What do you guys do to prepare for a competition in the day(s) before it? I mainly try to get enough sleep the week or so before, because I can never sleep the actual night before. I find sleep is a large cause of my mistakes, and caffeine just makes me feel nervous and twitchy.

realize that it doesn't matter how well i do so i don't really care when i make careless errors or don't pass or whatever

I agree about the sleep part, though I found I can get a good sleep the night before if I go to bed early enough (actually, this is probably the best preparation I can do before any contest). Also, I personally completely stop training a couple days before the contest. This makes me less nervous and more clearheaded. For long contests I like to bring water and a lunch, and really good expensive European chocolate.

See what works for you. Perhaps some exercise, like a run, before the contest (or even during a long contest, if you are allowed), could help to relax you. But probably the worst thing you can do is to study intensely the night before. Also, if you are writing an invitational contest, you should make sure that your teacher has set aside a quiet room for you to write it in. (I have heard from some of my peers that their teachers do not take these contests very seriously and don't make the proper arrangements.) If not, you can pressure your teacher to find a better room, or try to book a room yourself.

if you're having trouble falling asleep, try reading a really boring textbook

good one klebian.

the sleeping part i think is by far the best advice. i personally had some bad experience with this, not on purpose, it just happened. the night before a chapter mathcounts competition i was just coming back from a 5-day school trip to yosemite..i came home at like 12:00 and was sooo tired i fell asleep on my table...then i woke up, took a shower and went to bed at like 2:00. the next day was a terrible ordeal. the entire examination was torture...i couldn't concentrate, i didn't finish 75% of the problems and i completely lost track of time. my results reflected this lack of sleep. i like to think that it was just pure bad luck that we came home the exact night before chapter, and the fact that our bus like broke down several times during the 5 hour trip back.
soo..in conclusion.. a lesson to all....GET A GOOD NIGHT'S SLEEP!!!!

If you have trouble sleeping, run around and get yourself really tired the day beforehand, then eat a really big meal

[quote="Klebian"] try reading a really boring textbook[/quote]
That works for me.
I'd also try to just have a completely math-free day (of course, math classes at school don't count) before the competition.

Try to avoid thinking too hard the day before.  Time your meals well, and relax your brain (possibly watch tv or play video games, even).
I think every person has an ideal amount of sleep: not too much, not too little.  For me, it's eight and a half hours (determined through experimentation).  Get that much sleep, and time things so you wake up to leave enough time for the following:

Stretching, possibly a ten-minute easy jog
A good breakfast (with protein) and time to digest it
A good shower
A little green tea (inhale the vapors a little too)

This has been what I have done every year.  This year, I added the green tea part.  It worked very well.

[quote="Altheman"]Dude, thats insane. But if it works for you, then awesome.

I think that it is way more important to plan test taking strategy than to develop an elaborate preparation ritual.[/quote]

it is pretty elaborate.
the great tea part sounds cool though.  :)

This whole thing has kinda come into being as I've done tests and stuff through the years.

The relaxing your brain thing makes sense.  The good breakfast with protein thing and shower are common sense.  It's common sense to leave time for digestion and to wake yourself up.

The sleep thing came as a result of an experiment I did anyways to sleep optimally to conserve time.

The stretching came naturally as I run track.

The green tea makes sense.

The elaborate procedure I described is not something I follow exactly, but those general things are good to keep in mind.

What I tend to like to do is do some recreational mathematical puzzles for fun, 2-3 days before the contest just to keep my mind alert and still be relaxed at the same time as I'm not trying to learn new material. You should just do whatever that relaxes you.

I try to do really easy problems to warm up a day or two in advance. Make myself super confident. I like K8107's plan. It does make a decent amount of sense. Running isn't what I personally do, but getting yourself tired out the day before can certainly help getting some sleep. I usually forget about most contests aside from AMC/AIME anyway.


geometry
circumcircle
incenter
Euler
geometry unsolved

The incircle of a triangle ABC is tangent to BC,CA,AB at M,N and P, respectively. Show that
the circumcenter and incenter of tr.ABC and the orthocenter of tr.MNP are collinear.

So here is the official solution:
The incenter of tr.ABC and the orthocenter of tr.MNP lie on the Euler line of the triangle
MNP.The inversion with respect to the incircle of ABC maps points A,B,C to the midpoints
of NP,PM,MN (A',B',C'), so the circumcircle of A'B'C' maps to the nine-point circle of MNP which is
also centered on the Euler line of MNP. It follows that the center of circle ABC lies on the
same line.

Well,the problem is that i can't understand this part: It follows that the center of circle ABC lies on the
same line.
Can anybody help me.

The incenter of tr.ABC and the orthocenter of tr.MNP lie on the Euler line of the triangle
MNP.The inversion with respect to the incircle of ABC maps points A,B,C to the midpoints
of NP,PM,MN (A',B',C'), so the circumcircle of A'B'C' maps to the nine-point circle of MNP which is
also centered on the Euler line of MNP. It follows that the center of circle ABC lies on the
same line.

My memory said this is GOB's Theorism(?).

Do you know about Euler line ? tr ABC othocenter, circumcenter, centroid and nine point circle's center(of tr ABC).

So MXㅗNP, NYㅗMP, PZㅗMN. X,Y,Z each on NP,MP,MN

tr XYZ and tr ABC are similar. So center of similar(K) and tr ABC's circumcenter and tr XYZ's circumcenter lie on the same line.

K, tr ABC's incenter and trXYZ's incenter are collinear.

At last, tr XYZ's incenter is trMNP's orthocenter . tr XYZ's circumcenter is nine-point circle of MNP.

So, XYZ's incenter, XYZ's circumcenter and MNP's circumcenter(ABC's incenter) are colliner(by Euler line ).

Q.E.D

What is "center of similar?" and how you proov that tr.XYZ and tr.ABC are similar.
PP:I want a proof with inversion.

angle BPM=angle PNM=angle PXY. So, AB parralel with XY

Same way, BC//YZ , CA//XZ → tr ABC , tr XYZ similar.

AX intersects BY at K. KZ intersects AC,BC at P,Q

AX : XK= BY : YK , AX:XK=PZ:ZK, BY:YK=QZ:ZK

So, P=Q → It become point C

AX, BY, CZ intersect one point ; this is K

I can't speck English well. I don't know what is English name of K point.

Ok,but how you proov that K and tr ABC's circumcenter and tr XYZ's circumcenter lie on one line.


A cloud of particles with density $\rho$ comes closer to the star of mass $M$ and radius $R$ with speed $v$.Find $\frac{dm}{dt}$ of the star.

It's a bit unrealistic, but... Is the momentum the dust gives to the star negligible?

I get \[\frac{\text dM}{\text dt}= \rho R^{2}\pi \sqrt{v^{2}+\frac{2GM}{R}}.\]

Is this correct?

Here is the correct answer:

$\frac{dm}{dt}=\rho \pi v(R^{2}+\frac{2\gamma MR}{v^{2}})$

How did you get that? I did make several approximations, but I'm not sure whether all of them are valid...


Exponential equation
equation
algebra
roots
IMO Shortlist
IMO Longlist

Determine all positive roots of the equation $ x^x = \frac{1}{\sqrt{2}}.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Taking $\ln$ both sides, we have that

${%Error. "additional" is a bad command.
\displaystyle x\ln x=\frac{1}{2}\cdot\ln \frac{1}{2}} \\\Rightarrow \frac{x}{\frac{1}{2}}=\frac{\ln\frac{1}{2}}{\ln x}$

but since $a\geq b \Rightarrow \ln a\geq \ln b$, we must have one of $\displaystyle \max \left\{\frac{x}{\frac{1}{2}},\frac{\ln\frac{1}{2}}{\ln x}\right\}\geq 1 \geq \min \left\{\frac{x}{\frac{1}{2}},\frac{\ln\frac{1}{2}}{\ln x}\right\}$, so we must have $x=\frac{1}{2}$

The solution of $campos$ is not sufficient !:D  It's easy to see that $x_1=\frac 14,\ x_2=\frac 12$ are solutions .
There are not other solutions because of the function  $f(x)=x^x$  is decreasing on $]0,\frac 1e]$ and  increasing on $[\frac 1e,+\infty[$
 :cool:

:blush:  :(  i thought this can't be so easy..., isn't $f(x)=\ln x$ strictly increasing on $\mathbb{R^+}$?

i think i've found a solution... hope it's correct this time...

as in the previous post we have
$x^x=\frac{1}{\sqrt 2} \Leftrightarrow x \ln x= \frac{1}{2}\ln \frac{1}{2}$

now, let $x=\left(\frac{1}{2}\right)^k$, then,

$x \ln x= \frac{1}{2}\ln \frac{1}{2} \Rightarrow \left(\frac{1}{2}\right)^k\ln \left(\frac{1}{2}\right)^k=\frac{1}{2}\ln \frac{1}{2} \\\\\Rightarrow 2^k=2k$

but, since $f(x)=2^x$ is an exponential function and $g(x)=2x$ is a linear function, we must have $f(x)$ and $g(x)$ intersect in at most two points, namely $k=1$ and $k=2$... so we must have $x=\frac{1}{2}$ and $x=\frac{1}{4}$ are the solutions to the equation.

How do you know that $x$ must be a power of $1/2$? :blush:

As I see it, $k$ is not necessary an integer...

Derivating, we have that $(x^x)’ = x^x(\ln x + 1)$. Therefore, $x^x$ is decreasing when $0 < x < \frac{1}{e}$ and increasing when $x > \frac{1}{e}$. It follows that there can be at most 2 solutions. After some trial and error, we see that $\frac{1}{4}$ and $\frac{1}{2}$ are solutions and by the aforementioned reasoning, they are the only solutions. $\blacksquare$

We can see that $x^x$ is decreasing when $0 < x < \frac{1}{e}$, and increasing when $x> \frac{1}{e}$, I used trial and error, to get that $\frac{1}{4}$ and $\frac{1}{2}$, are the only solutions. $\square$


Let F1,F2 be first Fermat point, second Fermat point of triangle ABC, respectively. N is its nine point center, O is its circumcenter. Prove that four point F1,F2, N, O are cyclic.

This is called the Lester circle. You can find some references here: [url]http://mathworld.wolfram.com/LesterCircle.html[/url].


1) find all 6-digit numbers so that their digits are in strictly increasing order.
2) same question with digits in just increasing order.

I assume your questions are with and without repeats, respectively. Note that $ 0$ can't be used.

Since any group of $ 6$ digits has exactly one sequence in increasing order, the answers are $ \dbinom{9}6\equal{}84$ and $ \dbinom{9\plus{}6\minus{}1}6\equal{}3003$. The second one is balls and urns: if $ x_n$ represents the number of $ n$'s, then this is the number of solutions to $ x_1\plus{}x_2\plus{}\cdots\plus{}x_n\equal{}6$.

So there are 9 "types" of #s and we are taking any 6 of them because each set of 6 = 1 solution (hence balls and urns). the other is just 9 C 6

That's essentially EXACTLY what I posted a while ago.


trigonometry
trig identities
Law of Cosines

I'm not sure if this problem is completely correct, but if it is it would be really cool.
a,b,c,and d are positive numbers.
Let a+b=3, c+d=5, ab=cd. Let n be any number between -1 and 1. Find $ \sqrt {(a^2 \minus{} 2adn \plus{} d^2)(b^2 \minus{} 2bcn \plus{} c^2)} \plus{} \sqrt {(a^2 \plus{} 2acn \plus{} c^2)(b^2 \plus{} 2bdn \plus{} d^2)}$.

Well... if $ a,b,c,d$ are positive integers, then $ ab\equal{}2$, and there's no way for $ cd$ to be equal to that...

[hide="solution outline"]
\[ S = \sqrt {(a^2 - 2adn + d^2)(b^2 - 2bcn + c^2)} + \sqrt {(a^2 + 2acn + c^2)(b^2 + 2bdn + d^2)}
\]
Let $ \frac {a}{d} = \frac {c}{b} = \alpha$ and $ \frac {a}{c} = \frac {d}{b} = \beta$ then
\[ S = \frac {1}{bd}\sqrt {(\alpha^2 - 2\alpha n + 1)^2} + \frac {1}{bc}\sqrt {(\beta^2 + 2\beta n + 1)^2}
\]

\[ = \frac {1}{bd}\left\vert\alpha^2 - 2\alpha n + 1\right\vert + \frac {1}{bc}\left\vert{\beta^2 + 2\beta n + 1\right\vert
}\]
The rest is uninteresting algebra, but the punchline is
\[ S=15\]
for $ n\in[-1,1]$
[/hide]

Good job, JoeBlow.
Here is my solution:
[hide]Make a segment of length 3 broken into a and b, and one of length 5 broken into c and d. Intersect them at their breaking point, with the angle between a and d arccos n. Since ab=cd, the endpoints of the segments are concyclic. We can use the law of cosines to find the lengths of the sides of the cyclic quad formed by connecting the points. So by Ptolemy, the answer is $ 3\cdot5$=15.[/hide]


Hello, I am creating another game. I am calling it game 2, because I can't think of any other name for it. This game is very similar to a TV show that I heard about a while ago. I warn everybody thinking about joining, that this game might last a while. It depends on how it turns out. Basically there will be two teams, a person on each team (at the beginning). Each team will have its own quick topic that they can do whatever they want in. Basically each team wants to make their team as large as possible. They do this by trying to get all the new arrivals to join their team. The largest team will proceed to the next round. I will explain what happens later once we are farther into the game.

So sign-ups begin now. If there is any confusion, please ask any questions you have.

sigh

wat happened to game 1?

sighs

join group A

everone join group a!!

/me joins group B

Everybody join group B!

Okay...

I join group B.

Everyone join group B!

Oops!!! I just noticed that this game is almost exactly the same as game 1. Sorry guys. [size=150][b]Somebody please lock this thread!!![/b][/size]

And if there was a similar game to this, I would actually try it like if someone joins, you put the guy in a. if the next guy joins, you put him in b and so on.

you know,
you can lock thsi thread i tink......

kdage, this thread wil be lockd

nothing happens :)


Does anyone know how to find the 2 points of intersection in the first quadrant for the two equations below:

$ x \equal{} \frac {\minus{}y^3}{3} \plus{} 2y \plus{} \frac {1}{3}$

$ y \equal{} \frac {x^3}{3} \minus{} ln(3\minus{}x) \minus{} 1$

Any help is greatly appreciated.

I would claim that there exists no routine method in algebra
of coming up with the solutions for this problem.

One way of approximating the solutions could be to use the
free http://www.quickmath.com website and go to the solving
section in the equations area.

EQUATIONS  (Type in as)
x = -y^3/3 + 2y + 1/3
y = x^3/3 - log(3 - x) - 1 . . . . Note:  "log(X)" in this website IS notation for the natural logarithm of X.

Use two different colors for your curves.  (Suggestion.)

VARIABLES x,y

Independent x  :  runs from  1.8 to  2.05
Dependent y :  runs from .8 to  1.8

(I whittled down the x and y windows after going back and forth,
shaving off pieces outside of the intersection points.  These
windows were the best I could do at that moment, and for my
limited time.)

 OPTIONS  Tick marks (check)  Grid lines (check)
Axes
 None

 Automatic origin (check)

 Origin at (  ,  )

Aspect ratio
 One to one

 Golden ratio

 (check) Height : width =  2 : 3  . . . (This is a suggestion.)

Text output size :   20 points  . . . (I suggest to leave it at this.)

Looking at the intersections of the curves in Quadrant I,
I estimate the solutions to be about:

(1.82, 0.85) & (2.02, 1.75)

Note:  On the x-axis, there are four little tick marks between
the vertically numbered lines.  And on the y-axis, there are
three tick marks between the horizontally numbered lines as
the system outlined above with that window as given.

(EDIT)


topology
invariant
calculus
derivative
advanced fields
advanced fields theorems

I was wondering if there is a way to motivate the notion of smoothness without the drivative. To be more exact, can we do this using topological notions? For example, in a metric space, can we define a curve as smooth if we can "pin" the curve with two open balls of non-zero radius for every point on the curve? (By "pin", I mean the curve sits on the boundaries of both open balls at a point and the open balls are on opposite sides of the curve.) Is the definition I gave sufficient to define smoothness?

No. Smoothness is not homeomorphism-invariant (and distance is not a topological notion).
For example, consider the curves $ y\equal{}x^2$ and $ y\equal{}x^{\frac23}$. One is smooth while the other isn't, and they're related by the homeomorphism $ f(x,y)\equal{}(x,y^3)$.

Didn't we cover this pretty thoroughly when you asked essentially the same question two months ago?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=303685

Smoothness is not a notion that makes sense in an arbitrary metric space.

There are some algebraic definitions that don't really use the concept of derivatives.

Yes, but the algebraic definitions only make sense in algebraic situations, where everything is smooth almost everywhere anyway.  If you're referring to the definition I think you're referring to, the derivatives are still there - they just aren't being computed as limits.

The correct definition in the algebraic case (varieties) is the regularity of the local rings (after base change to an algebraicaly closed field). No derivations here.
For schemes, flatness comes into the game, too.

Yes, but the point of the definition by regularity is that it agrees with the geometric one that the Jacobian has maximal rank, which is - surprise - a statement about derivatives.

Right I was just pointing out that while it was originally motivated by differentiation (which is done formally), there are also intrinsic definitions.


USAMTS
analytic geometry
calculus
geometry
quadratics
perimeter

I deeply enjoyed number 3. I thought it was a great problem that coulds be approached in many ways. Also, the steps that built up to the solution for me were so logical and beatiful. I couldn't help being happy at the end when I had it all written down.

I thought 1 leaves a lot of space for extra exploration - which interested me.

I thought 2 and 4 to be pretty conventional.

5 was interesting due to its obvious but false solution, which, I, like many others, found first, only go back and realize that it isn't the case. Unfortunately it seems that I made some mistake in my solution to this problem.

[quote="Danbert"]What's your favorite problem from this round?

I'd have to say it was #5, with #4 a close second.
1,2 were too easy; 3 was easy but hard to write out well.

I used a neat coordinate geometry/differential calculus approach and ended up with 24.

It was fun.

 :D[/quote]

I also enjoyed #5, solved using a coordinate geometry and calculus, and got 24. :)

#2 problem was impossibly hard for me.  Maybe i didn't see something obvious.  It took so much time to solve that problem for me.  Each of other problems was a breeze.

[quote="mathfanatic"]I liked #5 and #2, which I also found harder than some have described.[/quote]

I didn't think #2 was all that hard... but #5 was.  I didn't manage to solve that until the day before it was due.

#4 gave me some problems before I saw an easy way.  At first, I had a series of 5 simultaneous equations of varying degree (:what?:) that would give me the side lengths of the square, but it was so nasty that not even a TI-89 would solve it...

[quote]#4 gave me some problems before I saw an easy way. At first, I had a series of 5 simultaneous equations of varying degree (What?!) that would give me the side lengths of the square, but it was so nasty that not even a TI-89 would solve it...[/quote]

Sounds familiar, although my method ended up with 3 simultaneous quadratics, which wasn't that hard to solve by hand.  I never did find an easier way (or is it more truthful to say that I never bothered to?  :lol: )

By the way, a TI-89 isn't always the best way to solve an equation.  I happen to have one of the new versions, and occasionally come across a problem that either (a) it refuses to solve at all, or (b) (very rarely) a problem which I can solve faster by hand that it does.

So what is the "popular" way to solve #2?

As I said, #2 has been impossibly hard for me.  I spent more time on that problem alone than other problems COMBINED.

For me, I took a radical approach.  I proved that all three sides of the right triangle (one-half of the isosceles triangle) must be integers, and went on from there.  It took 1 page just proving that all three sides must be integers.

When is the grade coming out anyway?

[quote="dts"][quote="mathfanatic"]#4 gave me some problems before I saw an easy way.  At first, I had a series of 5 simultaneous equations of varying degree (:what?:) that would give me the side lengths of the square, but it was so nasty that not even a TI-89 would solve it...[/quote]

I think that I initially did the same thing as you.  I think the problem was that our method wouldn't deliver one solution, but a great number of solutions.  By the way, I tried it on MAPLE, and I got nowhere--in fact, one method went on infinitely

[quote="Danbert"]By the way, a TI-89 isn't always the best way to solve an equation. I happen to have one of the new versions, and occasionally come across a problem that either (a) it refuses to solve at all, or (b) (very rarely) a problem which I can solve faster by hand that it does.[/quote][/quote]

I should just note, the only difference between the TI-89 Titanium and the TI-89 is hardware--the Titanium comes with a USB controller, more memory, etc.  In essence, the TI-89, TI-92, Voyage 200, and TI-89 Titanium are all the same.

I liked number 4 because I used one similar triangle equation to solve it.

I was amazed at how easy it was. Stunned. And I had 5 simultaneous equations before realizing the easy way and saying to myself, "DUH! I'm such an idiot!". Number 5 was a killer for me. I don't know calculus the least bit, so I tried 3-D plane tangencies which got me something close to 24, but not 24, which upsets me.

[quote="origamimasterjared"]Um, the question asked for "three isosceles triangles, no two of which are congruent, [b]with integer sides[/b]..."

I too spent more time on that one than on all the others....because I knew I could do it...and actually succeeded.

I used Hero's formula, set it equal to 6p, and algebra-d my butt off...  I could have used Pythagoras' theorem, but I decided Hero's was cooler :) Plus when I was trying to figure out the problem (one of the first seven times) I ended up proving Hero's formula by use of the Pythagoras Theorem and A=bh/2[/quote]

The ISOSCELES triangles have integer sides--that doesn't mean the right triangles have integer sides.  I proved that the right triangles have integer sides.

I liked #3 a lot because it offered so many different approaches. Number #1 was cool because, as tetrahedr0n mentioned, it left a good deal of space for extra exploration. Also, it didn't get bogged down in ugly equations or boring calculations.

After seeing the approach to problem 2 used in the Math Jam, I was once again reminded that geometry problems often have extremely beautiful solutions, IF YOU ONLY TAKE THE TIME TO LOOK. I have a bizarre tendency to turn beautiful geometry into complex, boring algebra. :P

#5 was fun -- I used the coordinate geometry/differential calculus approach. (Ah, the power of calculus.  :lol: )

numbers two and five gave me fits, but I think number 4 was perhaps the best overall problem, for its simplicity.  Number 1 was just way too easy and number three was a bit weird: proving addition, etc.

There is a relatively easy, non-rigorous method for #2 (i.e. not proving that you found all solutions).

Draw an altitude to the side of the triangle that is not equal to the other two sides, forming two congruent right triangles.  If these triangles have integer sides, our original triangle will have integer sides, which is what we want.  So, we use Pythagorean triples.  Let the lengths of the sides be 3x, 4x, and 5x, use the area-perimeter relation to solve for x, and boom, you have a valid triangle.

Then do it again with 4-3-5 and 5-12-13, and you're done.

It won't get you a commendation, but it will get you a 5.


ratio
geometry
geometric transformation
homothety
inequalities
triangle inequality

Given an equilateral triangle ABC, construct an interior point P such that PA:PB:PC=3:4:5

Since it is completely unnecessary that the given triangle should be equilateral, I will assume a general triangle. Construct 3 circles (A), (B), (C) centered at the triangle vertices A, B, C and with radii in the ratio $r_A : r_B : r_C = PA : PB : PC$. The locus of points P with a constant ratio of distances $\frac{r_A}{r_B} = \frac{PA}{PB}$ from the points A, B is a circle $(O_1)$ centered on the line AB and passing through the internal and external homothety centers $G_{12}, H_{12}$ of the circles (A), (B). To find these 2 homothety centers, erect normals to the center line AB of the circles (A), (B) at their centers A, B. The 1st normal  intersects the circle (A) at points A', A'' and the 2nd one the circle (B) at points B', B'', the points A', B' lying on the same side of the center line AB. The external homothety center $H_{12}$ is the intersection of the lines A'B', A''B'' with the center line AB and the internal homothety center is the intersection of the lines A'B'', A''B' with the center line AB. The center of the locus circle $(O_1)$ is then the midpoint of the segment $G_{12}H_{12}$. The homothety centers $G_{23}, H_{23}$ of the circles (B), (C) and $G_{31}, H_{31}$ of the circles (C), (A) and the locus circles $(O_2), (O_3)$ are constructed similarly. Assume that the circles $(O_1), (O_2)$ intersect at a point P (i.e., that the problem has a solution). Then they also intersect at one other point Q. Since the point P lies on the circle $(O_1)$, we have $\frac{PA}{PB} = \frac{r_A}{r_B}$. Since the point P lies on the circle $(O_2)$, we also have $\frac{PB}{PC} = \frac{r_B}{r_C}$. From transitivity of equivalence, it follows that

$\frac{PC}{PA} = \frac{PC}{PB} \cdot \frac{PB}{PA} = \frac{r_C}{r_B} \cdot \frac{r_B}{r_A} = \frac{r_C}{r_A}$

which means that the point P lies on the circle $(O_3)$ as well, i.e., that the circles $(O_1), (Q_2), (Q_3)$ concur at the point P. Similar reasoning leads to the conclusion that the 3 locus circles also concur at the point Q, i.e., they belong to the same pencil and their centers $O_1, O_2, O_3$ are collinear. We do not need the last fact, because the problem already has been solved - both the points P, Q have the distances from the vertices A, B, C in the desired ratio and we can pick the one lying inside the triangle $\triangle ABC$.

[quote="jelly119"]$\triangle{ABC}$ must be "[b]equilateral[/b]"!![/quote]
How can you say this when faced with a construction showing that a point P with the given ratio of distances $PA : PB : PC = 3 : 4 : 5$ from the vertices A, B, C can be found in a general triangle $\triangle ABC$, a construction, which certainly can be repeated for an equilateral triangle? Would measurements of the segments PA, PB, PC in the above figure of a general triangle persuade you that they really are in the ratio $PA : PB : PC = 3 : 4 : 5\ ?$

Actually, in an equilateral triangle ABC, we can construct an interior point P such that PA:PB:PC=$r:s:t$  if both the following conditions hold:
(1) $r,s,t$ satisfy the triangle inequality $r+s>t$  (assuming that $r\le s \le t$)
(2) The largest angle of the triangle with sides lengths $r,s,t$ is not larger than 120 degrees.


Prove
$ a) \frac{a\plus{}b}{2} \plus{} \frac{a\minus{}b}{2} \equal{} a$

$ b) \frac{a\plus{}b}{2} \minus{} \frac{a\minus{}b}{2} \equal{} b$

a) Multiply both sides by 2. This gives us $ a \plus{} b \plus{} a \minus{} b \equal{} 2a \implies 2a \equal{} 2a$.

b) Do the same thing.

Or you could just simplify the numerators because the denominators are the same. In the first one, you get $ \frac{2a}{2}\equal{}a$ and for the second one you get $ \frac{2b}{2}\equal{}b$.


Ok, so here's my problem, in a regular lotto 6/49 game, what is the probability that you will have exactly one number matching with the winning numbers?

BTW: If you don't know how to play Lotto 6/49 here's how it goes: You choose 6 numbers from 1 to 49 inclusive, (no repeats), and the winning numbers are also selected randomly in the same manner, so, you win prizes based upon how many numbers are matching. (Position of the numbers does not matter. For example if your numbers were 1, 2, 3, 4, 5, and 6, and the winning numbers were 5, 11, 14, 28, 37, and 44, you would have one number matching.

[hide]My teacher says the answer is $\frac{6}{49}$, whereas I think that it is $\frac{\dbinom{6}{1}* \dbinom{43}{5}}{\dbinom{49}{6}}$.... Please help :!: [/hide]

Thank you
 :maybe:  :(
- NAWAHD WERDNA

i believe you are right, if I understood the problem correctly

is your teacher like a ...gym teacher?...or an actual math teacher  :lol: no offense meant

Yes I agree with this answer. :) I can't even see how you get $\frac{6}{49}$ unless you think the name of the lottery is the answer. :D


algebra
polynomial
induction
algebra solved

hey
i need the official solution to this problem .i have a solution whic is 4 times the length of a model olympiad answer :D .here goes the problem
show that for any positive integer n
f(x) = (x 2  + x)^(2^n) + 1 is irreducible in Z(x).
i wud be grateful if ne1 provides a short and sweet solution
cheers

First we shall prove by induction that the polynomial f_n(x) = x^(2^n) + 1 is irreducible over Z[X].For n=0 it's clear , because it is linear.Assume that the result holds for n-1 and let's prove it for n.Let us use a lemma for this :

Lemma:Let f be a polynomial which is irreducible over Z[X].Then , if f(x^2) factors over Z[X] then it factors  in the form f(x^2) = (-1)^deg(k)*k(x)*k(-x) , where k is in Z[X].

Proof:Assume that f factors in linear factors over C[X] like this :
f(x) = (x-a1)(x-a2)...(x-am) , where m is the degree of  f. It cannot have double roots. Assume that it does , then let g(x) be the minimal polynomial of such a root over Z[X]. Then g^2(x) | f(x) in the ring Q[X] , and thus f(x) isn't irreducible in Q[X] , which is a contradiction with Gauss's Lemma , which says that a polynomial irreducible over Z[X] is irreducible over Q[X].Now , let bi and -bi be the roots of the equation x^2 = ai. Let g_i(x) be the minimal polynomial of bi over Z[X].We shall prove that deg(g_i)=m and g_i(x) = +/- g_j(+/- x) , for all i,j.First, since 0 isn't a root of g_i (wouldn't be irreducible anymore if it was) gcd(g_i(x),g_i(-x))=1. But g_i(x) * g_i(-x) = T(x^2) for some polynomial T in Z[X]. So T has the root a_i and it's in Z[X] , it's irreducible and it has leading coefficient 1 , so by the definition of the minimal polynomial , it's equal to f(x) , because f(x) is the minimal polynomial over Z[X] for each of its roots. Thus we conclude that deg(g_i)=m and for each i,j :
g_i(x) * g_i(-x) = g_j(x) * g_j(-x) . Since g_i is irreducible , it follows that it divides g_j(x) or g_j(-x) and this means that it's equal to one of them , except perhaps for a sign.

Now , let f(x^2) = u(x) * v(x). Then if there is a root a_i such that a_i is a root of u(x) and -a_i is a root of v(x) then it follows easily that f(x^2) factors in the desired form (where k=h_i(x)). If not , it means that u(x) is of the form u1(x)*u1(-x) and v(x) is of the form v1(x)*v1(-x) , or equivalently u(x) = u2(x^2) and v(x)=v2(x^2) , which means that f(x^2)=u2(x^2)*v2(x^2) , or f(x) = u2(x)*v2(x) , false because f is irreducible.

Getting back to the proof of the irreducibility of f_n , notice that f_n(x) = f_{n-1}(x^2) so it must factor as f_n(x)=g(x)*g(-x). The roots of f_n are {w*x_1,w*x_2,..,w*x_{2^(n-1)} , -w*x_1 , -w*x_2 , ... , -w*x_{2^(n-1)} } where w is a complex number such that w^(2^n) = -1 and x_1,x_2,.. are the roots of f_{n-1}. So g is in fact of the form (possibly by a reindexation of the roots x_i) g(x) = (x-w*x_1)(x-w*x_2) ... (x-w*x_k)(x+w*x_{k+1})(x+w*X_{k+2}) .. (x+w*x_{2^(n-1)}) . But then -g(wx) is a polynomial in Z[X] which has leading coefficient 1 and has the root x_1 , it has degree 2^(n-1) so it's exactly f_{n-1}(x) . That means that k=2^(n-1). So basically this means that g(x) = (x-w*x_1)(x-w*x_2) ... (x-w*x_{2^(n-1)}) . All we have to prove now is that g isn't in Z[X]. But that's pretty easy since the coefficient of x is w^{2^(n-1)-1} times an integer which is not an integer.

Now , we have proved that f_n(x) is irreducible. Let's prove that f_n(x^2 +x) is irreducible. Let k_i and -1 - k_i be the roots of the equation x^2 + x = x_i , where x_1,x_2,..,x_{2^n} are the roots of f_n. Then let's denote by h_i(x) the minimal polynomial of k_i over Z[X] and notice that (-1)^deg(h_i) * h_i(-x-1) is the minimal polynomial of -1-k_i over Z[X].

Let us prove that f_n(x^2+x) can only factor as the product of two polynomials of degree 2^n.

Case I : for all i , gcd(h_i(x) , h_i(-x-1))=1

In the same manner as before , h_i(x) * (-1)^deg(h_i) * h_i(-x-1) is a polynomial in x^2+x (say T(x^2+x)) and it's from Z[X] and has the root x_i and degree at most 2^(n+1) (because h_i(x)*h_i(-x-1) divides f_n(x^2+x) and thus deg(h_i) <= 2^n) and leading coefficient 1 so it must be equal to f_n(x^2+x). Thus T=f_n and for all i,j :
h_i(x) * h_i(-x-1) = h_j(x) * h_j(-x-1) . So h_i(x) is h_j(x) or h_j(-x-1) , possibly differing by a sign.
If f_n(x^2+x) = a(x)*b(x) , there is a root k_i such that -1-k_i is a root for b and X_i is a root for a. If there wasn't such a root , just like we did in the lemma we would get that f_n(x^2+x) = a1(x^2+x) * b1(x^2+x) , which is a contradiction with the irreducibility of f_n. So,in fact f_n(x^2+x) factors as :f_n(x^2+x) = g(x) * g(-x-1) , where g is a polynomial in Z[X] of degree 2^n.

Case II : there is an i such that h_i(x) and h_i(-x-1) have a common root.Thus they only differ by possibly a sign. If h_i(x)=h_i(-x-1) , then h_i=T(x^2+x) and since it divides f_n(x^2+x) we arrive at the obvious contradiction that f_n is reducible. If h_i(X)=-h_i(-x-1) then let's prove that this is impossible too. h_i is a minimal polynomial of a complex number (easy check that all roots of f_n(x^2+x) are complex) and its other roots are also roots of f_n(x^2+x) so it must have an even degree (since it's in Z[X]).But -h_i(-x-1) then has leading coefficient -1 , contradiction.

Now , it's pretty easy to notice that in Z_2[x] , f_n(x^2+x) = (x^2+x+1)^(2^n) , so in Z_2[X] ,

g(x) = (x^2+x+1)^{2^(n-1)} = x^{2^n} + x^{2^(n-1)} + 1 .

Now we're almost done.I got so close but I couldn't finish off from here.Here's how they do it in the official solution :

Let h(x) = g(-x-1) , and :
g(x) = a_0 + a_1*x + .. + a_{2^n} x^{2^n}
h(x) = b_0 + b_1*x + .. + b_{2^n} x^{2^n}

You compute the coefficient of x^{2^n + 2^(n-1)} + the coefficient of x^{2^(n-1)} in g(x)h(x) and show that it's 0 mod 4

After that by Lucas's theorem you notice that the same coefficient in f_n(x^2+x) is 2 mod 4 and we are done.

This is what I call a suppa-duppa proof! :D :D :D :D

good work iandrei!!

nice work andrei
my solution is xactly the same as the official solution then (the way u describe it,using lucas's theorem and stuff).though the problem is really monstrous ,i think u came up with a pretty neat solution ourselves :D
cheers mate

Hey , actually the true 'official' solution (the one given by the author) is pretty humiliating. It's short and ellegant . The one I was referring to is the one in the Mathematical Competitions 1998-1999 book.

Ok, so you notice the equality f_n(x^2+x) = (x^2+x+1)^(2^n) in Z_2[X] , so if it factors as a(x)*b(x) , then:

a(x) = (x^2+x+1)^p + 2u(x)
b(x) = (x^2+x+1)^(2^n-p) + 2v(x)

Then compute a(w)b(w) , where w^2 + w + 1 = 0. You get u(w)v(w)=1/2 , but that's impossible since u,v are in Z[X] and so using the fact that w^2=-w-1 , u(w)v(w)=A + B*w, where A,B are in Z.

Impressive , huh?

how cud ne1 come up with that?????!!!!! :D
really short and sweet


geometry
incenter
inradius
geometry solved

Prove that if inscribed circle in triangle $ABC$ contains his centroid then $\frac{ab+ac+bc}{a^2+b^2+c^2}=\frac{5}{6}$

Let G be the centroid, I the incenter and r the inradius of triangle ABC. Then, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=21703 , we have

$IG^2=r^2+\frac{1}{36}\left(5\left(a^2+b^2+c^2\right)-6\left(bc+ca+ab\right)\right)$.

Now, since the incircle of triangle ABC has center I and radius r, it is clear that the centroid G of triangle ABC lies on the incircle if and only if IG = r, what is equivalent to $IG^2=r^2$, and thus, as the above formula shows, to $5\left(a^2+b^2+c^2\right)-6\left(bc+ca+ab\right)=0$. But this is obviously equivalent to $5\left(a^2+b^2+c^2\right)=6\left(bc+ca+ab\right)$, i. e. to $\frac{bc+ca+ab}{a^2+b^2+c^2}=\frac56$. And the problem is solved.

  darij

Thanks, but is there any pure geometric solution?

Yes, there is  :D

The incircle touches the sides $a,b,c$ at the points $D,E,F$ respectively.
$M$ is the midpoint of $BC$.

The median from $A$ intersects the incircle at the points $G$ and $P$, where $AM = 3MG$

We know that
$AF=AE=s-a$ (where $s$ is the semiperimeter)
Also,
$BD = s-b$
$CD = s-c$

Suppose that $CD>BD$. Then $M$ is between $C$ and $D \Rightarrow  MD = CD-CM = s-c-\frac{a}{2} = \frac{a+b+c}{2} - c - \frac{a}{2} \Rightarrow$
$MD = \frac{b-c}{2}$

On the other hand, if $CD\leq BD$, then $MD = \frac{c-b}{2}$

At any case, we have $MD^2 = (\frac{b-c}{2})^2$

The power of $M$ with respect to the incircle is

$MD^2 = MG \cdot MP = \frac{1}{3}AM \cdot MP$

The power of $A$ is

$AE^2 = AG \cdot AP = \frac{2}{3}AM \cdot (AM-MP) = \frac{2}{3}AM \cdot AM - \frac{2}{3}AM \cdot MP \Rightarrow$

$AE^2 = \frac{2}{3}AM^2 - 2MD^2$

using the fact that $2AM^2 = b^2+c^2-\frac{a^2}{2}$ and $MD^2 = (\frac{b-c}{2})^2 \Rightarrow$

$AE^2 = \frac{b^2+c^2-\frac{a^2}{2}}{3} - \frac{(b-c)^2}{2} =$

$\frac{2(b^2+c^2-\frac{a^2}{2})}{6} - \frac{3(b-c)^2}{6} =$

$\frac{2(b^2+c^2-\frac{a^2}{2})-3(b-c)^2}{6} =$

$\frac{2b^2+2c^2-a^2-3b^2-3c^2+6bc}{6} =$

$=\frac{-(a^2+b^2+c^2)+6bc}{6}$
:spider:

But $AE^2 = (s-a)^2\Rightarrow$

$AE^2 = (\frac{b+c-a}{2})^2 = \frac{(b+c-a)^2}{4} = \frac{a^2+b^2+c^2+2bc-2ab-2ac}{4}$

and finally

$\frac{a^2+b^2+c^2+2bc-2ab-2ac}{4} = \frac{-(a^2+b^2+c^2)+6bc}{6}$

$3(a^2+b^2+c^2+2bc-2ab-2ac) = 2[-(a^2+b^2+c^2)+6bc]$

$3(a^2+b^2+c^2)+6bc-6ab-6ac = -2(a^2+b^2+c^2) + 12bc$

$5(a^2+b^2+c^2) = 6(ab+bc+ca)$

$\frac{ab+bc+ca}{a^2+b^2+c^2} = \frac{5}{6}$


[quote="delegat"]Nice solution. :blush:
I think that this was A3. in IMO Shortlist 2001.[/quote]
And this one has appeared repeatedly many times in other contest.for example,Korea 2005(if  i was right.)

he just forgot n under the first radical.in rest it is just cauchy

Thanks for explaination. It's clear now. But why did you said "Cauchy"?


probability
symmetry
induction
function
ARML
linear algebra
matrix

A bug is on the $0$ of a number line.  At any point in time, he may jump up $1$ or down $1$ on the number line.  What is the probability that he will at some point end up on $0$ again?

I've seen a solution to this using state diagrams (which you can use, of course) but is there a way this can be done with Markov chains?

He doesn't stop, right? Doesn't the probability automatically approach 1 by law of averages?

To be more rigorous... Define $P(k)$ as the probability that the bug will reach space 0 if he is on space $k$, such that $0 < P(k) < 1$ for all $k$. By symmetry $P(k) = P(-k)$. Now let the bug hop one space so that $k = \pm 1$. We have the following by the problem:
$P(0) = 1$
$P(k) = \frac{P(k-1)+P(k+1)}{2}$ for all integers $k \neq 0$
Let $P(1) = P(-1) = x$. You can use strong induction to show that $P(k) = P(-k) = k(x-1)+1$ for all positive $k$. From this, suppose that $x < 1$. Then there exists some $n$ such that $x \leq \frac{n-1}{n}$. Consider $P(n+1)$. It is equal to $(n+1)(x-1)+1 \leq (n+1)(\frac{n-1}{n}-1)+1 = \frac{-1}{n}$, which is absurd. So $x < 1$ is false. It follows that $P(1) = 1$, and thus $P(k) = 0 \cdot k+1 = 1$ for all $k$.

Aren't state diagrams and Markov chains two names for the same thing?

By the way, did you hear about the state diagram approach from a teacher at Stuy?

An interesting approach if the question is asking for the probability after taking $n$ steps: Using the generating function $(x+x^{-1})^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{-n+2k}$, we obtain that the constant terms tells us the number of ways one can get back to $0$ after $n$ steps.  Clearly, $n$ must be even since for odd $n$ the probability is $0$.  Therefore, the probability to get back to $0$ after $n$ steps is $\frac{\binom{n}{\frac{n}{2}}}{2^{n}}$.  I'll think about this later...my illness does not allow me to think efficiently.

[quote="Ravi B"]By the way, did you hear about the state diagram approach from a teacher at Stuy?[/quote]
Yes, why?

I once gave a talk on state diagrams at an ARML practice in NYC.  A couple of Stuyvesant teachers said they liked the talk, and one said he planned to explain the method to his students.

I don't have lecture notes for the talk, but you can find the list of problems here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38284[/url].

let p_n be the probability that if you are at (n,0), you will hit 0,  p_n=1/2(p_{n-1}+p_{n+1}), for n = N-{1,-1}, where p_1=1/2(1+p_2) and p_{-1}=1/2(1+p_{-2})

this is just a difference equation that can be solved...we want 1/2(p_1+p_{-1})

[quote="miyomiyo"]This does not really answer my question, does it?[/quote]

[quote="Wikipedia"]Suppose we draw a line some distance from the origin of the walk. How many times will the random walk cross the line? The following, perhaps surprising, theorem is the answer: for any random walk in one dimension, every point in the domain will almost surely be crossed an infinite number of times.[/quote]

Yes it does.  The probability is $1$.

[quote="Ravi B"]Aren't state diagrams and Markov chains two names for the same thing?[/quote]
I thought Markov Chains was the Matrix method.

Markov chains, stochastic matrices, and state diagrams are three representations of the same concept.  Each boils down to solving a system of linear equations.

Another question based on the original question: what is the expected number of moves that need to take place before the bug lands on $0$ again?

Clearly the path length will be even.

For the purposes of expected value, we can assume that the first jump is to the right, and thus the last jump must be to the left.

[hide="Conjecture"]
The number of ways to return to zero in exactly $2k$ moves is given by $f(k)=\frac{1}{k}\binom{2k-2}{k-1}$
[/hide]

Someone please verify/disprove

Yes -- it is the famous Catalan sequence (with the index shifted by 1 from the standard form, to account for the fact that you want to return [i]for the first time[/i] on the $2k$-th move).


Jane has a dime a nickel and a dollar.  Bob has twice more then the amount of money that Jane has.  How much money does Bob have?

[hide] $2.30 [/hide]

last time I checked this was basics....not classroom....

This is not too complicated...
1.00 + .10 + .05 = 1.15 dollars x 2 = [b]$2.30. There's your answer.[/b]

yeah I think it is,
[hide]Jane has 1.15, Bob has 2(1.15) or 2.30[/hide]

Amazingly...This doesnt seem to be as easy as you think it is.

NO ONE

Has gotten this right...

Care to have another go?  :P

[hide]twice more = 200% more or 300 % = $3.45[/hide]

[hide]oops, then the answer is 3.45[/hide]

lmao :rotfl:

I like that problem

 :spider:  :ddr:  :starwars:


It's very easy . From Minkowski's ineq , we have : $ \sum{\sqrt{x^2+1}} \geq \sqrt{(x+y+z)^2+9} $  ;)


Prove that
$\frac{\varphi(1)}{2^1-1}+\frac{\varphi(2)}{2^2-1}+...+\frac{\varphi(n)}{2^n-1}<2$,
where $\varphi(k)$ - number of such integer $a$, that $1\leq a\leq n$ and $gcd(a,k)=1$

Very nice problem  :)

I will prove a bit more (for $x>1$):
\[ \sum_{n=1}^\infty \frac{\phi (n)}{x^n-1} = \frac{x}{(x-1)^2}  \]

Proof:
Set $y=\frac{1}{x}$ and see $\frac{x}{(x-1)^2} = \frac{y}{(y-1)^2}$ and $\frac{1}{x^n-1} = - \frac{y^n}{y^n-1}$.

Note that $\frac{y^n}{y^n-1} =- \sum_{k=1}^\infty y^{kn}$ and also $\frac{y}{(y-1)^2} = \sum_{n=1}^\infty n y^n$.

Now the equality $\frac{y}{(y-1)^2} = \sum_{n=1}^\infty n y^n = \sum_{n=1}^\infty \sum_{d|n} \phi(d) y^n = \sum_{d=1}^\infty \sum_{k=1}^\infty \phi(d) y^{dk} = - \sum_{d=1}^\infty \frac{\phi(d) y^d}{y^d-1}$ follows.

Lambert series http://mathworld.wolfram.com/LambertSeries.html ? ;)


geometry
incenter
perimeter
inradius
inequalities
geometry unsolved

Prove $ \cat{GIH}> 90^\circ$
($ G$ is centroid, $ I$ is incenter, $ H$  is orthocenter)

As I remember, I 've found a geometric solution for this problem. Later, I knew another solution base on algebraic geometry. First, calculate $ GI^2, GH^2$ and $ IH^2$ based on $ p, r, R$ where $ p$ is half the perimeter of the triangle, $ r$ is the inradius and $ R$ is the circumradius.
And then all that we have to show is $ GH^2 > IH^2 \plus{} GI^2$ (The inequality will be very elegant).

I don't think you really mean "algebraic geometry". ;)

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=39447 .

  darij

Yeah. I just don't know what to call it.


superior algebra
superior algebra unsolved

How to find the automorphisms of $ \mathbb{Z}/n\mathbb{Z}$? If you have any articles pls give me.

Notice that the image of $ 1$ determines the entire automorphism. Then think about what are the possibilities for its image  :wink:

Yes, and the result will depend on whether you speak of rings or groups.


Consider a $ m$ by $ n$ matrix, with all entries real.
Prove that the span of the vectors formed by the columns of the matrix, in $ R^{m}$, has the same dimension, as the span of the vectors formed by the rows of the matrix, in $ R^{n}$.

this is evident from the reduced row echelon form of the matrix (row operations maintain dependence relations between columns and preserve the row space). i'm sure this has appeared on the forum before, so do a search.

Key word: rank.

The truth of this theorem implies that rank is a well-defined concept.


vector
probability
topology
function
integration
abstract algebra
algebra

Let $ Y$ be a random vector on a complete p.s. $ (\Omega,\mathcal F,P),\ \mathcal G\subset \mathcal F$
We know that there is a version $ P(\omega,A)$ of the conditional probability $ P(Y\in A\mid \mathcal G)$ s.t. $ P(\omega,\cdot)$ is a measure a.s. and $ P(\cdot,A)$ is $ \mathcal G$-measurable.

Now let $ f(x,y)\colon\mathbb R^k\times\mathbb R^d \to \mathbb R^m$ be measurable, $ B$ be Borel set in $ R^m$.

Is it true that $ g(\omega,x) \equal{} P(\omega,\{y: f(x,y)\in B\})$ is $ \mathcal G\otimes \mathcal B(\mathbb R^k)$-measurable?

In fact, I need only that $ \{(\omega,x): g(\omega,x) \equal{} 1\}\in \mathcal G\otimes \mathcal B(\mathbb R^k)$ for $ f$ linear and $ B$ open.

[quote="zhoraster"]
We know that there is a version $ P(\omega,A)$ of the conditional probability $ P(Y\in A\mid \mathcal G)$ s.t. $ P(\omega,\cdot)$ is a measure a.s. and $ P(\cdot,A)$ is $ \mathcal G$-measurable.[/quote]
I'm bit confused about what kind of "conditional probability" you are talking of here. More precisely, what are $ P(Y\in A\mid \mathcal G)$ and $ P(\omega,A)$? :?

This is not to say that you wrote anything wrong, just to say that my knowledge of the standard notation is not enough to figure out what is meant.

Well I'm a bit confused too  :huh: what is the space where $ Y$ is landing ?
I don't know if "a complete space" means in your mind that the probability space is complete  (measure theory) or if you mean that the arrival space of Y is complete space (topology theory).
If the r v $ Y$ takes its value on polish space then I agree with the existence of regular conditionnal PROBABILITY (not expectation that always exist no matter what).

Then the question is again quite unclear :

what does $ \omega,\{y: f(x,y)\in A\}$ mean ? does it mean that $ \omega$ s.t. $ f(x,Y(\omega))\in A$ ?

If so don't you want to prove the slightly different statement that

$ g(x,\omega) \equal{} P(f(x,Y)\in A|\mathcal{G})(\omega)$ is $ \mathcal{G}\otimes\mathcal{B}(\mathbb{R}^{k})$-measurable ?

Well if so, I think it's true but to get the proof I think one should express g the as a composition of different measurable functions, and it would be also appreciable to express every spaces with sigma fields to get things neat.

[quote="fedja"]
I'm bit confused about what kind of "conditional probability" you are talking of here. More precisely, what are P(Y\in A\mid \mathcal G) and P(\omega,A)?
[/quote]
$ P(Y\in A\mid \mathcal G) \equal{} E[1_{Y\in A}\mid \mathcal G].$

There is a function $ P(\omega, A)$ with properties given above s.t. $ P(Y\in A\mid \mathcal G) \equal{} P(\omega,A)$ a.s.

[quote="TheBridgeTwo"]Well I'm a bit confused too  :huh: what is the space where $ Y$ is landing ?
[/quote]
$ R^d$, forgot to say, sorry (though it is quite clear from the fact that $ f(x,y)$ is defined on $ \mathbb R^k\times \mathbb R^d$).
[quote]
I don't know if "a complete space" means in your mind that the probability space is complete  (measure theory)
[/quote]
yes

[quote]
what does $ \omega,\{y: f(x,y)\in A\}$ mean ? does it mean that $ \omega$ s.t. $ f(x,Y(\omega))\in A$ ?
[/quote]
No. It means what it says. For fixed $ \omega$, $ P(\omega,\cdot)$ is a measure on $ \mathbb R^d$. And $ \{y: f(x,y)\in A\}$ is some set, depending on $ x$, of course. And $ P(\omega,\{y: f(x,y)\in A\})$ is just $ P(\omega,\cdot)$ of this set.

[quote]
If so don't you want to prove the slightly different statement that

$ g(x,\omega) \equal{} P(f(x,Y)\in A|\mathcal{G})(\omega)$ is $ \mathcal{G}\otimes\mathcal{B}(\mathbb{R}^{k})$-measurable ?
[/quote]
This is exactly what I want. But there is a problem with your formulation because the conditional expectation can be changed on a set of measure $ 0$. So if we define them in a non-consistent way for different values of $ x$, there is no hope for measurability. In fact, I can give a counter-example.

That is why I need the regular version of conditional expectation.

OK, then (unless I still misunderstand something) for open $ B$ and continuous $ f$, it is trivially so. Indeed, the pre-image of $ B$ is open whence it is a countable disjoint union of semi-open parallelepipeds $ U_j\times V_j$. Thus your function is $ \sum_j 1_{U_j}(x)E(1_{Y\in V_j}|\mathcal G)(\omega)$ for a.e. $ \omega$ and every $ x$.

I guess the general Borel case follows by the usual tricks with monotone classes and MCT but, since you do not really need it, I leave it to someone else to figure out if it is, indeed, the case.

Well fedja I am not sure how to show explicitely that $ \sum_{j}1_{U_{j}}(x)E(1_{Y\in V_{j}}|\mathcal{G})(\omega)$ defines a probabilty measure in the case where x is not in any $ V_j$, because in that case $ g(x,\omega)$ is nill for any $ \omega$, isn't it ?

But if we had a measure $ \mu$ on space on which x is living, maybe we could define something like
 $ \sum_{j}\mu[1_{U_{j}}(x)].E(1_{Y\in V_{j}}|\mathcal{G})(\omega)$ as a candidate ?

Otherwise why did you wrote that "for open B  and continuous f, it is trivially so" I don't get it.

No, all those are not measures but functions. As I said, I may still misunderstand the notation but my idea was that for a fixed $ A\subset \mathbb R^d$, $ E(1_{Y\in A}|\mathcal G)$ means a $ \mathcal G$-measurable function $ P(\cdot,A)$ such that $ \int_B P(\omega,A)\,dP(\omega) \equal{} P(\{Y\in A\}\cap B)$ for all $ B\in\mathcal G$. $ P(\omega,A)$ is also a measure, but in the $ A$ variable, which is used to justify the identity
\[ g(\omega,x) \equal{} \sum_{j: x\in U_j}P(\omega,V_j)\]
for a.e. $ \omega$, namely, for all $ \omega$ such that $ P(\omega,\cdot)$ is a measure (which is equivalent to the formula I wrote but, may be, is expressed in a better notation).

I see no problem with [b]fedja[/b]'s argument.
Many thanks, by the way! It really helps.

Actually, I needed the measurability to apply the measurable selection theorem. Yesterday I proved it without the measurability -- using Efros notion of measurable random sets and Kuratowski-Ryll-Nardzewski theorem. There was nothing like easy, a lot of effort to prove measurability.

But this way it's much easier, and we can extend to Borel sets with the method of appropriate sets, and probably to measurable functions.

ok fedja I missed the point that only measurability was needed  :blush: I got confused and thought probabilty kernel was the desired property
Your approach is working very nicely.

to zhoraster
measurable selection theorem ? what is it ? I only know "measurable section theorem" could you state it as well as Kuratowski-Ryll-Nardzewski theorem and the definition of Efros measurable random set for my personal culture

thx

[b]Measurable selection theorem[/b] (in its simplest form)
Let $ (\Omega,\mathcal F, P)$ be a complete probability space, $ X$ be a Polish space.
Suppose $ A\in \mathcal F\otimes \mathcal B(X)$ and its projection to $ \Omega$ has probability $ 1$.
Then there is an $ X$-valued random variable $ \xi$ s.t. $ (\omega, \xi)\in A$ almost surely.

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

Now let (for simplicity) $ \mathcal C$ denote the family of closed subsets of $ X$.
The (closed) set-valued mapping $ F: \Omega\to \mathcal C$ is called (Efros) $ \mathcal F$-measurable if for any open $ V\subset X$
$ \{\omega: F(\omega)\cap V\neq \varnothing\}\in\mathcal F$.

Domain of $ F$ is $ \mathrm{dom} F \equal{} \{\omega: F(\omega)\neq \varnothing\}$

[b]Kuratowski-Ryll-Nardzewski theorem[/b]:
For a measurable closed set-valued mapping $ F$ there is a measurable [i]selector[/i], i.e. an $ X$-valued random variable $ \xi$ s.t. $ \xi(\omega)\in F(\omega)$ for all $ \omega\in \mathrm{dom} F$.

There is a stronger
[b]Castaing theorem[/b]
There exists a family $ \{\xi_n\colon \mathrm{dom} F\to X\}$ of measurable functions s.t. $ \{\xi_n(\omega)\}$ is dense in $ F(\omega)$ for all $ \omega\in \mathrm{dom} F$.

Ok thx zhoraster your selection tehorem what I know under the term section theorem

But I didn't knew either Efros definition nor the two theorems that you mention

regards

[quote="zhoraster"]But this way it's much easier, and we can extend to Borel sets with the method of appropriate sets, and probably to measurable functions.[/quote]
I think this following can avoid the heavy canon balls: We are done if we can show that $ g_C(\omega,x) \equal{} P(\omega,\{y: (x,y)\in C\})$ is measurable for every $ C\in\mathcal B(\mathbb R^k\times\mathbb R^d)$. This should hold just by taking $ f$ as identity on $ \mathbb R^m$ if $ m \equal{} k \plus{} d$. If this is correct then the claim holds for arbitrary measurable $ f$.

So let $ \mathcal D$ be the system of such $ C\in\mathcal B(\mathbb R^k\times\mathbb R^d)$ such that $ g_C$ is measurable. Then show that $ \mathcal D$ is a Dynkin system containing the system $ \mathcal E$ of sets of the form $ U\times V$ with $ U,V$ open in $ \mathbb R^k$ and $ \mathbb R^d$, respectively. Then $ \mathcal E$ is a $ \pi$-system and the claim follows by Dynkin's $ \pi$-$ \lambda$-theorem.

I quickly scribbled a reasoning on my piece of paper in front of me - and it seemed to be sort of straight forward as fedja already expected. Hope that it still holds if someone else will have cross-checked it. ;)


What is the sum of all integer values of $ x$ such that $ \frac{67}{2x \minus{} 23}$ is an integer?


$ 4x^2 \minus{} 3\equiv0\pmod7\implies4x^2\equiv3\equiv24\pmod7$

$ \implies x^2\equiv6\pmod7\implies\boxed{\varnothing}$.


algebra
polynomial
induction
function
number theory unsolved
number theory

find all polynominals w with real cofficients having the following property:
if x+y is a rational number, then so is w(x)+w(y)

The set of such polynomials is $W=\{ ax+b,\ a,b\in Q \}$. It is easy prove first by induction on $n=deg\ P, \ P(x)=a_{0}x^{n}+\dots+a_{n}$ that $a_{i}\in Q, \ i=1\dots n$, just let $x=y\in Q$.
Now suppose that $deg\ P\geq 2$. Let $x=q+\alpha,\ y=q-\alpha$, where $q\in Q$ and  $\alpha$ is transcendental. So, $P(x)+P(y)=G(\alpha)$, where $G$ has rational coefficients which depend on $q$ and $2\leq deg\ G\leq deg\ P$. But than $P(x)+P(y)$ can't be rational cause $\alpha$ is transcendental, so $deg\ P < 2$.

Let $z=\frac{x+y}{2},t=\frac{x-y}{2}$, then $w(x)+w(y)=2w(z)+2f(t^{2})$, were \[f(t^{2})=\sum_{k=2}^{n}\frac{a_{k}}{2}((z+t)^{k}+(z-t)^{k}-2z^{k})\]
if \[w(x)=\sum_{k=0}^{n}a_{k}x^{k}.\]
$f(t^{2})$ continiosly function with only rational means, therefore $f(t^{2})\equiv 0 \ (f(0)=0).$ It give $w(x)$ is additive function, therefore $w(x)=ax+b, \ a,b\in Q$.


If the consecutive integers from 50 to 1 were written as 5049484746..., what would be the 67$ ^{th}$ digit to be written?

WE write 68 digits of 2 digit numbers:50,49....34 numbers, so its like 26, so 2 is the 67th digit.

I believe the $ 67th$ digit is $ 1$. I just brute-forced it. I started counting at the '$ 5$' of $ 50$, then to $ 0,4,9,4,8$. I found, oddly enough, that the $ 10th$ digit is $ 6$, the $ 20th$ digit is $ 1$, the $ 30th$ digit is $ 6$, and it alternates like that. I assume that you can find the answer by modular arithmetic, but I'm not sure.

Can anyone show how to arrive at this answer?


Give $ a,b\in \mathbb N^*,gcd(a,b) \equal{} 1$Prove that for all $ n\in \mathbb N^*$ such that $ n\ge ab,$ Exist a pair $ (k;l)$such that two conditions $ k,l\in N$ and $ n \equal{} ka \plus{} lb$
I tried some numbers and saw it is true,but proving is too hard

[url=http://www.mathlinks.ro/Wiki/index.php/Chicken_McNugget_Theorem]Is this what you're going for?[/url]


I use Mac and i can't include eps images!!

pdflatex cannot use eps images. Either
1) convert eps to another format first such as pdf, png or jpg
or
2) compile to DVI using latex and then convert to pdf via ps


[[User:Temperal/The Problem Solver's Resource]]

I thought I should post this in the forum just so that people would know about it...

Anyways, it doesn't display properly in many browsers, I'm trying to fix that.

Number Theory $ \neq$ inequalities!!!
And I don't think it's usefull for IMO...

All right, I know number theory isn't inequalities, but I don't really know any non-inequality number theory...

if you think anything is incorrect, just edit it in. I'm not stopping you or anything, rather the opposite.


calculus
integration
limit
trigonometry
calculus computations

Find $ \lim_{n\to\infty} \int_0^{\frac{\pi}{2}} x|\cos (2n\plus{}1)x|\ dx$.

Integrating separately on each part of the integrate interval where $ \cos(2n \plus{} 1)x$ is positive or negative, we can define the integral as:

$ \int^{\frac {\pi}{2}}_0 x \left| \cos (2n \plus{} 1) x\right|\, dx \equal{} \frac {2 \pi (n \plus{} 1) n \plus{} \pi \minus{} 2}{2(2n \plus{} 1)^2}$

so the limit is easy, and it equals $ \frac {\pi}{4}$.

The integral equals $ \frac{1}{{{(2n\plus{}1)}^{2}}}\int_{0}^{\frac{(2n\plus{}1)\pi }{2}}{x\left| \cos x \right|\,dx}.$

Now,

$ \int_{0}^{\frac{(2n\plus{}1)\pi }{2}}{x\left| \cos x \right|\,dx}\equal{}\sum\limits_{k\equal{}0}^{n}{\int_{\pi k}^{\frac{(2k\plus{}1)\pi }{2}}{x\left| \cos x \right|\,dx}}\equal{}\sum\limits_{k\equal{}0}^{n}{\int_{0}^{\frac{\pi }{2}}{(x\plus{}k\pi )\cos x\,dx}}.$

The rest are straightforward calculations.


vector
linear algebra
matrix
anticommutative

First, I think there must be a type or something like this since I cannot believe this is the real statement. But anyway, here is what I did:
   First step: for any $ x_1,...,x_n$ we have $ (x_1A_1+...+x_n A_n)^2=-(x_1^2+...+x_n^2)I$. This can be observed just by expanding and grouping the terms $ x_i x_j (A_i A_j+A_j A_i)$, which are 0. From this step we conclude that these matrices are linearly independent and thus they span a vector space of dimension $n$. It also follows that any non-zero matrix in this vector-space is invertible. This gives us already a better bound: $n\leq 4$. Indeed, if not we regard the first column in each matrix from $A_1,...,A_5$. We have 5 vectors in $R^4$ thus they are dependent and that combination of matrices is non-zero (since they are linearly-independent) and non-invertible.
    Step 2: I think we even have $n\leq 3$. Let us look at the vector space of matrices with identical first row and column (that is $a_{12}=a_{21}, a_{13}=a_{31}, a_{14}=a_{41}$). This has dimension 13 over $R$ and I will prove it is separated from the space generated by our matrices. But if the matrix $x_1A_1+...+x_n A_n$ is in the above space, we regard its square. The element on the first position of its square will be nonnegative and on the other hand it will be $-(x_1^2+...+x_n^2)$, thus all $x_i$ are 0 and the two spaces have trivial intersection. From here it follows $n\leq 3$. Is there a huge mistake here?  :?

if the matrices can have complex entries, and their squares don't have to be $-I$, then the maximum size of a family of invertible, pairwise anticommuting matrices is $2v_2(n)+1$. when $n=2$, one example is the Pauli matrices.


Any body has a solution for this problem? Please help me to solve. I tried many times but I can't find out the elementary solution:
For all positive integer numbers $ n,m$ and all  possitive real numbers $ x,y$ Prove that:
$ (n\minus{}1)(m\minus{}1)(x^{n\plus{}m}\plus{}y^{n\plus{}m})\plus{}(n\plus{}m\minus{}1)(x^ny^m\plus{}x^my^n) \geq mn(x^{n\plus{}m\minus{}1}y\plus{}xy^{n\plus{}m\minus{}1})$

$ (m\minus{}1)(n\minus{}1)(x^{n\plus{}m}\plus{}y^{n\plus{}m})\plus{}(n\plus{}m\minus{}1)(x^{n}y^{m}\plus{}x^{m}y^{n})\geq mn(x^{n\plus{}m\minus{}1}y\plus{}xy^{n\plus{}m\minus{}1})$

<----->

$ nm(x^{n\plus{}m\minus{}1}\minus{}y^{n\plus{}m\minus{}1})(x\minus{}y)\plus{}(n\plus{}m\minus{}1)(x^{n}\minus{}y^{n})(x^{m}\minus{}y^{m})\geq0$
very easy .
$ x\equal{}y$


quadratics
algebra
system of equations
complex numbers

If $(x,y)$ is a solution for the system of equations below, find the maximum value of $x^2+y^2$

$2x^2+5xy+3y^2=2$
$6x^2+8xy+4y^2=3$

[size=59][color=blue]b-flat:  Fixed some typos[/color][/size]

I found that the [i]minimum positive[/i] value is [hide]$24$[/hide], but I did not find an upper bound on the numbers. In fact, without an upper bound I got that [hide]$x^{2}+y^{2}=24^{n}$[/hide] can be true, where $n$ is an integer.

I know that $x$ and $y$ have to be complex numbers (just solve the first equation as a quadratic in $x$ or a quadratic in $y$) so is there even a maximum? Wouldn't most restrictions on answers rely on the fact that the answers have to be real, thus allowing you to apply things like AM-GM. I don't know, I might be wrong.  :maybe:

drunner2007, how did you solve that systhem of equations? It is better to express the method for others, like me, to learn with you.


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AMC 12
Putnam

What is/are good strategy/strategies to improve score on AMC and AIME? Share your practice methods!

I'm pretty much reading the Art of Problem Solving books, over and over when I don't understand stuff, and doing some of the old tests that can be found in the contest sections. I am not timing myself yet, I'm just trying to do as many of the problems as I can. Accuracy is more important than speed at this point. I'll probably time myself in the month before the competition, and I am saving a few AIMEs that I have never done before on purpose so that I can have a true test-taking experience. Also, I am taking the AIME Problem Solving B class, which is really helpful.

What I think is the best strategy is to not look at the solution to a problem unless you have an answer, pretty much ever. Instead, I'll work on similar problems that are easy, or read the section about that type of problem in AoPS. You learn a ton more when you figure out a solution yourself than when someone else tells you the solution. If you are totally stumped, you could ask for a hint to get you going in the right direction on the forums.

You can take classes as well. They are very helpful and will prove to give good result ONLY IF you put enough time to work on the problems.

Depends on what your goals are - for USAMO, spend your time on the 21-22 easiest problems.  Check them over a lot, and then if you have extra time, do one of the last three and answer it if you have a GOOD hunch.  Worked for me...

[quote="PTynan89"]It's much more efficient to go for like a 120 on AMC 12 and a 14 on AIME, if you are slow but good.

But, if speed is your thing, then 132 - 147.5 and 8 - 10 would be a better goal.

Of course practice both still, but spend more time on whichever you need to improve more on (or whichever you can improve on the easiest/quickest).[/quote]

14 on AIME??

Well, it's okay if you don't get 14... as long as you make 12+ on AIME, you're probably going to make USAMO.

I'd say that 120-130 on AMC and 10-11 on AIME is good to make USAMO...

[quote="Phelpedo"]Have you ever taken the AMC12?[/quote]
Not officially unfortunately, but I have taken old ones though to most people that doesn't mean much. However, I usually do better under pressure because my brain is forced to work faster because I feel like I have no time.

the older AMC 12's (AHSME) are definitely harder than they are today

[quote="MysticTerminator"][quote="bpms"][quote="PTynan89"]It's much more efficient to go for like a 120 on AMC 12 and a 14 on AIME, if you are slow but good.[/quote]
I've never thought that the AMC is very time pressuring, I always was under the impression that 3 minutes is plenty of time for the problems, especially since I've fallen into what MysticTerminator calls "the Mathcounts Trap". I'm so used to a ton of time pressure that I do all the multiplying and other stuff in my head.[/quote]

HA! THANK YOU! SEE MATHCOUNTS IS EVIL HERE IS IRREFUTABLE PROOF[/quote]

Umm.... bpms just said that Mathcounts helped him do the problems faster... Which is probably a good thing...

He's never even taken the test so he doesn't really know what the time pressure is like...

But I do better under pressure usually (at least with Mathcounts)
Then again, maybe (probably) I'm just a naive middle schooler.

um the poitnof the AMC is about being slow rite I dont get it???

help me

[quote="jmerry"]I was at 1996 Mathcounts nationals: the Masters round included Reid Barton, Sasha Schwartz, and Po-Shen Loh, all of whom would later be on the IMO team (I don't remember who the fourth participant was).[/quote]
David Zhang, from Kansas. I don't think he ever made it to MOP, but he did qualify for the USAMO multiple times.

Well it's not really that you should instantly know what 15^2 is, it's just that you should have been practising (at least) long enough to be able to do that.

yeah also mathcounts is evil

y'know just to reiterate

[quote="Boy Soprano II"][quote]But then, MathCounts is fun in the way that it's pretty cool to watch on TV... it's fun to see those middle schoolers sweat  :D [/quote]
See, this is what irritates me a bit—MATHCOUNTS is the only thing the general public really sees about competition mathematics, and so they think it's this thing with buzzers and lightning-fast, when this is in fact the shallowest part of the competition.  By contrast, I think the media do not even mention the masters round, which is really the best part, and the masters round doesn't happen at most state or chapter competitions, which is a real shame, since most people don't even get to participate.  Yeah, it definitely has a place—it was my first math competition, and I got involved in it because my parents were just willing to take me to practice each week, and if I hadn't gotten involved then, who knows where I would be now?  But the nationals problems are simply too easy for the people competing there (wouldn't it be better to give them a test of about 10 much harder problems?) and it puts you in this very unwholesome attitude about speed being important, and gives you the completely wrong idea about math—you get into being nimble with arithmetic, instead of seeking elegance.

[quote]Even so, there's a remarkably strong correlation between the two contests- the same people win both.[/quote]
This may have been true in the past, but today it seems to be patently false.  I was wondering if this was true last year after I qualified for USAMO, and about half of the top twelve MATHCOUNTS people from last year didn't even make USAMO.  If I remember aright, most people in the top 57 didn't make USAMO, either.  Several people qualified who were not even at nationals, and Delong and Eric Larson (who pretty much beat everybody at red MOP this year) were not in the top 57, if I remember aright.[/quote]
Eric Larson got 53rd place

[quote="Boy Soprano II"][quote="Phelpedo"]Actually, Eric is a two time top-57er (but he's been around 53? both times). And there is quite a strong correlation of success, even in recent results, if you call '03 and '04 recent. Adam Hesterberg, Arnav, Brian Lawrence, Shaunak Kishore, and Greg Gauthier, all moppers, were in the '03 countdown. So while not everybody who does well in MathCounts does well in the USAMO, there is a pretty high correlation.[/quote]
I don't find a this much overlap a high correlation at all, in the context—I think 50% is pretty low.  And Gregory Gauthier didn't make MOP this year, whereas many people who didn't do as well as he in Mathcounts did.  And only about 6/21 of the MathCounts champions up to 2004 have yet made the IMO team.  Compare that, for instance, to the number of IMO team members in the past decade or so who made IMO without having ever been to MOP before (yay Zeb!).

And why should you instantly know what $15^{2}$ is?[/quote]

Interesting, the website had Gauthier listed.

In any case, I think 50% is fairly high when you consider the difference between MathCounts and USAMO. It's almost like the difference between geography bee and USAMO :wink:  Though I agree of course that people who are really good would obviously have no trouble handling the mathcounts problems, it's just that the speed and the simplicity tend to make results a bit random.

um dude that is exactly our point, that they are two vastly different contests

and one sucks

guess which one

hint it is not the USAMO

[quote="Boy Soprano II"]Yeah, there were some typos on the AMC site regarding who was at MOP.  Gregory Gauthier was not at MOP this summer.

Okay, I see your point, though.[/quote]

You know what I like to call those?

LIES!

[quote="mustafa"][quote="Karth"][quote="Boy Soprano II"]Yeah, there were some typos on the AMC site regarding who was at MOP.  Gregory Gauthier was not at MOP this summer.
[/quote]

I wasn't at mop, but I know that Fredrick Johnson of Paideia wasn't there either. He would be a black mopper if he were at mop  :wacko:[/quote]

Wait, so did he decide not to go, or did he get sick and go to the hospital during the USAMO, or what?[/quote]

It was a mistake.. he didn't make black mop. (At least i don't think so...)

I'd say the best strategy for taking the tests is getting all the problems right that you can...preparation is a different story

yes, and while we're talking about strategies, were you aware AoPS is looking to hire a mathematics educator

we would fill this job excellently


calculus derivative trigonometry limit algebra function domain

Three students attempt to solve the following problem:

"If f(x) = $ sin^{-1}$$ \frac{2x}{1+x^{2}}$ +$ c$.$ tan^{-1}x$, find the value of  the constant c for which the derivative of f with respect to x is zero for all values of x."

Student A  claims that c can have any real value. There is actually no need to calculate the derivative -the fact that it is zero for all x implies that f(x) is constant.And sincce f(0)=0 we must have f(x) =0 for all x.

Students B and C,on the other hand calculate the derivative. Student B finds one value of c satisfying the conditions of the question while student C finds two values.

Who is correct?

we can have at most one value, since otherwise $ \tan^{\minus{}1}$ would be a constant function. Differentiation yields $ c\equal{}\minus{}2$ (if I did not produce an error)

[quote="Michael Niland"][color=darkred]Consider the function $ f: \mathcal R\rightarrow\mathcal R$, where $ f(x) = \arcsin\frac {2x}{1 + x^{2}} + k\cdot\arctan x$ and $ k$ is a constant. Find the value of  the constant $ k$ for which $ f'(x) = 0$ for any $ x\in\mathcal R$[/color].[/quote]
[color=darkblue]Since $ f( - x) = - f(x)$ (odd) can suppose w.l.o.g. $ x\ge 0$.

Observe that $ \{\begin{array}{c}f(0) = 0 \\
 \\
f(1) = (k + 2)\cdot\frac {\pi}{4} \\
 \\
f(\sqrt 3) = (k + 1)\cdot\frac {\pi}{3}\end{array}\implies k\in\emptyset$. Can try and $ \lim_{x\to\infty}f(x) = k\cdot\frac {\pi}{2}$.

Prove easily (elementary, without calculus) $ g(x)=\arcsin\frac {2x}{x^{2} + 1} = \{\begin{array}{ccc}2\cdot\arctan x & \mathrm{for} & 0\le x\le 1 \\
 \\
\pi - 2\cdot\arctan x & \mathrm{for} & 1 < x\end{array}$.

[b]The enunciation is no correctly. At first you must require the domain $ D'$ of the derivative-function $ g'$, i.e. the construct of it.[/b] Prove easily that $ D' = [0,1)\cup (1,\infty )$ and the point $ (1,\frac {\pi}{2})$ is angularly. Exactely, $ \{\begin{array}{c}g'_{l}(1) = 1 \\
\ g'_{r}(1) = - 1\end{array}$.[/color]

[hide]They are all incorrect.

The value c = -2 will make f'(x) = 0 for all $ x\in (\minus{}1,1)$. The value c = 2 will make f'(x) = 0 for all $ x\in (\minus{}\infty,\minus{}1)\cup (1,\infty)$. But there is no value of c that will make f'(x) = 0 for all x.

Note, in fact, that this function is not differentiable at x = \pm 1 for any value of c.[/hide]

[quote="gauss202"][b]They are all incorrect [/b] ....[/quote]

Sorry, [b]Gauss-2002[/b], you don't read with attention ! Since the function $ f$ is odd I supposed w.l.o.g. $ x\ge 0$ !!

forgetting for a moment the stipulated derivative taking in the claim, make the unique substitution $ x=\tan w,-\frac{\pi}{2}<w<\frac{\pi}{2}$ and calculate $ f(x)=\sin^{-1}[\sin 2w]+c\cdot\tan^{-1}[\tan w]=\left\{\begin{array}{cc}(2+c)w &\mbox{if }|w|\leq\frac{\pi}{4}\Longleftrightarrow |x|\leq 1\\[0.1cm]\pi+(c-2)w &\mbox{if }w>\frac{\pi}{4}\Longleftrightarrow x>1\\[0.1cm]-\pi+(c-2)w &\mbox{if }w<-\frac{\pi}{4}\Longleftrightarrow x<-1\end{array}\right.$

confirming gauss202


Ascertain the real numbers $x,y,z$ for which $xz\ne 0$ and
$z^2=xy+3xz\ ;\ y^2\ge 4xz\ ; \ 4x^2+15xz\ge 4yz\ .$[hide="Answer"]The set of the fundamental solutions is $S=\{(1,-2,1)\ ;\ (1,4,4)\ ;\ (4,7,-2)\}\ .$[/hide]

We divide all the expressions by  $z^2$  and we set $\frac{x}{z}=a ,\frac{y}{z}=b$
Now we have he following system to solve :

$1=a b +3 a\\ \\ b^2\ge 4a\\ \\ 4a^2 +15a\ge 4b$

But from the first one we take $a=\frac{1}{b+3}$ and now we have to consider two possibilities:

$\bullet \ (b+3)>0$ which after simplifying lead us to the following inequalities

$(b-1){(b+2)}^2\ge 0$

and

$(b-1){(2b+7)}^2\le 0$

The possible values of  $b$  are   ${1,-2,}$.

$\bullet \ (b+3)<0$which after simplifying lead us to the inequalities

$(b-1){(b+2)}^2\le 0$

and

$(b-1){(2b+7)}^2\ge 0$

The only value of  $b$  for this system is  ${-\frac{7}{2}}$.

Let check them now.

If $b=1$ then $y=z$ and the first equation gives $y=0$ (which is impossible because $y=z$ and $z\ne0$ ) or $y=4x$.

If $b=-2$ then $y=-2z$ and the first equation gives $z=x$

If $b={-\frac{7}{2}}$ then $y=-\frac{7}{2} z$ and the first equation gives $z=-\frac{1}{2} x$

Finally we have the following solutions for the system

$(w,4w,4w) , (w,-2w,w) ,(w,\frac{7}{4} w,-\frac{1}{2}w) , w\ne 0$

I edited my post.Too many mistakes....  :blush:
I hope it is ok now  :D

These are infinite solutions!

Virgil, your last two "fundamental solutions" are redundant/repetitive on each other.

Ramanujan only lists the 3 types.

O.K. I corrected. Thanks, Box of Rocks !


Two distinct prime numbers a and b with a>b satisfy

3b^2 - ab - a^2 = 0.

Prove that a-b is the square of an integer.

[quote="socrates"]See that $a^2+ab-3b^2=0 \ or \  a^2+ab=3b^2  \ or \  a(a+b)=3b^2 $
so $a$ divides $3b^2$
Since $a>b$ and a,b are primes we have $(a,b)=1 $ so $a$ does not divide $b^2$
So $a/3$ or $a=\pm1,\pm3$
[/quote]

According to socrates' reply, $a|3$ but it is a prime so $a=3$

But $a>b$ and $b$ is a prime hence $b=2$

$3b^2 - ab - a^2 = $

$3\cdot 2^2 - 3\cdot2 - 3^2 = $

$12 - 6 - 9 = -3$ , not zero!

Now, I'm confused  :(

[quote="socrates"]See that $a^2+ab-3b^2=0 or  a^2+ab=3b^2 or a(a+b)=3b^2 $ so $a$ divides $3b^2$
Since $a>b$ and a,b are primes we have $(a,b)=1 $ so $a$ does not divide $b^2$
So $a/3$ or $a=\pm1,\pm3$

This is my approach. Check it in case i have made some mistake :)  :)[/quote]

Actually , from a(a+b)=3b^2 ,  b must have divide either a or a+b but (a,b)=(a,a+b)=1 hence b can only be 1 which contradict with the question .....


Hello everybody. In the process of trying to typeset my USAMTS solutions this year, I ran into a very picky problem. I cannot find out how to include the fancy "P" and "Q" symbols used in problem 5 (both letters should look somthing like $\Re$). I have run into a site that talks about AMS fonts (and one of these fonts does have the appropriate symbols), but I don't understand how to include/use them. Can anybody help me?

Thanks!

\mathcal{P} becomes $\mathcal{P}$ :D
\mathcal{Q} becomes $\mathcal{Q}$ :D

You can type it drectly onto the solutions.tex template. But if you're working on a document from scratch, you have to include one of the following packages:
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}
(The solutions.tex template already has these packages)

aah... mathcal... sounds cool!

thanks grapes.


topology
symmetry
linear algebra
advanced fields
advanced fields unsolved

Let $ R$ be the Riemann Curvature Tensor of some Riemannian manifold $ (M, g)$.
At a point $ p$, we may view $ R$ as a $ (1,3) \minus{}$tensor, that is, a linear operator:

$ R_p$ : $ (T_pM)^{*} \times (T_pM)^3 \rightarrow \mathbb{R}$.

Now, I've got a reference that says it can be viewed as a linear operator on the space of
two-forms at a point, which I understand to be the exterior product $ T_pM^{*} \wedge T_pM^{*}$,
and that this operator is self-adjoint.

How do I prove it?

By "raising and lower indices," one can regard $ R_p$ as a linear map \[ R_p: T^*_p M \otimes T^*_p M\longrightarrow T^*_p M\otimes T^*_p M.\]  (Or more abtractly, the metric gives an isomorphism between $ T_p M$ and $ T_p^* M$, which allows us to think of $ R_p$ in this way.)  One can show that the Riemann curvature tensor has the following symmetries: $ R_{ijkl}\equal{}\minus{}R_{jikl}$ and $ R_{ijkl}\equal{}R_{klij}$.  From these it follows that $ R_{ijkl}\equal{}\minus{}R_{ijlk}$.  The first and third equations tell us that

\[ R_p: T^*_p M \wedge T^*_p M\longrightarrow T^*_p M\wedge T^*_p M\]

and then the second tells us that $ R_p$ is self-adjoint.

There's a point here that I couldn't follow: how does one uses the symmetry $ R_{ijkl} \equal{} R_{klij}$
to prove self-adjointness?
It seems that the symmetry needed is $ R_{ij}^{kl} \equal{} R_{kl}^{ij}$, and I can't prove it using the
one above...  :|

Thinking about linear algebra on $ \mathbb{R}^n$, recall that a linear transformation $ A: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ defines a bilinear form $ (u,v)\equal{}Au\cdot v$.  Under this identification (which comes from the inner product), a self-adjoint linear transformation corresponds to a symmetric bilinear form.

Back to curvature: The equation $ R_{ijkl}\equal{}R_{klij}$ says that curvature $ R_p$ can be thought of a symmetric bilinear form on $ T_p M\wedge T_p M$ (which, again using the metric, is the same as $ T^*_p M\wedge T^*_p M$).


Is there any way to denote the smallest positive real number? It's obvious it isn't 0, but any larger wouldn't be the smallest. So, is there any notation for this?

There is no sense in denoting something that doesn't exist. You may want to take a look at "hyperreal numbers", nonetheless.

The real numbers are not [url=http://mathworld.wolfram.com/WellOrderedSet.html]well-ordered[/url].


$\ Subiectul\ nr\ 1$
   $\ a)$ $\ Rezolvati\ in\ Z\ ecuatia: \ x^2+\ y^2+\ 2(x+y)$=$\ 3$.
   $\ b)$ $\ Sa\ se\ afle\ x\ natural\ stiind\ ca\ 165_x\ divide\ 561_x\ ,x\ fiind\ baza\ de\ numeratie.$

  $\ Subiectul\ nr\ 2$
    $\ a)$ $\ Aratati\ ca\ numarul\sqrt{n^2+7n+11}\ este\ irational,\ n\ fiind\ natural.$
    $\ b)$ $\ Aflati\ prima\ zecimala\ a\ numarului\ de\ mai\ sus.$

  $\ Subiectul\ nr\ 3$
        $\ Fie\ triungfhiul\ ABC\ cu\ BC=4a,a>0.$ $\ Pe\ paralela\ dusa\ prin\ A\ la\ BC\ se\ considera\ punctele\ D,E\ astfel\ incat\ AD=AE=a,\ BD\ intersectat\ cu\ CE\ este\ O\ in\ interiorul\ triunghiului.$$\ Aratati\ ca:$
    $\ a)$ $\ Punctul\ O\ este\ centrul\ de\ greutate\ al\ triunghiului\ MDE,\ unde\ M\ e\ mijlocul\ lui\ (BC).$
    $\ b)$ $\ Daca\ BD\ intersectat\ cu\ AC\ e\ P\ si\ CE\ intersectat\ cu\ AB\ e\ N,\ atunci\ NP\ paralela\ cu\ BC.$

  $\ Subiectul\ nr\ 4$
        $\ In\ exteriorul\ unui\ patrulater\ convex\ ABCD\ de\ semiperimetru\ p\ se\ construiesc\ triunghiurile\ echila-terale\ ABM,\ BCN,\ CDP,\ DAQ$.$\ Demonstrati\ ca:$
    $\ a)$ $\ ABCD-\ dreptunghi\ daca\ si\ numai\ daca\ MP+NQ=\ p(1+\sqrt 3).$
    $\ b)$ $\ ABCD-\ patrat\ daca\ si\ numai\ daca\ aria(MNPQ)=\frac{\ p^2(2+\sqrt 3)}{4}$.

    $\ Autorii\ problemelor\ sunt: \ Constantin\ Dragomir,\ Mariana\ Saraolu,\ Ion\ Ghica,\ Cristinel\ Mortici$

Dar subiectele de la a6-a si a8-a  le are cineva?

Andrei, daca vrei sa vezi si subiectele de la celelalte clase,, du-te pe Asociatia Scoala cu Ceas, pe care o accesezi de pe http://www.epsilon.ro :)


Its a little difficult to read the measurements, could you tell us what they are?

2,2

4,0

0,2

-2, 0

that was for #2

1) the length from E to F is 10

[quote="now a ranger"]2. Find the radius of the circle which circumscribes polygon ABCD.
1. Trapezoid ABCD is circumscribed aroudn a circle as shown. Points E and F are midpoints of AD and BC respectively. What is the perimeter of ABCD? click on the picture to see it more clear.

thanks[/quote]

For 1 does the circle touch E or F?

The perimeter for problem 1 depends on how big the radius is of the inscribed circle.

I put all the sides of the trapezoid in terms of the radius of the circle which I called $r$.

I won't go through all the steps, but I'll give you the sides of the trapezoid.

$AB = 5+\frac{1}{2}r$
$CD = 15-\frac{1}{2}r$
$BC = 2r$
$AD = \sqrt{5r^{2}-20r+100}$

Also $r$ must be less than $5$


Daca a,b,c sunt laturile unui triunghi atunci a^2,b^2,c^2 sunt laturi de triunghi ?

Doar daca trinughiul cu laturile $a,b,c$ are unghiurile ascutite.


OMG!!  I'm really excited.

So plan is for each of the 6 days before this comes out, I'm going to read each of the six Harry Potter Books (mainly cause I don't remember what happened).

And then I'll start reading the seventh book at midnight.

When does it come out?  I'm a little behind...

[quote="Ignite168"][quote="bubka"]Deathly Hallows, huh? Where did you get the name? And that is probably still a year or so away.[/quote]

easter egg on her website

so me and 3 of my friends are going to a local bookstore at 9pm so we can be first and line and then we're each going to read a different chapter starting from the last one and then spoil it as soon as we get to the two deaths[/quote]
LOL

@hspotter:you can read 600+ pages a day?(do you have that much freetime?) :what?:
or is it gonna be out during the summer?

[quote="Elemennop"][quote="junggi"]
@hspotter:you can read 600+ pages a day?(do you have that much freetime?) :what?:
or is it gonna be out during the summer?[/quote]

I think it's coming out towards winter 2007...and even if he has school, HP books aren't terribly challenging reads; if he spends all afternoon and time in classes reading he'd be able to get through it  pretty easily.[/quote]
eh a whole year away
and wouldn't he have homework to do, and his teachers would get pissed, etc

I thought the date was set for July 31st (imagine that)

there is no date yet and im pretty sure it would come out in summer considering all the past books and trends.

i got through hp5 and hp6 in about 5-6 hours each of nonstop reading so it's def possible (but hp7 is going to be longer)

[quote="Ignite168"][quote="bubka"]Deathly Hallows, huh? Where did you get the name? And that is probably still a year or so away.[/quote]

easter egg on her website

so me and 3 of my friends are going to a local bookstore at 9pm so we can be first and line and then we're each going to read a different chapter starting from the last one and then spoil it as soon as we get to the two deaths[/quote]
Dude, that is so going on bash.org . Best. Idea. Ever.

Check whether it gets on at : http://www.bash.org/?723513

just sacrifice sleep for reading time  :D

[quote="Ignite168"]
so me and 3 of my friends are going to a local bookstore at 9pm so we can be first and line and then we're each going to read a different chapter starting from the last one and then spoil it as soon as we get to the two deaths[/quote]

how do you read a different chapter each if you start from the last one? And there is for once nothing to spoil - the two deaths are obvious.

And I heard somewhere that the triple seven is a baseless rumour.

Anyway I for one will be in no hurry. I will get hold of it from a friend during the first week and read it during the next.

[quote="Fraxia"]Yeah, it's probably going to come out sometime this summer, 7/7/07 is a date I saw passed around a bit.

Can't wait  :lol:[/quote]

Doubt it - the movie comes out less than a month from that date, and it'd be marketing suicide to release the book and movie concurrently; there's alot of lost advertisement and profit if they don't move the dates apart.

I'm shooting for Christmas season 07.

Actually I heard that the movie will be given a months leeway.
But it won't be later than August, I assure you.

[quote="Elemennop"][quote="junggi"][quote="Koda"]And there is for once nothing to spoil - the two deaths are obvious.
[/quote]
oh yea? im clueless[/quote]

harry and you know oo[/quote]
eh

07/07/07...I can hardly wait....
99% Harry will die....because it's a well fuelled rumour that his scar is a horcrux....It'll be a real twist if Harry finds out that to kill Voldy he hass to kill himself...
Hagrid also is a likely candidate...
Oh by the way:
1.Deathly Hallows is an anagram of All H Die Slowly where H could stand for Horcruxes...
2.Godric's Hollow, the village in which Harry's parents used to live..is named after Godric Gryffindor
3. Ron and Hermione WILL hook up.
4. It's most likely that Draco Malfoy turns good.
5.Snape used to love Lily Potter
6.Snape is on Dumbledore's side...that's why he killed him.
If u guys want me 2 I can give u evidence for each of these...but not now. Next time...
----more to come later..keep watchin-----

wow VERY NICE JOB spoiling EVERYTHING for us clueless idiots(eh)

What kheg_k says looks retarded, don't listen to him.
bubka has no idea what he's talking about either.

The book is NOT finished yet so there has NOT been a release date set.
The movie comes out on 7/13/06, it'd be highly unlikely for them to release the final book before the movie for marketting purposes.

[hide="Possible Minor Spoilers"]
-Harry is going to visit Godric's Hollow which is said in HP6.
-R.A.B is most likely Regulas Black, due to the fact that while they were cleaning out the desk in HP5, they found a heavy unopenable locket.
-The last word of HP7 is currently scar.
-There are no Quidditch matches in HP7.

Appearances
-JKR told the filmmakers of HP5 to include Kreacher because he's extremely important meaning that Kreacher will likely have some role.
-JKR said that there was something extremely important about why Dumbledore had James' Invisibility Cloak.
-Something extremely important about Lily is going to be revealed in HP7.
-Harry's eyes resembling his mothers is extremely important.
-We will learn something about Peter Pettigrew, Dumbledore, and his family.
-Something about Petunia will be revealed.
-Umbridge, Rita, and Krum will both reappear.
-The two way mirror and Sirius' flying motorbike will reappear.
-It's very important that Dumbledore had a gleam of triumph in his eye at the end of HP4 when Voldemort said he restored his body using Harry's blood.
-Someone from Harry's grade will become a teacher.[/hide]

Hmm 7/7/07 sounds sort of nice, but I think 7/31/07 is a lot nicer  :wink:

I've been told that the titles are usually released a year and a half before the book.  If this continues, then something like 7/31/08 is more likely than 7/7/07, as cool as the latter would be.

[quote]99% Harry will die....because it's a well fuelled rumour that his scar is a horcrux....It'll be a real twist if Harry finds out that to kill Voldy he hass to kill himself...[/quote]
A big part of me says that Harry can't die, but I also have a sneaking feeling that he will.  The scar (or just Harry himself) being a horcrux is plausible, and there's even some textual evidence to support it.  That does raise the question of who will kill Voldy if Harry has to die first.

[quote]Hagrid also is a likely candidate...[/quote]
Agreed.  As is Neville, IMHO.

[quote]3. Ron and Hermione WILL hook up.[/quote]
They pretty much already have.  I think book 7 can take for granted that they are a couple by the time it starts.

[quote]4. It's most likely that Draco Malfoy turns good.[/quote]
Nah, Draco is irrelevant after failing to kill Dumbledore.  I don't think he'll even have that big of an impact on events in book 7, even though he's likely to remain a major character.

[quote]5.Snape used to love Lily Potter[/quote]
Plausible, and consistent with a lot of other things.  I find this theory quite tempting, especially since it helps me answer the question of Snape's loyalties.

[quote]6.Snape is on Dumbledore's side...that's why he killed him.[/quote]
I disagree.  I think Dumbledore planned on dying, since he's the secret-keeper for a whole bunch of people.  But I think Snape is working for no one but himself.  The reason he appears to be playing both sides is that he really is playing both sides.  Or, if #5 above is true, he could just be out for revenge.

[quote="Ignite168"]The movie comes out on 7/13/07...[/quote]
The [b]fifth[/b] movie come out on 7/13/07.

[quote="Sly Si"][quote]4. It's most likely that Draco Malfoy turns good.[/quote]
Nah, Draco is irrelevant after [b]failing[/b] to kill Dumbledore.  I don't think he'll even have that big of an impact on events in book 7, even though he's likely to remain a major character.[/quote]
He [b]did[/b] kill him.

[quote="i_like_pie"]

[quote="Sly Si"][quote]4. It's most likely that Draco Malfoy turns good.[/quote]
Nah, Draco is irrelevant after [b]failing[/b] to kill Dumbledore.  I don't think he'll even have that big of an impact on events in book 7, even though he's likely to remain a major character.[/quote]
He [b]did[/b] kill him.[/quote]
when?
i dont think someone can die twice....

Yeah, Draco was supposed to, but couldn't do it, which shows a lot of stuff about him that I won't go into.  But basically, he can't cut it as a Death Eater.  And I don't think he'll want to join the good side, which leaves going into hiding or some other form of irrelevance.

why are you all ruling out the possibility of snape being on the good side still

Dumbledore probably told Snape to kill him, for a reason I still haven't figured out. Snape intended not for this to happen, but he had to make the unbreakable vow to show his questionable (and with good reason) loyalty to Voldemort. Obviously he wanted to save his own life.

I can't quote the exact page no. but there's one part where Hagrid overhears a conversation between Dumbledore and Snape..read it and it seems as if Dumbledore is convincing Snape to kill him...
Also...tell me...if Snape wasn't on the good side...then why is it written that before taking the unbreakable vow..his hand quivers??(I think the last pg of the chapter-Spinner's End) Why is he hesitant to kill Dumbledore if he is on Voldemort's side??
About the Snape and Lily thing...
We know that Snape joined Voldemort the day the Potters died right?
This is the theory someone sent me...
"Snape tells voldemort-"I will tell u the last part of the prophecy only when u promise me that u wont kill Lily"
Voldy(I like that name!) says-"Yeah, sure" but then he is forced to kill her...and Snape thinks Voldy betrayed him...that's why dumbledore says that he can't tell Harry why Snape turned good...
Also...Harry can NEVER meet Snape's eyes. It's because when Snape looks at Harry, he sees the same green eyes Lily had... :rotfl:
Thirdly...if Snape hated Lily then he would modify his memory(which harry saw) such that Lily is portrayed as the villain.(Sllughorn says ppl's memories are based on how much they like/dislike the person there..that's why he blurs out Tom Riddle) But in the memory, Lily sticks up for him and he remembers that...which shows that he doesn't hate her."
That theory seems quite plausible to me...but it's just an email...
I have a feeling Harry is a horcrux coz look at what Dumbledore tells Harry regarding Parseltongue in book 2. When harry asks dumbledore why only him and voldemort can speak it, notice dumblesore's reply...
I'll post some more later...

youre reasoning makes no sense at all.

1) his hand never quivered
2) snape was NOT hesitant to kill dumbledore at all
3) snape has met harry's eye many times as to use leglimency on it and i doubt that snape would betry voldy over something as childish as that

uhh no? lol
in the pensieve, Lily was like I'd watch my pants if I were you... I don't think that was a compliment

His hand DID quiver...I'll tell u...[quote]'And should it prove necessary...if it seems Draco will fail....'whispered Narcissa([b]Snape's hand twitched within hers but he could not draw away[/b] 'will you carry out the deed that the Dark Lord had ordered Draco to perform?[/quote]
Snape HAD to look confident of killing Dumbledore...he had to be convincing...if the death eaters knew he was on Dumbledore's side then he would be dead by now...he HAD to put on a good act.
I have absolutely no evidence about the meeting-eyes thing because that isn't my theory.
Moogra- u completely misunderstood...see...WHY did Slughorn modify the memory in which Tom Riddle asks him about Horcruxes? Because he hated him and wanted to forget him right? But Snape remembers Lily in a good way...as sticking up for him and telling James to let him down...she only told him to wash his pants because he insulted her...before that if u notice..she was supporting him. He wouldn't want to remember her being kind to him if he hated her, would he?

[quote="kheg_k"][quote="Ignite168"]What kheg_k says looks retarded, don't listen to him.
[/quote]

Huh? Since when have I said anything in this conversation?[/quote]

Only since now :rotfl:

Harry (or any part of him) is not a horcrux and there's tons of overwhelming evidence pointing to the fact that he can't be a horcrux, plus it's a bad theory, so yea

he's not a horcrux

But I somehow still feel he is going to die... :|

1 <3 Harry Potter!!!

I found out about the title from mugglenet (the most awesome HP site other than JKR's) and went on her site and did the little hangman thing and it was so cool. :)

For speculation about the Deathly Hallows...listen to mugglecast episode 69 or pottercast's deathly hallows book title episode (haven't listened to the second one, but I'm assuming thats what it's about!)

Doesn't mugglenet rock?

I think the release date should be 7/07/07 just cause 7 is a powerful magical number and it looks cool. Look here for more reasons: http://www.mugglenet.com/editorials/editorials/edit-damonk01.shtml *there's a debate on episode 71, which i just finished listening to!

and here for more speculation about the title: http://www.mugglenet.com/editorials/editorials/edit-sephia02.shtml


probability
logarithms
geometry
expected value

http://www.slate.com/id/2215503/

I'll add one way I deal with the St. Petersburg Paradox in a probability class. Assume the payoffs go $ 1,2,4,8,\dots.$ Now, how big is the bank? That is, what is the largest possible payoff the other partner could ever make (or would ever make)? If the bank has size $ N,$ then the expected value of the game is roughly $ \frac12\log_2(N)$ in dollars. The other guy has a billion dollars? than the expected payoff is about $ \$15.$ A trillion? Then it's about $ \$20.$ If the other guy would walk away after $ \$1000,$ then the expected payoff is about $ \$5.$

The correct title to your post could also be "Richard gets misquoted in Slate".  For example, the way of dealing with St. Petersburg that you mention is exactly what I explained to him, or tried to.  I'm not sure where he got that other stuff.  I think I didn't do a good job because I assumed he had more background in the area than he did.  I figured that by telling him that you get into credit risk issues because I don't have infinite money would make the issue clear, but I should have expounded on that in greater detail.  (Also, I'll note that I wouldn't ask this question in a quant interview because it is so well-known.)

[b] @Richard [/b]: Does one stand a chance in getting a job as a "Quant" if one managed to pursue an M.S. in computer science?

Sure.  For a pure quantitative position, a PhD would probably be a better bet, though.

How about a M.S. degree in Actuarial Science from a reputed University, coupled with Associateship from Society of Actuaries? Does it stand a chance in getting a "Quant" job?

[b] @ Richard [/b] : Do you mean to say a PHD in Computer Science (say from SUNY,Buffalo) could land me a quant job?

Since you are experienced, can you tell me does a PHD (in Computer Science) land me in some software/quant job in highly reputed firms?

Thanks!