutative algebra.

I'm sorry, I don't know what a commutative algebra is. Is there some other way of proving it?

All I mean is that powers of $ A$ commute with each other, and so do linear combinations of powers of $ A$.  Have you actually computed $ PQ$ and $ QP$?

[quote="kurt.godel"]If $ A$ is a real skew symmetric matrix, show that $ B$ is orthogonal, where $ B \equal{} (A \plus{} I)(A \minus{} I)^{ \minus{} 1}$ (inverse only on the second matrix).

I came as far as showing that $ BB^T \equal{} PQ^{ \minus{} 1}P^ { \minus{} 1}Q$, where $ P$ is $ A \plus{} I$, and $ Q$ is $ A \minus{} I$. But I need to show some kind of commutativity to prove this. So close ...[/quote]
 The fist  if $ C \in GL_n{(\mathbb{R})}$ then $ {(C^{ \minus{} 1})}^{T} \equal{} {(C^{T})}^{ \minus{} 1}$ ,$ A \minus{} I \in GL_n{(\mathbb{R})}$ then $ \minus{} (A \plus{} I) \in GL_n{(\mathbb{R})}$ or $ (A \plus{} I) \in GL_n{(\mathbb{R})}$
Use it $ B^{T} \equal{} (A^T \minus{} I)^{ \minus{} 1}(A^T \plus{} I) \equal{} ( \minus{} A \minus{} I)^{ \minus{} 1}( \minus{} A \plus{} I) \equal{} (A \plus{} I)^{ \minus{} 1}(A \minus{} I)$
Then $ BB^T \equal{} (A \plus{} I)(A \minus{} I)^{ \minus{} 1}(A \plus{} I)^{ \minus{} 1}(A \minus{} I) \equal{} (A \plus{} I)(A \plus{} I)^{ \minus{} 1}(A \minus{} I)(A \minus{} I)^{ \minus{} 1} \equal{} I$


calculus
geometry
Harvard
college
function
linear algebra
real analysis

Please post in this format on what you have taken/ plan to take:

[u][b]My plans:[/b][/u]
[b]10th grade:[/b] AP Calculus BC
[b]10th grade summer:[/b] Multivariable Calculus
[b]11th grade fall: [/b]Intro to Linear Algebra
[b]11th grade spring:[/b] Linear Algebra
[b]11th grade summer:[/b] Intro to Ordinary Differential Equations
In 12th grade, I had an extra period open at school so I took 2 classes:
[b]12th grade fall:[/b] Ordinary Differential Equations
                          & at the same time
[b]12th grade fall: [/b]Intro to Partial Differential Equations
[b]12th grade spring:[/b] Partial Differential Equations
                          & at the same time
[b]12th grade spring:[/b] Real Analysis

Now, some questions:
1. In how much detail are your classes taught compared to a top-rate college, such as MIT/Caltech?
2. Do you have to take intro's to _______?
3. What one is normal to take after Calculus: Multivariable Calc, Linear Algebra, or ODE?

Thanks and feel free to share!

I'll answer for my son A+MATH, because he is off at his summer program. Speaking only of CURRICULAR math, he is in a five year course with this sequence:

1st year: algebra 1 and algebra 2
2nd year: geometry and math analysis [precalculus]
3rd year: university honors calculus I
4th year: university honors calculus II
5th year: university honors calculus III, and optional advanced topics

You can get more detailed syllabus information about my son's program, UMTYMP, on [url=http://www.itcep.umn.edu/umtymp/]its Web site[/url], and I can tell you that Ivy League admissions officers are VERY familiar with this program, as several [url=http://www.itcep.umn.edu/annual/aluminfo.html]alumni of UMTYMP[/url] have gone to each of the top math schools in the United States over the years. Harvard flatly doesn't grant credit toward Harvard graduation for credits from another university gained by someone of high school age, but Harvard does take into account previous college credits earned by applicants (which a HIGH percentage of all admitted applicants have) for placement into Harvard courses. I think UMTYMP would serve as sufficient preparation for Harvard's [url=http://abel.math.harvard.edu/~laurent/math25/math25.html]Math 25[/url] course, and UMTYMP supplemented with other locally available courses would be ample preparation for Harvard's legendary [url=http://www.math.harvard.edu/~siu/math55b/index.html]Math 55[/url] course. But, no, I don't think my son's curricular math course will be a match for the college courses in the top colleges in the United States, but then again he will be only of junior high or senior high age when he is taking them. He has completed the first year of UMTYMP at age eleven.

Now, as for what else he might take, we are still investigating that, but he will likely take a lot of AoPS online courses and more summer courses like the JHU CTY course he is taking right now. One curricular math sequence is not enough. Falling into the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]calculus trap[/url] doesn't guarantee a good math education.

Hope this helps! Good luck on your ambitious plans.

I took AP Calculus BC in ninth grade, Differential Equations in Tenth Grade, and will take either linear algebra or multivariable calc during my current eleventh grade year (I already know both but will not be allowed to skip them).  However, most of my math I have not learned in classes.  After Calc BC, I started doing ALOT of reading on my own.  I would reccomend that if you REALLY want to learn to do math, and to solve problems in clever ways and such, that you read on your own.  The books I would reccomend are A Course of Modern Analysis by Whittaker and Watson, an Introduction to the Theory of Numbers by Hardy and Wright, An Introduction to Algebraic Structures by Landin, The Theory of Functions by Titchmarsh, Mathematics of Classical and Quantum Physics by Byron and Fuller, Special Functions by Andrews, Askey, and Roy, and Real Analysis by Haaser and Sullivan.  All of these books are great (especially a Course of Modern Analysis, which is worth its weight in gold) and will teach you ALOT.  I find that the best way to learn mathematics, or anything for that matter, is to find out how you learn most effectively, and then teach it to yourself.


sec^4x-tan^4x=1+2tan^2x

tan^3a-1/tan^a-1=tan^2a+tana+1

Every method I did leads to a dead end

well..what exactly did you try?
the method im using requires a straightforward application of the trig identities. gets kinda complicated though.  :oops:

[hide="hint"]$ 1 \plus{} \tan^2x \equal{} \sec^2x$.[/hide]

EDIT:Might as well prove:
[hide="#1"]Factor:
$ \text{LHS} \equal{} (\sec^2x \minus{} \tan^2x)(\sec^2x \plus{} \tan^2x) \equal{} 1(\sec^2x \plus{} tan^2x).$.
We know $ 1 \plus{} 2\tan^2x \equal{} \sec^2x \plus{} \tan^2x$.
$ \sec^2x \plus{} \tan^2x \equal{} \sec^2x \plus{} \tan^2x$.
[/hide]

[hide]
The lack of parenthesis makes your questions unclear, so I just follow the most reasonable interpretation of the identities.
1.
$ \sec^4x \minus{} \tan^4x
\equal{} (\sec^2x \minus{} \tan^2x)(\sec^2x \plus{} \tan^2x)
\equal{} \sec^2x \plus{} \tan^2x
\equal{} \sec^2x \minus{} \tan^2x \plus{} 2\tan^2x
\equal{} 1 \plus{} 2\tan^2x$
2.
cubic difference factorization: $ x^3 \minus{} 1 \equal{} (x \minus{} 1)(x^2 \plus{} x \plus{} 1)$, apply it to the identity.

...No need of any tirg. identities....[/hide]

Actually, you did use $ 1\plus{}\tan^2x\equal{}\sec^2x$ :wink:

Well, I usually consider it a corollary of the Pythagorean Theorem...

Well...I guess, although it is an identity.


calculus
integration
trigonometry
Euler
real analysis
real analysis unsolved

Find
$ \int_0^{ \plus{} \infty } {\frac{{\cos \left( {x^n } \right) \minus{} \exp \left( { \minus{} x^m } \right)}}
{x}dx}$
If m ,n > 0

hello, for $ m\equal{}n\equal{}1$ i have got
$ \int_{0}^{\infty}\frac{\cos(x)\minus{}e^{\minus{}x}}{x}\,dx\equal{}0$
Sonnhard.

That's an irrelevant result.

Can you compute the general integral?


See http://www.mathlinks.ro/Forum/viewtopic.php?t=46496 .

  darij


function
integration
calculus
calculus computations

Hey,Could someone help me to calculate the fourier transform of

$ \frac{J_1(a\sqrt{x^2\plus{}b^2})}{\sqrt{x^2\plus{}b^2}}$

where $ J_1(.)$ is the first order bessel function of the first kind.

i.e.

calculate

$ \int_{\minus{}\infty}^{\infty} \frac{J_1(a\sqrt{x^2\plus{}b^2})}{\sqrt{x^2\plus{}b^2}} \exp(\minus{}i2\pi{wx})dx$

Thank you!!


Let $ n$ be an integer such that $ n>3$. Suppose $ a,b\in\mathbb N$ satisfy $ 2^n\equal{}10a\plus{}b$ and $ b<10$. Find the number $ k$ such that $ k|ab$ for all $ a$ and $ b$.

In other word, let $ S\equal{}\{(a,b)\in\mathbb N^2| 10a\plus{}b\equal{}2^n, b<10, n>3, n\in\mathbb N\}$. Find $ k\equal{}\gcd\{ab|a,b\in S\}$.

[hide]6

$ (3\minus{}1)^{n}\equal{}10a\plus{}b$

when $ n\equal{} 4k\plus{}1, b\equal{} 2$, and the last term of $ RHS \equal{} \minus{}1$
        $ n \equal{}4k\plus{}2, b\equal{} 4$, and the last term of $ RHS \equal{} 1$
         $ n \equal{}4k\plus{}3, b\equal{} 8$, and the last term of $ RHS \equal{} \minus{}1$
         $ n \equal{}4k, b\equal{} 6$

So when $ n \ne 4k$, $ a$ is always divisble by $ 3$

[/hide]


geometry
power of a point
radical axis
angle bisector
geometry proposed

It is given a triangle $ ABC$. Let $ X$ is a point on the straight line $ BC$ such that $ C$ is between $ B$ and $ X$.
The incircles of $ \triangle ABX$ and $ \triangle ACX$ intersect in points $ P$ and $ Q$. Prove that $ PQ$ passes through a fixed point, which doesn't depend on the position of $ X$.

PS Look only into the cases when the incircles intersect themselves .  :wink:

$ (I)$ is incircle of $ \triangle ABX,$ touching $ BX, XA$ at $ D, E.$ $ (J)$ is incircle of $ \triangle ACX,$ touching $ CX, XA$ at $ U, V.$ $ B', C', X'$ are midpoints of $ AB, AC, AX.$ $ DE$ cuts the midline $ B'X' \parallel BX$ of $ \triangle ABX$ at $ K.$

$ \overline{KX'} \equal{} \overline{X'E} \equal{} \overline{XE} \minus{} \overline{XX'} \equal{} \frac{_1}{^2}(\overline{BX} \minus{} \overline{AB})\ \Longrightarrow$
$ \overline{B'K} \equal{} \overline{B'X'} \minus{} \overline{KX'} \equal{} \frac{_1}{^2}(\overline{BX} \minus{} (\overline{BX} \minus{} \overline{AB})) \equal{} \frac{_1}{^2}\overline{AB}$

Then $ \angle XBK \equal{} \angle BKB' \equal{} \angle B'BK$ $ \Longrightarrow$ $ BK \equiv BI$ bisects the angle $ \angle ABX.$ [hide]This gives another proof of  [url]http://www.mathlinks.ro/viewtopic.php?t=215418[/url]. [/hide]. Similarly, $ UV$ cuts  the midline $ C'X' \parallel CX$ of $ \triangle ACX$ at $ L$ and $ CL \equiv CJ$ bisects the angle $ \angle ACX.$ Then $ DKLU$ is a parallelogram. Whether the incircles $ (I), (J)$ intersect or not, their radical axis is the midparallel $ m$ of $ DE \parallel UV$ or $ DK \parallel UL$ (both perpendicular to the angle bisector $ XJI$), cutting the fixed segment  $ KL$ at its midpoint $ M.$


geometry
incenter
geometric transformation
reflection
homothety
circumcircle
trigonometry

Dear Mathlinkers,
let ABC a triangle, A' the foot of the A-altitude of ABC, I the incenter of ABC, X the point of intersection of the perpendicular issue from I with BC, X' the symmetric of X wrt AI and Fe the Feuerbach's point of ABC.
Prove synthetically that the circle through Fe, X' and A' pass through the midpoint of AI.
Any reference?
Sincerely
Jean-Louis

Dear Jean-Louis Ayme,

it seems you're investigating the Feuerbach point quite actively...  :)
Here is a solution to your problem, synthetic as always, I hope you like it.

In a [url=http://www.mathlinks.ro/viewtopic.php?p=1219763#1219763]recent topic[/url] on a problem of yours, I proved that $ D,X',F_e$ are collinear, where $ D$ is the midpoint of $ B$ and $ C$. You can find both proof and reference to a problem in Mathemtical Reflections there.

[b]Proof:[/b] We call $ D,E,F$ the midpoints of $ BC,AC,AB$ and $ S$ the intersection of $ AI$ and $ BC$.  We denote $ M$ for the midpoints of $ A$ and $ I$.

Because $ IX'SX$ is a cyclic quadrilateral, it follows that $ \measuredangle X'SD \equal{} \measuredangle X'IX \equal{} \measuredangle OAA'$. Now, if we call $ N$ the midpoint of $ H$ and $ A$, it is well known that $ N$ lies on the nine-point circle of $ \Delta ABC$. A homothecy centered at $ H$ with factor $ 2$ maps $ N$ to $ A$, it maps $ D$ to the antipode of $ A$ in the circumcircle of $ \Delta ABC$, and it maps $ A'$ to the reflection of $ H$ in $ BC$, which evidently lies on the circumcircle of $ \Delta ABC$. We can now see that $ ND$ is parallel to $ AO$, which proves that $ \measuredangle OAA' \equal{}\measuredangle DNA' \equal{} \measuredangle DF_eA'$. We conclude that $ F_eX'SA'$ is cyclic.

Now, call $ A_1$ and $ A_2$ the projections of $ A$ onto $ BI$ and $ CI$. Clearly, $ A_1A_2$ is parallel to $ BC$ (angle chase). Now, call $ H'$ the orthocenter of $ \Delta AA_1A_2$, and $ X''$ the antipode of $ X$ in the incircle of $ \Delta ABC$. It is crystal clear that $ M$ is the circumcenter of $ \Delta AA_1A_2$. We have that $ AH'\equal{}2.AM.\cos{A_1AA_2} \equal{} AI.\cos{\frac{B\plus{}C}{2}} \equal{} AI.\sin{\frac{A}{2}} \equal{} r$. Of course $ r$ is the radius of the incircle as always. This shows that $ X''$ is the reflection of $ H'$ in $ M$, and hence it lies on the Euler line of $ \Delta A_1AA_2$. Now, it follows from Hatzipolakis' theorem that $ MX''$ passes through $ F_e$.

Because $ F_e$ is the external center of similitude of the incircle and NPC of $ \Delta ABC$, it follows that $ F_eX$ passes through the midpoint of arc $ A'D$ in the NPC of $ \Delta ABC$. It follows that $ F_eX''$ is the external angle bisector of $ \angle A'F_eD$, which is still equal to $ \angle A'AO$. It follows that $ \measuredangle MF_eA' \equal{} \measuredangle ASA'$. This shows that $ F_eSA'M$ is cyclic. We already proved that $ F_eX'SA'$ is cyclic, so we are done.

[b]Comments:[/b]

As to avoid any misinterpretations, the proof of $ X''$ being the De Longchamps point of $ \Delta AA_1A_2$ is due to [u]treegoner[/u], and can be found in Hyacinthos message 10913. In fact, using some earlier results of [u]Darij Grinberg[/u], he proved Hatzipolakis' theorem.

For those who don't know Hatzipolakis' theorem: Don't worry. This theorem was found in 2004 by Antreas Hatzipolakis, and first proven by Darij Grinberg. You can find it on Hyacinthos. The theorem states that the Euler line of $ \Delta AA_1A_2$ passes through the Feuerbach point of $ \Delta ABC$.

Two recent topics, all related to this configuration:
http://www.mathlinks.ro/viewtopic.php?t=219830
http://www.mathlinks.ro/viewtopic.php?t=219598

[b]Figure:[/b]

[img]http://img360.imageshack.us/img360/650/aymefeuerbach2kr1.png[/img]

Let $ V$ denote the foot of the A-angle bisector, $ M$ the midpoint of $ BC,$ $ D$ the midpoint of $ AI$ and $ D'$ the projection of $ D$ onto $ BC.$ From the right trapezoid $ AIXA',$ it follows that $ DD'$ bisects $ \angle DAX$ $ \Longrightarrow$ $ DX'$ and $ DA'$ are the reflections of $ DX$ across $ AI$ and $ DD',$ respectively. Hence

$ \angle A'DX' \equal{} 2\angle VDD' \equal{} 2\angle VAA' \Longrightarrow$

$ \angle A'VX' \plus{} \angle A'DX' \equal{} 2(\angle VAA' \plus{} \angle AVA') \equal{} 180^{\circ}$

$\Longrightarrow  D,X',V,A'$ are concyclic $ (\star)$

Since $ (X,Y,V,A) \equal{} \minus{} 1,$ where $ Y$ is the tangency point of the A-excircle $ (I_a)$ with $ BC$ and $ M$ is also midpoint of $ XY,$ we have $ MX^2 \equal{} MV \cdot MA'.$ But the inversion with center $ M$ and power $ MX^2$ takes the 9-point circle $ (N)$ into the common external tangent of $ (I),(I_a),$ differerent from $ BC,$  which is the reflection of $ BC$ about $ AV$ $\Longrightarrow$ $ F_e \in MX'.$ Power of $ M$ to $ (I)$ is $ MX^2 \equal{} MV \cdot MA' \equal{} MX' \cdot MF_e.$ Because of $ (\star),$ it follows that the five points $ D,X',V,A',F_e$ are concyclic and the proof is completed.

Dear Mathlinkers,
1. let (1) be the incircle of ABC, A" the midpoint of AI
(2) the circle with diameter A"X , (Garitte's circle it goes through Fe; see http://perso.orange.fr/jl.ayme vol. 4 Symmetric of OI  p. 14)
(3) the circle passing through A', A" ,X'
M the meetpoint of AI with BC and N the second point of intersection of AI with (2).
2. The tangent to (1) at X' goes through M.
3. By a converse of the pivot theorem applied to triangle X'MN with (1), (3), (2),
(3) goes through Fe.
Sincerely
Jean-Louis


inequalities
LaTeX
inequalities unsolved

if X1+X2+X3+...+Xn = X1^-1 + X2^-1 + ... +Xn^-1,

    Prove that (n-1+X1)^-1 + (n-1+X2)^-1 +... + (n-1+Xn)^-1 is smaller than 1.

  P.S. somebody who knows to type sigma or inequalities?

I think you forget about $ x_i>0$.
If $ x_i\equal{}\minus{}1$ for all i, then $ \sum_i x_i\equal{}\sum_i x_i^{\minus{}1}$, but $ S\equal{}\sum_i (n\minus{}1\plus{}x_i)^{\minus{}1}\equal{}1\plus{}\frac{2}{n\minus{}2}>1$.

[quote="bjh7790"] P.S. somebody who knows to type sigma or inequalities?[/quote]
We use LaTeX for typing/creating such formulas.
In art of problem solving, just put two dollar signs around the text you want to be LaTeX'ed. Like this:
[code] Show that $a^2+b^2+c^2\ge ab+bc+ca$ for $a, b, c\in\mathbb{R}$[/code]
Would look like this: Show that $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$ for $ a, b, c\in\mathbb{R}$
 :)
Commands:
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php
Symbols:
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
[quote="bjh7790"]Why no new replies? I want to know the solution.... [/quote]
Sorry that I cannot see the solution. :(


$ xy \plus{} yz \plus{} zx \equal{} 1$ then prove that:

$ \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x} \ge \frac {5}{4}$

I think $ x,y,z\geq 0$ because if $ x,y,z<0$ then $ LHS<0$.
And RHS must be $ \frac52$
We have:
$ LHS^2\equal{}\frac{1}{(x\plus{}y)^2}\plus{}\frac{1}{(y\plus{}z)^2}\plus{}\frac{1}{(z\plus{}x)^2}\plus{}\frac{4(x\plus{}y\plus{}z)}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\geq \frac{25}{4}$
Because apply the Iran 96 inequality we have $ (xy\plus{}yz\plus{}zx)(\frac{1}{(x\plus{}y)^2}\plus{}\frac{1}{(y\plus{}z)^2}\plus{}\frac{1}{(z\plus{}x)^2})\geq \frac{9}{4}$
And $ \frac{(x\plus{}y\plus{}z)}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\equal{}\frac{(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\geq 1$
Equal when $ (1;1;0)$ and permution!


How did u guys solve number 17 in the amc, care to help, plz?

Would you care to post the problem?  I haven't seen it.


geometry
3D geometry
sphere
trigonometry
analytic geometry
function
parameterization

My friend and I were thinking. If Parametric Form can be a simple way to describe curves in 2d then why not in 3d? I understand you cannot describe planes or spheres or anything like that, but whould this ever be useful? I was wondering just to see if I should look into this more.

Some of the curves we figured out how to show in parametic form were circles, sprials, and some other curves with neat effects.

Equations for circle:
$x=\sin \theta$
$y=\cos \theta$
$z=c$ where c is a constant

Equation for a slinky or sprial:
$x=\sin \theta$
$y=\cos \theta$
$z=\theta$

I think this might be useful, so if you find any good appilactions for the 3d parametric form please tell me.

o.0

It actually looks practical in modeling...

Perhaps you would be interested in looking up and studying [url=http://www.google.com/search?q=cylindrical+coordinates&sourceid=mozilla-search&start=0&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official]cylindrical coordinates[/url]

In 3D, parametric form is just like parametric in 2D with an extra component.
In 2D, you define a curve with equations $x=f(t), y=g(t)$.
In 3D, you add another function for $z$: $x=f(t), y=g(t), z=h(t)$.

The coordinate system you discovered is actually one of two polar systems in 3D, called cylindrical coordinates. A point $(x, y, z)$ in rectangular coordinates can be converted to cylindrical coordinates $(r, \theta, z)$ with these relationships: $r^2=x^2+y^2$, $\tan\theta=\frac{y}{x}$.
The other polar system is called spherical coordinates. A point $(x, y, z)$ can be converted to spherical coordinates $(\rho,\theta,\phi)$, where $\rho^2=x^2+y^2+z^2$, $\tan\theta=\frac{y}{x}$, and $\cos\phi=\frac{z}{\rho}$.

The stuff at MathWorld is excessive but you can find pictures there to better improve your understanding. It's great that you experiment with this kind of stuff. I reccommend you do some more self-discovery before grabbing a book and reading everything you can about it.

Actually, just for the record, you can expand parametric equations in 3-D to describe surfaces, as well.  You just need to add a second parameter.  For example, if we take your equation for the circle, but make c another parameter which takes real values between 0 and 1, we get a portion of a cylinder of height 1, the center of whose bases lie on the z-axis.

(I really like parametrics, by the way -- lots of fun stuff to be done.  Try making a spiral that goes along a cone or a sphere, for a good challenge.)

Thank you for your comments. I am becoming very interested in this subject now and I will look into it. I will also see if I can understand the stuff on math world about it, spherical, and cylindrical coordinate systems.

edit: Oh, I was just wondering what a good way to abstractly describe this in other higher dimensions would be as well. I was thinking about the spherical coordinates and how in 4d you might have a sphere transending through the hyper-space and how that coordinate system might take into effect. I suppose Wyou could have a hyper-spherical coordinates too. here it is defined similarly to spherical coordinates except with another angle, say $\alpha$.

Well you have given me alot to think about (and something to do after my finals this week  :lol:) So, thanks again.

Hmm ... I'm not sure I have a good enough grasp on what 4-D looks like to say what hyper-spherical coordinates would be like.  You can always do parametric equations, though.  Yay, parametric equations!

By the way, here's another challenge: find the parametric equations for a Mobius strip.


Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $ 13$ visible numbers have the greatest possible sum. What is that sum?
[asy]unitsize(.8cm);

pen p = linewidth(.8pt);
draw(shift(-2,0)*unitsquare,p);
label("1",(-1.5,0.5));
draw(shift(-1,0)*unitsquare,p);
label("2",(-0.5,0.5));
label("32",(0.5,0.5));
draw(shift(1,0)*unitsquare,p);
label("16",(1.5,0.5));
draw(shift(0,1)*unitsquare,p);
label("4",(0.5,1.5));
draw(shift(0,-1)*unitsquare,p);
label("8",(0.5,-0.5));[/asy]$ \textbf{(A)}\ 154 \qquad \textbf{(B)}\ 159 \qquad \textbf{(C)}\ 164 \qquad \textbf{(D)}\ 167 \qquad \textbf{(E)}\ 189$

[hide]
When we fold up the cube, we will have $ 32$ opposite $ 1$, $ 16$ opposite $ 2$, and $ 8$ opposite $ 4$.  For the bottom two cubes, we hide two opposing faces, so we hide $ 8$ and $ 4$ since they have the smallest sum.  On the top, we hide only $ 1$ face, so we hide $ 1$.  Thus the sum is $ 3(32\plus{}16\plus{}8\plus{}4\plus{}2\plus{}1)\minus{}2(8\plus{}4)\minus{}1\equal{}164$.  Answer is $ C$.
[/hide]

[hide]The two bottom cubes have 4 numbers showing. These numbers are either $ 1\plus{}2\plus{}32\plus{}16\equal{}51$, $ 1\plus{}4\plus{}32\plus{}8\equal{}45$ , or $ 2\plus{}4\plus{}16\plus{}8\equal{}30$. Thus the maximum sum of the numbers on the bottom 2 cubes is $ 51\cdot2\equal{}102$.
The top cube has all but 1 number showing. Thus, the maximum possible sum of the 5 numbers on the top cube is $ 32\plus{}16\plus{}8\plus{}4\plus{}2\equal{}62$. Thus the maximum possible sum of all 13 numbers is $ 102\plus{}62\equal{}\boxed{164 \Rightarrow C}$.[/hide]

I know you are about to strike here again @sugar_rush

please don't bump 13 year old problems, and if you are posting solutions for practice, use the aops wiki's edit feature.

The age of the thread should have nothing to do with it. If your solution is significantly different than anything found in the thread, you should post. If you have a significantly better version or a LaTeX-ed version of someone else's solution that's hard to read, you should post. If you have a question about someone else's solution in the thread, you should post.

sugar_rush's posts don't always meet these criteria. I'm just replying to clear up that novel posting about contest problems in the contests forum is not discouraged.

Edit: It's actually pinned, and the rules are even nicer than I thought: novelty is not necessary.
[quote=v_Enhance]
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
[/quote]


Could there be a forum added with discussions regarding math books or at least the AoPS books? If not, where should I post my questions about math books?

For questions about the AoPS Books this is the right forum to ask. For general questions about other math books, you can use the dedicated forum: http://www.mathlinks.ro/Forum/index.php?f=319


[quote="test20"]I have just learnt that IMO 2011 will take place in the Netherlands.[/quote]

Great. And where does it take place in 2010?  :lol:

I have read that it is India, but I'm not sure the source is the best  :wink:  :(

[quote="Arne"]I'm very happy that the IMO is coming to the Netherlands :)[/quote]Hey, Tasar is going to be there so I'm happy to have proven him wrong when he stated he will not be seen at the IMO again :P

Is IMO 2010 indeed going to be in India or is this a joke?

[quote="Arne"]I will also be at IMO 2011, I guess, since I live very close to the Netherlands :)[/quote]And Peter maybe ;)

Also I have no idea where IMO 2010 will be. I have heard Kazakstan, you now say India. It will be decided soon and I will put that information on the main page :)

The main page is really under the spotlight :)

[quote]Hey, Tasar is going to be there so I'm happy to have proven him wrong when he stated he will not be seen at the IMO again[/quote]

Well, I said "there is a good chance" I will be there. I'll still have to lobby to get myself a job as coordinator or some such. Or I could just pass by, as it should be close to my home (the Netherlands being the size it is :-)), or at least my parents place (for if all goes well I'll get a post-doc position in some exotic place next year, so I'll not be in the Netherlands anymore). Anyway I can't be 100% certain about what I'll do in four years, though I'd be happy to be proven wrong.

[quote="Valentin Vornicu"][quote="Arne"]I will also be at IMO 2011, I guess, since I live very close to the Netherlands :)[/quote]And Peter maybe ;) [/quote]Even IMO 2009 (Bremen) is within hours of here, so if other people go I might come along. Time to meet the ML people that do not come to the IMC. :)

Sorry I was wrong.

As you may have reas already, Kazakstan will be the host of IMO 2010. :wink:

IMO 2011 will be in Amsterdam, Netherlands. [url=http://olympiads.win.tue.nl/imo2011/]Here[/url] is the preliminary web page.


We have elements $ W,X,Y,Z$ with atomic numbers $ N, N+1, N+2, N+3$ respectively. Their natural boiling points are $ 58,-152, 688, 1380$ (in $ ^{\circ}C$) respectively. Which of the following compound is the most stable?

$ A) W_{2}X$
$ B) W_{2}Y$
$ C) W_{2}Z$
$ D) XZ$
$ E) XY_{2}$

[hide="Answer"]The elements W, X, Y, and Z are consecutive in the periodic table. Looking at their boiling points, we can conclude that, at room temperature, W is liquid or solid, X is a gas, and Y and Z are liquids or solids. The only way for this to happen is the following: W is an halogen, X is a noble gas, Y is an alcaline metal, and Z is an alcaline-earth metal. The only noble gas that forms reasonably stable compounds is Xenon, but only with fluorine or oxygen, both of which are also gases at room temperature. In our four elements, only one is a gas, so we should not expect any compund with X to be stable: this reasoning eliminates possiblities A), D), and E). Hence, we must consider ionic solids formed by halogens and metals of the s block: the most stable of these (the ones with greater lattice enthalpy) are those who combine small, highly charged ions. So, in our four elements, it is expected that the most stable compound is the one formed from the halogen (W) and the alcaline-earth metal (Z). As the metals of group 2 form doubly charged cations, and halogens form mononegative anions, the compound we are looking for is therefore $ ZW_{2}$, C).

When I first saw these four elements and their boiling points, my intuition told me that W appeared to be bromine. Consulting the boiling points in a table, this is indeed the case: W is bromine, X is Kripton, Y is Rubidium, and Z is Strontium, so our compound is strontium bromide, $ SrBr_{2}$.
[/hide]


inequalities
calculus
integration
function
abstract algebra
trigonometry
logarithms

Let $ x$ be a real number;$ n$ be a natural number;$ n \geq 2$
Prove that
$ 1\plus{}\sum_{i\equal{}1}^{n} \frac{cosix}{i} >0$

try this hint:
at first try to find closed form of the sum (it can be represented as integral of some function)
for example: note that $ \sum _{1}^{N}e^{(ik)x} \equal{} (\frac {e^{ix(N \plus{} 1) }\minus{}1}{e^{ix} \minus{} 1} \minus{} 1)$
now if you try to integrate it by $ x$ and after take $ Im$ part you will get something.then is easy how to estimate this integral .

[hide="Ignore this since it is wrong."]This is false. The Dirichlet kernel is not uniformly bounded either above or below.

We can write $ 1 \plus{} \sum_{k \equal{} 1}^n\frac {\cos kx}{k} \equal{} \frac12 \plus{} \frac {\sin\left(\left(n \plus{} \frac12\right)x\right)}{2\sin\left(\frac {x}2\right)}.$

Try looking at the value of this for $ x \equal{} \frac {3\pi}{2n \plus{} 1}$ and asking what happens as $ n\to\infty.$[/hide]

[quote="Kent Merryfield"]

We can write $ 1 \plus{} \sum_{k \equal{} 1}^n\frac {\cos kx}{k} \equal{} \frac12 \plus{} \frac {\sin\left(\left(n \plus{} \frac12\right)x\right)}{2\sin\left(\frac {x}2\right)}.$

[/quote]
In  this time I don't agree with you kent.
this is false since if we take $ x \to 0$ then
$ 1 \plus{} \sum_{k \equal{} 1}^n\frac {\cos kx}{k} \approx 1 \plus{} \sum^{n}1/k \approx \ln n$
and
$ \frac12 \plus{} \frac {\sin\left(\left(n \plus{} \frac12\right)x\right)}{2\sin\left(\frac {x}2\right)}\approx n \plus{} 1$
but $ \ln n \neq n \plus{} 1$

but I can assume that :
$ 1 \plus{} \sum_{k \equal{} 1}^n\frac {\sin kx}{k} \approx  \frac12 \plus{} \frac {\sin\left(\left(n \plus{} \frac12\right)x\right)}{2\sin\left(\frac {x}2\right)}.$
when $ x \to 0$
 :roll:

Oops. I was thinking this was the Dirichlet kernel but I was ignoring the $ k$ in the denominator. Sorry for the mistake.

Now I believe that the inequality is true.

This is what I think :
suppose $ N \equal{} 2l \minus{} 1$
Lets' try to find minimal value of the sum :
$ S_{N}(x) \equal{} \sum_{k \equal{} 1}^{N}\frac {\cos kx}{k}$ on the interval $ [0,\pi]$
since:
$ S'_{N}(x) \equal{} \sum_{k \equal{} 1}^{N} \minus{} \sin kx \equal{} ...calculating... \equal{} \frac {\sin (N \plus{} 1)x/2\sin Nx/2}{\sin x/2}$
hence we can note that at the point $ x_{k} \equal{} 2k\pi/(N \plus{} 1)$ we have local minimum and at the point $ x_{k} \equal{} 2k\pi/N$ local maximum.
 thus all $ x_{k} \in [0,\pi]$
let's try to investigate the numbers
$ S_{N}(x_{k})$ for $ 1\leq k < l$
let's show that minimal value holds only for point $ x_{l} \equal{} \pi$
really :
$ S_{N}(x_{k}) \minus{} S_{N}(x_{k \plus{} 1}) \equal{} \minus{} \int_{x_{k}}^{x_{k \plus{} 1}}S'_{N}(x)dx \equal{} ..calculating .. \equal{} \int_{x_{k}}^{x_{k \plus{} 1}}\sin^{2}lx \frac {1}{\tan (x/2)}dx > 0$
thus $ S_{N}(\pi) \equal{} \minus{} 1 \plus{} 1/2 \minus{} 1/3 \plus{} ... \minus{} 1/N \to \minus{} \ln 2$
and the same result will be for $ N \equal{} 2l$ since
$ |S_{N} \minus{} S_{N \plus{} 1} | \leq \frac {1}{N \plus{} 1}$


advanced fields
advanced fields theorems

Can anyone explain to me what is entailed in axiomatizing a binary relation?

If I have some candidate set of axioms, how do I know they are consistent?

Thanks.

You probably want to find a model for your [url=http://en.wikipedia.org/wiki/Axiomatic_system]axiomatic system[/url]. A binary relation on a set $ X$ is just a subset of $ X\times X$: if you can define a subset that satisfies all of your axioms, then they are consistent.

Do you have a particular binary relation in mind?


Pete's father never knows what the date is,

so Pete decided to make a calendar for him out of two cubes.

He wrote a digital looking number on each side of each cube,

and put one cube next to the other in a clear container.

From the front faces of the two cubes you can read a 2-digit number

indicating today's date.

(For example the 9th day of a month is indicated by 09.)

You may switch, turn, and move the cubes whichever way you want to,

and this way you are able to show the date of every day of every month.

 What numbers did Pete put on the sides of the cubes if we know that

one of the cubes has the numbers 3 , 4 , and 5  on it?

The digital numbers Pete used look like these:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

[hide]One of the cubes has 1, 2, 3, 4, 5, 6. The other has 0, 1, 2, 7, 8, 9.[/hide]

Move to HSB then delete this post please. Also, could a mod clean up the grammar and spelling?

[hide]We know that we have to have all nine digits on the cubes, for days like the 10th, 11th, 12th....19th. This will cover all of the days except for the days with double digits. These days are the 11th and the 22nd. This means that both cubes need the digits 1 and 2, and the other 8 digits can be split up in any way that we like. A sample solution would be 012345 and 126789.[/hide]

This probably should go into high school basics because I can do it without taking half an hour. :rotfl:

I've seen this problem before. I like it. Maybe because it's easy. :D

$30000^{th}$ post in HSB.

[quote="mustafa"]30001st post in HSB :D[/quote]

[quote="i_like_pie"][quote="mustafa"]30001st post in HSB :D[/quote]

 :roll:  :roll: [/quote]

Please do not spam with posts such as these.

Thanks & locked.


Let n \geq 1 be an integer, A1,..An positive real numbers and B1....Bn
positive real numbers smaller than 1.
Prove that

1/(A1/B1+..+An/Bn) + 1/((A1/(1-B1)  +...+(An/(1-Bn)) \leq 1/(A1+..An)

This is my solution. I hope it is correct.

We can rewrite the inequality as

(A1+..+An)(A1/B1(1-B1) +..+ An/Bn(1-Bn)) \leq (A1/B1 +..+ An/Bn)
(A1/(1-B1) +..+ An/(1-Bn))

Now RHS - LHS =  \sum (i<j) AiAj[1/Bi(1-Bj) + 1/Bj(1-Bi) - 1/Bj(1-Bj) -

1/Bi(1-Bi)] = \sum (i<j)AiAj[Bj(1-Bj) + Bi(1-Bi) -Bj(1-Bi) -Bi(1-Bj)]/BiBj(1-

Bi)(1-Bj) = \sum (i<j)AiAj (BiBi +BjBj -2BiBj)/BiBj(1-Bi)(1-Bj) \geq 0


Given $ a,b,c>0$.Prove that:
$ \frac{5}{4}. \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge \frac{a^2}{a^2\plus{}ab\plus{}bc}\plus{} \frac{b^2}{b^2\plus{}bc\plus{}ca}\plus{} \frac{c^2}{c^2\plus{}ca\plus{}ab}\plus{} \frac{1}{4}$

We can use cauchy and S.O.S to prove it.:)

By CS and Am-Gm, we are easily to prove the stronger and nicer:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \geq\ \frac {a^2}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ca \plus{} ab}$
 :)
@Secrets: [hide]Tiec qua', em ko di trai he` Hue'. Vui lam' day'![/hide]
@2424: [hide]Hiz, anh vua` di Sam` Son ve` lai danh' bai` thua het tien` oai`. Dung' la` do? tinh` den bac. Sang' nay luc ve` chang? con` tien` ma` di xe om ve` nha` nua~ :-<[/hide]


If $ 4: x^2 \equal{} x: 16$, what is the value of $ x$?

We have the ratio \[ \frac{4}{x^2}\equal{}\frac{x}{16}.\] Cross multiplying gives \[ x^3\equal{}64\implies x\equal{}\boxed{4}.\]

Set up ratios and solving is the slow way.

Just note that $ 4^2\equal{}16 \implies x\equal{}\boxed{4}$.


[quote="grafitti123"]Given a set $ \{1,2,3,....n\}$ of natural numbers, find the number of triangles possible with distinct sides.[/quote]

Use the fact that in a triangle with sides $ a,b,c$ that

$ a \plus{} b > c$

$ b \plus{} c > a$

$ c \plus{} a > b$

[hide="Hint"]If $ S_n$ is the set of triangles with distinct integer side-lengths $ \leq n$, then $ S_n \subset S_{n\plus{}1}$.
[hide="If that was not clear"]Consider using recursion[/hide][/hide]

[hide]Try $ a \equal{} x \plus{} y$ and extend.[/hide]


function
algebra
polynomial
induction
real analysis
real analysis unsolved

Suppose $n$ is a nonnegative integer and

$f(x) = c_{0}e^{r_{0}x}+c_{1}e^{r_{1}x}+\cdots+c_{n}e^{r_{n}x}$,

where $c_{i}$ and $r_{i}$ are real numbers. Prove that if $f$ has more than $n$ zeros in $\mathbb{R}$, then $f(x) \equiv 0$.

the result is clear if $r_{i}\in \mathbb{Q}$ because you can write it as:
$f(x) = e^{\frac{Ax}{q}}P(e^{\frac{x}{q}})$
where $P$ is a polynomial of degree $n$ and $A$ a constant. For the general case, just make an approximation of $r_{i}$ with rational numbers.

Hmm, I did not understand how exactly you can guarantee that you still have real roots after approximation, but probably I'm blind now. Anyway, there is a very simple approach using induction and Rolle's theorem.

you're not blind, but I was an idiot and posted an answer without thinking  :blush:  Rolle kills the problem immediately  :)


Ying earns a commission of 3% on all sales in excess of $ \$20,000 $.  In one month, she sells $ \$56,000 $ of merchandise.  What is the number of dollars in her commission for that month?


Let $S$ be a finite set of points in the plane.

A linear partition of $S$ is an unordered pair ${A,B}$ of subsets of
$S$ such that $A \cup B = S$, and $A \cap B = NULL$ and $A$ and $B$ lie on opposite sides of some straight line disjoint from $S$ ($A$ or $B$ may be empty).

Let $LS$ be the number of linear partitions of $S$.

For each positive integer $n$, find the maximum of $LS$ over all sets $S$ of $n$ points.

This is Putnam 2006 B3 and, as such (1) has already appeared on the forum, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122454]here[/url] and (2) is totally inappropriate for the Intermediate forum.


I found this while surfing the web, and I can't figure out how to solve without a calculator or trial and error.

If there is a good way to solve it, please post.

2LW + 2WH + 2LH = 192
LWH=144
H=1+W

Btw, this is an old ACT question.

[quote="undefined117"][hide]
i got W = 3, H = 4, and L = 12

$2\frac{144}{H}+2\frac{144}{L}+2\frac{144}{W}= 192$

$\frac{144W}{W+W^{2}}+\frac{144}{W+W^{2}}+\frac{144}{W}= 96$

$\frac{144W^{2}+144W+144(W+W^{2})}{W^{2}+W^{3}}= 96$

$2 \cdot 144W^{2}+2 \cdot 144W = 96W^{2}+96W^{3}$

$192W^{2}+288W = 96W^{3}$

$0 = 96W^{3}-192W^{2}-288W$

$0 = W^{3}-2W^{2}-3W$

$0 = W(W^{2}-2W-3)$

$W = 0$ or $0 = (W-3)(W+1)$

$W = 0$ or $W = 3$ or $W =-1$

$W = 0$ doesn't work because then $LHW \neq 144$
$W =-1$ doesn't work because then $H = 0$ so $LHW \neq 144$

$W = 3$ is the only one that works, so $H = 3+1 = 4$ and $L = \frac{144}{3 \cdot 4}= 12$.
[/hide][/quote]

I was wondering wouldn't it be simpler if the equation was $\frac{144}{1+W}+\frac{144}{W+W^{2}}+\frac{144}{W}= 96$

instead of the one you have given in the second line? because then as $W^{2}+W$ can be factorised to $W(W+1)$ we can multiply both sides by $W^{2}+W$ which does not involve $0$ as an answer.  This is a suggestion but I may be wrong.

max1546, could you then provide a solution using the equation in your suggestion? i just posted a solution off the top of my head and it may not be the most efficient one..

[quote="max1546"]OK..I'm lost now I forgot how to do it.
Can anyone explain how we got $L=W+W^{2}$?
[hide="If we know L in terms of W"]If we know it we can just put them all into the equation given.
$W(W+1)(W(W+1))=144$
It simplifies to $W(W+1)=12 or-12$
For $W(W+1)=12$ the discriminent is negative, meaning no real roots.  And due to the first equation given, it is clear that all numbers have to be real.
Therefore the only equation is $W(W+1)=-12$
This comes out that $W=-4 or3$
When we use $W=3$, it is true.
When we use $W=-4$, it is false as it does not fit the criteria of all the given equations.[/hide]

I was actually thinking of using undefined's way of solving.
Mine is similar to yours.

[hide="prev solution"]$LW=144/H$, $LH=144/W$, $WH=144/L$
Therefore
$2*144/H+2*144/L+2*144/W=192$
$144/H+144/L+144/W=96$
$144/(W+1)+144/(W(W+1))+144/W=96$
$144W+144+144W+144=96W^{2}+96W$
This simplifies to $W=-1 or W=3$
And it comes to simply $W=3$[/hide][/quote]

ohh.. I never thought about using $LWH = 144$
that seems easier.
idk but I forgot how I got my $L = W+W^{2}$ :huh:

So does anyone have a quick way of getting $L=W+W^{2}$?
Actually i found 2more ways to do this problem.
[hide="1"]Make an equation with the power of 4 and then use synthetic division to find the values of $W$.[/hide]
[hide="2"]Write down the factors of 144.  Then as H=1+W, find all pairs of factors that are 1 apart, then try calculate the third factor.  The pair has to be either both positive or negative.  As none of the values can be 0.  However, the issue of negative values can easily be ruled out as:
for the first equation to be the largest, the values of the negatives have to be as small as possible, which comes to -8, -9.  However, even though the third value is 4, the sum is always smaller than 192.  So we only have to consider the positive values.[/hide]
I actually found the second method faster, but the first method would be the most simple one.


This is a question I saw from a contest page, don't remember which contest. But it was interesting in that I couldn't solve it although I had the distinct feeling that I SHOULD be able to. Here it is:

You have five circles that are all tangent to each other and to two lines. The radius of the smallest circle is 8, and the radius of the largest circle is 18. What is the radius of the third (middle) circle? Sorry, I don't know how to represent this in LaTeX, so I'm hoping this can be visualised.

Gut instinct says 13, but please give full solutions with your answers =)

Thanks in advance!

First of all, do you mean three circles and two lines, five things in total? Also, I'm not sure how you can have all those figures mutually tangent (without coinciding and such).

Edit--Now that I read it again, I think I know what you mean.. kind of like two lines with five circles sandwiched between them...

If 'a ' be the angle between the two lines,

it can be proved that the radii of the circles are in GP with common ratio

  K = (cosec(a) +1)/(cosec(a) - 1)

 So r_5 = r_1* K^4   => K^4 = r_5/r_1  = 18/4 = 9/4   => K^2 = 3/2

and r_3 =  r_1* K^2 = 8 * 3/2 = 12 units

You used 4 instead of 8 in one step... but yeah that's generally right

I'm not sure exactly how you showed that there is a constant K, but my approach was to look at three circles instead of five.  Labelling their radii a, b, and c, it shouldn't be too hard to show that ac = b^2.  From there you easily generalize for n circles.

[quote="probability1.01"]You used 4 instead of 8 in one step... but yeah that's generally right

I'm not sure exactly how you showed that there is a constant K, but my approach was to look at three circles instead of five.  Labelling their radii a, b, and c, it shouldn't be too hard to show that ac = b^2.[/quote]

True , I used 4 instead of 8 in 18/4..

Anyway the constant can be arrived at by a little geometric consideration..

In fact the result can be generalised for ANY no. of circles.

Here's a little correction in the value of the constant though the same doesn't have much significance as far as this problem is concerned.

K = (cosec([b]a/2[/b]) +1)/(cosec([b]a/2[/b]) - 1)

where a is the measure of the angle between the two intersecting st. lines


The area of a square with side $ x$ cm is 5 cm$ ^2$. What is the number of square centimeters in the area of a square with side $ 2x$ cm?

We have $ x^2\equal{}5 \implies x\equal{}\sqrt{5}$. Thus, $ (2\sqrt{5})^2\equal{}\boxed{20}$.

Alternately, the ratio of the length of the larger square to the length of the smaller square is $ \frac{2x}{x}\equal{}\frac{2}1$.  Therefore, the ratio of their areas (larger square to smaller square) is $ \left(\frac{2}{1}\right)^2\equal{}\frac{4}{1}$.  Thus, the area of the larger square is $ 5\cdot 4\equal{}\boxed{20}$


inequalities
geometry
perimeter
inequalities unsolved

Prove that if $ y,z,t $ are non-negative numbers smaller than 1 and such that $ y+t\leq 1$ then the following inequality:
  $\sqrt{y^2+z^2}+\sqrt{t^2+1}+\sqrt{(1-z)^2+(y+t)^2}\geq (1+\sqrt{5})(y+zt) $

Well maybe a geometric interpretation [u]might[/u] work...

For example if we take the points $O(0,0)$, $A(t, 1)$, $B(-t, 1)$ and $C(y, z)$ ...
Then the left hand side is just the perimeter of $\Delta OBC$...
and for the right hand side maybe something with a dot product?

Dunno, it's just an idea, maybe it's not the right idea at all...


induction
number theory unsolved
number theory

WLOG assume that $x_1 > x_2 > ... > x_n$ and $y_1 > y_2 > ... > y_n$ and also $x_1 \geq y_1$.
Prove by induction:
$n=1$: trivial
So assume $n>1$ and let it be proven for $n-1$.
Now $(y_1+1)^{y_1+1} > y_1^{y_1+1} > \sum_{k=1}^{y_1} k^k > \sum_{k=1}^n y_k^{y_k} = \sum_{k=1}^n x_k^{x_k} > x_k^{x_k}$.
Thus $y_1+1 > x_1$, thus $x_1=y_1$, so the term $x_1^{x_1} = y_1^{y_1}$ cancels and we are done by induction hypothesis.


Find the number of all sequences $ (x_1,...,x_n)$ of letters $ a,b,c$ that satisfy $ x_1\equal{}x_n\equal{}a$ and $ x_i \not\equal{} x_{i\plus{}1}$ for $ 1 \le i \le n\minus{}1.$

[quote="moldovan"]Find the number of all sequences $ (x_1,...,x_n)$ of letters $ a,b,c$ that satisfy $ x_1 \equal{} x_n \equal{} a$ and $ x_i \not \equal{} x_{i \plus{} 1}$ for $ 1 \le i \le n \minus{} 1.$[/quote]

Let $ P(k)$ be the number of sequences for $ n\equal{}k$

Then for $ k \ge 2$, $ P(k\plus{}1)\equal{}2^{k\minus{}1}\minus{}P(k)$, where $ 2^{k\minus{}1}$ is number of the sequence $ (x_1,x_2,...,x_k,x_{k\plus{}1})$ satisfies almost all of the condition except that $ x_{k}$ may be $ a$.

So for $ n \ge 3$ odd , $ P(n)\equal{}2^{n\minus{}2}\minus{}2^{n\minus{}3}\plus{}2^{n\minus{}4}\minus{}...\minus{}2^2\plus{}2\equal{}\frac{2^{n\minus{}1}\plus{}2}{3}$
For $ n \ge 4$ even, $ P(n)\equal{}2^{n\minus{}2}\minus{}2^{n\minus{}3}\plus{}2^{n\minus{}4}\minus{}...\plus{}2^2\minus{}2\equal{}2^{n\minus{}3}\plus{}2^{n\minus{}5}\plus{}...\plus{}2\equal{}\frac{2^{n\minus{}1}\minus{}2}{3}$
For $ n\equal{}1,2$, $ P(n)\equal{}0$.

[hide="Minor Error"]For $ n \equal{} 1$: $ (x_1) \equal{} (a)$ satisfies the required conditions, so $ P(1) \equal{} 1$. [/hide]


For which $n\in\mathbb N$ there does exist $n$ points in plane satisfying these conditions. There are $n-1$ numbers $d_{1},d_{2},\dots,d_{n-1}\in\mathbb R^{+}$ that for each $i$ there exist exactly $i$ pairs of points that their distance is $d_{i}$.

[quote="Omid Hatami"]For which $n\in\mathbb N$ there does exist $n$ points in plane satisfying these conditions. There are $n-1$ numbers $d_{1},d_{2},\dots,d_{n}\in\mathbb R^{+}$ that for each $i$ there exist exactly $i$ pairs of points that their distance is $d_{i}$.[/quote]

$n-1$ numbers : $d_{1},d_{2},\dots,d_{n}$  ???  :maybe:


calculus
integration
trigonometry
calculus computations

I think that this substitution works: $ x^{6}=y$.

I got: \[ \frac{(x^{6}+2)\sqrt{x^{6}-1}}{9}\]

$ \int\frac{x^{11}}{\sqrt{x^{6}-1}}\, dx$

$ u = x^{6}-1 \implies u+1 = x^{6}$
$ du = 6x^{5}\, dx \implies \frac{1}{6}du = x^{5}\,dx$

The integral can be rewritten as:

$ \int x^{6}(x^{6}-1)^{-\frac{1}{2}}x^{5}\, dx$

Just substitute in (get the integral in terms of u), distribute, and integrate term by term (and then get the resulting expression in terms of x, of course).

[quote="mdk"]$ \int\frac{x^{11}}{\sqrt{x^{6}-1}}dx$[/quote]

I suggest the trigonometric substitution $ x^{3}= \sec \theta$
This gives $ 3x^{2}dx = \sec \theta \,\tan \theta \,d\theta$
And the integral becomes
$ \frac{1}{3}\int{\sec^{4}\theta }d\theta = \frac{1}{3}\tan \theta+\frac{1}{9}\tan^{3}\theta$
Getting back to x, we get the integral as $ \frac{(x^{6}+2)\sqrt{x^{6}-1}}{9}$

[quote="mdk"]there are other things you could do as well, but this is nice.[/quote]
Can you please elaborate a bit about the other things?

I think you can do $ x^{3}=cosh(u)$, $ x^{3}=tan(\theta)$, and even integration by parts.


Let a1,a2,...an in R. Prove that there exist k  in {0,1,...n} such that:
           abs {(a1+a2+...+ak)-(a(k+1)+...+an))} <= max  abs ai  (1<=i<=n)


Accordingly, there are 4[b].[/b]3[b]x[/b]1[b]0[/b]([b]With nineteen[/b] zeroes[b].[/b])[b] different positions of Rubik[/b]'[b]s[/b] Cube[b].How is such position be feasible,calculateddly[/b]? How is such calculation be made?

                                                                         Respectfully,
                                                                           A.Eddington

http://akbar.marlboro.edu/~mahoney/courses/Spr00/rubik.html
Groups of permutatations...


I just wondered; when is the next time the AIME Problem Series will be held? It still says Summer 2005 on the Online School page...

there are 2 AIME Problem Series, A and B.  AIME Problem Series B is offered this fall/winter, and I'm assuming AIME Problem Series A will be offered in the spring.

The date of the next problem series can be found on the enrollment page.

We believe that we will offer both AIME Problem Series during the Winter, but we will first see how enrollments are shaping up for the AIME B Problem Series.

[quote="chess64"][quote="MCrawford"]We believe that we will offer both AIME Problem Series during the Winter....[/quote]

If that happens, will there be one in the Spring?[/quote]

We offer the AIME Problem Series classes only during the Winter (essentially to end around two weeks before the AIME) and during the Summer.  The first is obviously the best time for preparation and we just wouldn't have time to start one between the AIME and the Summer and probably wouldn't have the enrollment.

[quote="Omar"]Another question:  How long does the AIME course last? How about the AMC 12/ AMC 10 classes?[/quote]

All Problem Series (and subject classes) are 12 week courses.

The Problem Series classes are all 90 minutes each.

Information about the courses can be found in the Online School area of the site.


Hello,

Could someone please tell me what the LaTeX code for the "Not Equal to Sign" (= with a / through) is?

Thanks.

[size=134]\neq[/size] should produce $ \neq$.

I use \ne as in $ 1\ne 2$. It's probably the same as \neq in $ 1\neq 2$, but one letter shorter.

I’ve just found out that [size=134]\not=[/size] also works: $ \not\equal{}$

Handy if you forget everything else. :D


algebra
polynomial
real analysis
real analysis unsolved

$ f \in C^{1}[0,1]$Show that for any $ \epsilon > 0$ there is polynomial $ P(x)$ such that $ \parallel{}f \minus{} P\parallel{}_{C_{1}} < \epsilon$, where $ \parallel{}*\parallel{}_{C_{1}}$
we define as:
\[ \parallel{}f\parallel{}_{C_{1}} \equal{} |\sup_{x\in[0,1]} f(x)| \plus{} |\sup_{x\in[0,1]} f^{'}(x)|
\]

Hint: approximate $ f'$ by a polynomial, then integrate.


function
topology
real analysis
real analysis unsolved

Suppose that $ f$ is a function and $ c$ is a number less than $ 1$ such that
\[ |f(x)\minus{}f(y)|\leq c|x\minus{}y|\]
for all $ x$ and $ y$.

Prove that $ f$ has only one fixed point.

Some nice problems you have there, Erken... :)

(Of course, $ c$ must be positive (or at least non-negative, but $ c \equal{} 0$ is a trivial case) for the problem to actually make sense.)

Anyway, let's first note that $ f$ can not have two or more fixed points. Let's assume the opposite: $ (\exists x_0,y_0\in\mathbb{R})(x_0\neq y_0 \wedge f(x_0) \equal{} x_0 \wedge f(y_0) \equal{} y_0)$. Of course, then we'd have $ |x_0 \minus{} y_0| \equal{} |f(x_0) \minus{} f(y_0)|\leq c|x_0 \minus{} y_0|$. Since $ x_0\neq y_0$, we have $ c\geq 1$, which is not allowed by the conditions of the problem.

Also, it is easy to notice that $ f$ is continuous: just take $ y$ as a constant and let $ x\rightarrow y$. We get $ |f(x) \minus{} f(y)|\leq c|x \minus{} y|\rightarrow 0$.

Now, let's assume $ f$ has no fixed points. Since $ f$ is continuous, either $ f(x) > x$ for all $ x\in\mathbb{R}$ or $ f(x) < x$ for all $ x\in\mathbb{R}$. WLOG we take $ f(x) > x$. We can now take $ y \equal{} 0$. We know that $ f(y) > 0$. Now, we will prove that $ x > 0 \Rightarrow f(x)\leq cx \plus{} f(0)$. Assume otherwise: $ f(x_0) > cx_0 \plus{} f(0)$ for some $ x_0 > 0$. Now we have $ |f(x_0) \minus{} f(0)| \equal{} f(x_0) \minus{} f(0) > cx_0$. Also, $ c|x_0 \minus{} 0| \equal{} cx_0$. That is an obvious contradiction per conditions of the problem. Now, we know that $ x > 0 \Rightarrow f(x)\leq cx \plus{} f(0)$. However, since $ cx \plus{} f(0) < x$ for $ x > \frac {f(0)}{1 \minus{} c}$, we also know that $ x > \frac {f(0)}{1 \minus{} c} \Rightarrow f(x) < x$. That is contradictory since we assumed $ f(x) > x$ for all $ x\in\mathbb{R}$. So, $ f$ has exactly one fixed point.

This is, of course, a particularization of the Banach fixed point theorem, also known as the contraction mapping lemma. That is, suppose $ f: X\to X$ where $ X$ is a complete metric space, and there exists $ c\in (0,1)$ such that $ d(f(x),f(y))\le cd(x,y)$ for all $ x\,y\in X,$ then $ f$ has a unique fixed point.

The standard proof: existence is the meat of the proof; uniqueness is easy. To see uniqueness, suppose $ f(x)\equal{}x$ and $ f(y)\equal{}y$ Then $ d(f(x),f(y))\equal{}d(x,y)\le cd(x,y)$ which rearranges to $ (1\minus{}c)d(x,y)\le 0$ which implies $ d(x,y)\le 0,$ hence $ x\equal{}y.$

For existence: pick an arbitrary $ x_0.$ Let $ x_1\equal{}f(x_0)$ and in general $ x_{n\plus{}1}\equal{}f(x_n).$ Show that this is a Cauchy sequence - hence a convergent sequence (since we assumed $ X$ complete). Show that its limit is a fixed point for $ f.$

That proof would certainly work in $ \mathbb{R}.$


limit
logarithms
algebra unsolved
algebra

let $\lim_{n\rightarrow \infty}a_{n}=0$ and $\lim_{n\rightarrow \infty}\sum_{i=1}^{n}a_{i}\in \mathbb{R}$

can we say anything about :

$\lim_{n\rightarrow \infty}(a_{n+1}+\ldots+a_{2n})$

Try $a_{n}= \frac{1}{n^{2}}$, the limit will be $0$.
Try $a_{n}= \frac{1}{n}$, the limit will be $\log (2)$.
Try $a_{n}= \frac{1}{\sqrt n}$, the limit will be $\infty$.

sorry , we have $\lim_{n\rightarrow \infty}\sum_{i=1}^{n}a_{i}\in \mathbb{R}$ ,too.

I want to know is there any sequence like $\{a_{n}\}$ that $\lim a_{n+1}+\ldots+a_{2n}\neq 0$
with perivoius properties.

It is obviosly \[\lim_{n\to \infty}\sum_{k=n}^{2n}a_{k}=\lim_{n\to \infty}(S_{2n}-S_{n-1})=0.\]


I've got a chessboard $n\times n$, $n>1$.
The question is: how many rectangles are there such that its sides are parallel to the sides of the board?

This is my reasoning:

I may choose the two lines containing the sides in $\binom{n+1}{2}$ ways. Then I may choose two of the vertices, in one of the selected lines in $\binom{n+1}{2}$ ways. Finally the answer is $\binom{n+1}{2}^2$ because this way the remaining vertices are fixed.

I'd like to know if that  is right and another solution would be fine too.

Thanx

I believe that is correct. :lol:

[size=75][color=green]Mod comment: It is :)[/color][/size]


trigonometry
inequalities
calculus
derivative
real analysis
real analysis unsolved

[b]Prove/or disprove that [/b]  $\frac{\sin{\varphi}}{\varphi}\le \frac{2+\cos{\varphi}}{3}\; \; \; ,\; \; \; \forall \varphi >0 .$

Note that for $\varphi>3,\,\frac{\sin\varphi}{\varphi}<\frac13\le\frac{2+\cos\varphi}3.$

Thus it is more than sufficient to prove this inequality on the interval $[0,\pi].$

Cross-multiply this inequality and subtract. Define

$f(\varphi)=\varphi(2+\cos\varphi)-3\sin\varphi.$ Note that $f(0)=0.$

We must show that $f(\varphi)>0$ for $\varphi\in(0,\pi].$

Take derivatives:

$f'(\varphi)=-\varphi\sin\varphi+2-2\cos\varphi.$ In particular, $f'(0)=0.$

$f''(\varphi)=-\varphi\cos\varphi+\sin\varphi.$ In particular, $f''(0)=0.$

$f'''(\varphi)=\varphi\sin\varphi>0$ for $\varphi\in(0,\pi).$

Thus, $f''(\varphi)$ is increasing and hence positive on $(0,\pi].$

Thus, $f'(\varphi)$ is increasing and hence positive on $(0,\pi].$

Thus, $f(\varphi)$ is increasing and hence positive on $(0,\pi].$

As a sidelight - what does this look like near zero?

$\frac{\sin\varphi}{\varphi}= 1-\frac{\varphi^2}6+\frac{\varphi^4}{120}+O(\varphi^6)$

$\frac{2+\cos\varphi}3=1-\frac{\varphi^2}6+\frac{\varphi^4}{72}+O(\varphi^6)$

[color=darkblue][b][u]Congratulations, Kent Merryfield ![/u][/b] The your proof is very nicely and completely.[/color]

I saw one thing I should have mentioned to make the argument complete. I should make a pont that $f(0)=0.$ For the moment, the edit is easy to spot, with the font difference.


I dont remember the exact words of this question, but it is one question i suggest everyone (including the ninth and tenth standards) should break their heads on. It is very beautiful... :roll:
I am posting it in the Physics forum also...

2 points A and B move in the same plane with velocities $ u$ and $ v$ respectively, constant in magnitude.($ v>u$). A follows a rectilinear path independent of B. (It moves linearly with constant velocity). B moves such that the velocity vector $ \vec v$ is always "aimed" at A. Initially, the particles are such that $ \vec u \perp \vec v$ and the particles are separated by a distance $ d$. After what time will the points converge?

I have done this and I dont suppose 9th and 10th graders will be able to solve it. I could solve it in 12th only. Anway I wont be posting the solution asi guess hell already knows it and I will let the 11thies think.

I was offline for some days,so i am sorry for late reply

[hide="solution"]
in the fig,we have
$ x_B\equal{}x_A\equal{}x$(say) (here,

then we have,$ \frac{dx_a}{dt}\equal{}vcos{\theta}$

so,$ \int dx_A\equal{}v \int cos{\theta}dt$

integrating,putting limits as 0 to x for RHS and 0 to T for LHS,we get

$ x\equal{}v \int cos{\theta}dt$

also,we have
x=uT(for B)

so,$ uT\equal{}v \int cos{\theta} dt$
$ \Rightarrow \int cos{\theta}\equal{}\frac{uT}{v}$

now,
$ \minus{} \frac{ds}{dt}\equal{}v\minus{}ucos{\theta}$(using vel. of app.)

$ \Rightarrow \minus{}ds\equal{}(v\minus{}ucos{\theta})dt$
$ \Rightarrow \minus{}\int ds\equal{}v\int dt \minus{} u\int cos{\theta}dt$(limit is fron 0 to T in LHS and d to 0 in RHS)
$ \Rightarrow d\equal{}vT\minus{}u\cdot \frac{uT}{v}$  (as $ \int cos{\theta}\equal{}\frac{uT}{v}$)
$ \Rightarrow d\equal{}vT\minus{}\frac{u^2}{v}$
$ \Rightarrow d\equal{}\frac{(v^2\minus{}u^2)T}{v}$
$ \Rightarrow T\equal{}\frac{vd}{v^2\minus{}u^2}$

method is right as i had solved it earlier,but calculation may be wrong as i haven't checked answers right now,will check answer.

[/hide]

Very Good. You didnot define your $ \theta$ but anyways, you understood the crux of the problem, so that doesnt matter.

$ {\theta}$ is the angle made by vel of A at time t=t(actually it is straight line,i failed to draw properly)with x axis


[quote="Svejk"]Prove that
\[\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\leq \sqrt[3]{2(a+b)(\frac{1}{a}+\frac{1}{b})}\]
,$\forall$ a,b>0[/quote]

Substitute $a=x^{3}$, $b=y^{3}$
Then we have $\sqrt[3]{2(x^{3}+y^{3})\left(\frac{1}{x^{3}}+\frac{1}{y^{3}}\right)}\geq \frac{x}{y}+\frac{y}{x}$
$\iff (x^{3}+y^{3})\left(\frac{1}{y^{3}}+\frac{1}{x^{3}}\right)(1+1)\geq \left(\frac{x}{y}+\frac{y}{x}\right)^{3}$ which is true by Hölder's Inequality


calculus
integration
calculus computations

hello, by substituting $ t\equal{}\frac{1}{x^2}$ we have $ \minus{}\frac{1}{2}dt\equal{}\frac{1}{x^3}dx$ and our integral is $ \minus{}\frac{1}{2}\int\frac{1\minus{}t}{\sqrt{t^2\minus{}2t\plus{}1}}dt$.
By substituting $ z\equal{}t^2\minus{}2t\plus{}1$ we have $ \frac{1}{4}\int\frac{dz}{\sqrt{z}}\equal{}\frac{1}{2}\sqrt{z}\plus{}C$. Last but not least is our antderivative
$ \frac{1}{2}\frac{\sqrt{x^4\minus{}2x^2\plus{}1}}{x^2}\plus{}C$.
Sonnhard.


vector
geometry
parallelogram
rotation
geometry unsolved

Let a convex polygon $H$ be given. Show that for every real number $a \in (0, 1)$ there exist 6 distinct points on the sides of $H$, denoted by $A_1, A_2, \ldots, A_6$ clockwise, satisfying the conditions:

[b]I.[/b] $(A_1A_2) = (A_5A_4) = a \cdot (A_6A_3)$.
[b]II.[/b] Lines $A_1A_2, A_5A_4$ are equidistant from $A_6A_3$.

(By  $(AB)$ we denote vector $AB$)

This seems like a very easy problem for a Vietnam TST (which probably implies that I made a mistake somewhere).

Take an arbitrary point that is not on any diagonal or side. Call this point $ X$. We will create many $ A_1A_2...A_6$ so that they satisfy condition II, and then from these show that there is one that also satisfies condition 1.

We now take an arbitrary line through $ X$, and label the endpoints $ A_3,A_6$. Label one side of the line $ P$ and the other side $ Q$. I claim that in $ P$ and $ Q$ are unique distances away from the arbitrary line so if 2 lines are taken at these distances, they intersect the boundary of $ H$ in segments $ A_1A_2$ and $ A_4A_5$ respectively s.t. they are parallel to the arbitrary line through $ X$ (trivial by how they were cut off) and of length $ a\cdot (A_6A_3)$. (Here, the line intersecting the boundary can be on an edge, in which case, for example, $ A_1A_2$ is not unique, but the distance to the line through $ X$ will be).

We will show this is true for $ P$ (a similar argument holds for $ Q$). Existence is simple, take a segment of length $ a \cdot (A_3A_6)$ on $ A_3A_6$ with one vertex on $ A_3$, and translate it parallel to itself with the vertex that was initially at $ A_3$ confined to the boundary of $ H$ in $ P$. Then it is easy to show that the other vertex of the line segment must eventually coincide with the boundary of $ H$ also. For uniqueness, if it was not true, there would be a parallelogram in $ P$ with 2 sides parallel to $ (A3A6)$ which had length $ a\cdot (A6A3)$. It is easy to show that this is not possible based on the fact that $ H$ is a convex figure (easier when aided with a figure).

WLOG $ A_1A_2$ is farther from $ A_3A_6$ than $ A_5A_4$. We rotate the line through $ X$. The length of $ (A3A6)$ is continuous as we rotate $ X$ around the full circle, and the unique distances, which were discussed before, are also changing continuously. This is easy to see when the line rotates through a section so that it is not parallel to a side, and when it is parallel to a side, a diagram, if I included one, would facilitate and easy argument that it is also continuous in this case.

Initially, $ A_1A_2$ is farther from $ A_3A_6$ than $ A_5A_4$. After rotating 180 degrees, we have $ A_1A_2$ having switched places with $ A_5A_4$, so that $ A_1A_2$ is now closer than $ A_5A_4$ to $ A_3A_6$. Since the distances away were continuous, there must have been a time that they were equal distances away.

These arguments are in fact rigorous despite requiring a diagram, and any argument requiring a diagram are easy to fill in (but too time consuming to include without such a diagram).

Hope I am right :)

Cheers,

Rofler


i want to verify if 2 vectors in vector space $R_{R^n}$ are liniar independent
say these vectors are $\left(u_1,u_2,...,u_n\right)$ and $\left(v_1,v_2,...,v_n\right)$

\[ W= \left( \begin{array}{cccc} u_1 & u_2 & ... & u_n \\ v_1 & v_2 & ... & v_n \end{array} \right) \]

if $rank(W)=2$ then u,v are linearly independent?
i dunno this exactly
if i can understand this i will also post a problem related to this...
i need to know if what i wrote is true

All linear algebra problems should be in the College Playground forums ;)

The rank of a matrix is the dimension of the span of the columns; it is also the dimension of the span of the rows. The dimension of a subspace is the number of elements in a maximal independent set, so the dimension of the span of a set of vectors is the number of vectors in the largest independent subset of that set.

Hence if a matrix has two rows and those rows are independent, the rank of the matrix is 2, and if the rank is 2, then the two rows must be independent.

[quote]"The rank of a matrix is the dimension of the span the columns"[/quote] [b]i dont agree[/b]

It is very foolish to disagree with one of the basic theorems of linear algebra.

its very weird to go around making people fools.
i just dont understand what you are trying to say in the first post of yours in the
first paragraph.
its really very weird what you're saying there.
what is pan of columns first ?

I was missing the word "of" (now fixed), but what is there to disagree with? Rank can be either the dimension of the column space or the dimension of the row space. It is a fundamental theorem that you do get the same count both ways.


Let u(n) is an infinite sequence such that
          u(n) = (1+[[(-1)^n]*n!]/u(n))*u(n-1) for n>1.
    u(1) is a 'good beginning' if there exists only one u(n) defined as above. A = {u(1)/u(1) is a 'good beginning'}
     a) Prove that |A| >0.
     b) Does A have finite or infinite elements ?
         :D  I have a solution to the first one. But the second one is very difficult in my opinion, and I have no solution yet. I hope you can do it. :D

I made a very stupid mistake when I put this problem in a wrong place. It should br in Alg :( ebra.


Hi all,

Currently my output profiles only consist of the three basic ones, namely LaTeX -> DVI, LaTeX -> PDF and LaTeX -> PS.  I would like to have a output profile: LaTeX -> DVI -> PDF.  How do I go about creating this?  Thanks!

Cheers,
ruifang86

First of all, it is not clear to me why you might want it given that LaTeX=>PS=>PDF is already there and can take care of all picture related problems that LaTeX=>PDF challenges you with sometimes. Anyway, to define a new profile, go to the "build" menu and click on "define profiles". Then either click on "wizard" (it is assumed to run you through all the steps one by one) or select "copy" and copy one of the existing profiles to the profile with your new name (it'll spare you a lot of time for typing in directories, command parametes, etc.). Then select your new profile and start editing it.

You do not need to touch the LaTeX part and the viewer can be copied exactly from the LaTeX=>PDF profile; your main concern is postprocessor. You need something like dvi2pdf.exe (I don't even know if such thing exists :P). Most likely you'll not need to touch the arguments but you need to enter the correct path to the "executable". After that just click OK and you are done.

I ddn't see LaTeX=>PS=>PDF in ruifang86's list of profiles which is perhaps what they want. It allows for use of eps pictures and PSTricks etc. It uses DVI so, as fedja says, there's no need for another profile going via DVI.
You can import LaTeX=>PS=>PDF from [url=http://www.latex-community.org/viewtopic.php?f=31&t=302]here[/url] (latex-ps-pdf.zip is in the second post) - all you need to do is change the paths.

[quote="stevem"]I didn't see LaTeX=>PS=>PDF in ruifang86's list of profiles [/quote]
Oops, you're right  :blush: (though I still wonder how came: you get it with the standard distribution of the TeXnic center, don't you? :?).

I believe you only get that profile if Ghostscript has already been installed.


f:N cu val in Z pt care f(1)=1 si nf(n-1) +(n-1)f(n)=0

We have $f(n)=-\frac{f(n-1)n}{n-1}$, so by induction $f(n)=(-1)^{n-1}n$.

Hope I correctly understood the problem.


search
function
number theory open
number theory

let a,b are 2 real positive numbers.Consider 2 sequences [a],[2a],...,[ka] and [b],[2b],...,[kb] which [x] is an integer such that x+1 >=[x]>=x.Prove that 2 sequences contain all distinct positive integers when ands onlywhen a,b not in Q set and 1/a +1/b = 1

This is Beatty's theorem which already appeared on the forum.

can you show me this theorem in this site

Search function -> take a search for "Beatty" -> done (please search as often as possible) [hide=":D"]well, lets hide it...:
http://www.mathlinks.ro/Forum/viewtopic.php?t=22782
http://www.mathlinks.ro/Forum/viewtopic.php?t=25764
http://www.mathlinks.ro/Forum/viewtopic.php?t=26352
http://www.mathlinks.ro/Forum/viewtopic.php?t=98365[/hide]


USAMTS AMC AIME geometry LaTeX articles function \/closed

[quote="theone853"]I posted this in the previous thread for giltches, but it probably would be better if i posted this here. In the USAMTS forum, the title is: here." class="maintitle">USA Mathematical Talent Search. Also it says there are no posts in Fun and Games factory, but once someone posts, that problem will probably go away.[/quote]Solved. :)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=134642#134642]This[/url] post should go here. Sorry for that.

In another thread:

[quote="Kent Merryfield"]Define a sequence $\{u_n\}_{n=0}^{\infty}$ by $u_0=u_1=u_2=1,$ and thereafter by the condition that
$\det\left(\begin{array}{cc} u_n & u_{n+1} \\ u_{n+2} & u_{n+3} \end{array} \right)=n!$
for all $n\ge 0.$ Show that $u_n$ is an integer for all $n.$ (By convention, $0!=1$.)

[Note: I can't get the browser to read the "&" properly, and hence the matrix won't post properly for me. Can anyone fix it?][/quote]

That's not a local problem of a browser.

Kent should really be telling us of this problem here, not somewhere random on the board  :blush: ... however I will dig into it (both problems). Meanwhile it's great to see not so many problems.

Ken actually thought that this was a local problem of his PC, I think... Anyway it's not too important.

  darij

[quote="mecrazywong"]It seems that the "About" (http://www.mathlinks.ro/Forum/about.php) button at the upper right corner of the page is invalid since I first visited Mathlinks. Well...it maybe the problem of my browser, since no one seems to have noticed it.[/quote]It's not a problem, it's not finished yet. Will be solved.

If it's not a problem, then how can it be solved?  :?

  dg

[quote="Myth"]Bug: I watch topics for replies on MathLinks site. But suddenly I received "Topic reply notification", which points to the same topic on aops site.[/quote]Noted, I'll fix it soon.

[quote="rrusczyk"]Please post any glitches you see in the new Forum (erm, except for the missing posts :blush:)[/quote]

The AIME problem series forum is missing too.

Edit: How come the orange "beta version" link leads to the second geometry problem ever posted on mathlinks?

bug 1: all latex images in reply preview have addresses, which belong to aops, so then I will see my reply on ML I will need download images again, because this time they will belong to mathlinks site.
bug 2: $$ ... $$ and \[ ... \] produce non-centered equations.

Bagatelle: The "Hit Counter" on the ML portal doesn't work. Heretic question: Do we need a hit counter at all?

  darij

Okay now a small recap of solved bugs:
- latex bugs solved (only a small disfunctionality of the copy to clipboard button remains, which untill I better understand how windows works it will be hard to solve :) )
- emailer bugs solved
- viewtopic bugs (in firefox the new / reply topic buttons were on top of each other instead of one next to the other, the hidded people were displayed as hidden when they were online and offline when they were offline, quote button being in the left hand side of the posting row, source field not appearing)

Still to solve moderator control panel (new) bugs and some viewing bugs.

Anything else I forgot and you guys observed?

[quote="darij grinberg"][quote="Valentin Vornicu"]source field not appearing)[/quote]

unfortunately not completely: the source field appears in the threads, but not in the topic lists e. g. http://www.mathlinks.ro/Forum/index.php?f=46 (scroll down for the older messages!)...

concerning all the rest: great work!

  dg[/quote]That's sort of a cache problem ... very strange. Normally it should be there, as it is with the new topic.

Well you can now use the moderator control panel ;)

[quote="Kent Merryfield"][quote]Well you can now use the moderator control panel[/quote]

Yes, I can and I have.

But some odd things about moderating: I see "edit" and "delete" buttons under every post in the threads on "College Playground" that I now moderate, but I don't see those buttons in the "Southern California ARML" hidden forum that I was already moderator for. I also seem to have a "delete" button under posts in places where I'm not a moderator.

--

Let me give the matrix thing one more try:

$M=\left(\begin{array}{cc}1&2\\3&4\end{array}\right)$

That's frustrating. It happens in every browser I try (currently in Firefox), it happens with both of the computers I post from, it happens whether I have the "Disable HTML in this post" checked or unchecked. I don't know what else to try.

Before the crash/changeover it happened sporadically but there usually seemed to be some way to get out of it. Now there's no way out of it.[/quote]We'll fix the & problem don't worry about it too much :) It's because of an un-escaped & somewhere.

As for the delete thing let me look into it ...

[quote="Myth"]I have discovered bug while deleting thread.
I was asked whether I really want to delete topic. I said "No". After that page reloaded and asked me the same, and so on.[/quote]Oops, improper redirect sentence. Will be fixed soon :)

[quote="jmerry"][quote="Valentin Vornicu"]Finally after days of work, I've managed to bring back the collapsible categories, better and improved (board wise). [/quote]

At the moment, they don't work right. Any sections that are collapsed by default cannot be expanded, and are accessible only through the jump box (since the section title is not a functional link).

Last night, it worked when I was logged in; today it's broken whether I'm logged in or not.
The sections that are expanded by default can be expanded and contracted freely.[/quote]This is not a bug. It's a default view set of artofproblemsolving.com. We do realize that most of the students coming through that url are US students looking for start oportunities, therefore they would not be interested in either Advanced stuff, neither College.

Most of the people interested in the last two enter throught the mathlinks.ro url.

At this point I am working into finding a way to make the advanced and college sections collapsed on the aops url without messing up cookies.
Untill then you are free to join the usergroup view all cats to avoid this problem.

Ok dealt with that the cookie way. No usergroups are required anymore.

[quote="rcv"]A regular user posts a new topic.  The topic shows a little "X" symbol below the topic.  But when the user clicks this link to delete his post, he gets an error "You are not a moderator of this forum".

I think this worked, previously, but I'm not completely sure, so I can't be certain this is a legitimate "bug".  Regular users can still delete replies (as long as nobody has responded), but they can't delete the original post in a thread, even if nobody has responded.[/quote]It's not a bug, it's designed for people to not be able to delete the first post in a topic. These are secondary effects ... will think about it.

The matrix & bug is solved. You can use arrays now in $\LaTeX$.