Sorry if this has already been explained...

I've always sent my solutions by email, and now we can't do that this year. So I was wondering, the web upload thing says "a single PDF file". What if you used "\includegraphics"? Do we not have to send in the .jpg file that we put into our pdf? How about usamts.tex? Do we not have to send that anymore, either?

Again sorry if this has been explained, I must have missed it!

When you build your .pdf file from the .tex and some images, the images are included in the .pdf file, so you do not need to include it in your submission. However, you must keep any images in the same folder as the .tex on your computer.

really? i thought that \includegraphics took the files from your computer... if i just send the pdf file, where will it get the picture file to put in the pdf?

When you build the .pdf file, it takes the images and the tex and compiles them into a .pdf file. Once this file is built, it has all the images you included. But, if you change the image, then you have to rebuild the file for the new image to appear in the .pdf.

Oh, okay. Now that I think about it, it makes sense... its a pdf, so...

Thanks!


What is the greatest product obtainable from two integers whose sum is 1998?


Find the following limit without using L'Hôpitla's rule:

 $\lim_{x \to a} \frac{\sqrt{2a^3x-x^4} -a \sqrt[3]{a^2 x}}{a-\sqrt[4]{ax^3}}$


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

I thought it'd be nice to make a definitive list of people who made USAMO from AOPS. So post your AMC score / AIME / USAMO index, as well as your name and the school/state your from if you want.

Me, I got a 126 on the AMC 10A, 129 on the AMC 12B, and a 9 on the AIME. I'm from Lexington High in Massachusetts.

139.5 on the AMC 12A and 10 on the AIME II. I'm a sophomore at Choate Rosemary Hall in Connecticut.

Congrats to everyone!  :lol:

136 on the AMC 12 B, 10 on the AIME I, and I DON'T GIVE OUT PERSONAL INFO OVER THE INTERNET!!! :)

123 AMC 12A, 124 AMC 12B, 10 AIME II. Junior at Ladue Horton Watkins High School in St. Louis.

139.5 AMC 12B
10 AIME I

12th grade.
Benjamin N. Cardozo HS, New York
Massachusetts Institute of Technology Class of '09.

congrats to all. it seems like many on the list are frequent aops visitors.

150 AMC10A, 144 10B, 12 AIME I => USAMO "index" 270

I go to Exeter.

Finally it's posted. I'm a freshman with a 9 on the AIME. I just made the cut.

I'm from California.

Um 119.5 <--- Bad score. AMC 12A. My school doesn't offer any B's.

9 - AIME I

Freshman at Henderson High in West Chester, PA.

I bombed AIME but I still kind of made it(I think my name's up there by a mistake), but anyway I had 129 AMC12A, 9 AIMEI, lucked out cuz I'm in 9th grade

btw I go to Acton-Boxborough Regional High School of Massachusetts.

AMC12B: 134.5
AIME I: 14
Index: 274.5

Don Mills C.I., Toronto, Canada.

143 (12B), 11 (AIME-I), 253 index.
I'm from Minnesota, and given that, you should be able to guess who I am.

[quote="tekno10m"]Sophomore
133.5 AMC12B
12 AIME I
253.5 index

From Massachusetts.

MA is definitely the smartest state in America - 16 USAMO qualies... boo ya! More qualies than any state except CA, but their population is 6 times as large.[/quote]

IL has 17... :P (I'm the soph from IMSA, btw)

Anyways, congrats to all!!!

Frosh, Choate Rosemary Hall
150 AMC10A 9 AIME II

Yay! Choate has two qualifiers! infinity percent of last year!

whoa, this is my 14th USAMO, and if all goes well, this will be my 12th time at MOP (and I'm obviously not kidding as I'm 15, so I was obviously alive when I started winning)

[quote="h_s_potter2002"][quote="SirErnest"]146.5 on AMC10A.
7 on AIMEII.
Freshman from Indiana.[/quote]

Not to be mean, but are you sure that you made USAMO?  Because you didn't qualify through index, floor score, or being highest in your state.  Maybe it's a typo or something.[/quote]

There's no freshman qualifier from Indiana. Perhaps he made a mistake. No worries though - 7 is an excellent score!

IL
11th Grade
Newman Central Catholic HS
(Sauk Valley Community College)
AMC 12B - 136
AIME II - 11 (I got 14 of the AIME I in 3 hours though)

I'm a little disappointed. If I would've known all of the answers while I was taking the test I would've got them all right. Like duh, guys! I'm happy I made it.

I got a 149.9 on the AMC 12 A and a 14.9 on the AIME. I made USAMO for the 4th time. JK.  ;)

not enough canadians posting... so allow me...

AMC12B 128.5, AIME 11 -> index 238.5
made USAMO for the 2nd time... but it's only the 2nd year canadians are eligible...

By the way, we Canadians are the luckiest when it comes to math competitions... we get to write APMO, CMO and USAMO... plus all the other canadian and american contests including ARML... aren't you jealous lol?

no, i am actually really really glad i live and was born in the United States.

canada is lame

and United States have more ppl, meaning we have better chance of having a better MOP team... no offense ther though, I still love stuff about canada too :lol:

AMC 10B: 130.5
AIME II: 9

I'm a IL freshman, so I just barely made the floor value...

[quote="JP"]are all the canadians taking AMC and stuff?[/quote]

I think it's safe to say that not [b]all[/b] Canadians are taking the AMC :D

AMC 12A: 127.5
AMC 12B: 133.5
AIME I: 11
Index: 243.5
MA sophomore.

[quote="yif man12"][quote="cn2_71828182846"][quote="yif man12"]I was counting Yasha, even though he lives in NC now.  :D

Well, I know two other qualifiers in CT have AoPS accounts, even though they don't post much anymore and since dbkhoa made USAMO, that makes 4 of 6.[/quote]So which CT residents are you not counting?[/quote]

Joshua Horowitz and Charles Nathanson, but by way you're pushing the issue, it sounds like you're one of the two or know one of them very well.  :)

Also, no one else cares about this, but Choate has 2 qualifiers this year while Deerfield's only got 1  :D[/quote]

hi

AMC12A - 146.5
AIME - 10

[quote="yif man12"]Cool, another CT dude.[/quote]

haha, sorry, thought my ID was kinejoshua. I get these confused sometimes.

point being, I'm Joshua Horowitz.

hi.


Let $ ABCD.A_1B_1C_1D_1$ be a cubic whose side is $ a$.Let $ N,M$ be two points on $ AB_1$ and $ BC_1$ respectively such that the angle between $ MN$ and the plane $ (ABCD)$ is $ 60^0$.Prove that ${ MN\ge 2a(\sqrt{3}}-\sqrt{2})$.When does equality occur?

[quote="aadil"]what do you mean by a cubic?[/quote]
with $ AB\equal{}AD\equal{}AA_1\equal{}BC\equal{}....$

[quote="every"]Let $ ABCD.A_1B_1C_1D_1$ be a cubic whose side is $ a$.Let $ N,M$ be two points on $ AB_1$ and $ BC_1$ respectively such that the angle between $ MN$ and the plane $ (ABCD)$ is $ 60^0$.Prove that ${ MN\ge 2a(\sqrt {3}} - \sqrt {2})$.When does equality occur?[/quote]
This problem is in M&Y,a contest on math magazine in Vietnam,please don't help him!
Mod should lock this topic!


What is the smallest value of the expression $\sqrt{x^2 + 1} + \sqrt{(y - x)^2 + 4} + \sqrt{(z - y)^2 + 1} + \sqrt{(10 - z)^2 + 9}$, where $x$, $y$, and $z$ are arbitrary real numbers?

Source: Australian Math Competition, S28, 2001

Nope.. but there's a sneaky way to do this one. It's actually a multiple-choice question and the options are:

$(A) 7$
$(B) 13$
$(C) 4 + \sqrt{109}$
$(D) 3 + \sqrt{2} + \sqrt{90}$
$(E) 149$

Your answer was (C) and I'm interested to know how you got it. :)

I worked from the first square root, min value when x=0, and thats 1. Then i just worked up the line to the end...I'm trying to think about this more now  :lol:

EDIT: Ok so my next stab is $3+\sqrt{2}+\sqrt{9}$, which occurs when $x,y,z=1$  :blush: I know it doesnt exactly use any mathematical principals...

Aidan, can you post solutions or atleast answers to these questions that have gone unanswered for some time?

The answer is $\boxed{13}$?

Can't be 7. Can't be 149. $4+\sqrt{109}$ is wrong and every other possible answer is larger than it 'cept 13...

Whoops, I made a mistake for option (E). It should be $\sqrt{149}$.  :blush: I'll post the solution shortly.

[quote="aidan"]Solution:

[hide]Ok, consider the points $(0, 0), (x, 1), (y, 3), (z, 4), (10, 7)$.

Distance between $(0, 0)$ and $(x, 1)$ = $\sqrt{(x - 0)^2 - (1 - 0)^2} = \sqrt{x^2 + 1}$

Similarly, distance between $(x, 1)$ and $(y, 3)$ = $\sqrt{(y - x)^2 + 4}$,
distance between $(y, 3)$ and $(z, 4)$ = $\sqrt{(z - y)^2 + 1}$,
distance between $(z, 4)$ and $(10, 7)$ = $\sqrt{(10 - z)^2 + 9}$

The sum of all these distances is $\sqrt{x^2 + 1} + \sqrt{(y - x)^2 + 4} + \sqrt{(z - y)^2 + 1} + \sqrt{(10 - z)^2 + 9}$.

To minimize the value of the expression, we want all the points to be collinear. The minimum distance is then $\sqrt{(10 - 0)^2 + (7 - 0)^2} = \sqrt{149}$.[/hide][/quote]
wow! I like this! tnx   ;)
I couldn't find the Source: Australian Math Competition, S28, 2001 , where is it?

Yeah, I don't think AM-GM is applicable in this case. Anyway, the solution I posted is the only solution Australia's mathematicians offer.

Err, you can't find the source anywhere online because Australian MC questions are very copyrighted. The "royalty fees" you have to pay for access to the questions are pretty high.

Visit http://www.amt.edu.au/ to see what the competition is about.

Ok, here's a hint.
[hide]
The given expression represents the sum of the distances from $(x,y)$ to 4 different points.
When is the sum of the distances minimized?
(From here, you can use the same idea as the previous problem.)
[/hide]

[hide="question"]
EDIT: Now that I think of it more, would the circumcenter be the minimal distance, or the intersection of the diagonals, centroid maybe??
[/hide]

frt gave a very good hint.. maybe you should list out all the points and figure it out from there. Not too hard!

I know what are the points...maybe they need to be on the same circle?

[quote="aidan"]Solution:

[hide]Ok, consider the points $(0, 0), (x, 1), (y, 3), (z, 4), (10, 7)$.

Distance between $(0, 0)$ and $(x, 1)$ = $\sqrt{(x - 0)^2 - (1 - 0)^2} = \sqrt{x^2 + 1}$

Similarly, distance between $(x, 1)$ and $(y, 3)$ = $\sqrt{(y - x)^2 + 4}$,
distance between $(y, 3)$ and $(z, 4)$ = $\sqrt{(z - y)^2 + 1}$,
distance between $(z, 4)$ and $(10, 7)$ = $\sqrt{(10 - z)^2 + 9}$

The sum of all these distances is $\sqrt{x^2 + 1} + \sqrt{(y - x)^2 + 4} + \sqrt{(z - y)^2 + 1} + \sqrt{(10 - z)^2 + 9}$.

To minimize the value of the expression, we want all the points to be collinear. The minimum distance is then $\sqrt{(10 - 0)^2 + (7 - 0)^2} = \sqrt{149}$.[/hide][/quote]

That was a nice solution.  cheers to Australian Mathematicians.

[quote="robinhe"][quote="lkryptonicl"]can u use AM-GM for this?[/quote]

I don't see how that would help.

That would only create a bigger mess unless there is a special way to implement it.[/quote]

AM-GM usually works when stuff cancels out, mostly.  There are other applications as well.

Yeah, that's true. If you doubt it, use triangle inequality... :)

[quote="srulikbd"][quote="mathgeniuse^ln(x)"][quote="aidan"]Anyway, I offer another similar question:

What is the smallest possible value the expression $\sqrt{x^2 + y^2} + \sqrt{(x - 1)^2 + y^2} + \sqrt{x^2 + (y - 1)^2} + \sqrt{(x - 3)^2 + (y - 4)^2}$ can be?

Source: Australian Math Competition, S29, 2004[/quote]

[hide="My solution"]Let us consider the coordinate system.

Then, we shall have the points be (0,0), (1,0), (0,1), and (3,4).

So now let us draw a quadrilateral for these points.

The point where it minimizes these distances is the intersection of the diagonals.

So, the intersection is (3/7,4/7).

So we see that the distances from each of these points and summed up will produce 5+$\sqrt{2}$

So the answer is

$5+\sqrt{2}$[/hide][/quote]
is it true? how do I proof that The point where it minimizes these distances is the intersection of the diagonals?  :)[/quote]

Consider the fact that if you chose a point on a segment.  The distance from that point to one endpoint, to that of another endpoint is always constant, and hence it will be the minimum value.  But as you move the point off the segment it becomes larger, so you want it on the diagonal.

Same idea goes with the other diagonal, and hence wherever they intersect should be it.

Oh, yeah, I also forgot that this only works when the say sum of the distances not sum of the squares of the distances.  When it is sum of the squares you use the same method except no square roots and rearrange and complete the square.

I understand now  :)  let's try to find a pattern:
N collinear points, where need to be a point in order the sum of the distances from this point will be minimum?
[hide="solution"]let's number the points-1 2 3...N. if N is even, the new point need to be between the two middle points, or in numbers-between N/2 to N/2+1, on the straight line. if N is uneven, the new point need to be on the middle point-(N+1)/2  [/hide]
what about 3 points that aren't collinear? (I have no idea)


What is the absolute value of the difference of the square of the cube of 2 and the cube of the square of 2?


Let be $ x,y,z\ge 0$ . Prove that

$ \sum_{cyc}\sqrt{x(x\plus{}y)(x\plus{}z)}\le \sqrt{(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\plus{}3xy\plus{}3yz\plus{}3zx)}$

problem is easy, just use ineq Cauchy_Schwar
$ (\sum_{cyc}\sqrt{x(x+y)(x+z)})^2$ $ \leq$ ($ x+y+z$)($ \sum(x+y)(x+z)$)=$ {(x+y+z)(x^{2}+y^{2}+z^{2}+3xy+3yz+3zx)}$


algebra
polynomial
real analysis
real analysis unsolved

Let $ p=\sum\limits_{k=0}^n a_kX^k\in R[X] $ a polynomial such that all his roots  lie in the half plane $ \{z\in C| Re(z)<0 \}. $ Prove that $ a_ka_{k+3}<a_{k+1}a_{k+2}, $ for every k=0,1,2...,n-3.

I think it related to :"Routh-Harwitz" theorem ,applied for determining the stability of polynomials, in "Barbeu"'s book or "ParOslov"'s book , it mentioned as a proposition.


Let $ A, B$ be two $ m\times n$ matrices and $ c,d$ two $ n\times 1$ columns such that the systems $ Ax\equal{}c$ and $ By\equal{}d$ have the same set of solutions. Prove or disprove; $ A$ and $ B$ are row equivalent.([i]E.Khanmohamadi[/i])

I think it is true. The solution space for each is the 'affine' subspace of a particular solution plus the subspace of all solutions to the homogeneous system. We know that two matrices are row equivalent if and only if they have the same solution space. If we call the affine subspace associated with A 'a' and that of B 'b', let us take any element n in B and forming the set S of all n - x, where x is in a, is the solution space of A. Similarly the set of all n - y where y is in b, is the solution space of B. It follows from the equality of a and b that the solution spaces are equal as well.


induction
ratio
limit
real analysis
real analysis unsolved

Let $(a_n)_{n\req 1}$ be sequence of positve numbers such that $a_n> 1, a_{n+1} \geq 2a_n -1, \forall n \in \mathcal l N$
Prove that:
$a) x_n=\frac{1}{a_1}+\frac{1}{a_1a_2}+...+\frac{1}{a_1a_2...a_n}$ converges and let $\lim(x_n)=x$.
$b) \alpha _n$ is monotonic and find it`s limit where $x=x_n+\frac{\alpha_n}{a_1a_2...a_n(a_{n+1}-1)}$

The first point is simple. Just pove by induction that $a_n$ is increasing. Then the series $x_n$ converges by comparision with the geometric serie with ratio $\frac 1 a_1$.

That was my approach to the first point  simply by noticing that $\lim_{n\to\infty}a_n=\infty$ and applynig some series convergence cryterias. What bothers me is calculating the second limit.


$a,b,c$ are reals with $a+b+c=1$. Prove or disprove that

$\dfrac{a}{b^2+b}+\dfrac{b}{c^2+c}+\dfrac{c}{a^2+a}\geq\dfrac{3}{4(ab+bc+ca)}$

It was given to me by my prof a month ago.

true.
let $f(x)=\dfrac{1}{x^2+x}$, simple checking yields f is convex. Now by jensen,
$a f(b) + b f(c) + c f(a) \ge f(ab+bc+ca) = \dfrac{1}{(\sum ab)^2+(\sum ab)}$. Since $\sum ab \le \dfrac{(a+b+c)^2}{3} = \dfrac{1}{3}$, the inequality follows.

[quote="silouan"]How? If I take at the position of $\sum ab$ the $\dfrac{1}{3}$ I get LHS$\geq 6$

Please explain it to me.[/quote]
I dont' understand what you mean.
You have $x \le \dfrac{1}{3}$ then $\dfrac{1}{x^2+x} \ge \dfrac{1}{\dfrac{1}{3}x + x} = \dfrac{3}{4x}$ isn't it?

I was confused by the notatation of $\sum$ sorry

[quote="Arne"]Why don't you just ask him whether he created it or not?[/quote]

he created but I don't know if someone before him craete the same one.

[quote="silouan"]Does anyone have seen this ineq on a book or at the internet?
If no then it is my prof construction :lol:[/quote]

Who is your prof?
Nice ineq and creation!
 :)

My prof is mr . Christos Patilas. He  wrote helping  books for the B' and C' of the lyceum :)

Guys, please try to discuss this kind of things using PM's...


How many 3-letter words can be formed from the standard 26-letter alphabet, if the first letter must be a vowel (A, E, I, O, or U)?

[quote="Blanc"]The question is a little ambiguous in that not all of the combinations are actually words.[/quote]

Assuming we are simply talking about 3 part blocks of letters, there are $ 5\cdot26\cdot26\equal{}\fbox{3380}$ combinations.

[quote="Blanc"]The question is a little ambiguous in that not all of the combinations are actually words.[/quote]

[quote="Maxima"]Assuming we are simply talking about 3 part blocks of letters...[/quote]

In problems like these, we assume "word" to mean nothing more than "a collection of letters". Fortunately for us, we're playing Alcumus and not Scrabble. ;)

[hide="Hidden Solution"]This is a very straightforward multiplication problem, as Maxima showed:

Since the first letter must be a vowel, we have only $5$ options for the first letter. There are no restrictions on the second and third letters, so we have $26$ possibilities for each. Therefore, the total number of possible words is $5 \times 26 \times 26 = \boxed{3380}$.[/hide]

Tim

[quote="gershmel"]this problem seams like it should be 9, because you would subtract 3 from 30 for the kids that have neither, giving you 27. then you have to find a number whos sum of the number plus half of it equals 27. which is 18 and 9. so theres 9 blue eyed kids. the 6 that are both shouldnt have any part in the solution. idk how to word what im saying, it just seems rong[/quote]
What does this have to do with this post?

The question now reads:

How many 3-letter "words" can be formed from the standard 26-letter alphabet, if the first letter must be a vowel (A, E, I, O, or U)? (We aren't talking about English words per se; something like EQX is perfectly valid here.)


this time it's about Econ 101. who has taken econ 101 with trimarchi? i heard her exams are pretty easy and they are quite similar to the past exams. give me some information about this course with trimarchi.


A shipment of 8 DVD players is found to have two which are broken.  The broken ones are not labeled however and the players need to be checked again until the broken two are found.  Assuming the players are checked one at a time, what is the probability that the first broken one is found sometime in checking the first five players, and the second broken one is found on the sixth player checked?  Express your answer as a common fraction.

Um...

[hide]The probability of broken, normal, norm., norm., norm. is (2/8)*(6/7)*(5/6)*(4/5)*(3/4). There are 5 places where you can find the broken, so multiply that by 5. Finally there'll be a one third chance that the sixth one is broken. Multiply it out to get [b]5/28[/b].[/hide]

By the way, its equipment, no extra t in the middle.

[hide]
There are 2 different broken DVD players.  Call them A and B.
The probability that A is chosen in the first 5 and B is not is

$\frac 18 \cdot 67 \cdot 56 \cdot 45 \cdot 34=\frac{3}{56}$
Then multiply by 5 for order to get $\frac{15}{56}$.

The probability that B is chosen next is $\frac 13$.

So the probability is $\frac 13\cdot\frac{3}{56}=\frac{1}{56}$

Then you have to multiply by 2 to account for B chosen in the first 5 and A as number 6.

So the answer is $\frac{1}{28}$[/hide]

You forgot to multiply the first choice by 2, and you multiplied 1/3 with the wrong one.
So then after that you should get 5/28.

Biff,

You multiplied by 5 to get 15/56, but later you reverted back to 3/56.  I think that's your only error.

frost


1) is correct.
This is because you have to do the exponent operation before the negation operation (by consulting BODMAS if you use that system!).

$(-4)^{2}= 16$
$-4^{2}=-16$
We always do exponents before the negation sign. Think of it this way:

$-1 \cdot 4^{2}= 16$

I think buzzer11 actually means: $-1\cdot 4^{2}=-16$
!!!

[quote="sieuhaha"]Please tell me which answer is correct 1 or 2
1) - 4² =  -16

2) - 4² =  16

                 Thanks[/quote]

[hide="answer, I think"]The correct one is number 1[/hide]


We have a standard deck of 52 cards, with 4 cards in each of 13 ranks.  We call a 5-card poker hand a full house if the hand has 3 cards of one rank and 2 cards of another rank (such as 33355 or AAAKK).  What is the probability that five cards chosen at random form a full house?

Erm...

The total number of possible outcomes is $ \binom{52}{5}\equal{}2598960$.

To make a full house, we need a rank for the $ 3$ cards.  We can pick this in $ 13$ ways.  We need $ 3$ of $ 4$ cards from that rank.  This can be done in $ 4$ ways.  We need a rank for the other two cards.  This can be done in $ 12$ ways.  We need $ 2$ of $ 4$ cards from that rank, which can be picked in $ 6$ ways.

Total number of full houses is $ 13 \cdot 4 \cdot 12 \cdot 6\equal{}3744$, so probability is $ \frac{3744}{2598960}\equal{}\boxed{\frac{6}{4165}}$.

I'm getting both ways. Like I think you could do it both ways are technically right, but they get different answers........ weird. They both make sense.

Eh, how does the first way make sense to you?

First way: [quote="akhil0422"]\[ \frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}\cdot\frac{4}{49}\cdot\frac{3}{48} \implies \frac{288}{311875200} \implies \boxed{\frac{1}{1082900}}\][/quote]

There are not just $1$ rank to choose from for the 3 cards and not just $1$ rank to choose from for the pair.... So you're off by a factor of $13\cdot 12$ there. And then another $\binom{5}{2}=10$ to arrange everything. And this gives $\boxed{\frac{6}{4165}}$ if I'm not mistaken.


calculus
integration
geometry
3D geometry
sphere
special factorizations
calculus computations

I am currently working on these practice problems my professor assigned

1.

Set up a double integral for the volume of a solid inside both the sphere $ x^{2}+y^{2}+z^{2}=3$ and the paraboloid $ 2z=x^{2}+y^{2}$. Then convert the integral to polar form and evaluate.

What I did was this:

$ \int_{-3}^{3}\int_{-\sqrt{9-x^{2}}}^{\sqrt{9-x^{2}}}\sqrt{3-x^{2}-y^{2}}-\frac{x^{2}+y^{2}}{2}dy dx$

which I converted to polar as being:

$ \int_{0}^{2\pi}\int_{0}^{3}(\sqrt{3-r^{2}}-\frac{r^{2}}{2})dy dx$

Although I am not sure if my limits of integration are correct on the polar one. If so, then it shouldn't be difficult to evaluate...

2.

Find the surface are of the portion of the sphere $ x^{2}+y^{2}+z^{2}=9z$ that lies inside the paraboloid $ x^{2}+y^{2}=4z$

The reason I am having trouble with this one is because I am curious about isolating the z in the equation with the sphere to use the formula

$ \int \int_{D}\sqrt{[f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}+1}\dA$

I was hoping someone could shed some light on how to start the problem

3.

Given $ \int_{0}^{2}\int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int_{0}^{\sqrt{4-x^{2}-y^{2}}}f(x,y,z) dzdydx$
Change the order of integration to $ dydxdz$.

If I interpret the triple integral correctly, it shapes out a quarter-sphere (only that which lies on the positive x quadrants and above the z=0 plane). I know it is not a volume integral but to change the order of integration, it helps me to map out my boundaries
I tried to map this and the answer I got was

$ \int_{0}^{2}\int_{-\sqrt{4-z^{2}}}^{\sqrt{4-z^{2}}}\int_{-\sqrt{4-x^{2}-z^{2}}}^{\sqrt{4-x^{2}-x^{2}}}f(x,y,z) dydxdz$

I hope that is correct, but I am not a great visual thinker so I often have trouble with my 3D interpretations...

Thanks for any help. It is really appreciated

For #1, the circle $ x^{2}+y^{2}=3$ has radius $ \sqrt{3}$, not 3.  But be careful where you're integrating over.  If you picture a paraboloid with its vertex at the center of a sphere, they will intersect at a circle above the $ xy$-plane so with radius less than $ \sqrt{3}$.  To find the radius, solve the system $ x^{2}+y^{2}+z^{2}=3$ and $ x^{2}+y^{2}= 2z$.

The only problem with your conversion is that you still have $ dydx$.

I'm still looking at #2, but is this what you were referring to?
$ x^{2}+y^{2}+z^{2}-9z = 0$
$ x^{2}+y^{2}+z^{2}-9z+(\frac{9}{2})^{2}= (\frac{9}{2})^{2}$
$ x^{2}+y^{2}+(z-\frac{9}{2})^{2}= \frac{81}{4}$

#3 your interpretation of the given limits seems correct.  I believe you made a mistake in the $ dx$ limits when you changed order.  Remember, $ x$ was always positive in the original one.  (and a typo in the upper $ dy$ limit  :))

For number one, r should then go from $ 0$ to $ \sqrt{2}$
The $ dydx$ should be the Jacobian transformation $ r dr d\theta$
Solving the system, I get that $ x^{2}+y^{2}= 2$, so it intersects above the xy plane in a circle centered at the origin of radius $ \sqrt{2}$

Yes, for number two I'm not sure how to solve it because I thought I had to isolate the $ z$ before solving for my first order partials $ f_{x}$ and $ f_{y}$ to plug into the surface area equation I posted.
Is that what I have to do?

For number 3, aside from the silly typos, then my integration limit for $ dx$ should start from $ 0$.

Thanks

I believe so.  I hope I'm not overcomplicating things, but here is how I did #2, hidden by steps so you can check as you go.

[hide]  After completing the square as shown above, begin with a sketch.  The sphere is resting on the $ xy$-plane and the paraboloid comes up and intersects at a circle.  To find where they intersect, we solve the system $ x^{2}+y^{2}+z^{2}=9z$ and $ x^{2}+y^{2}=4z$.

[hide]  Subsitituting $ 4z$ for $ x^{2}+y^{2}$ in the first equation yields $ 4z+z^{2}= 9z$, so $ z^{2}-5z = 0$.  Then $ z=0$ or $ z=5$.  $ 0$ is trivial (the origin) so we're interested in $ z=5$.  [Since $ 5 > \frac{9}{2}$ we can picture that the region in question is a little less than the half sphere.  I use this to check and make sure my answer is less than $ \frac{1}{2}(4\pi(\frac{9}{2})^{2}$).]  Putting $ z=5$ back into either equation yields $ x^{2}+y^{2}= 20$

[hide]  Solving for $ z$ in the sphere equation yields $ z=\sqrt{\frac{81}{4}-x^{2}-y^{2}}+\frac{9}{2}$ so we can now calculate the partial derivatives.  $ z_{x}= \frac{-x}{\sqrt{\frac{81}{4}-x^{2}-y^{2}}}$ and similarly $ z_{y}= \frac{-y}{\sqrt{\frac{81}{4}-x^{2}-y^{2}}}$.

  Applying the surface area formula, we have $ \int_{D}\int \sqrt{\frac{x^{2}}{\frac{81}{4}-x^{2}-y^{2}}+\frac{y^{2}}{\frac{81}{4}-x^{2}-y^{2}}+1}\ dA$.  Writing $ 1$ as $ \frac{\frac{81}{4}-x^{2}-y^{2}}{\frac{81}{4}-x^{2}-y^{2}}$, the integral simplifies to $ \int_{D}\int \sqrt{\frac{\frac{81}{4}}{\frac{81}{4}-x^{2}-y^{2}}}\ dA = \frac{9}{2}\int_{D}\int \sqrt{\frac{1}{\frac{81}{4}-x^{2}-y^{2}}}\ dA$.

  If you want to think rectangularly first and then convert thats fine, but I always jump straight to polar/spherical/cylindrical when the application is obvious.  Since the region $ D$ we are integrating over is $ x^{2}+y^{2}= 20$ (see above), the integral becomes $ \frac{9}{2}\int_{0}^{2\pi}\int_{0}^{\sqrt{20}}\sqrt{\frac{1}{\frac{81}{4}-r^{2}}}rdrd\theta$.

  Now we solve with a subsitution of $ u = \frac{81}{4}-r^{2}$ so $ \frac{-1}{2}du = rdr$.

$ \frac{9}{2}\int_{0}^{2\pi}\int_{0}^{\sqrt{20}}\sqrt{\frac{1}{\frac{81}{4}-r^{2}}}rdrd\theta$ = $ \frac{9}{2}(2\pi)\int_{\frac{81}{4}}^{\frac{1}{4}}\sqrt{\frac{1}{u}}(\frac{-1}{2})du = \frac{9\pi}{2}\int_{\frac{1}{4}}^{\frac{81}{4}}u^{-\frac{1}{2}}du = \frac{9\pi}{2}[2u^{\frac{1}{2}}]_{\frac{1}{4}}^{\frac{81}{4}}= 9\pi(\frac{9}{2}-\frac{1}{2}) = 36\pi$.

Checking, $ 36\pi < \frac{1}{2}(4\pi(\frac{9}{2})^{2}) = \frac{81}{2}\pi$, so I think we're good.

[/hide][/hide][/hide]

Thanks so much for the help. Once I realized that I could just put the equation in terms of z like that, the rest was tedious calculations. I really appreciate the help. Thanks for taking the time.


We have the graph $y=x^{z}$.

For each z, define $S_{z}=\sum_{x=1}^{n}x^{z}$.

Compute $S_{z}$ for all integers z.

$S_{3}= (\frac{n(n+1)}{2})^{2}$

But I think he's looking for a generalization.

The more natural generalization is what gets referred to sometimes as the "hockey stick formula": if we denote
\[\binom{x}{k}=\frac{x(x-1)(x-2)\cdots (x-k+1)}{k!}\]
(where $k$ is nonnegative integer), then for any positive integers $n$ and $m$,
\[\sum_{k=0}^{n}\binom{k}{m}=\binom{n+1}{m+1}\]
(The index of 0 in the summation can be replaced by any integer less than or equal to $m$.)
From this, finding a sum of powers results from writing $x^{m}$ in terms of $\binom{x}{j}$'s for $0\leq j\leq m$. This can be used to give a formula in terms of recursive sequences, but it's not much of a closed form.

OK. Bad question.

New one.

Dr. Dervative derived a new notation:

$S_{z}^{n}=\sum_{x=1}^{n}x^{z}$

For all naturals n and z.

Prove that:

$S^{n-1}_{z}<\frac{n^{z+1}}{z+1}<S^{n}_{z}\forall z<5$


Find the values of $\alpha$,0<$\alpha$<$\pi$ and $\alpha$ different $\frac{\pi}{2}$,such that
$f(x)=4x^2-4x-tg^2\alpha$, f:R->R,has minimum value -4.


function
algebra
polynomial
number theory solved
number theory

Let N=p_1*p_2*..*p_n , p_i are prime numbers (3<p_1<p_2<..<p_n)
      prove that 2^N + 1 has least 4^n divisors

This one is Imo Short List 2002, pb. N-3.

Pierre.

This is a simple problem,
I think we can replaced 2 by p-1 where p is a prime and  2,p,p1,p2,...,pn are distince primes  as you see in the following proof!

Here goes,
  From the fact that
             gcd(2^p+1,2^q +1)=3 for all odd primes p<>q
   and     gcd(2^p+1,9)=3 for all prime p>3
and the multiplitive of the divisor d(n) fuction we just have to show that
             d(2^p+1) \geq  4  for all p is odd prime >  3
If not then d(2^p+1) <4 so from the decomposite 2^p+1 =p1^e1 *...*pr^er  where e1,...,er are positive integer
we got (1+e1)...(1+er)<4
so r<2 and then r=1 and e1<3.But 2^p +1 is not prime and is divisible by 3 so we have 2^p+1 =3^e with e>1.Since e1<3 so we got e=2.But p>3 so this contracts!!!!!!!!
Now join all the above remarks we have
         d(2^N+1) >= d(2^p1+1)....d(2^pn +1)>= 4...4= 4^n
This completes the problem!

I think there also was a solution using cyclotomic polynomials, does anyone knows about it?

ie: considering 2^N+1 as a product of cyclotomic polynomials evaluated at 2.

In fact you can manage to prove that for each polynomial P(x) there exist a number m, sucht that $2^{p_1p_2p_3 \cdots p_n}+1$ have at least $P(n)$ distinct prime divisors...


Do you mean $a_{2}a_{3}$? And that part is not included in the $\sum$, right?

[quote="Phelpedo"]If I square something like $(a_{1}+a_{2}...+a_{n})$, I get
\[\sum_{cyc}a_{1}+2(a_{1}a_{2}+2a{}_{2}{}_{3}...) \]
is there any name for the kind of "symmetric" sum that appears in the second part? I know it's not cyclic sum or symmetric sum.[/quote]

Isn't it a cyclic sum though since you are just cycling through consecutive numbers including n and 1.

Plus I believe you mean a_{1} ^2, right.

But when you square $(a_{1}+...+a_{n})^{2}$, the nonsquare terms go through every possible $a_{i}a_{j}, i \neq j$.

(This is written as

$\sum_{i, j \le n, i \neq j}a_{i}a_{j}$

But I am not sure if it has a name.)

[quote="t0rajir0u"][quote="nutz_for2.718281828"]uhh.. the sum of the cubes of the first n natural numbers is the square of the sum of the natural numbers... maybe that can help.[/quote]

Totally beside the point.  The poster wants a [b]name[/b] for the sum

$\sum_{i \neq j}a_{i}a_{j}$[/quote]
nutz for e enjoys posting repeated solutions to problems that have been solved many times and providing useless info.


when we have a mixture of gaseous substances then what do we mean when we say volume percentage

The volume percentage of a gas X in a gaseous mixture sample of total volume $ V_{T}$ is defined as

$ percentage(V)_{X}\equal{}\frac{V_{X}}{V_{T}}\times 100$.

where $ V_{X}$ is calculated from the number of moles present in the sample (which can be found by suitable analytic techniques) of volume $ V_{T}$, assuming ideal gas behaviour.

Here's a relevant exercise for the application of that concept:

A 5.0 L air sample containing H2S at STP is treated with a catalyst to promote the reaction,

$ H_{2}S(g)\plus{}O_{2}(g)\longrightarrow H_{2}O(g)\plus{}S(s)$.

If 3.2 g of solid S were collected, calculate the volume percentage of H2S in the original sample.

well what actually i have problem with is when we say volume percentage of $ X$ in $ V$ is $ V_{x}$ do we mean that in a given room of volume $ V$ the ellement $ X$ will ocuupy just $ V_{x}$

Of course not: if X is a gas, then it will occupy all the volume available to it - that's a basic property of gases. The volume percentage of a gas is just a concentration unit, its just a mean of reporting the concentration of a gas in a given sample, like molar fraction or partial pressure.

thanks wonderful.that was what confusing me  :D

is there any meaning to thsi problem
a mixture of $ O_{3}$  and $ O_{2}$ occupies 600ml at NTP if the mixture weighs $ 1 gm$ what is the volume occupied by $ O_{3}$
[hide="ans to ur question carcul"]if i understood this right
now the mass of $ S$ collected implies there is $ \frac{1}{10}$ mole of it so $ \frac{1}{10}$ mole of $ H_{2}S$ which occupy $ 2.24 l$ under STP so the volume percentage is $ \frac{2.24}{5}100$ $ \equal{}44.8$ percent

if i got u right the $ V_{x}$ is that volume of the fixed mass of gas that exerts the sam epressure as the given mixture under identical conditions of temp.[/hide]

The answer to my question and your reasoning are correct. Regarding your problem, it makes sense: you have two equations on two unknowns, so it can be solved.

but obviously doesn't $ O_{3}$ also occupy $ 600 ml$

[quote="Carcul"]Here is the answer:

[quote="Pardesi"]if i got u right the $ V_{x}$ is that volume of the fixed mass of gas that exerts the sam epressure as the given mixture under identical conditions of temp.[/quote][/quote]

 :D  :rotfl:


If you have a word of length N, repeated letters allowed, how many words of arbitrary length can be formed? (words don't actually have to be real words) I know there's a easy solution to the case with only N-letter words, but i'm not sure if there's a formula for total number of words.

I have some doubts there is the closed formulae.

i couldn't find one either - i spent quite a while working on the answer  :(
I suppose the answer would just be a really ugly sum of different cases. What i'm hoping though, is that all the cases somehow add up and simplify.


analytic geometry
algebra
polynomial
algebra unsolved

A point in the plane with a cartesian coordinate system is called a [i]mixed  point[/i] if one of its coordinates is rational and the other one is irrational. Find all polynomials with real coefficients such that their graphs do not contain any mixed point.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78454[/url] says even a bit more (it's not hard to get out of this that only the linear polynomials with rational coefficients work).

$ax+b$, with $a,b\in \mathbb{Q}$, $a\neq 0$.

This is a beautiful problem.

First, I claim that all such polynomials have rational coefficients. This isn't too bad: Say that the polynomial was $f$, and $\deg f = j$. Now apply Lagrange Interpolation on any $j+1$ rational points, and we're good for this part.

Now say that $\deg f \ge 2$. I claim that $f$ passes through a mixed point. First of all, multiply all the coefficients of $f$ by a suitable integer so $f$ becomes an integer polynomial. Subtract $f(0)$ from $f$, so $f(0) = 0$. Now take $-f$ if $f$ had a negative leading coefficient. For what follows, consider the domain $x > A$, where $A$ is the largest root of $f'(x)$. It is clear that $f$ is monotonic and hence bijective in this domain. Now let $a$ be the leading coefficient of $f$. Assume that $f$ passes through no mixed points. Then, it must be true that for all intergal $c$ larger than $f(A)$, there exists a unique $x > A$ such that $f(x) = c$. Moreover, by the hypothesis, this $x$ must be rational. In fact, by the rational root theorem, $x$ is a integer multiple of the fraction $\dfrac{1}{a}$! Let the sequence $k_c$ for all integers $c > A$ be this exact $x$. Since $f(x)$ is strictly increasing over this domain, so must $k_c$. Observe that \[ k_{c+1} - k_c \ge \frac{1}{a} \] since they are obviously different, and $k_{c+1} > k_c$. Also observe that \[ (c+1) - c = 1.\] By the mean value theorem, there exists a $u$ in the domain $[k_c, k_{c+1}]$ such that \[ f'(u) = \frac{f(k_{c+1}) - f(k_c)}{k_{c+1}-k_c} = \frac{1}{k_{c+1}-k_c} \le a\]

However, since the degree of $f$ is larger than $1$, and $k_c$ tends to infinity, $f'$ is also monotonically increasing past a certain point. So we just need to take a sufficiently large $c$ to get contradiction in the above equation. So the only polynomials which have a chance of working are $f(x) = ax + b$, where $a, b \in \mathbb{Q}$.  But it is trivial that these polynomials work, so we're done.

Quite a nice problem.

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots +a_{1}x+a_{0}$ be a polynomial which goes through no [i]mixed points[/i]. The key claim is the following

[b][color=#f00]Lemma.[/color][/b] $P$ must only have rational coefficients.

[i]Proof:[/i] We'll only use that $P(\mathbb{N})\in\mathbb{Q}$ as otherwise $P$ would go through a [i]mixed point[/i]. Recall that if $g\in\mathbb{R}[x]$, $\Delta_t g(n):=g(n+t)-g(n)$ and $D_k(g)=\Delta_{2^{k-1}}(D_{k-1}(f))$ where $D_0(g)=g$, then $\deg(\Delta_{k+1})+1=\deg(\Delta_{k})$ for $0\leq k\leq \deg g$ and $D_{\deg g+1}(g)$ is the zero polynomial. Hence, since $P(n)\in\mathbb{Q}$ for all $n\in\mathbb{N}$ we have that:
\[\mathbb{Q}\ni D_{n}(P)=D_{n}\left(\sum\limits_{i=0}^{n} a_{i}x^{i}\right)=\sum\limits_{i=0}^{n} D_{n}\left(a_{i}x^{i}\right)=D_{n}(a_{n}x^{n})=c\cdot a_{n}\]
where $c$ is some nonzero constant as $D_{n}(a_{n}x^{n})$ is a constant polynomial, but not the constant zero. Also, we know that:
\[c=\frac{1}{a_{n}}D_{n}(P)=\frac{1}{a_{n}}\sum\limits_{k=1}^{2^{n}} \varepsilon_{k} P(x+k)=\sum\limits_{k=1}^{2^{n}} \varepsilon_{k} (x+k)^{n}\Longrightarrow c = \sum\limits_{k=1}^{2^{n}} \varepsilon_{k} k^{n} \in \mathbb{Z}\]
where $\varepsilon_{k}\in\{+1,-1\}$ and depends on $D_{n}(P)$. Therefore $c\cdot a_{n}\in \mathbb{Q}$ and $c\in \mathbb{Z}\Longrightarrow a_{n}\in\mathbb{Q}$. Now, we must have $P_{1}(\mathbb{Q})\in \mathbb{Q}$ where $P_{1}(x)=P(x)-a_{n}x^{n}$ and applying the same procedure implies that $a_{n-1}\in\mathbb{Q}$, $a_{n-2}\in\mathbb{Q}$ and so on, we get inductively that every coefficient must be rational.

If $P$ is a constant polynomial, then all inputs of $x$ spit out a rational, hence there are mixed points $P$ goes through and if $P$ is linear, then it doesn't go through any [i]mixed points[/i] as $P(x)=a_{1}x+a_{0}$ for $a_{1}, a_{0}\in\mathbb{Q}$ and $a_{1}\neq 0$ as $P(x)\in\mathbb{Q}\iff x\in\mathbb{Q}$. We'll show that if $\deg P\geq 2$, then $P$ can't give irrational outputs for all irrational inputs. WLOG we can assume that $P$ has integer coefficients because we can multiply the polynomial by some $C\in\mathbb{N}$ such that $Ca_{i}\in\mathbb{Z}$ $\forall i$ and this wouldn't change if $P$ goes through any [i]mixed points[/i].

WLOG let's also assume that $a_{n}>0$ (if that's not the case just negate the polynomial). Now we must have that $P(x)-n=0$ only has rational roots for all integers $n$. Let $n=p+a_{0}$ for any prime $p>\max\{a_{n}, |P(0)|+|a_{0}|\}$. Since $P(0)-p-a_{0}<0$ and $P(x)-p-a_{0}>0$ for a large enough $x$ (as the leading coefficient of $P$ is positive), $P(x)-p-a_{0}=0$ has a real solution due to IVT and this root must be rational to avoid contradiction with the assumption that $P$ doesn't go through [i]mixed points[/i]. Now, by the Rational Root Theorem, it follows that if this root is $\frac{b}{c}$ where $b,c\in\mathbb{Z}$ and $\gcd(b,c)=1$, then $b\mid p$ and $c\mid a_{n}$ since $-p$ and $a_{n}$ are respectively the constant term and leading coefficient of $P(x)-p-a_{0}=0$. Note that the two equations $P(x)-p_{1}-a_{0}=0$ and $P(x)-p_{2}-a_{0}=0$ clearly can't share a root as that would imply that $p_{1}=p_{2}$. Now $\frac{b}{c}$ is either of the form $\frac{1}{d}$ where $d\in\mathbb{Z}$ and $d\mid a_{n}$ or $\frac{p}{d}$ where $d\in\mathbb{Z}$ and $d\mid a_{n}$. Since there are finitely many rational numbers of the first form we must have that for infinitely many primes $p$, $\frac{b}{c}=\frac{p}{d}$. Now, as there are finitely many choices for $d$, it follows that for infinitely many primes $p$ we have that $\frac{b}{c}=\frac{p}{D}$ for a fixed integer $D$, which divides $a_{n}$. Hence $P\left(\frac{p}{D}\right)-p-a_{0}=0$ for infinitely many primes $p$. This implies that the polynomial $P(X)-DX-a_{0}=0$ has infinitely many roots and is therefore the zero polynomial, but $\deg(P)\geq 2$, contradiction.

[hide=Remark]The proof of the lemma is nicer (and quicker) in #4 above, however, I think my proof is instructive in a way as the same idea also appears in [url=https://artofproblemsolving.com/community/c6h2840836p25162375]BMO Shortlist 2021 A4[/url] as well as a couple of other problems I've seen.[/hide]


I have heard some people using the words "I claim" or some form of that with the word "claim" in it in proofs. What does it mean to "claim" something? Or, am I just making this up? Sorry if this sounds like a stupid question; I am sortof curious.

Well I suppose that when you write "I claim..." then you are just stating something and then you go on to prove it. For example, you may be doing some problem and you end up with some solutions. Then you could say "I claim these are all the solutions" and your next few lines would be proving that. Nothing that special though..

"I claim" just means "The next thing I'm about to say is a theorem for which I will present a proof."

A brief example:

Given that a:^2: + b:^2: = 0, find all real solutions (a, b).

I claim that (0, 0) is a unique solution.
Proof:  if a is not 0, a:^2: > 0.  b:^2: :ge: 0 for all real b, so a:^2: + b:^2: > 0.  Thus a = 0.  By identical reasoning, we must have b = 0.  Thus, our claim, that (0, 0) is the only solution, is proven.  Q.E.D.


Let  $ a_0\equal{}a_1\equal{}1$ and  $ a_{n\plus{}2}\equal{}a_{n\plus{}1}\plus{}(n\plus{}1)a_n$, $ n\equal{}0,1,2,\ldots$. Prove that for each odd number  $ k\in {\mathbb N}$, the number $ k$ divides $ a_k\minus{}1$


[quote="perfect_radio"]Find the sum of all numbers of three digits written in base $10$, which have both even and odd digits.

[i]Petre Bătrăneţu[/i][/quote]
[hide] First find the sum of all 3-digit numbers. Then subtract the numbers with only even or only odd digits. The sum of all 3-digit numbers: $\frac{999*1000}{2}-\frac{99*100}{2}=494550$.
If we were to list out all 3-digit numbers with only odd or even digits, there would be 25 1's, 25 2's, 25 3's, etc. in the hundreds place, since there are 25 combinations for the other 2 digits. That sum is $25*100*\frac{9*10}{2}=112500$. There are also 25 0's, 25 1's, 25 2's, etc. in the tens and hundred places. Therefore, the sum of all 3-digit numbers with only odd or even digits is $112500+11250+1125=124875$.
$494550-124875=\boxed{369675}$. [/hide]


1. What is Trost's Reagent?

2. What are its synthetic values (i.e. use in synthesis).

I saw the term when reading about the Total Synthesis of Periplanone B.

1. Trost's reagent is cyclopropyldiphenylsulfonium ylide.

2. Interesting question, since I happen to have one ACR article where Barry Martin Trost describes the applications of that reagent. In summary: introduction of cyclopropyl group, synthesis of spiropentanes and cyclobutanones are some of its applications.

could you draw its structure i cant infer from its name???and whats its use???


can somone help??

Question:
I need to get the number of sequences of length 20 consisting of 8 X chars, 7 Y chars and 5 Z chars in which no two Z chars appear adjacently, and X,Y do not appear adjacently.

Please Please Be specific
thanks
Alisha

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193739
Why have you double posted?  :huh:
And also i rem seeing JBL post a reply to this though i cant find the link  :mad:


function
calculus
integration
limit
real analysis
real analysis unsolved

[quote="xxxxtt"]Prove that every convex function on a closed,finite interval is integrable.[/quote]

if $ f$ is convex on $ [a,b]$ then is continuous on $ (a,b)$ so is Riemann integrable on every $ [c,d]\subset(a,b)$. The improper integral however might be not convergent  e.g $ f[0,1]\to \mathbb{R},\ f(x)=\frac{1}{x}, x\neq 0, and\ f(0)=0$ then $ \int_{0}^{1}f(x)dx=\infty$

cckek, you're wrong. That function ain't convex.

Let $ f : [a,b] \to \mathbb R$ be a convex function. A [b]well-known[/b] result states that $ f$ is left- and right-differentiable in every $ x \in (a,b)$, hence $ f$ is continuous on $ (a,b)$.
If $ f$ is decreasing on $ (a,b)$, then $ f$ has a limit in $ b$. Otherwise, there are $ a < u < v < b$ such that $ f(u) < f(v)$. For every $ b > p > q > v$ we have that $ \frac{f(p)-f(q)}{p-q}\geq \frac{f(u)-f(v)}{u-v}> 0$, so $ f$ is increasing on $ (v,b)$, so it has a limit in $ b$.
In either case, $ \lim_{x \to b^{-}}f(x)$ exists. If it were infinite, then the contradiction with the convexity of $ f$ on $ [a,b]$ would be quite blatant (if, for example, it's $ \infty$, then $ f$ would be increasing on some $ [v,b]$, which would make $ f(b)$ a real number bigger than any other real number).
Similarly, $ f$ has finite limit in $ a$, so we're done.

Hmm, u'r right, and this even can be seen on my example's graph:D


0317785b = 2866.68 + b

it's driving me nuts and it's been a while since I have done formulas

Maybe it's just me but what does the 0 on LHS and . on RHS mean?
Also is b a digit of the number for LHS or do you mean that it is multiplied?


Hi. If I am given a problem which gives me two sets of vectors spanning two subspaces of a linear space, how would I go about finding a basis for the intersection of the two subspaces?

For example:

In R4, let S denote a subspace spanned by (1, 1, 3, -1), (1, 0, -2, 0), (3, 2, 4, -2) and let T be a subspace spanned by (1, 0, 0, 1), (1, 1, 7, 1). Find a basis for the intersection of S and T.

Is Ramanujan's method correct? I does not make sense to me. What is the reasoning behind it?

That method doesn't work. It produces $ S^{\perp}\cap T^{\perp}$. Note that the dimension count is wrong for "generic" choices of bases.

Spanning sets for the two spaces are an inconvenient starting point for getting at the information we want. Given two bases, we can combine them to find the sum of the two spaces, but intersections are harder. One way around that is to go to the orthogonal complement: $ S\cap T\equal{}(S^\perp \plus{} T^\perp)^\perp$. (Ramanujan's method gives $ (S\plus{}T)^\perp$ instead)

A method for this: Let $ A\equal{}\begin{bmatrix}1&1&3&\minus{}1\\1&0&\minus{}2&0\\3&2&4&\minus{}2\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}1&0&0&1\\1&1&7&1\end{bmatrix}$. Solve the systems $ Ax\equal{}0$ and $ Bx\equal{}0$; let the columns of $ C$ span the solution space of $ Ax\equal{}0$ and the columns of $ D$ span the solution space of $ Bx\equal{}0$. Define $ M\equal{}\begin{bmatrix}C&D\end{bmatrix}$. As a row space, $ S$ is the set of solutions of $ xM\equal{}0$. Find a basis for it.

jmerry how can you prove that $ S\cap T \equal{} (S^\perp \plus{} T^\perp)^\perp$? I've proved that the intersection is contained in $ (S^\perp \plus{} T^\perp)^\perp$ but I am stuck the other way around. Any hints? I also tried to prove that they have the same dimension without any success though.

Since $ S^\perp\subseteq S^\perp\plus{}T^\perp$, $ S\equal{}(S^\perp)^\perp\supseteq (S^\perp\plus{}T^\perp)^\perp$. The same holds for $ T$. Since both $ S$ and $ T$ contain $ (S^\perp\plus{}T^\perp)^\perp$, their intersection $ S\cap T$ does.

This is a version of DeMorgan's laws. In the lattice structure, vector space sums look enough like set unions, vector space intersections look enough like set intersections, and orthogonal complements look enough like set complements that the same arguments work.

KUNG-WEI YANG Western Michigan University in a one-page article "A Basis for the Intersection of Subspaces" introduces another method for finding a basis for intersection of two subspaces. I asked one of my students to prepare it for a  $ 15$ min lecture in my class. (This article is available from jstor. I'll put the main idea of that article on the forum next week unless someone else does it sooner.)

After a rather long absence I want to keep my promise and explain the method used in the above article.

Let $ W$ be an $ n$ dimensional vector space and let $ U, V$ be two subspaces of $ W$ of dimensions $ r, s$ and bases $ \mathcal{B}_U\equal{}\{u_1,\dots, u_r\},\ \mathcal{B}_V\equal{}\{v_1, \dots, v_s\}$, respectively. Now, define $ A$ to be the $ n\times (r\plus{}s)$ matrix with columns $ u_1,\dots, u_r, v_1,\dots,v_s$, and let $ B$ be the reduced row echelon form of $ A$, i.e., $ B\equal{}\text{rref}(A)$.

Suppose $ B$ has $ t$ non-pivotal columns. Then according to the formula $ \text{dim}(U\plus{}V)\plus{}\text{dim}(U\cap V)\equal{}\text{dim}(U)\plus{}\text{dim}(V)$,
we have $ t\equal{}\text{dim}(U\cap V)$.

Suppose $ [a_1,\dots, a_r, *]^{t}$ is one of the $ t$ non-pivotal columns. We claim that the $ t$ vectors of the form $ a_1u_1\plus{}a_2u_2\plus{}\dots\plus{}a_ru_r$ form a basis for the intersection of $ U, V$.

[hide="How to prove it"]Show that [i]each[/i] element of $ \mathcal{B}_V$, is obtained this way:

$ v_k\equal{}a_1u_1\plus{}a_2u_2\plus{}\dots\plus{}a_ru_r$+ (a linear combination of "previous" pivotal v_j's), where $ [a_1,\dots, a_r, *]^{t}$ is the $ (r\plus{}k)$-th column of $ B$.

The above identity completes the proof. Because it implies that $ (a_1u_1\plus{}a_2u_2\plus{}\dots\plus{}a_ru_r)\in U\cap V$ and that these vectors together with the $ (s\minus{}t)$ pivotal $ v_j$'s span $ V$.[/hide]


Mafia
geometry
geometric transformation
search
FTW
blogs
Support

You are in a room. You do not know why you are in the room, but you are pretty sure you don't want to be anywhere near the room. What will you do?

Please leave a possible command in your post. I will try to update this every day. I will select a command from the posted ones and then describe the resulting situation. Note that everyone is controlling the same character, not your own individual ones. Credit for the idea goes to a site called MSPA, but without the pictures. I hope this works.  :) Go!

EDIT: [Command will be selected from first three commands. I wasn't sure how many posts there would be. So commands after that will not and cannot be accepted. You are allowed to comment about existing commands and the situation. If I have a hard time deciding, I may look at the comments.]
EDIT #3: [One command per person between my posts, please! Please try to not take all the commands, either.]

[quote="randomguy64"]Look for a way out.[/quote]

Win.

Will you please describe the room a little more?

And exactly which "MSPA" is this?

/look around room, but also look behind objects I can't see behind.

Command chosen: Look around room.
The room is about the size of a room. In other words, it is rectangular, about 24.5763 feet wide and 32.1452 feet long. It is hard to tell how high the ceiling is because of an optical illusion painted on it. On one wall is a window with light coming in. It looks to be about 3:52 pm. You don't know how you know this, or, thinking about it, the dimensions of the room.. There is a door that is clearly locked. There is half a key on the floor. It clearly does not fit the door. There is also the number 58, in red, on one of the walls, which are blue. It occurs to you that a lot of this information is absolutely useless. The clock on the wall says 3:52 pm. Under it are two light switches, which is odd, or even, as there is only one light in the room. It's off. The light switches are in opposite positions. There are no objects you can't see behind, except for the wall, etc. You cannot look around these objects!

\jump to see if ceiling is higher than i can reach
\use key to break window

Chosen: \jump to see if ceiling is higher than i can reach
You jump towards the optical illusion. It's hard to aim at something that messes with your vision like that. You miss. It quite obviously is higher than you can reach. You hit the floor. Hard. How did you jump that high? Your infinite health bar has been depleted by 300 HP, which, obviously, does nothing at all. You feel invincible, until you contemplate the possibility of taking infinite damage. You get up.

You are now in exactly the same position as before, except now you have more knowledge about the height of the ceiling, which probably is, like much of the other information, completely useless.

/me asks self: Is this similar to Crimson Room?

I use the key to try to break a window. (basically what Phire said a few posts back).

[quote="randomguy64"]/tries all possible combinations of positions with the light switches and observes what happens.[/quote]

again.

@aleph0: It's a game where you try to get out of a crimson room. Look it up.

§§§§§
those are my favorite symbols
▐▐▐▐  ▐▐
and that is a piano!

i typed A alt+0160 B on a mac and thats what happened.
But      now     i'm     on     a     PC

[color=blue]The last few posts haven't been related to this game :(, is anyone still playing?[/color]

[color=#0000FF]Just when I thought it had been abandoned...
[hide=" "]but now I have to dig up all of my files about it.[/hide][/color]

[quote="aleph0"][color=#0000FF]Just when I thought it had been abandoned...
[hide=" "]but now I have to dig up all of my files about it.[\hide][/color][/quote]

hmmm.......

[color=#00BF00]Goes back to every room and observes everything in each room[/color]

uh ... i'm pretty sure this was LONG dead.

WE DONT CARE!

WHO EVER THE MOD IS,
give us a hint!

please?

:(

I'm the mod, and this is long dead.

Edit: I request a lock, because this should be long dead.

Well, this game was fun. Let's grow wings and fly out of the place! Yay! We win! =D

It's up to you, mod.

Yeah, please don't let this die.  The only reason people lost interest is because they got stuck.
I'm sure you can make a solution to that problem...

Due to several people requesting me to lock this thread, I shall lock this thread. Good night.


Hello. I am new here, I would like to know if someone could help me out with this problem.
   In Babba´s island they use an alphabete with two signs a and b, and every finite sequence of signsis called a word. For every set P of six words of four signs each, we define the set  $ N_p$ as the set whose words do not have any of the words (sub-words) of set P. Demonstrate thaif$ N_p$ is finite hence its words can not exceed ten signs length.

Hi..if you want I can copy the solution from a book... (maybe you want a more original solution)
but I will not have time until tomorrow
By the way..who are you? :P

Daniel

Thank you a book would do. I am practicing for the national ad I am in the last level but I have never practiced before, quite a complicated situation.

Well I´ll be in the national round..so why don´t you contact me by msn?
kohendaniel@gmail.com

I´ll help you :D

Daniel

Thank you a lot. I have added you to my msn. But if you could give me the answer to the problem I have presented at the top it would be the best.

could you upload or write the answer here daniel? Please I need it. Thank you.

Here it goes
Suppose that $ N_p$ has a word of size $ 11$ We will show that contains words of any size.
We consider all the sets of three consecutive letters of the word of lenght $ 11$, there are $ 9$ sets. Since there are $ 2^3\equal{}8$ word of lenght $ 3$ there must be at least a word of lenght $ 3$ that appears more than one time. Let´s call it $ t\equal{}c,d,e$
We consider the subword of the word of lenght $ 11$ (let call it p) which goes form the first letter of p from the first time $ t$ appears.
then we have  $ c,d,e,s....c,d,e$ We copy $ s....c,d,e$ after $ e$ and  it is easy to see that satisfies the condition of the problem..
Repeating this we obtain word of infinite lenght.Absurd
The reasoning is valid even when the two subwords $ c,d,e$ are overlapped inside $ p$. You can do a similar reasoning and you obtain the same thing as in the previous case

Daniel


geometry
circumcircle
modular arithmetic
trigonometry
incenter
power of a point
radical axis

Construct the circles $(O_{1}),(O_{2}),(O_{3})$ with diameters the sides $BC, AC, AB$ of the acute triangle $ABC$, if we denote as $U, V, W$ the cut points between the radical axis of the inner Apollonius circle $(O_{0})$ of $(O_{1}),(O_{2}),(O_{3})$ and the circumcircle $(O)$ of $ABC$ with the lines $BC, AC, AB$ respectively. From $U$ draw the tangent $UX$ to the arc $BC$ of the circumcircle $(O)$ that not contain $A$, similarly drawn the tangents $VY$ and $WZ$ to the circumcircle $(O)$. Prove that $AX,BY,CZ$ concur.
I can't prove that: If $X', Y', Z'$ are the touch points of $(O_{0})$ with $(O_{1}),(O_{2}),(O_{3})$ respectively, then $AX',BY',CZ'$ concur.

This is a very interesting configuration, but the concurrency of the straight lines $AX,BY,CZ$ is true for any line cutting the extentions of the side-segments $BC,CA,AB.$

[b]Lemma.[/b] $A',B',C'$ are three points on the arcs $BC,CA,AB$ of the circumcircle $(O)$ of $\triangle ABC.$ Tangents of $(O)$ at $A',B',C'$ cut $BC,CA,AB$ at $A_0,B_0,C_0.$ Then $A_0,B_0,C_0$ are collinear iff $AA',BB',CC'$ concur.

Since $\angle BA'A_0=\angle BAA'$ and $\angle AA'A_0=\angle CAA'$ $\pmod\pi,$ it follows that

$\frac{A_0B}{A_0C}=\frac{A'B}{A'C} \cdot \frac{\sin \widehat{BAA'}}{\sin \widehat{CAA'}}= \left ( \frac{\sin \widehat{BAA'}}{\sin \widehat{CAA'}} \right)^2.$   Similarly, we have the expressions

$\frac{B_0C}{B_0A}=\left ( \frac{\sin \widehat{CBB'}}{\sin \widehat{ABB'}} \right)^2 \ , \ \frac{C_0A}{C_0B}=\left ( \frac{\sin \widehat{ACC'}}{\sin \widehat{BCC'}} \right)^2$

$\Longrightarrow \frac{A_0B}{A_0C} \cdot \frac{B_0C}{B_0A} \cdot \frac{C_0A}{C_0B}=\left ( \frac{\sin \widehat{BAA'}}{\sin \widehat{CAA'}} \cdot \frac{\sin \widehat{CBB'}}{\sin \widehat{ABB'}} \cdot \frac{\sin \widehat{ACC'}}{\sin \widehat{BCC'}}  \right)^2$

By Menelaus' theorem and trig-Ceva we conclude that $A_0,B_0,C_0$ are collinear if and only if $AA',BB',CC'$ concur.

[quote="juancarlos"]I can't prove that: If $X', Y', Z'$ are the touch points of $(O_{0})$ with $(O_{1}),(O_{2}),(O_{3})$ respectively, then $AX',BY',CZ'$ concur.[/quote]
Pairwise circles $(O_1),(O_2),(O_3)$ cut at the feet $A',B',C'$ of the A-,B- and C- altitude of $\triangle ABC.$ Inversion through pole $A$ with power $\overline{AC} \cdot \overline{AB'}=\overline{AB} \cdot \overline{ AC'}$ transforms $(O_1)$ into itself and $(O_2),(O_3)$ into the lines $CC',BB'.$ Thus $(O_0)$ is taken into the circle $(O_a)$ tangent to $CC',BB'$ and internally tangent to $(O_1)$ at the inverse $X''$ of $X'$ $\Longrightarrow$ $X'' \in AX'.$

Let $H$ be the orthocenter of $\triangle ABC.$ In the figure $\triangle HBC \cup (O_1) \cup (O_a),$ it's well-known that the angle bisector of $\angle BX''C$ goes through the incenter of $\triangle HBC$ and the incenter of triangle bounded by $BC,CB',BC',$ i.e. incenter of $\triangle ABC,$  for  a proof see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=303520]R,S and T are collinear, very nice![/url]. Thus, $IX''$ goes through the midpoint $V_a$ of the semicircle $BC'B,$ i.e center of the inner Vecten square with side length $BC.$ Similarly, $IY''$ and $IZ''$ go through the centers $V_b,V_c$ of the inner Vecten squares $CA,AB.$  In this configuration, $AX'',BY'',CZ''$ concur at Kimberling center $X_{1336}$ $\Longrightarrow$ Lines $AX',BY',CZ'$ concur at $X_{1336}.$

Dear Mathlinkers,
A thought for the beloved Juan Carlos Salazar who died the 30 march 2088 and was a great contributor to Mathlinks.
See: http://perso.orange.fr/jl.ayme  vol. 2  In memorian Juan Carlos Salazar.
Sincerely
Jean-Louis


graph theory
combinatorics proposed
combinatorics

Let $ G$ be an infinite graph such that for any countably infinite vertex set $ A$ there is a vertex $ p$, not in $A$, joined to infinitely many elements of $ A$. Show that $ G$ has a countably infinite vertex set $ A$ such that $ G$ contains uncountably infinitely many vertices $ p$ joined to infinitely many elements of $ A$.

[i]P. Erdos, A. Hajnal[/i]

[quote="dgrozev"]I think König's infinity lemma has nothing to do with this problem since the main point here is that G contains [u][b]uncountably[/b][/u] many vertices of some kind. Of course it is possible to use the König's lemma somewhere in the proof but it cannot itself guarantee the uncountability of something.
By the way, as stated, this problem is not true since we can give as a counterexample a complete graph with countably many vertices. So, I was curious to look at the book "Contests in Higher Mathematics", but the statement there was the same as here! Then I looked at the solution and there it was used essentially the fact that for any countably infinite vertex set $A$ there is a vertex $p$, [color=#FF0000]not belonging to[/color] $A$, joined to infinitely many elements of $A$.
As it appears the marked in red text should be inserted in the problem statement.

[color=red]Original post edited. -mod[/color][/quote]
another version of infinity lemma no this version.
if i find a link about infinity lemma i will post it


Given that the polynomial $ x^2\minus{}ax\plus{}16$ has only positive integer roots, find the average of all distinct possibilities for $ a$.

First, we factor $ 16$ as $ 2^4$. By Vieta, we know that $ a$ is the sum of the solutions. Thus $ a$ could equal $ 2\plus{}8$, $ 4\plus{}4$, or $ 16\plus{}1$. Thus the average is $ \frac{10\plus{}8\plus{}17}{3}\equal{}\frac{35}{3}$..... I think.....


can someone tell me about the math version of Poincare Conjecture,
Thanks

In mathematics, the Poincaré conjecture  is a theorem about the characterization of the 3-sphere, which is the hypersphere that bounds the unit ball in four-dimensional space.

The conjecture states:

Every simply connected, closed 3-manifold is homeomorphic to the 3-sphere.

An equivalent form of the conjecture involves a coarser form of equivalence than homeomorphism called homotopy equivalence: if a 3-manifold is homotopy equivalent to the 3-sphere, then it is necessarily homeomorphic to it.

Originally conjectured by Henri Poincaré, the theorem concerns a space that locally looks like ordinary three-dimensional space but is connected, finite in size, and lacks any boundary (a closed 3-manifold). The Poincaré conjecture claims that if such a space has the additional property that each loop in the space can be continuously tightened to a point, then it is necessarily a three-dimensional sphere. The analogous conjectures for all higher dimensions were proved before a proof of the original conjecture was found.

After nearly a century of effort by mathematicians, Grigori Perelman presented a proof of the conjecture in three papers made available in 2002 and 2003 on arXiv. The proof built upon the program of Richard S. Hamilton to use the Ricci flow to attempt to solve the problem. Hamilton later introduced a modification of the standard Ricci flow, called Ricci flow with surgery to systematically excise singular regions as they develop, in a controlled way, but was unable to prove this method "converged" in three dimensions. Perelman completed this portion of the proof. Several teams of mathematicians verified that Perelman's proof was correct.

The Poincaré conjecture, before being proved, was one of the most important open questions in topology. In 2000, it was named one of the seven Millennium Prize Problems, for which the Clay Mathematics Institute offered a $1 million prize for the first correct solution. Perelman's work survived review and was confirmed in 2006, leading to his being offered a Fields Medal, which he declined. Perelman was awarded the Millennium Prize on March 18, 2010. On July 1, 2010, he turned down the prize, saying that he believed his contribution in proving the Poincaré conjecture was no greater than Hamilton's. As of 2020, the Poincaré conjecture is the only solved Millennium problem.

On December 22, 2006, the journal Science honored Perelman's proof of the Poincaré conjecture as the scientific "Breakthrough of the Year", the first time this honor was bestowed in the area of mathematics.

sorry for 5 year bump


AoPS had one of these topics some time ago...  Since reviving 9-page topics is bad, here's a new topic.

So what is the status of Taiwan?

I am highly interested in political topics like these, however it's always risky (just yesterday valentin advised me to throttle back a bit on a topic of mine in the physics forum (which i did))

ccy, are you a chinese , living in china (i get the impression there are several chinese here who live in usa)

Well uhm don't know much about it, but i saw a documentary that i watched with great interest when british reporter simon reeve travelled to nonexisting countries (like nagorno karabach, trans dneister,...)

he went to prc and roc

the people he talked to said :
in china : recruits for the army are motivated to join by posters about the threat coming from taiwan, they still seem to hope to get it back, and are willing to take action if they declare themselves independent

in taiwan : the old generation consider themselves chinese, they still hope to take over the mainland, which i think will be hard
the young people consider themselves taiwanese and feel the two nations should stay as they are

I also would like to ask one or two questions :

in the documentary he is shown bombs near a beach pointed at the mainland, and loud speakers that can yell propaganda, he is also taken on a boat trip with the coast guard, where they intercept tourists that get a kick from getting close to taiwan, and order them back
Is that on taiwan?  or was it islands closer to the mainland, that politically belong to taiwain?
Because it seems the street of formosa is wider than the canal between UK and France/Belgium, and i can tell you, i've never seen england from the beach

Second question : on the wikipedia site about [url=http://en.wikipedia.org/wiki/Three_Gorges_Dam]link[/url]
the three gorges dam it says taiwan has claimed it is capable of destroying that dam
uhm, are tensions that high?

I would say most countries recognize Taiwan as a province of China, even if the media doesn't support that. Considering the fast development of China, it is probably in Taiwan's best interest to stay a province.

but does taiwan benefit from any of that? just how isolated are they?

perhaps relevant, what is this talk of a jingtai expressway,not only technologically hard, but why would the chiense want to connect beijing to taipei?

[quote="fredbel6"]just yesterday valentin advised me to throttle back a bit on a topic of mine in the physics forum (which i did))[/quote]Good for you. Anyway Taiwan vs POC is a classic on AoPS - MathLinks and at times it's reminding me of Romania vs Moldovia issue (which is rather similar).

[quote="fredbel6"]Romania Moldova issue :?   In those documentaries i spoke of it seemed that eastern part of moldova used to belong to ukrainee?  Why is there an issue?[/quote]Whaaat?! That is a joke. Ukraine, like Hungary did not even exist when the Romanians were living on these lands.

Romanians are the descendents of the Romans that colonized Dacia province when it was conquered by Emperor Traian in 106 A.Ch. Because the land was one of the most fertile in the whole Balkan area (and much closer and thus the crops were much easier to bring to Rome and to the Germanic fronts than the crops from Egypt) it was heavily colonized.

Many many centuries afterwards the slaves and the hungarians came from Asia and settled down nearby, claming lands like Transilvania and Moldovia. Transilvania was conquered at around 1100 A.Ch. and was freed only after World War I. Moldovia resisted the attacks of the turks for centuries, and the Russian Empire afterwards. After World War II, the Soviet Union claimed half of Moldovia and made it into a Soviet Republic which they attached to USSR. After 1991, that half became a stand-alone country.

Ukraine really hasn't got anything to do with it. The issue was if Romania should unite (or rather take back) with Moldovia.

[b][color=darkblue]Question[/color][/b] : Ubi patria est ?
[u][color=violet]Answer Nr.1[/color][/u] : Ubi bene, ibi patria.
[u][color=green]Answer Nr.2[/color][/u] : The answer Nr.1 is false.

[color=red]DANTON[/color] : Est-ce qu'on emporte la patrie a la semelle de ses souliers ?
[color=red]F. G. KELLER[/color] : Achte jedes Menschen Vaterland, aber das deinige liebe.
[color=red]L. J. JAURES[/color] : Les patries, en leur mouvement magnifique de la nature a l'esprit,
de la force a la justice, de la competition a l'amitie, de la guerre a la federation,
ont a la fois toute la force organique de l'instinct et toute la puissance de l'idee.
[color=red]V. H. HEINE[/color] : Es ist eine eigne Sache mit dem Patriotismus, mit der wirklichen
Vaterlandsliebe. Man kann sein Vaterland lieben und achtzig Jahr' dabei alt
werden, und es nie gewusst haben; aber man muss dann auch zu Hause geblieben sein.

[quote="Valentin Vornicu"]

Romanians are the descendents of the Romans that colonized Dacia province when it was conquered by Emperor Traian in 106 A.Ch. Because the land was one of the most fertile in the whole Balkan area (and much closer and thus the crops were much easier to bring to Rome and to the Germanic fronts than the crops from Egypt) it was heavily colonized. [/quote] never knew that, silly me, the name romania must come from somewhere

[quote]Many many centuries afterwards the slaves and the hungarians came from Asia and settled down nearby, claming lands like Transilvania and Moldovia. Transilvania was conquered at around 1100 A.Ch. and was freed only after World War I. Moldovia resisted the attacks of the turks for centuries, and the Russian Empire afterwards. After World War II, the Soviet Union claimed half of Moldovia and made it into a Soviet Republic which they attached to USSR. After 1991, that half became a stand-alone country. [/quote]

Wow, i had this simplistic view that moldova was part of ussr since rise of communism, but that the part east of dniester was an autonomous region of ukraine.  After world war II russia changed it so it was joined with moldova instead of ukraine, after cold war, they became an independent country.  Wikipedia also says that romania annexed parts of moldova with the axis??  Why would romania join up?  Or has wikipedia been vandalised..again

[url=http://en.wikipedia.org/wiki/Transdniester#Autonomous_Moldavian_Republic_in_Soviet_Ukraine]link[/url]

Forgive my ignorance and curiosity.  Most people in belgium barely know where moldova is, it is considered poor and uninteresting (it is among the poorest, together with albania , in europe, am i not mistaken?)  However it seems transnistria is highly important as huge remnants of weapons during cold war are stacked there, which may play a key factor in terrorism etc...
Some european countries are only talked about when there is another eurosong competition, my info is pasted together from a handful of wikipedia pages, some documentaries and some people i know who went there
a friend of mine went to romania twice to work voluntarily on the construction of a school : interestingly he told me that on that other side of romania, at the hungarian border, both nations clearly as he said 'pick' on people from the other nation, is there any reason for that?

[quote="fredbel6"]never knew that, silly me, the name romania must come from somewhere[/quote]Well yes, Romania - Rome - Romans - Romanians. Get the picture? :) Heck, I never studied Latin, but I can read and understand around 60% of any Latin text! :)

[quote="fredbel6"]Wow, i had this simplistic view that moldova was part of ussr since rise of communism, but that the part east of dniester was an autonomous region of ukraine.  After world war II russia changed it so it was joined with moldova instead of ukraine, after cold war, they became an independent country.  Wikipedia also says that romania annexed parts of moldova with the axis??  Why would romania join up?  Or has wikipedia been vandalised..again

[url=http://en.wikipedia.org/wiki/Transdniester#Autonomous_Moldavian_Republic_in_Soviet_Ukraine]link[/url]
[/quote]Let's get one thing straight. Moldova means the land from the Carpatian Mountains up to Dnister. What's east of Dnister is not Moldova, and I frankly do not know to whom that teritory really belongs. In simpler ways Moldova is a part in Romania now and the other part is called [url=http://en.wikipedia.org/wiki/Bessarabia]Basarabia[/b]. It was part of Romania up until World War II when the Russians took it away.

[quote="fredbel6"]a friend of mine went to romania twice to work voluntarily on the construction of a school : interestingly he told me that on that other side of romania, at the hungarian border, both nations clearly as he said 'pick' on people from the other nation, is there any reason for that?[/quote]Well besides that fact that for centuries the hungarians occupied Romanian teritory, tried to make all Romanians peasants, by taking away their schools and churches (Romanians are ortodox, while Hungarians are Catolics), and a lot of other bad stuff, I think it's a similar situation with Flanders and <the other place> (sorry but I forgot the name :) ) in Belgium.

The love (for anything) hasn't voice and don't make noise. You feel and you live profoundly the love. To say (declare) always "I love you" is very much alike to the strip tease. So is and with the love for the your people (nation) which formed you. \[ \boxed {\ \mathrm {To\ be\ freely\ \Longleftrightarrow\ to\ be\ indebted\ to\ anybody.}\ } \] The only and eternal dept is the civic duty and the moral responsibility for your people to the last instant of the your existence.

[b]Please V. Vornicu (if you can and/or you wish), traduce in engleza si apoi sterge textul in romaneste.[/b]

[b]P. Doumer[/b] : Exista pentru tine un bun si unic mijloc de a sluji umanitatea: acela de a te stradui pentru inaltarea patriei tale.
[b]J. J. Rousseau[/b] : Cine nu are patrie, inceteaza de a mai fi.

Going back to the original question...

Answering it literally, Taiwan is obviously functionally independent of China right now.  The real questions are, whether it "should" reunify with China in the future, and whether it will or not.

As a grandson of a Chinese WW2 bomber pilot and other members of Chiang Kai-Shek's "baggage train", I unsurprisingly think it "should".  However, people like me comprise only ~20% of Taiwan's population.  The vast majority has mixed feelings about the Kuomintang's 1949 intrusion, since the KMT repressed them during the subsequent 40 years.  They, very reasonably, want Taiwan to be governed in a manner that focuses foremost on the interests of Taiwanese, and many of them are not confident that this would happen under reunification.  Independence would seem a safer bet... except that China's leadership would be politically compelled to retaliate.  This won't change unless/until the Communists lose power.

To achieve the best outcome under these constraints,

- It is China's responsibility to demonstrate that China is worth unifying with.  The government needs to show that it'll never make big mistakes like those that characterized the first 30 years of Chinese Communist rule ever again, and that it understands how to fully modernize China.  It also needs to show with its management of Hong Kong that the practice of "one country, two systems" offers enough local autonomy to satisfy rational Taiwanese.

- It is Taiwan's responsibility to be politically realistic in its actions and demands, to accept that China does constrain it and focus its efforts on negotiating shrewdly with China rather than offensively defying it.  It is the responsibility of Taiwanese to ensure that the governments they elect in the future are more competent than that of Chen Shui-bian.  Chen's incompetence to date can be forgiven to some extent thanks to inexperience, but the electorate needs to show that they'll punish a failure to learn from mistakes, even when it's a Taiwanese that will suffer.

I've outlined what I hope for.

I don't know if it will happen.  I am all too conscious of the sad results of public choice theory.

My apologies for once again dragging the discussion away from its original topic (although I agree with the previous poster that as an empirical matter Taiwan is independent of the PRC and the interesting question here is the un-asked one: what Taiwan's relationship with China [i]should be[/i], rather than what it is.)

It is worth pointing out that in Transylvania, currently (and likely to remain) part of Romania, there are large sections where the population is majority-Magyar (Hungarian).  While Valentin counts the area as part of "historic Romania" or "Greater Romania," Hungarians consider it as part of "historic Hungary" or "Greater Hungary."  Transylvania has been in the past attached to the Hungarian Kingdom (including as part of the Austro-Hungarian Empire), to Romania, and even been briefly independent (when much of what is now Romania and Hungary was controlled by Turks).  During the Ceausescu regieme, there was a concerted effort to wipe out the Hungarian communities in Transylvania, to the extent that it garnered the attention of the international community as a human rights disaster (not that the Ceausescu era was exactly wonderful for Romanian-Romanians, of course).

I say this not to suggest that Hungary has somehow a "better" claim to Transylvania than Romania, but rather to point out that there are two sides to every story -- indeed, in Central and Eastern Europe, it often seems as if there are at least four or five sides.  For Americans, who have done such a good job of eliminating and minimizing those others who would claim portions of our territory, it often seems strange that histories of a thousand years or more are called upon when discussing current political arrangements.  The fact of the matter is that the two major concepts of "a people," the idea of the nation-state (probably due to the French Revolution) and the idea of the historic people (perhaps best associated with the German/Nazi Volk concept) can't be brought into agreement in much of Europe.  Since the world has chosen the nation-state as its unit of political significance, there will inevitably be groups associating themselves with the volk-concept who lay claim to land which they once held or in which they maintain a presence.

(As one final note, I should say that in the lead-up to the joint Hungarian-Romanian governmental session that happened last year, Gyurcsany Ferenc, the prime minister of Hungary, made a formal declaration stating that Hungary was no longer trying to regain control of Transylvania.  The basic Hungarian version of the story is that it was unfairly taken from them at the end of World War 1, in what we in the U.S. call the Versailles Treaty but which they refer to as the Treaty of Trianon.)

[quote="JBL"]It is worth pointing out that in Transylvania, currently (and likely to remain) part of Romania, there are large sections where the population is majority-Magyar (Hungarian).  While Valentin counts the area as part of "historic Romania" or "Greater Romania," Hungarians consider it as part of "historic Hungary" or "Greater Hungary."  Transylvania has been in the past attached to the Hungarian Kingdom (including as part of the Austro-Hungarian Empire), to Romania, and even been briefly independent (when much of what is now Romania and Hungary was controlled by Turks).  During the Ceausescu regieme, there was a concerted effort to wipe out the Hungarian communities in Transylvania, to the extent that it garnered the attention of the international community as a human rights disaster (not that the Ceausescu era was exactly wonderful for Romanian-Romanians, of course).
[/quote]This is a joke right? As told before the Romans conquered Transylvania from the Dacs population (capital was at Sarmisegetusa, city in Transylvania) in 106 A.Ch. (and it was under the Roman Empire until the retreat of Marcus Augustus at 276 A.Ch. - or similar date anyway). Almost 1000 (!!!) years later the Huns (led by Attila) settled in the Panonic fields, and conquered Transylvania at around 1200-1300. Attached is thus not the correct word to use here.

Secondly, there are only two counties (which are in the inner south-eastern border of Transylvania land) where the population is Hungarian in majority (out of around 17 that compose Transylvania), namely Covasna and Harghita.

Thirdly, there was no concerted effort to wipe out anything during Ceausescu regime. He did a lot of bad stuff, but wiping out Hungarians is not on the list.

[quote="Valentin Vornicu"]This is a joke right? As told before the Romans conquered Transylvania from the Dacs population (capital was at Sarmisegetusa, city in Transylvania) in 106 A.Ch. (and it was under the Roman Empire until the retreat of Marcus Augustus at 276 A.Ch. - or similar date anyway). Almost 1000 (!!!) years later the Huns (led by Attila) settled in the Panonic fields, and conquered Transylvania at around 1200-1300. Attached is thus not the correct word to use here.[/quote]

I am slightly upset by your reaction.  One reason is that it is incredibly typical of nationalist disregard for both truth and nuance.  The factual errors are relatively minor: Attila led the Huns through Europe in the 5th century AD.  The Magyar, or Hungarian, people arrived in the Carpathian basin in the late 9th or early 10th century AD (the official date in Hungary is the year 896) and at some point conquered and settled parts of what is now Transylvania.
But let's seriously consider what you're saying: the Romanians get the area in the 2nd or 3rd century and hold it for 1000 years.  Then the Hungarians come and conquer it and hold it for 500 years (non-consecutively as there is a period of 100 to 200 years where most of both Hungary and Romania, including Transylvania, were effectively Turkish).  Somehow, in your eyes, 1000 years are sufficient to establish a permanent claim to a piece of territory, but 500 years don't even merit the use of the word "attached?"  (And what about the claim of Turkey?!)  I presented a complicated historical situation: Transylvania has been ruled by Romanians, Turks, Hungarians and Transylvanians.  The region has been, as I correctly stated, attached to a variety of different pricedoms, kingdoms and nations, including the Kingdom of Hungary and a Romanian nation.  For you, this significant, interesting 2000 year history seems to boil down to "we got there first!"  I'm unimpressed.

[quote="Valentin Vornicu"]Secondly, there are only two counties (which are in the inner south-eastern border of Transylvania land) where the population is Hungarian in majority (out of around 17 that compose Transylvania), namely Covasna and Harghita.

Thirdly, there was no concerted effort to wipe out anything during Ceausescu regime. He did a lot of bad stuff, but wiping out Hungarians is not on the list.[/quote]

I should re-state: there was no ethnic cleansing campaign as in modern-day Serbia, Bosnia and Kosovo.  However, he did make a serious effort to wipe out the Hungarian culture.  Romania had an official state policy starting almost immediately after WWI of resettling ethnic Romanians into Transylvania.  Hungarian-language education was banned, eliminated or discouraged at various times.  Ceausescu himself was personally convinced that the Hungarian population (acting as agents of the Hungarian state, of course) was plotting against him and Romania.  You can read about the history of Hungarians in Romania in greater detail at this [url=http://www.minority2000.net/Gr-75/t84gb.htm]this Minorities in Europe website.[/url]

A side-note: I imagine that you pointed out that the two Hungarian-majority provinces are in the the south-east in order that I should notice that they are surrounded entirely by Romania and should thus seem Romanian.  I think this was something of a strategic error, however: the Hungarians in that region are known (in Hungarian, at least) as Szekely -- the world literally means "frontier guard."  There existence so far to the interior establishes a Hungarian claim to the territory, not a Romanian one.

A second, more important side-note: I don't care who owns Transylvania.  If it were still part of Turkey, that would be essentially fine with me.  I have no investment in the Hungarian or the Romanian historical, ethnic, cultural, linguistic, military or other claims to the territory.  I do have an investment in the truth, and in people recognizing that questions of this nature are complicated.  The situtation in Eastern Europe is much more complicated than that of Taiwan (although of less international significance currently), but everyone contesting claims in situations such as these should recognize the historical complications that make simple political answers so difficult and should form their views accordingly.

[quote="JBL"]But let's seriously consider what you're saying: the Romanians get the area in the 2nd or 3rd century and hold it for 1000 years.  Then the Hungarians come and conquer it and hold it for 500 years (non-consecutively as there is a period of 100 to 200 years where most of both Hungary and Romania, including Transylvania, were effectively Turkish).  Somehow, in your eyes, 1000 years are sufficient to establish a permanent claim to a piece of territory, but 500 years don't even merit the use of the word "attached?"[/quote]First of all the Romanians "didn't get the area". What are Romanians? Romanians are the population resulted by the intermixing of the huge number of Romans (colonizers brought by Traian to replace the dead Dacii from the long war) with Dacii (which ruled the teritory for hundreds of years before that also).

And yes, being first and staying there for more than 1000 years does make a claim to  the land. Egypt was conquered almost all the time since Alexander the Great. Does it not mean that the land belogs to the Egypteans? Or should the French, British, Turks (that held that territory for hundreds of years) have claims to it?

Furthermore, even if the territory itself was under political ruling of Hungary and afterwards the Austriac Empire, it was populated at all times by Romanians (in majority). All these (and some more even) justify the claims.

[quote="JBL"]A side-note: I imagine that you pointed out that the two Hungarian-majority provinces are in the the south-east in order that I should notice that they are surrounded entirely by Romania and should thus seem Romanian.  I think this was something of a strategic error, however: the Hungarians in that region are known (in Hungarian, at least) as Szekely -- the world literally means "frontier guard."  There existence so far to the interior establishes a Hungarian claim to the territory, not a Romanian one.[/quote]There is nothing to hide that Transylvania had a political border with Moldovia (!) which the Hungarian Kingdom (and then the Austriac Empire) wished to protect, and did so by amassing a huge number of troops there, and certainly doesn't bring any arguments in favor of any Hungarian claim to the land.

My goodness, you really aren't saying anything deeper than "we got there first!"  I am disappointed you have such disagreement with the statement "history is complicated."  I won't be posting on this topic in this thread any more, I think.

Basque, North-Irish, North-Italian, Cyprus-Greek or Turkish a.s.o. Come on to talk over these ! Leave Taiwan, is enough !

Uh, can a moderator please split this thread into the obvious two pieces?

yes it appears an occasional curiosity of mine has opened a pandora's box
sorry
well valentin is here, i bet he'll do it?

[quote="JBL"][quote="Buffalo"]Ignore the historical facts, Is it well accepted that the Hungarians are descendants of the Huns(Attila)?[/quote]
No, it's not accepted.  The Szekler people, one of the Hungarian groups living in Transylvania, claim to be, but their claim does not seem to be supported by the historical record.[/quote]
I think I've used the wrong words, what I really want to ask is, do Hungarians consider themselves descendants of the Huns? And if you walk on a street in Europe, randomly pick somebody(skip him if he's an expert in history) and ask "Do you think that the Hungarians are descendants of the Huns?", what will the answer usually be?
Your reply suggests at least some of the Hungarians will say yes.

You didn't use the wrong words, exactly, but I interpreted them in a different sense than you meant them:  I assumed "accepted" to mean "accepted as actual history, by historians who have studied the situation."  Certainly, many people, if asked that question, would be likely to say "yes."  However, that question is of the sort that would lead to an unusually large number of "yes" answers anyhow, because it provides no other options and the single provided option sounds reasonable.  Among Hungarians, I do not know how much of the population believes they are descendents of the Huns, but it is probably a non-trivial number.

Yes, there should be a Taiwan thread, and a Hungary thread. (This comment can be put into either thread when the threads are split.   ;)   )

this is messed. I voted for province of china.

but if I was taiwanese, I would vote independent country or china is province of taiwan, which is quite odd... because china is way bigger


Παιδια με συγχωρειτε που δινω το προβλημα σε λινκ αλλα η ΕΜΕ κατι εχει κανει και δεν γινεται να αντιγραψει κανεις τα προβληματα :cursing:  :furious: .

Απο το λινκ με ενδιαφερει το 1ο ερωτημα του 2ου Προβληματος(combinatorics). Ισως να κανω λαθος αλλα δεδομενων των ελαχιστων πληροφοριων που δινονται οι απαντησεις θα μπορουσαν να ειναι πραγματικα παρα πολλες.

Μπορει καποιος να μου πει και να μου εξηγησει την απαντηση...

Το λινκ: http://www.hms.gr/eme/modules/wfsection/article.php?articleid=1295

Νομίζω πως έχεις δίκιο ότι μπορεί να υπάρχουν πολλές λύσεις. Ίσως όμως το πρόβλημα να εννοεί το εξής:

Αν για να επισκεφθείς το Α υποχρεούσαι να πας στο Β, και για να επισκεφθείς το Β υποχρεούσαι να πας στο Γ, τότε για να επισκεφθείς το Α υποχρεούσαι να πας και στο Γ.

Αρχικά νόμιζα πως με αυτόν τον περιορισμό θα υπήρχε μοναδική λύση αλλά δυστυχώς βρίσκω δύο λύσεις.

Χμ, καταλαβα θα το ξανακοιταξω.

Για να μην ανοιγω καινουργιο τοπικ να ρωτησω και κατι αλλο απο ΙΜΟ 2005.

In a mathematical competition, in which 6 problems were
posed to the participants, [b]every two of these problems were solved by more
than 2/5 of the contestants[/b]. Moreover, no contestant solved all the 6 problems.
Show that there are at least 2 contestants who solved exactly 5 problems
each.

Το μπολντ κομματι εννοει.
α) Οτι αν παρουμε 2 οποιαδηποτε προβληματα εστω τα Α και Β τοτε αυτοι που ελυσαν και τα 2 ειναι μεγαλυτεροι απο 2/5 των διαγωνιζομενων
ή
β) Οτι αν παρουμε 2 οποιαδηποτε προβληματα εστω τα Α και Β τοτε αν προσθεσουμε αυτους που λυσαν το Α με αυτους που λυσαν το Β τοτε το αθροισμα θα ειναι μεγαλυτερο απο τα 2/5 των διαγωνιζομενων?

Thanks a lot :)  :lol:

Έτσι όπως το διαβάζω καταλαβαίνω το (α). Θα κοιτάξω την άσκηση αργότερα.


[img]http://img185.imageshack.us/my.php?image=17275168jg3.png[/img]

don't think that worked, but the image is at that location.

I'm not a moderator, but it's kind of tacky to post screenshots of book problems. Among other things, it's not searchable.

I'm not sure which problem you are asking about, but the funny thing is that if it is number 12, then it is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=512566286&t=222413]here[/url] (and we also know that is Hatcher's book).


$ x + y\geq z$   $ \forall x,y,z\in$ $ R^ +$
$ f(x + y - z) + f(2.{\sqrt {xz}) + f(2.\sqrt {yz}) = f(x + y + z)}$
$ f: R^ + \rightarrow R^ +$


Let ABC be a triangle and lengths of sides are a,b,c and area of ABC be S.Let A'B'C' be another triangle that lengths of sides a+b/2,b+c/2 and c+a/2.Find the minumum area of A'B'C' triangle as a fuction of S.

http://www.mathlinks.ro/Forum/viewtopic.php?t=55494

  darij


I get all nervous and triple-check every computation I do right after I do it and I panic and I can't think of ways to solve problems and my mind wanders off to my happy place and it's just really crazy...I still do well, but only like 70% of my full potential.

I like to imagine that I am nervous - or I really have to to become stressed in order to make me more aware - something I learned in health class. I like to take things seriously, and I promise myself lots of fun after I put in 100% effort during competition.

must be frenzied.  Therefore, i dont go to the bathroom between sprint and chapter, even if i have to go.  It helps me think faster.

I am usual too nervous and end up never performing as well as I should.


algebra
polynomial
number theory
greatest common divisor
number theory solved

Find all n for wich it exists m such that m2= -9 (mod n).

Dear,

I am wondering what leads you to the problem,but the problem reminds me to the following well-known problem:(It may be a problem from Korean MO)
 Problem: Prove that if 2^n-1 | m^2 +9 then n is 2^k for some k

Compare to your problem, it turn out that it is more than better prob.After one hour of thingking,i found the solution as below:

  Solution:We will prove that alll such n is in the form
       3^a * product(p^k) where a is an integer which 0 \leq a  \leq 2 and p is 4t+1 primes
Here goes, Assume that n is a positive integer number fpor which n|m^2+9.Take any p, a prime divisor of n, so we have p|m^2+9 .This implies -9 is QR of p and so p=3 or p<>3 and (-9/p)=1 where (a/b) is the Jacobi symbol! So we have (-1/p)=1 which implies p==1(mod4) .So
     n=3^a*p1^k1 *.......* pr^kr
where pi are 4t+1 primes
Now if a>2 then this implies m is divisible by 3 and so cause a>2 the 3 is a divisor of (m/3)^2+1 which is imposible cause 3 is not 4t+1 prime

Now we have to prove that  all such n satisfies the condition of the problem .In deed by from the fact that the polynomial the equation x^2+1 ==0 (mod p_t) has an root x_t for which (x_t,p)=1 and cause (2,p_t)=1 ,so THE equation x^2+1 ==0 (mod p_t^k_t) has a solution wich we denote by y_t.Now by the Chiness Remainder Theorem there exists an integer x such that
          x== y_t (mod p_t^k_t) for t=1,...,r
So we have x^2+1 ==0 (mod p1^k1*...*pr^kr)
Chose m=3*x we have m^2+9 ==0(mod n).This completes the proof!!!

Applies this result we can solve the Korean problem: with n replaced by 2^n-1 the number has to has 4t+1 prime divisors and 3 only.Write n=2^b*m where m is odd.If m>1 then 2^n -1 is divisible by 2^m -1 which has form 4t+3 so it must has an prime divisor wich has the form 4t+3.Of cause 3 does not divides by 2^m -1 cause m is odd so this contracts to the fact that 2^n -1 has only 3 or prime divisors in 4t+1
So n has to be 2^k.

An last we also find all such n, ie... 2^n -1 | m^2+9 for some m
Indeed, at first we have to show that 2^(2^k) - 1  is divisible by 3 but  9  for k>0.Indeed 2^k=6t+2 or 6t+4, then  2^6==1(mod 9) so 2^n -1 ==3 or 6 (mod 9) .
Now we prove that if p is an odd divisor of 2^n-1 p>3 and n=2^k ,k>1, then p ==1(mod4)
Indeed, by Fermat Theorem p| 2^(p-1) -1 then p | 2^( gcd(
p-1,2^k)) -1 .Since the fact that p>3 so gcd (p-1,2^k)> 2.So we must have gcd (p-1,2^k ) >=4 .And this implies p ==1(mod 4).
This completes all the....work.Now i have to drink something for relax!

A quick observation: I think the first problem pluricomplex mentioned, the one which asks us to find all n s.t. there is an m s.t. 2n-1|m2+9, was, I think, also used in a Romanian TST.

It is a problem from IMO short-list 1998.

Pierre.

[quote="pluricomplex"]We will prove that alll such n is in the form
3^a * product(p^k) where a is an integer which 0 \leq a \leq 2 and p is 4t+1 primes[/quote]

I think you are missing some of the solutions, since -9 is a square mod 2, but not mod 2^k for k \geq 2, we can do everything that you said, and see that n is of the form 2a*3b*product(p^i)
with 0 \leq a \leq 1 and 0 \leq b \leq 2

Oh god, i forgot the case! Yes you true, cause for any odd k we have k^2==1(mod8) so n is of the form 2^a*3^b*product(p^i)
with 0<=a<= 1 and 0<=  b<=  2
The proof still works!
Thanks for your corrections!


find all  polynominal $ f(x)$ such that  ;  $ f(x)f(2x^2) \equal{}f(2x^3\plus{}x^2)$
find all  polynomial $ f(x)$ such that  ;  $ f(x)f(2x^2) \equal{}f(2x^3\plus{}x)$

[quote="Hong Quy"]find all  polynominal $ f(x)$ such that  ;  $ f(x)f(2x^2) \equal{} f(2x^3 \plus{} x^2)$
find all  polynomial $ f(x)$ such that  ;  $ f(x)f(2x^2) \equal{} f(2x^3 \plus{} x)$[/quote]
2)$ f(x)\equal{}(x^2\plus{}1)^n$
1)Try to find out an root with low degree it can find out all solution.


Today my bio teacher and I got into a small quarrel. We were talking about the 4stage hypothesis and how it was predicted that life came from inroganic compounds such as NH3 etc. But she claimed that methane was inorganic.... at which point i swiftly stated that it wasnt as it contained the vital C that makes it organic. IM just checkin to make sure, who was right? also, is it true that although C defines organic nature, CO and CO2 are one of the only exceptions?

$ \ce{CH4}$ is organic. There are other exceptions too like $ \ce{CO3^{2\minus{}}}$ and $ \ce{SCN\minus{}}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211765]This[/url] will also help you.  :)

Methane? Inorganic? That's not right. Definitely not right.

[quote="feynman_wannabe"] my bio teacher and I got into a small quarrel. We were talking about the 4stage hypothesis and how it was predicted that life came from inroganic compounds such as NH3 etc. But she claimed that methane was inorganic.... [/quote]well both seem correct...

ur teacher wasn't wrong in saying that life originated from inorganic matter --and well CH4 was one of them earliest compounds :(
well cant think clearly..... :?:

the definition is a little controversial.  some say that an organic has to contain at least 1 carbon-carbon bond...

and that is cent pc wrong
the same way as the earliest defination(compounds obt frm living organisms :P )

In summary, knowing the exact definition of an organic compound is not important: what is really important is to understand the chemical behaviour of molecules.