Ok, I really don't know in which forum I should post this question so I'll ask here
thanks you :)

I would like to know, if someone know how to draw a tree with LaTe
I'm trying to do some huffman compresion algorithms sheme on my blog
but unfortunately I cant find how to draw a tree with the basic package of LaTeX...

maybe if an administrator is reading he could try to figure out how to add
other package ?

http://www.essex.ac.uk/linguistics/clmt/latex4ling/trees/parsetree/

there's some interesting package to draw tree... so I dont know what could be done
anyway... thanks :)

Nib,

The appropiated forum for this post is the [url=http://www.mathlinks.ro/Forum/index.php?f=123]LaTeX Forum[/url]. On the other hand, if you want, you could use pictex. Finally, I don't think they would add this package to the forum, because probably it can be a little heavy on the compilation part, hence it could be a not that good idea.

The XY-pic package is a fairly easy way to draw nice graphs.

More information here: [url]http://tug.org/applications/Xy-pic/[/url]

If you're using TeXnic Center, it should download the package automatically for you.

You can't use it on the forum, though.  (As djimenez pointed out, if we start adding packages to the forum, then it would become very slow to compile.)

Ok, thanks both for your answers you are totally right :)


calculus
integration
algebra
function
domain
analytic geometry
trigonometry

Well the problems I'm stuck on is these:

1)Let S be the unit disk centered at the origin. Consider the mapping $ u \equal{} \frac {x}{x^2 \plus{} y^2}$ and $ v \equal{} \frac { \minus{} y}{x^2 \plus{} y^2}$
Find the image of the set S\{$ \vec 0$} under the mapping.

I have no idea how to do this.. I try to solve for u and v and keep getting the original expression....
I don't the formula for the unit disk and no clue what the S\{$ \vec 0$} means.

Help...

2) Let T be the triangle with vertices (0,1) (1,0) and (0,0). By making an appropriate change of variables, convert the integral:
$ \iint_{T} f(x \plus{} y)\, dx\,dy$ into an integral on the interval [0,1].

Does this mean convert the triangular domain into a square? Cause that seems to be hard.. I tried u=x+y, but couldn't get a v value...

Finally, a moment of inertia problem:
3) Let $ \Omega$ = [0,1]x[0,1] represent a 2-d distribution of a certain material of density $ \rho \equal{} \rho(x,y)$. Establish a formula using integrals of the moment of inertia about the line $ y \plus{} x \equal{} 0$

The part I'm confused on is how to find the moment of inertia about the line $ y \plus{} x \equal{} 0$. It gets me confusing answers like $ r \equal{} \sqrt {(x_{ij} \plus{} x)^2 \plus{} (y_{ij} \minus{} y)^2}$, where r is the distance from the line, which I'm pretty sure is wrong since it doesn't come out cleanly.

I don't understand 1 either.  I didn't cover multivariable calculus-moment of inertia together so I can't help you with 3.  I can help with 2.  You don't need to convert anything to a square.  Draw a picture, and you'll see that the left-hand $ x$ limit is $ x\equal{}0$ and the right one is $ x\equal{}1$.  The lower $ y$ limit is $ y\equal{}0$ and the upper one is $ y\equal{}1\minus{}x$.  Then the integral becomes
$ \int_{0}^{1}\int_{0}^{1\minus{}x}f(x\plus{}y)\mbox{ d}x\mbox{ d}y$
and that's it.

Isn't that just setting the bounds. And how is that change of variables? I think that's what the problem is asking for... I think it wants a thing with u and v and jacobian...

1) $ S\setminus\{\vec 0\}$ means the unit disk $ S$ with point $ (0,0)$ (denoted as $ \vec 0$ here) removed. Simply put, it's the unit disk without the center. A suggestion: use polar coordinates $ x\equal{}r\cos t$, $ y\equal{}r\sin t$. In terms of these coordinates $ S$ is the set of all points with $ 0<r<1$ (or $ 0<r\le 1$ if $ S$ includes the boundary circle). Express $ u$ and $ v$ in terms of $ r$ and $ t$.

3) The distance from a point $ (x,y)\in \Omega$ to the line $ x\plus{}y\equal{}0$ is $ \frac{x\plus{}y}{\sqrt{2}}$. Think of the perpendicular line passing through $ (x,y)$ and where it crosses the line $ x\plus{}y\equal{}0$.

Well your question said nothing about changing systems or variables.  Usually, when an integral is converted into one on the unit square, its from a parallelogram, in which case you use change of variables.  This doesn't appear to be the case.

@JRav:
Sorry, I didn't type it in my post, but on my wkst it says you have to use "an appropriate change of variables". I'll edit the post..

@mlok:
1) Well, I did express u and v in terms of r and t. You get $ u \equal{} \frac {cos(t)}{r}$ and $ b \equal{} \frac { \minus{} sin(t)}{r}$ However, I think the problem asks for me to draw a picture. Is it just a square with r, t coordinates since polar coordinates turn circular to squarish?

3) I'm not sure about how you derived it. It seems correct, but I derived it in 2 parts one for if the point is on the line $ y \equal{} x$ and one if it isn't. Is there an easier way?

Peter Shi's second exam is a beast isn't it?  :ninja:

Can anyone help?

For #1, I get the unit circle on $ u$ and $ v$ as the image that is made, but that wouldn't really make sense would it?


At a plant nursery, it costs $ \$37$ for four geraniums and three daises or $ \33$ for three geraniums and four daisies. What is the number of dollars in the cost of one geranium and one daisy?

Sum it, you get 7d+7g=70, so g+d=10
Where 3 stands should be 33


logarithms
calculus
calculus computations

Hi everyone! I have the following problem, since I don't know how to compute $ \frac{\partial^2 z}{\partial x^2}$  :blush: :

Let $u=\sqrt {x^2+y^2} $ and $v=\ln\sqrt {x^2+y^2}$. Transform the following equation:

\[y^2\frac{\partial^2 z}{\partial x^2}-x^2\frac{\partial^2 z}{\partial y^2}+\frac{x^3}{x^2+y^2}\frac{\partial z}{\partial x}=0\]

That clearly isn't the right transformation- your two new variables aren't independent. Before you can do anything, one of them needs to become an angular variable (or at least not a pure radial variable).


Find all polynomials $ P(x)$ of the smallest possible degree with the following properties:
$ (1)$ The leading coefficient is $ 200;$
$ (2)$ The coefficient at the smallest non-vanishing power is $ 2;$
$ (3)$ The sum of all the coefficients is $ 4;$
$ (4)$ $ P(\minus{}1)\equal{}0, P(2)\equal{}6, P(3)\equal{}8.$

(3) equavalent to $ P(1)\equal{}4$.
Constuct Lagranj's polinom $ P_0(x)$, suth that (3),(4):
$ P_0(x)\equal{}(x\plus{}1)(x\minus{}2)(x\minus{}3)\minus{}2(x\plus{}1)(x\minus{}1)(x\minus{}3)\plus{}(x\plus{}1)(x\minus{}1)(x\minus{}2)\equal{}2x\plus{}2$.
$ P(x)\equal{}2x\plus{}2\plus{}Q(x)P_1(x), P_1(x)\equal{}(x\minus{}1)(x\plus{}1)(x\minus{}2)(x\minus{}3)\equal{}\minus{}6\plus{}5x\plus{}...$.
Therefore smallest degree when $ Q(x)\equal{}200x\plus{}b$, then
coefficient $ a_1\equal{}2\minus{}1200\plus{}5b\equal{}2\to b\equal{}240.$


I thought, you understood that it was incorrect to send problems from the current BWM!
So, don't try to post it again.


AMC USA(J)MO USAMO inequalities algebra polynomial

If you're just beginning to work olympiad problems, what is the best way to spend 4.5 hours and pick up some partial credit on the three USAMO questions each day? For instance, if you don't really know how to proceed on the harder problems, should you just try to dumbass one of the easier ones?

[quote] For instance, if you don't really know how to proceed on the harder problems, should you just try to dumbass one of the easier ones?[/quote]

Could you explain what you mean by this question, particularly the last part?

Yakko

[quote="paladin8"]He means that if there is a question that you have no idea how to solve, should you try to "dumbass" it, basically making up something that sounds remotely like a solution to get some extra points I suppose.[/quote]

Nope. "Dumbassing" an inequality means basically expanding everything, cancelling whatever you can, and hoping for things to simplify to an easy problem or a full solution.

Haha. Well in my language it has a different meaning :P

[quote="MithsApprentice"]

Nope. "Dumbassing" an inequality means basically expanding everything, cancelling whatever you can, and hoping for things to simplify to an easy problem or a full solution.[/quote]

Unfortunately, this doesn't really work anymore.  Almost every (polynomial) inequality in the last 5+ years on high-level exams like USAMO, APMO, IMO has been either stronger than AM-GM (e.g USAMO 03 which was as strong as Schur) or required some sort of trick (USAMO 04/APMO 05) before any sort of progress could be made.

Any fool can solve a problem which is equivalent to Muirhead with a bit of patience and no ingenuity; solving an inequality that is stronger than Muirhead actually requires skill.

And thank goodness it doesn't work. I don't see the point of an inequality problem that could be solved with Algebra I knowledge of polynomial expansion.

But such a horrid method is prone to mistakes and takes a long time.

Indeed. However, faced with the prospect of obtaining 7 points on the USAMO, many people would not mind. Nowadays, though, it doesn't work.


conics ellipse geometry rectangle perimeter calculus

Rectangle ABCD has area 2008. An ellipse with area $ 2008\pi$ passes through A and C and has foci at B and D. What is the perimeter of the rectangle?

Being x ,y rectangle sides we have 2a=x+y  2c=(xx+yy)^.5.So b=(xy/2)^.5
Ellipse area=pi*a*b
Rectangle area=x*y
After simple calculus we have P=16*(251)^.5

and since when was HSB dealing with calculus..?


function
topology
complex analysis
set theory
complex analysis theorems

In my complex analysis course we are currently considering product representations of entire functions and canonical products. One thing that seems implicit in the text, but isn't stated, is that any non-zero entire function has only countably many zeros.

What about an analytic function on any open set? Only countably many zeros?

Is the statement true, is it difficult to prove? I imagine this follows from the fact that zeros are isolated, but I still have no idea how to prove such a statement. My background in set theory and topology is a little bit too weak for the course I am taking.

Consider a partition of the plane into squares of side length $1$.   There are only countably many squares in this partition.  An analytic function with uncountably many zeros has infinitely many zeros in one of the partitions, since a countable union of finite sets is countable.

Infinitely many zeros in a bounded set implies an accumulation point of zeroes, and an accumulation point of zeros for an analytic function implies that that function is zero everywhere.

For a function that is analytic only on an open set $U\subset\mathbb C$ this argument must be adjusted, since zeros may accumulate at boundary points. In this case you can consider  the sets $K_{n}=\{z\in U\colon |z|\le n,\text{ and }\mathrm{dist}(z,\partial U)\ge 1/n\}$. Each $K_{n}$ is a compact subset of $U$, which implies that there are only finitely many zeros in $K_{n}$. But $\bigcup_{n}K_{n}=U$.

(This also must be adjusted to account for connectedness or lack thereof. If $U$ is not connected, $f$ may vanish on some of its components.)

[quote]What about an analytic function on any open set? Only countably many zeros? [/quote]

Also true, at least for a [i]connected[/i] open set, for reasons very similar to blahblahblah's argument.

At heart, blahblahblah used that the entire complex plane is $\sigma$-compact. That is, it is the countable union of compact sets. Can you show that any open subset of $\mathbb{C}$ is $\sigma$-compact? If you can, then you can copy blablablah's proof.

Except for one thing - he did quote what is essentially the uniqueness of analytic continuation: that an analytic function with zeros on a set with a limit point is zero "everywhere." If you're going to quote that theorem, you'd better be working on a connected set.

[color=green][mlok beat me to the reply.][/color]


i'm planning to either got to Hopkins.... or Wilbur Cross where mostly everyone's going... :D in my school...

I think we have 67 students going into 22 (should've been 21, one guy decided not to go to Cheshire Academy) different high schools.

Hamden High
Notre Dame
Fairfield Prep
Hopkins
Choate
Wilbur Cross
Career
Hillhouse
HSC
Co-Op
New Haven Academy
Metro
Hyde
Xavier
Lyman Hall
North Haven High
ECA

And a lot more I can't remember

if i am going to a public school for high school, i am planning to go to: [list]ludlowe high school
fairfield warde (in fairfield)[/list]
if i am going to a privat school for high school, i am planning to go to: [list]john hopkins
choate rosemary hall[/list]

[quote="Ignite168"]John Hopkins is a college and are you going to pick one as you can't go to 2 schools..[/quote]

hopkins HIGH SCHOOL in new haven

u see, choate and hopkins have an entry exam
the choate exam is harder btw

if i dont make the choate exam, ill go to hopkins
if i dont make the hokins exam, ill go to public school

plus, i wont go to choate or hopkins if my parents rnt willing to pay .

fairfield has two high schools. Since the districting in my town is all funky, i dont kno which high school i am going into

Umm Choate and Hopkins don't have separate "exams" so one isn't harder. I took the ISEE for both and the ISEE was pretty easy considering I did it when I hadn't slept.

Unless you mean that Choate prefers the SSAT. Hopkins accepts that too and I think the SSAT is technically easier as you can take it as many times as you want and submit the best score. The ISEE you can only take once in the application window. Anyway, Choate looks at a lot more than your test scores, you have to write essays and stuff.

You can apply for financial aid, the reason I'm not going to Wilbur Cross is because Hopkins gave me enough financial aid to convince me to go to their school.

The SSAT isn't really choate's entry exam, it's a generic exam for most secondary schools. It's really easy, but has little weight in whether or not you get accepted (I would hope...).

Choate also accepts the ISEE. Hopkins accepts the SSAT (wish I knew  :mad: ) so I don't see how one can be harder than the other.

[quote="Timecat8"]yeah High school choosing is soooooooooooooo hard.. i mean doi you want to go to a public or private school..... do you want to go to a high school.. in New haven or not... i mean lol.... it take forever to get a real choose....[/quote]

Really? I made my decision about 1 minute after I opened the financial aid letter  :P (Seriously)

[quote="Timecat8"]ignite your going to Hopkins right............... did you have to get into a  Malone program..... to get finacial aid?[/quote]

I got both. You don't have to do anything special for the Malone program though, it's automatically done.

Err, I didn't "apply" for the Malone thing. The letter said it was automatically chosen through the admissions process. And I would say it isn't a full scholarship seeing as I got it WITH financial aid and still have to pay a little bit.

On the website somewhere it says 7-9th graders who are eligible get a regular scholarship, but 10-12th have to put in work hours (library, school store etc)to get their scholarship. Are Malone Scholars exempt from putting in work hours?

Err, I have no idea. I wouldn't mind putting in work hours at all if I had to. Anyway, I'm pretty sure all Financial Aid and Scholarships are re evaulated year by year.

[quote="Timecat8"]SO A QUESTION FOR THE MALONE PROGRAM FOR HOPKINS........................... YOU DON'T HAVE TO DO ANYTHING SPEICAL....[/quote]

I replied to all the questions about Hopkins in the PM you sent me which should cover that.


function
real analysis
real analysis theorems

[quote="rustam"]I'm searching for the proof of the theorem:
The fuction is integrarble as a Reimann iff the Lebesgue measure is 0.
Where i can find it?[/quote]

You mean if the measure of the set of points of discontinuity is zero.
You can find it in any advanced real analysis book.It is quite standard.

Nick


linear algebra
matrix
algebra unsolved
algebra

We conside the set M which contains all matrices with 3 rows and 3 columns and all the elements in the set {-1;1}.

Prove that:
1)if $B$ in M is invertible then its inverse is not in M
2)if $A$  is a matrix in M then $A^{2006}$ has all the elements not zero.


algebra
function
domain
number theory
greatest common divisor
Ring Theory
complex numbers

$x^5=(y+i)(y-i)$ and if an irreducible in $Zi$ divides $y+i$ and $y-1$ it is $i+1$ or $i-1$, but taking norms, we would find out that $2$ divides $x$, and therefore $2^5$ divides $y^2+1$, contradiction. So we must have:

$y+i=(a+bi)^5=a^5-10a^3b^2+5ab^4+i(5a^4b+b^5-10a^2b^3)$ so we have $5a^4b+b^5-10a^2b^3=1$ so $b=+-1$, there we find possible values for $a$, and therefore, we find $y$, and $x$. But i dont want to do that work!

ok, here goes a little of work: if $b=-1$, we would have that $5$ divides $2$, therefore, $b=1$ and it gives integer values for $a=0$ and thats it!

Pascual i cant understand your solution. im not that advanced¡¡ :blush:  could explain with detail please¡¡¡ what is that expression you have for x5(5 is expeonent) and that z letter with the i below it. what does it mean??'

Pascual works in the euclidean domain $\mathbb{Z}[i]=\{a+bi;(a,b)\in \mathbb{Z}^2\}$, where $i=\sqrt{-1}$.
In this ring, you can use arithmetic properties (unique factorization etc.).

$(x-1)(x^4+x^3+x^2+x+1)=y^2$
$gcd\{x-1,(x-1)(x^3+2x^2+3x+4)+5\}\rightarrow\{1,5\}$

1case:
$(x-1)=k^2$
$(x^4+x^3+x^2+x+1)=l^2$

for $x>3$ we have $(x^2+\frac{x+1}{2})^2>x^4+x^3+x^2+x+1>(x^2+\frac{x-1}{2})^2$
and
$x^4+x^3+x^2+x+1=(x^2+{x \over 2})^2$ haven't solutions

so the only solution second equation is:
$(x,l)=(3,11)$
but it isn't solution our main equation

2 case:
$(x-1)=5k^2$
$(x^4+x^3+x^2+x+1)=5l^2$

Can anybody simply solve this case i have no idea

[quote="Pascual2005"]
$x^5=(y+i)(y-i)$ and if an irreducible in $Zi$ divides $y+i$ and $y-1$ it is $i+1$ or $i-1$[/quote]
$p+qi=(i+1)\left(\frac {p+q} 2+\frac {q-p} 2i\right)=(i-1)\left(\frac {q-p} 2-\frac {p+q} 2i\right)$, but what if $p$ is even and $q$ is odd (or vice versa)? So, if $y$ is even than we can't be sure that $y+i=(a+bi)^5$.

Also, I forgot to mention, solutions are integers (not only positive integers).

I dont understand your question,  i just said: if $p$ divides $x+i$ and $x-i$, then it divides their difference $2i$, And since $i$ is a unit it divides $2$, but then its norm is $2$, so taking norms it follows that $2$ divides $x$. Nothing is wrong!

blang where can i learn more of what pascual was using to solve the problems  it seems useful. he uses complex numbers right??
 help me please :blush:

[quote="Pascual2005"]I dont understand your question,  i just said: if $p$ divides $x+i$ and $x-i$, then it divides their difference $2i$,[/quote]
Oh yes, the difference is $2i$. Sorry, my mistake, no hard feelings.

[quote="blang"]To marko avila : try google with "gaussian integers".[/quote]
Can you point the definite source from where we can learn it?

Bojan! there is no problem, i just was confused because i didnt knew what was wrong! Thanks for the information in the file!

According to the Catalan Theorem, we know that the equation x ^ n+y ^ m = 1 has only two solutions: the trivial solution (x = 1, n = 0, y = m = 0) and the solution (x = 3, n = 2, y = 2, m = 3), which implies that the equation x ^ 5 = y ^ 2 + 1, n "has only the trivial solution.


Find all pairs of integers (x, y) satisfying
$1+x^{2}y = x^{2}+2xy+2x+y$.

I'm pretty sure it's just (0,1), (-1,-1), (1,-1). my way is terrible tho, I just made it a function of x and then used calculus to examine behavior

[quote="Hokkage"]Here's what I did although I didn't finish...
\begin{eqnarray*}1+x^{2}y&=&x^{2}+2xy+2x+y\\ x^{2}y-2xy-y+2&=&x^{2}+2x+1\\ y(x^{2}-2x-1)+2&=&(x+1)^{2}\\ y&=&\frac{(x+1)^{2}-2}{(x^{2}-2x-1)}\\ &=&\frac{(x+1)^{2}-2}{(x-1)^{2}-2}\end{eqnarray*}
I don't know where to go from here though (proving that for y to be an integer x must be 0, 1, -1) although I don't think it should be too hard...[/quote]

Why don't we....

[hide]$y=\frac{x^{2}+2x-1}{x^{2}-2x-1}=1+\frac{4x}{x^{2}-2x-1}$
It should be a lot easier to work cases from here
x has to be between $-1-\sqrt{2}and 3+\sqrt{10}$ in order for $\frac{4x}{x^{2}-2x-1}$ to be greater than one, or less than negative , sooo plug in x=-2, -1...5,6 and we find:
$(-1,-1);(0,1);(1,-1);(2,-7);(3,7)$[/hide]


Let $ a,b,c$ be the sidelengths of a triangle such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \sqrt {\frac {a}{a^2 \plus{} b^2}} \plus{} \sqrt {\frac {b}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {c}{c^2 \plus{} a^2}} \ge \sqrt {2} \plus{} 1.$
Equality holds if and only if $ a \equal{} b \equal{} 1/2,c \equal{} 0$ and its cyclic permutations.

Ps. I saw this inequality from the book of Pham Van Thuan and also found that the proof in that book is wrong. So I post it here and wait for a nicer proof than mine.

Assume $ a \equal{} \max\{a,b,c\}$ and let $ f(a,b,c) \equal{} LHS$ then \[ \frac{1}{3}\leq a\leq \frac{1}{2}, f(a,b,c)\geq f\left(a,\frac {1}{2},\frac {1}{2} \minus{} a\right) \equal{} g(a) \geq g\left(\frac {1}{2}\right)\]  :)


How many solutions are there to the equation:
$ a\plus{}b\plus{}c\plus{}d\equal{}13$
given that a $ \neq$ b $ \neq$ c $ \neq$ d

Unless you say that they all have to positive, and they must be integers, then there is an infinite number of solutions

If this is the case, then we have

1 2 3 7 - 4 choose 2 or 24 ways to rearrange.
1 2 4 6 - 4 choose 2 or 24 ways to rearrange.
1 3 4 5 - 4 choose 2 or 24 ways to rearrange.

This gives us a total of 72 solutions.

Well they are Balls and Urns so $ a,b,c,d \in \mathbb{Z}$

I think the number of distict solutions is $ P(13) \equal{} F_{13 \minus{} 5} \equal{} 21$ where $ F_ n$ is the $ n^{th}$ Fibonacci number

But I have yet to figure out why

Why are we all getting different answers?  :ninja:

[hide="blech"]Assuming they are positive integers...

There are $ \binom{16}{3} \equal{} 560$ ways to partition 13 into 4 positive integers, by balls and urns.

We will never have a situation where $ a \equal{} b \equal{} c \equal{} d$ because then $ 4a \equal{} 13\implies 4|13$, contradiction.

Suppose $ a \equal{} b$ and $ c \equal{} d$, so we have $ 2a \plus{} 2b \equal{} 13\implies a \plus{} b \equal{} \frac {13}{2}$, a contradiction because $ a$ and $ b$ are integers. So there are no solutions in this case

Suppose $ a \equal{} b$ but $ c\neq d$ so that $ 2a \plus{} b \plus{} c \equal{} 13\implies b \plus{} c \equal{} 13 \minus{} 2a$.
There are $ n \minus{} 1$ solutions to $ b \minus{} c \equal{} n$ by simple balls and urns (and noticing that $ b$ and $ c$ are greater than 0).
Notice that $ a\leq 5$. If $ a\geq 6$, then $ 2a\geq 12\implies b \plus{} c\leq 1$, a contradiction.
So for each $ a\in[1,5]$ we have $ 13 \minus{} 2a \minus{} 1 \equal{} 12 \minus{} 2a \equal{} 2(6 \minus{} a)$ solutions.
So $ 2\sum_{a \equal{} 1}^{5}6 \minus{} a \equal{} 2(30 \minus{} 15) \equal{} 30$.

Suppose $ a \equal{} b \equal{} c$, so that $ 3a \plus{} d \equal{} 13\implies d \equal{} 13 \minus{} 3a$. Note that $ a\in[1,4]$.
So there are 4 solutions in this case.

Now, we could also have $ b \equal{} c$, $ c \equal{} d$, or $ a \equal{} d$ instead of $ a \equal{} b$ in the first case, so the 30 solutions must be multiplied by 4. So we have 120 solutions here.

In the second case, we could also have $ b \equal{} c \equal{} d$, $ a \equal{} c \equal{} d$, or $ a \equal{} b \equal{} d$ instead of $ a \equal{} b \equal{} c$, so we have 16 in this case.

Finally, $ 560 \minus{} 120 \minus{} 16 \equal{} \boxed{424}$[/hide]

Edit: Fixed that $ \binom{15}{3}$ thing. I'm still not sure it's right though.

Edit 2: Fixed the 4*4=16 thing.

Edit 3: They can be 0!!!!

Edit 4:
Brute force
[hide="blech II"]
If order doesn't matter, the solutions are:
(0, 1, 2, 10)
(0, 1, 3, 9)
(0, 1, 4, 8)
(0, 1, 5, 7)
(0, 8, 2, 3)
(0, 2, 4, 7)
(0, 2, 5, 6)
(0, 3, 4, 6)
(1, 2, 3, 7)
(1, 2, 4, 6)
(1, 3, 4, 5)

There are 11. Each one can be rearranged in $ 4!\equal{}24$ ways.
So $ 24*11\equal{}\boxed{264}$.
[/hide]

Shouldn't that be $ \binom{16}{3}$ partions

also in the second case when $ a \equal{} b \equal{} c$ then there are $ 4\cdot 4 \equal{} 16$ solutions, not $ 12$.

Also since $ a \neq b \neq c \neq d$ and you have counted all permutations of every set of solutions shouldn't your answer be divisible by $ 4!\equal{}24$

Actually I think we've both made a mistake there.
Each integer needs to be greater than or equal to 1, so we can't have 2 dividers next to each other.
So I think it should actually be $ \binom{12}{3} \equal{} 220$.

EDITz0rz:
The integers can be equal to 0 so it's just traditional balls and urns, so $ \binom{16}{3}\equal{}560$.

Oh, you're right, it's 16.
Forgot to count the $ a \equal{} c \equal{} d$ case.

[quote="OMGusernametaken"]Unless you say that they all have to positive, and they must be integers, then there is an infinite number of solutions[/quote]

Wow, I forgot to specify!!!  I was thinking that a,b,c and d were all integers greater than or equal to zero.

Oh, wow, equal to zero.
Well that just poops on everything I said before.

I shall edit now  :D.

Edit: I wrote a Python script that tells me that the real answer is 264.  :huh:

We can assume the numbers are a, a+x, a+x+y, and a+x+y+z, where $ a\ge 0$, x,y,z >0
Then $ 4a \plus{} 3x \plus{} 2y \plus{} z \equal{} 13$
$ 4a \plus{} 3(x \minus{} 1) \plus{} 2(y \minus{} 1) \plus{} (z \minus{} 1) \equal{} 7$
$ 4a \plus{} 3m \plus{} 2n \plus{} p \equal{} 7$
So it's the number of ways 7 can be made of 1s, 2s, 3s, and 4s (partition).
If a=1, then we have (m,n,p)=(1,0,0) (0,1,1) (0,0,2)
If a=0 m=2 then (n,p)=(0,1)
If a=0 m=1 then (n,p)=(2,0) (1,2) (0,4)
If a=0 m=0 then (n,p)=(3,1) (2,3) (1,5) (0,7)

So 11 possibilities, each can be permuted 24 ways for a total of 264.


I have a question in my textbook which is confusing me and would appreciate any feedback:
'The random variable $ X$ has probability density function
$ f(x) \equal{} 4x^k$    $ 0 <\equal{} x <\equal{} 1$, $ 0$ otherwise,
where $ k$ is a positive integer. Find the value of $ k$'

Now, I integrated f(x) to find that $ k \equal{} 3$ but what I don't understand is how $ f(x)$ can be $ 4x^3$ over the range $ 0 <\equal{} x <\equal{} 1$. For example, when $ x\equal{}1, f(x) \equal{} 4$ but how can this value for $ x$ have a probability of 4? Am I missing something? Please put me straight.

A probability density function has the property that an event $ x \in [a, b]$ has probability

$ \int_{a}^{b} f(x) \, dx$.

As you can see, this implies that the probability that $ x \equal{} 1$ is actually $ \int_{1}^{1} f(t) \, dt \equal{} 0$ rather than $ f(1)$.

I think I see. My confusion stems from thinking that the y-axis of the pdf is probability.


geometry
circumcircle
projective geometry
radical axis
geometry unsolved

let $ ABCD$ be quadrilaterale inscribed in a circle.

$ \|\begin{array}{c} E\in AB \cap CD \\
F\in AC \cap BD \end{array}$

let $ (C_1)$ and $ (C_2)$ respectively the circumcircle of $ \triangle AFD$ and $ BFC$
and   $ \|\begin{array}{c} H\in (C_1) \cap (C_2) \end{array}$  with    $ H$ # $ F$

show that $ \triangle EHF$ is right

DI/IC=HD/HC=(AD*BH*AD)/(BC*AH*BC)=(AD^2 * BD^2 *HC)/(BC^2 * AC^2 *HD)

DI/IC=(AD*BD)/（BC*AC）=EA/EC * ED/EA=ED/EA

So E,D,I,C are harmonic division and it is easy to prove HI is the bisector of angle DHC
Thus we are done

$ I\in CD\cap HF$, $ J\in AB\cap HF$. $ AD$, $ HF$, $ BC$ are concurrent at the radical center, $ R$, of the three circles. It is well-known that $ E,A,J,B$ form a harmonic range. Then the pencil $ R(E,A,J,B)$ is harmonic, intersecting with $ CD$ gives $ E,D,I,C$ harmonic. Then $ \angle IHD\equal{}\angle FHD\equal{}\angle FAD\equal{}\angle CAD\equal{}\angle CBD\equal{}\angle CBF\equal{}\angle CHF\equal{}\angle CHI$. And the result follows.


A square measures 6 centimeters on each side. A 1.5-centimeter square is cut from each corner. The sides arc folded up to form a box. How many cubic centimeters are in the volume of the box formed? Express your answer as a decimal to the nearest tenth.

If you cut a 1.5-cm square in each corner of a 6-cm square and form a box... The box will have a dimension of 6 - 2(1.5), 6 - 2(1.5), and 1.5...

So the volume would be V = lwh = 3(3)(1.5) = 13.5  :|


geometry
circumcircle
cyclic quadrilateral
perpendicular bisector
geometry unsolved

[quote="outback"]Let $ ABCD$ be a cyclic quadrilateral, with its circumcenter centered at $ O$. Angles $ \angle B,\angle C$ are obtuse. Let $ E \equal{} AB\cap CD$. Let $ P$ be the foot of the perpendicular from $ E$ to $ BC$. Let $ Q \equal{} EP\cap AD$. Let $ R$ be the foot of the perpendicular from $ E$ to $ AD$. Let $ S \equal{} ER\cap BC$. Let $ K$ be the midpoint of $ QS$. Show that $ E,K,O$ are collinear.[/quote]
$ BC\cap AD\equal{}F;BD\cap AC\equal{}M;SQ\cap RC\equal{}N$
Easy to see $ M,N,F$ colinear
so $ EK,EO\perp MN$
so $ E,K,O$ are collinear
:D

[quote="outback"]Let $ ABCD$ be a cyclic quadrilateral, with its circumcenter centered at $ O$. Angles $ \angle B,\angle C$ are obtuse. Let $ E \equal{} AB\cap CD$. Let $ P$ be the foot of the perpendicular from $ E$ to $ BC$. Let $ Q \equal{} EP\cap AD$. Let $ R$ be the foot of the perpendicular from $ E$ to $ AD$. Let $ S \equal{} ER\cap BC$. Let $ K$ be the midpoint of $ QS$. Show that $ E,K,O$ are collinear.[/quote]
[hide="Solution"]
Let the circumcenters of $ \triangle EBC$ and $ \triangle EAD$ be $ X$ and $ Y$ respectively.  Notice that since $ \triangle EBC$ and $ \triangle EDA$ are similar, we see that $ ES$ and $ EP$ are isagonal conjugates with respect to $ \angle BEC$, so since $ EP$ is an altitude of $ \triangle EBC$, we have that $ ES$ contains the circumcenter of $ \triangle EBC$.  Hence, $ X$ lies on $ ES$ and similarly, $ Y$ lies on $ EQ$.  Now, notice that $ X$ and $ O$ both lie on the perpendicular bisector of $ BC$, so $ OX\perp BC$.  Also, $ EY\perp BC$.  Similarly, $ OY\perp AD$ and $ EX\perp AD$, so $ OYEX$ is a parallelogram.  Then, $ EO$ passes through the midpoint of $ XY$.  Since $ \triangle EBC\sim \triangle EDA$, we have that $ \frac{EX}{XS}\equal{}\frac{EY}{YQ}$, and therefore, $ XY\parallel SQ$.  This gives us that $ EO$ passes through the midpoint of $ SQ$, which is $ K$.  This yields that $ K$, $ O$, and $ E$ are collinear, as desired. [/hide]

[quote="outback"]Could you please give more detail, such as why is this?

[quote="Mr Green"]
Easy to see $ M,N,F$ colinear

[/quote][/quote]
Because,
$ M,N,F$ are conjugate with $ E$
so $ M,N,F$ lie on Polar of $ E$
:D


Let $ ABC,\ A'B'C'$ be the triangles such that $ \angle A+\angle A'=180^\circ , \angle B=\angle B'$.
Prove that $ aa'=bb'+cc'$, where $ a,\ a'$ denotes the side lengths of $ BC,\ B'C'$ etc. respectively.

[hide]
The relation becomes
$ \sin A\sin A'=\sin B\sin B'+\sin C\sin C'\Leftrightarrow \sin^{2}A=\sin^{2}B+\sin C\sin C'$
But $ C=\pi-(A+B),C'=\pi-(A'+B')=\pi-A'-B=A-B$
We obtain $ \sin^{2}A-\sin^{2}B=\sin (A+B)\sin (A-B)$ which is true in every triangle.
[/hide]


THIS IS A RANDOM TAG
rotation
conics
videos
group theory
..

This is asking if u like him as a person. And i said no because he's a freak.

I don't know him as a person.

He seems odd, but I can't really make a judgement having never met the guy.

His music is the best though!
I still regard being able to moonwalk as one of the greatest accomplishments of my life.

i love some of his songs(like we are the world) but do not quite agree with some of his life style.

[quote="blahblahblah"][quote="djimenez"][quote="denizuzun"]i hate him[/quote]

That is a waste of energy!

Nevertheless, the guy is a phenomenon, a circus freak basically. Eventhoug, fifteen years ago and before he a few good sogs (rithm, in general not lyrics)[/quote]

You're grossly underrating just how good he was musically.  'Thriller' didn't sell millions upon millions of copies by being novel - it was just REALLY good.[/quote]

Thriller (and Billy Jean, and Bad, and many others) sold millions because most of the people do not care about lyrics, just how nice is the rithm, or would you consider an excelent lyric this?

[quote="Billie Jean"]
Billie jean is not my lover
She’s just a girl who claims that I am the one
But the kid is not my son
She says I am the one, but the kid is not my son
[/quote]

And for example, I haven't checked lately, but at least up to 2001, Nsync had the best seller disk in hystory, more than Thriller, nevertheless, I do not consider that they are really good, just people in general don't have good taste  :D

Michael jackson is one of the few fears in my life, including the first chapter in the 4th Harry Potter book, and the dream I had of a cartoon charectar going through a tunnel.

i always thought micheal jackson was weird.

Michael Jackson's is one of the greatest musicians ever to grace the planet.  While I may not agree with his "appearance" or his "personal life," I do enjoy his music and his dancing (Smooth Criminal).

i know nothing about michael's personality. but his "neverland" ranch just scares me to death...

[quote="djimenez"][quote="blahblahblah"]

You're grossly underrating just how good he was musically.  'Thriller' didn't sell millions upon millions of copies by being novel - it was just REALLY good.[/quote]

Thriller (and Billy Jean, and Bad, and many others) sold millions because most of the people do not care about lyrics, just how nice is the rithm, or would you consider an excelent lyric this?

And for example, I haven't checked lately, but at least up to 2001, Nsync had the best seller disk in hystory, more than Thriller, nevertheless, I do not consider that they are really good, just people in general don't have good taste  :D[/quote]

I don't know what you're looking at.  NSYNC's best selling album sold roughly 11 million copies, making it the 58th best selling record of all time.  Thriller sold 26 million copies, and for a long time was the best selling album of all time.

[quote="Ragingg"][b]Michael Jackson's is one of the greatest musicians ever to grace the planet.[/b]  While I may not agree with his "appearance" or his "personal life," I do enjoy his music and his dancing (Smooth Criminal).[/quote]

Wow, it is scary. First because I tend to think that Jackson is an interpreter, not an author. His music probably is not really [b]his[/b] music, because probably (I haven't checked) he has never really written the music or the lyrics of a song. And well, that is a subjective and not objective statment ever (yours as well as mine), but I honestly disagree with you.

[b]BTW blahblahblah[/b], the list of the RIAA lists the sellings inside the US, it does not count the sells outside the US!! I am a little lazy to look for the source I found few years ago right now,  but I will look for it (or it may be that I am just remembering wrong, nevertheless, I know that I saw a different list than the one the RIAA has published).

jackson was fine as a black male singer, but once he turned into a white women, that's where i draw the line

[quote="djimenez"][quote="stupidkid"]I think Jacksons got serious problem, I just dont see how someone could be attracted to another person of the same gender and 20 years younger. But thats only my opinion.[/quote]

Well, the acusations are about abuse to a kid around 35 years younger!

Pedophilia is considered by most of the Society (myself included) as a mental illness. I think that if you are 45 years old, it is kind of natural that you like a person 25 years old (twenty years younger), but both are adults, and to be honest, most of the people look great on their twenties, they are totally grown up (at least phisically), nevertheless, I can not understand how is that an adult sexually like a 10 or 12 years old kid or younger. It is just too dirty.[/quote]

I heard on tv that the reason isn't because they want to and think it's cool or anything.  It's just like being gay, it's just natural to them.  I guess thats kind of scary, and I feel kind of bad for them.

Hmm i dont think i ever wanna meet him in person but you gotta love Thriller...

[quote="h_s_potter2002"]I heard on tv that the reason isn't because they want to and think it's cool or anything.  It's just like being gay, it's just natural to them.  I guess thats kind of scary, and I feel kind of bad for them.[/quote]
Well.... that is a very soft argument. I mean, I am not a psicologist, but I tend to think that pedophilia is clinically considered an illness, homosexuality is not (even when it was for a long time). Nevertheless, at least I am sure that in all the civilized world, pedophilia is a crime and homosexuality is not!

[quote="alkachi"]in first  i was a fun but now after that he do  i HATE him :mad:[/quote]

you don't know what he did. the trial is not even close to finish yet

A black boy turned into a white woman..... whether any of the trial stuff is true or not, I  don't know nor want to comprehend the mess.

But then again, isn't it discrimination just because he's a little different than everyone else? But, where do you draw the line?

He has chosen this way of life, it's not like it's something he can't help. And he puts himself out in front of the media....

Hmmm.... I don't know. I still think he's a freak though.

[quote="elementsofmathkid"]He has chosen this way of life, it's not like it's something he can't help. And he puts himself out in front of the media....[/quote]Well, if the acusation is true, it does not really matter if he can or cannot help it, PEDOPHILIA IS A CRIME!, and if it is the case that he is a pedophilian and he can not help it, he should not be allowed to be a mile near of any under age person!

i wonder how much his plastic surgery cost!!

[quote="Derek"]I like his songs. But I agree his face is a little bit messed up.[/quote]Hehe, if he goes to jail, just imagin how is he going to be without all that Crayola people that works for him!!

Well, it's a thread made in time when the public was extremely polarized by the Jackson process, and take into account the age of participants in this poll...

I was shocked to hear the news this morning, although the media prepared us for this months ago when writing about MJ's health problems.

 :(

[quote="Jermaine Jackson"]“May Allah be with you, Michael, always,” [/quote]

[youtube]RYrUQItmW4s[/youtube]

Michael Jackson was a really good entertainer. RIP.

[youtube]Uqxo1SKB0z8[/youtube]

He certainly had good music, but I voted no for several reasons.
1) He almost dropped his son off a balcony in Germany
2) He was almost convicted of serious felony assault charges (he's probably guilty)
3) Many of his "surgeries" show an extremely low self esteem

However, whether he is a "freak" shouldn't really have anything to do with it.  You have to look at the evidence, and it shows why people shouldn't like him.


I remember seeing a formula for the sum of the first n terms of a fibonacci-type sequence (not necessarily 1,1,2,3,5...) but don't remember what it was...can anyone help me out?

To add from $F_0$ to $F_n$ in a Fibonacci-like sequence, the formula is simply $F_{n+2}-F_{0}$: see the wikipedia article here: [url=http://en.wikipedia.org/wiki/Fibonacci_sequence]en.wikipedia.org/wiki/Fibonacci_sequence[/url]

did you know the fibbonaci sequence was made up by some guy named fibbonaci wondering about the breeding habits of rabbist?
we start with a baby rabbit,
o.
then it grows up.
O
then it reproduces, but it doesn't go away.
Oo
Every time the same rules are followed: old rabbits reproduce and baby rabbits grow up:
OoO
OoOOo
OoOOoOoO
OoOOoOoOOoOOo
if you continue the pattern there are lots of fun patters to try and find. I read this all ina a book about phi. I forget why it was relevant to phi.

it's probably the least boring math book in existance :P .

The Number Devil (i forgot who it is by) is a book where this kid goes on adventures in his dreams with the number devil doing math. It's really good, and really famous. You really should read it.

The Formula for the nth Fibonacci number is my signature, you can copy it from there.

Latexed for clarity:

$\frac{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}$

[quote="Hexahedron"]Shouldn't it be

$\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}}{\sqrt{5}}$

because the way you had it, F(0)=0, which is false.[/quote]

from the signature, that would be right.

By the way why does it work? Do you need induction to prove it?

That would probably do the trick...it's also pretty easy to intuit I suppose

a   n   a+n   a+2n    2a+3n   3a+5n ....

Actually looking at what I just typed, shouldn't it be A[size=59]n+2[/size] - A[size=59]1[/size]?

Yeah, actually, that makes sense.  I saw the n+1 thing in AOPS Volume 2, chapter 16, but I guess it's an error.

Time to change the siggy, NY_HEEGer37.

Btw, this thing is called Binet's Formula for the Fibonnaci sequence.

yeah, but I checked online, and it says F(0) = 0, so...i'm confused


geometry
geometric transformation
reflection
circumcircle
geometry proposed

If you like geometry problems with very nonstandard solutions, try this one:

Let ABC be a triangle, and x, y, z three lines through its vertices A, B, C. Denote by x', y', z' the reflections of the lines x, y, z in the angle bisectors of the angles CAB, ABC, BCA, respectively. Prove that the triangle formed by the lines x, y, z has the same "fourth side" as the triangle formed by the lines x', y', z'.

Here, the [i]"fourth side"[/i] of a triangle is defined as $\frac{\Delta}{R}$, where $\Delta$ is the area and R the circumradius of the triangle.

This is a slightly corrected version of the AMM (American Mathematical Monthly) problem #4470, part [b](1)[/b], by Victor Thébault. The proposed solution (AMM volume 60, no. 4, april 1953, pages 268-269) uses some formulas. I have a rather short and, in my opinion, surprising solution - but it is not quite a geometry solution, honestly. Can you find it?

  Darij

I have been scaring people once again, as there is a rather simple synthetic solution. I think I am going to write a note about that in some time. One more interesting thing Armpist notified me about is that we can take x, y, z to be the parallels to BC, CA, AB through A, B, C; then, x', y', z' are the tangents to the circumcircle of triangle ABC at the points A, B, C, and thus we see that the fourth side of the tangential triangle of triangle ABC equals twice the fourth side of the triangle ABC. This is the crucial part in the solution of http://www.mathlinks.ro/Forum/viewtopic.php?t=68468 .

  darij


geometry
3D geometry
number theory unsolved
number theory

Is it possible to find two integers $ x$ and $ y$ such that the sum of their cubes is 6? What if $ x$ and $ y$ are allowed to be rational?

[hide="Hint for integers"]
$ x^3\minus{}y^3\equal{}6$
$ x\equal{}y\plus{}p$
$ 3p y^2\plus{}3p^2 y\plus{}p^3\equal{}6$
$ p|6$
[/hide]

[hide="Hint for rationals"]
$ \frac{a^3}{b^3} \minus{} \frac{c^3}{d^3}\equal{}6$ with $ (a,b)\equal{}(c,d)\equal{}1$

$ (ad)^3\minus{}(bc)^3\equal{}6(bd)^3$
$ (ad)^3\equal{}b^3 (c^3\plus{}6d^3) \Rightarrow b|d$
$ (bc)^3\equal{}d^3 (a^3\minus{}6b^3)  \Rightarrow d|b$

$ d\equal{}b$
$ a^3\minus{}c^3\equal{}6b^3$
[/hide]

If i understood well...
The cubes of integers are congruent to 1, 0 or -1 mod 9.
So, the sum of 2 cubes is congruent to 2, 1, 0, -1, -2 mod 9, but 6 is congruent to 6(or -3) mod 9.
So, the answer for the first part is no?

Is this correct?

Yes, that's right.
@mszew Are you sure your hints are right? I think the problem says [i]sum[/i] of cubes equaling 6.

Well it remains true that b=d if @mszew changes signs: - to +.

I mean on the rational part.

the problem is equivalent to finding integer (not all zero) solutions of $ x^3 \plus{} y^3 \equal{} 6z^3$
my guess is using $ \mathbb{Z}[e^{\frac {2i\pi}{3}}]$
probably none exist

Again by mod 9 i get that z is divisible by 3, and x^3 is congruent to 1 mod9, and y^3 is congruent to -1 mod 9 or the other way.
Help!  :spider:

x^3 + y^3 = 6z^3
this isn't very hard, factor the LHS, then prove that z^3 divides one of the factors


do u mean for the last one that it was split between professionals and farmers, not laborers?

[quote="vidyamanohar"]A study has determined that the occupation of a boy, as an adult, depends upon the
occupation of his father. If the father is a professional, 80% of son's became professionals
and the remaining 20% are split between farmers and laborers. If the father is a farmer, 50%
of son's became farmers, 30% became professionals, and 20% became laborers. If the father
was a laborer, then 60% of son's became laborers and the remaining 40% was split between
professionals and farmers.

In the long run, what proportion of the population will become farmers?[/quote]

Thank you.  I corrected the mistake.

[quote="cincodemayo5590"]I can see two methods of doing this. But first is this correct?:
[hide] $30%
$[/hide][/quote]

unfortunately, I am unable to see the solutions.

I am getting the following message

"Unparseable or potentially dangerous latex formula. Error 6"

A tip: If you can't see LaTeX code, then click Quote on that message - it will be repeated in the typing window, so you'll see what's in there.

[quote="cincodemayo5590"]I can see two methods of doing this. But first is this correct?:
[hide] $30%
$[/hide][/quote]

thanks for the hint by Farenhajt
I dont have the answer with me.  Pls provide the solution.

I don't think there's enough information here.  First we need to know the proportions we start with for the fathers's occupations...

actually i think that it is possible, the society would tend toward a certain percentage after an infinite amount of time in this case so that no matter what we start with we end up coming to what it should be.

[hide="hint"]  In an equilibrium position, suppose the percentages of the three types of people are $a$, $b$ and $c$.  Then, what would the percentages be in the next generation?[/hide]

[hide] ok so I got these equations:
p + f + L  = 100
20p= 30f +20L
50 f =10p +20L
40L = 10p + 20f

so 30 p = 80f
and 70f = 60 L

so 8/3f +f + 7/6f = 100

f = 600/29

ok so im not sure if thats right, i think i may have messed up somewhere[/hide]

Umm, f can't be greater than 1, because then there would be more farmers than total population...

[hide]I got $\frac{4}{15}$, but I'm not sure if it's right, because I ended up just assuming that they started divided evenly...[/hide]


trigonometry AMC AIME USA(J)MO USAMO geometry MATHCOUNTS \/closed

Yes, it's appropriate for the AMC, but it doesn't cover all of the AMC-12 material; you'll need Volume 2 for that (for example, trig and logarithmic manipulation are in Volume 2).

Silverfalcon, you should definitely buy AoPS volume 1 now.  It will fill in some of your gaps in knowledge (just to mention one, number bases).  And buy the solutions manual too.  You can buy volume 2 either now or later.

[quote="Rep123max"]Might as well buy both now. Isnt there a small discount to buy them both.[/quote]

Yes, it is slightly less expensive to buy both at once.

[quote="Silverfalcon"]How hard is AoPS Art of Problem Solving Book Volume 1 is?
Is that book appropriate for AMC?[/quote]

It's the only thing I used for AMC/AIME (and I haven't even completely finished AoPS II yet...in fact, I'm not even halfway thru by chapter count)  :oops:

...and it's gotten me into AIME 3 years now and USAMO this yr...

Volume 1 is too easy for USAMO.  There are parts of Volume 2 that would be useful - but for strict USAMO prep, you'd probably be better off with Zeitz or Engel.

Volume 2 would be a great thing to use prepare - it has lots of IMO problems, which are good things to use to prepare for the USAMO (in my opinion).

I'm curious...I just found that my brother has a copy of AoSP and well it's probabaly pretty old....it was the second edition from the year 1995 and well the order form on the back has prices that are cheaper than the order form that this site has...so is it possible to still use the order form in the book of the book for cheaper prices?...and the book my brother has doesnt have the solution manuel so if i buy just the solution now would it be different than the book that i have?

In response the question about prices, those prices are no longer valid.  You can go to the Books Order page (click order on the sidebar) for the current prices.

However, the current solution manual should match the book nearly exactly - there have been almost no changes in the book since the second edition, just fixing of typos and a few little corrections.

Can motivated but (temporarily, I hope) not strong 11 year old succeed in self studing the AOPS books? Is the first book starts completely from first principles?
(for instance, does one needs to know what a variable is?)

I would really like to hear your opinions, tnx!

My 10-year daughter is working with me through (slowly) all the exercises and problems of Volume 1.  She would not be able to self-study the book -- she needs me for assistance when stuck.

It would help if a student has had some exposure to algebra before starting Volume 1.  My daughter had seen some algebra before, such as solving a linear equation, so that wasn't a problem.

I think Ravi B is right.  Volume 1 could be self study for a student that already understands the basic principles of algebra and geometry, but a student without that background would need to work with someone more knowledgeable.

You might just try and see which chapters you can follow.  You can always ask questions in these forums.

A sequel? Then it should be pretty close to the Olympiad class level.
Maybe an administrator can tell you...

Art and Craft of Problem Solving is a great bridge from medium AIME level math up to a great deal of Olympiad level math.

[quote="Karth"]What is the best way to prepare for the AIME? Is there a course you can take or a book you can buy? I tried doing an AIME once and only got 2 out of 15  :([/quote]

The Art of Problem Solving Volume 2 is the book that we sell that is very in-line with AIME level math.  Paul Zeitz's Art and Craft of Problem Solving is a little more advanced -- good once you can get half or more of the problems.

Our introductory level subject books (the Counting/Probability book is already available) take students up to a level of around half the AIME problems.

We host classes here, including an AIME Problem Series, that help students learn to apply the math they know plus more advanced curriculum to difficult problems such as those on the AIME.

[quote="Karth"]So, erm if we read all those books, will we be almost guaranteed to get a really good score on the AIME?  Like maybe 10+? And how would you recommend reading the books? Like for example, I think that the AoPs Volume 1 is pretty simple, but Volume 2 has stuff I don't really know... and which books would help you get past the USAMO?  :D[/quote]

Nobody is guaranteed a good score by reading a book.  You get the most out of the books by working as many of problems as you can on your own and thinking for a long time about the harder ones.

Once you're at the level at which you can work a lot of the hardest problems in AoPS 2 and ACoPS, there is a book called Problem Solving Strategies that can help with USAMO students.

There are other book recommendations in our Resources guide.

[quote="Karth"]Is qualifying for the AIME, USAMO, etc. only possible in high school (9 - 12th grade)? Or is it possible to do so in college? (Considering it's the AMC [i]12[/i])...[/quote]

Only high school.

ha.. I just read this topic and realized it's 2 years ago! :lol:

It's amazing how in the beginning of this thread that I was preparing to buy AoPS Vol. 1 and now I'm thinking of buying Art and Craft of Problem Solving.

Speaking of that, will I benefit from buying Art and Craft of Problem Solving by Zeitz?

[quote="Karth"]Around what grade would a typical person (well, us, I guess...) go and get the Art and Craft of Problem Solving book? Like for example, I'm in tenth grade, and find AoPS 1 to be very easy and AoPS 2 to be a bit challenging...[/quote]

It's not a matter of grade level, but rather a matter of your experience.  When you can solve 1/3 - 2/3 of the problems on the AIME and you're getting into proofs -- that's the level when the Zeitz book will help you most.

[quote="cooljoe"]Which is better: an AoPS course or the AoPS books? Also, is it imperative that one buy Intro to Counting and Probability if you get a 2 on the AIME?[/quote]

I'll note that the surveys for the Intro C/P course were the best we've ever had for any course -- and it was the first time the course has been taught with the book.

Our intro books are written with students who are trying to qualify for the AIME in mind, so yes, we believe they will help.


trigonometry
vector
calculus
derivative
geometry
calculus computations

I am told to solve this:

$f'(x)=0$

when:

$\cos(x)-2\sin(x)$; $xe[0;2\pi]$

So first of i derivate it:

$f'(x)=-\sin(x)-2\cos(x)$

This obviously is right. But how do i then solve this:

$-sin(x)-2cos(x)=0$

$\ sin^2(x)+cos^2(x)=1$ so you have to solve the sistem:
$\ a^2+b^2=1$
$\ a+2b=0$
Then $\ b^2=\frac15$ and now it's clear .

Another way:

$-\sin x-2\cos x=0$
$\sin x = -2\cos x$
$\frac{\sin x}{\cos x} = -2$
$\tan x = -2$
$x = \mathrm{arctan}\;(-2)$

[quote="Sporally"]OK, that i don't get honorstly. I know that $\sin^2(x)+\cos^2(x)=1$

So when we have $-\sin(x)-2\cos(x)=0$ we can say it like $-a-2b=0$. But how do you go on from there? That i don't find that obvious...[/quote]
$\ a^2=4b^2$ so $\ 5b^2=1$

Honorstly, i don't get your point Thor. Sorry. Maybe something i haven't learned, but i don't see how you get $a^2=4b^2$.

[u]@ TheBB[/u]
That's true. Didn't think that far, but that is a law i know...

Thx for helping both of you..

from $f'(x)=0\Longleftrightarrow (2,1)\cdot (\cos x,\ \sin x)=0,$ the vector $(\cos x,\ \sin x)$ is an unit vector which is perpendicular to the vector $(2,1).$ Thus $(\cos x,\ \sin x)=\pm \frac{1}{\sqrt{5}}(1,-2).$

Oh my, now i feel even more confused. Not your fault, i just need to stick with things like $\tan(x)=\frac{\sin(x)}{\cos(x)}$. That i know what is :)

Anyway, i have yet another task. I don't know if it is because of the current time of day here in Denmark, but this one isn't too easy for me right now.

$\tan(x)=\sin(x)+0.5$

Can $x$ be isolated in an almost as easy way as we got $x$ isolated in the other task?

Can someone plz help me with this last one?

$\tan(x)=\sin(x)+0.5$

[u]Sporally[/u], you know the calculus of derivatives but you don't know to solve a very  easy trigonometrical equation: $\sin x+2\cos x=0\ !$ Sorry, in my opinion, here is a contradiction.

[quote="Virgil Nicula"][u]Sporally[/u], you know the calculus of derivatives[/quote]
There might be something wrong in what you said..... :|

I have now worked with derivatives for a long time and i think i know how it works now. Might not be the best in this area, but it is getting better all the time. The reason i can't do this equation is that i've never been too good working with $\sin$, $\cos$ and $\tan$. I just looked back in the textbook but didn't find an example where both $\tan$ and $\sin$ was in. If it was $\tan(x)=0.5$, then i wouldn't have any problems with it. But can't recall what to do when you have two of them in one equation. If you can help me on this one i won't ask you again about one of these problems.

[hide="Only for Spix."][quote="spix"] There might be something wrong in what you said..... :|[/quote]
No, nothing wrong ! I remember some words (from my math-teacher): in gymnasium, we learn to step; in lycee, we learn to swim; in university, we learn to take one's flight ! My dear Sporally meets with difficulties in the gait. Please Spix, help him, if you can. Give him some addresses where he can find the answers for his  problems. I don't know. In this section don't exist ! There are and another sections here, on mathlinks. Thanks.[/hide]

No spamming, i just like to write the last post in every thread i've started so that i know i have read all the posts... But don't worry, i have solved this problem now and lately my understanding of $\sin$, $\cos$ and $\tan$ has increased though they might still give me problems from time to time.

Solve $\tan x=\sin x+0.5$

Let $\cos x=a,\ \sin x=b,$ we have $\frac{b}{a}=b+\frac{1}{2}\ (b\neq 0) \ \Longleftrightarrow 2b(1-a)=a.$ Since $a\neq 1,$ yielding $b=\frac{a}{2(1-a)}.$ Substitute this for $a^{2}+b^{2}=1,$ and arranging we get $4a^{4}-8a^{3}+a^{2}+8a-4=0.$ since $a\neq 0,$ dividing both sides by $a^{2},$ we have $4\left(a^{2}-\frac{1}{a^{2}}\right)-8\left(a-\frac{1}{a}\right)+1=0.$ Set $a-\frac{1}{a}=t,$ yielding $\pm 4t\sqrt{t^{2}+4}-8t+1=0\Longleftrightarrow \pm 4t\sqrt{t^{2}+4}=8t-1.$ Squaring both sides, we have $16t^{4}+16t-1=0.$ Let $2t=u,$ we obtain $u^{4}+8u-1=0.$
Regrettably we can't solve this equation by hand.

[u]@ Kunny[/u]
How do you go from $4\left(a^{2}-\frac{1}{a^{2}}\right)$ to $\pm4t\sqrt{t^{2}+4}$ by claiming that $t=a-\frac{1}{a}$?

$x\not\in \frac{\pi}{2}Z\ ,\ \boxed{\tan \frac{x}{2}=t}\Longrightarrow \tan x=\frac{2t}{1-t^{2}}\ ,\ \sin x=\frac{2t}{1+t^{2}}\ .$
$tan x=\sin x+\frac{1}{2}\Longleftrightarrow \frac{2t}{1-t^{2}}=\frac{2t}{1+t^{2}}+\frac{1}{2}\Longleftrightarrow \boxed{t^{4}+8t^{3}-1=0}\ .$

[u]@ Kunny[/u]
Hold the phone. What rules allow you to claim that $\left(a+\frac{1}{a}\right)^{2}=\left(a-\frac{1}{a}\right)^{2}+4$?

[quote="Virgil Nicula"]1. Solve the following equation : $1+\tan x=2\sin 2x\ ,\ |x|<\frac{\pi}{2}\ .$
2. Why $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\ ?$[/quote]
The equation has the period $T=\pi\ !$

[b][u]The first method.[/u][/b] $\sin x+\cos x=2\sin 2x\cos x$ $\Longleftrightarrow$ $\sin x+\cos x=\sin 3x+\sin x$ $\Longleftrightarrow$
$\cos x=\cos \left(3x-\frac{\pi}{2}\right)$ $\Longleftrightarrow$ $4x-\frac{\pi}{2}\in 2\pi Z$ or $2x-\frac{\pi}{2}\in 2\pi Z$ $\Longleftrightarrow$ $\boxed{\ x\in S\equiv \left\{-\frac{3\pi}{8},\frac{\pi}{8},\frac{\pi}{4}\right\}\ }\ .$

[b][u]The second method.[/u][/b] $(1-\sin 2x)+\tan x=\sin 2x$ $\Longleftrightarrow$ $\cos x(\cos x-\sin x)^{2}+\sin x=\sin 2x\cos x$ $\Longleftrightarrow$$\cos x(\cos x-\sin x)^{2}=\sin x\left(2\cos^{2}x-1\right)$ $\Longleftrightarrow$ $\cos x(\cos x-\sin x)^{2}=\sin x\left(\cos^{2}x-\sin^{2}x\right)$ $\Longleftrightarrow$
$\cos x-\sin x=0$ or $\cos x(\cos x-\sin x)=\sin x(\cos x+\sin x)$ $\Longleftrightarrow$
$\tan x=1$ or $\cos^{2}x-\sin^{2}x=2\sin x\cos x$ $\Longleftrightarrow$ $\tan x=1$ or $\tan 2x=1$ $\Longleftrightarrow$
$x\in \frac{\pi}{4}+\pi Z$ or $x\in \frac{\pi}{8}+\frac{\pi}{2}Z$ $\Longleftrightarrow$ $x\in S\ .$

[b][u]The third method.[/u][/b] ${\boxed{\tan x=t}\Longrightarrow}$ $1+t=\frac{4t}{1+t^{2}}$ $\Longleftrightarrow$ $t^{3}+t^{2}-3t+1=0$ $\Longleftrightarrow$
$(t-1)(t^{2}+2t-1)=0$ $\Longleftrightarrow$ $t\in \left\{1,-1+\sqrt2,-1-\sqrt 2\right\}$ $\Longleftrightarrow$ $x\in S\ .$
[b]Remark.[/b] $\tan \frac{\pi}{8}=\frac{\sin \frac{\pi}{4}}{1+\cos \frac{\pi}{4}}=\frac{\sqrt 2}{2+\sqrt 2}=\frac{1}{\sqrt 2+1}=-1+\sqrt 2\ .$

I THINK i've got it now. And this obviously also is a math rule you use. $A^{2}-B^{2}=(A-B)\cdot (A+B)$, right?


[img]http://epsilon.uscoders.net/ML_Problem2.png[/img]

Thank you.

If $ r \equal{} \frac ab \not\in G$, then for any $ n\in N$, $ \frac{a}{bn}\not\in G$.


function
calculus
integration
calculus computations

i know that
[tex] \sum_{i=1}^ni^2 = n(n+1)(2n+1)/6[/tex]

but how can we simplify this summation?:

[tex] \sum_{i=1}^nx_i^2 [/tex]

That's not a meaningful question, since $x_i$ is simply a very general name for a very general function of $i.$ Replace the summation signs by integral signs, and the question would look like this:

"
I know that $\int_0^bx^2\,dx=\frac{b^3}3$

but how can we simplify the integral $\int_0^b[f(x)]^2\,dx$ ?
"

As I said, not a meaningful question.


algebra
function
domain
inequalities
combinatorial geometry

Let $x;y$ be real numbers satisfying $|2x-y| \leq 3$ and $|x-3y| \leq 1$.Find max of:
$A=x^{2}+y^{2}+xy$

[hide="I suspect that..."] $A(x, y)$ is convex in both variables, and therefore takes on its extrema at the vertices of the bounding polygon that constitutes its domain.  I am not sure, though. [/hide]

[hide="Well, if my rather large guess is right..."] The vertices of the domain are

$(\pm 2, \pm 1), \left( \pm \frac{8}{5}, \pm \frac{1}{5}\right)$

We have $A = 7$ at the first two points and $A = 2.92$ at the others.  So I believe the maximum is $A = \boxed{7}$. [/hide]

http://www.ece.tamu.edu/~hpfister/courses/elen689-2007/convexfun.pdf

Page 61 (15th page of the PDF) has a corollary of Jensen's Inequality that seems to imply the hypothesized result, i.e. a convex function attains its maximum at one of the points in the convex hull of the domain. As for the proof, well I didn't read through it all.


Im currently on the "Chain Rule" section and i've been understanding everything well so far as far as deviation goes. But I kinda stumbled against an algebraic example, and im not sure how they got that answer.

H(X)= (X^2 + X + 1)^2 = (X^2 + X + 1)(X^2+X+1)

=X^4 + 2X^3 + 3X^2 + 2X + 1

Now i know how to calculate the deviation H'(X)= 4X^3 + 6X^2 + 6X + 2

But how do you get this answer. X^4 + 2X^3 + 3X^2 + 2X + 1

Thanks.

thanks. i think my problem was not knowing how to add the exponents, meaning x^4 + x^4 = 2x^4.

appreciate the help.


If $ f: X \rightarrow Y$ is a map and if $ g$ and $ h$ are paths in $ X$ with $ g(1)\equal{}g(0)$, then $ f(g \cdot h)\equal{}fg \cdot fh$.

I was told that this follows directly from the definition of path composition. This is where $ f: X \rightarrow Y$ so the first is a path in $ Y$ and the second is a path in $ Y$. Is this completely justified by this, or do I need to justify this more than what is above?

First of all it has to be $ g(1)\equal{}h(0)$ and not $ g(0)$ and second, $ f$ has to be continuous.
To prove the equality it suffices to write down the path parametrization of both paths.


For the numbers $ a$, $ b$, $ c$, $ d$, $ e$ define $ m$ to be the arithmetic mean of all five numbers; $ k$ to be the arithmetic mean of $ a$ and $ b$; $ l$ to be the arithmetic mean of $ c$, $ d$, and $ e$; and $ p$ to be the arithmetic mean of $ k$ and $ l$.  Then, no matter how $ a$, $ b$, $ c$, $ d$, and $ e$ are chosen, we shall always have:

$ \textbf{(A)}\ m \equal{} p \qquad \textbf{(B)}\ m \ge p \qquad \textbf{(C)}\ m > p \qquad \textbf{(D)}\ m < p \qquad \textbf{(E)}\ \text{none of these}$

[hide]$ m=\frac{a+b+c+d+e}{5}, k=\frac{a+b}{2}, l=\frac{c+d+e}{3}, p=\frac{3a+3b+2c+2d+2e}{12}$.
\begin{align*}60(m-p)&=60(\frac{a+b+c+d+e}{5})-60(\frac{3a+3b+2c+2d+2e}{12})\\&=12a-15a+12b-15b+12c-10c+12d-10d+12e-10e\\&=2c+2d+2e-3a-3b\end{align*}
We can choose certain numbers for $ a, b, c, d, e$ to make this positive, negative, or 0, whichever we wish.  So we cannot definitively say any of A, B, C or D, so the answer is $ \boxed{E}$.[/hide]


geometry
rectangle
calculus
integration
analytic geometry
rotation
derivative

Let $z_1, z_2, z_3, z_4$ be four complex numbers
Let $\Gamma = [ z_1, z_2, z_3, z_4, z_1 ]$ be a rectangle with center at $z = 0$ .
Show that
\[ \oint_{\Gamma} \frac{\mbox{dz}}{z} = \pm 2 \pi i  \]

I know very little complex analysis, but isn't this just an application of the residue theorem?

Also, what do you mean by the center of a triangle?

Well it seems more like a pre-residue problem that shall be solved by Cauchys formula (or it's even just a step before that).
At least by the integral formula we can assume that we have a circle around the origin.

(But one question: a triangle out of four points¿)

I said the residue theorem because $\frac{1}{z}$ behaves "badly" (at least, it seems to me) around $0$. And, yes, triangle out of four points... I didn't notice that

perfect_radio: What do you mean by the residue theorem ?
(Could you post a link or a statement ?).
ZetaX: I'm sorry for the "triangle stuff" it's just some bad habit...  :P
I'm just beginning complex analysis, and have seen a proof of the above statement.
Homever, it was not satysfying, nor detailed enough. Hence I posted my question here, hoping to see a rigorous proof...
I've already got through the Cauchy theorems...

I've posted residue theorem (but without exact definitions) in http://www.mathlinks.ro/Forum/viewtopic.php?p=317935#p317935 .
In http://mathworld.wolfram.com/ResidueTheorem.html it's shown that you need something like the above to even prove it (thus it shouldn't be used ;) ).

So, how do you exactly prove {x}'s statement?

I'd do it first for a circle $C$ (with radius $r=|z_1|=|z_2|=|z_3|=|z_4|$). That's a simple calculation.

Now $\int_\Gamma \frac1z\,dz-\int_C\frac1z\,dz=\int_{\Gamma_1}\frac1z\,dz+\int_{\Gamma_2}\frac1z\,dz +\int_{\Gamma_3}\frac1z\,dz=\int_{\Gamma_4}\frac1z\,dz$, where $\Gamma_i$ is the union of the segment from $z_i$ to $z_{i+1}$ and the arc outside it, traced counterclockwise. By Cauchy's theorem, each $\int_{\Gamma_i}\frac1z\,dz=0$.

It's also possible to do the integral directly as a line integral. Making the simplifying assumption that $\Gamma$ is oriented along the coordinate axes and that $z_1=a+ib$ is in the first quadrant (if these aren't true, just rotate), we have

$\int_\Gamma\frac1z\,dz$
$=\int_{-a}^a\frac1{x-ib}\,dx+\int_{-b}^b\frac1{a+iy}\,i\cdot dy+\int_a^{-a}\frac1{x+ib}\,dx+\int_b^{-b}\frac1{-a+iy}\,i\cdot dy$
$=\int_{-a}^a\frac1{x-ib}\,dx+i\int_{-b}^b\frac1{a+iy}\, dy-\int_{-a}^{a}\frac1{x+ib}\,dx-i\int_{-b}^{b}\frac1{-a+iy}\, dy$
$=\int_{-a}^a\frac1{x-ib}-\frac1{x+ib}\,dx+ i\int_{-b}^b\frac1{a+iy}-\frac1{-a+iy}\,dy$
$=\int_{-a}^a\frac{2ib}{x^2+b^2}\,dx+ i\int_{-b}^b\frac{-2a}{-a^2-y^2}\,dy$
The derivative of $\arctan\frac xc$ is $\frac{\frac1c}{\frac{x^2}{c^2}+1}=\frac{c}{x^2+c^2}$, so
$\int_\Gamma\frac1z\,dz=\left.2i\arctan\frac xb\right|_{x=-a}^{x=a}+ \left.2i\arctan\frac xa\right|_{x=-b}^{x=b}=4i\left(\arctan\frac ba+\arctan\frac ab\right)$
$=2\pi i$.

Why does the integral remain the same upon rotation?

Just check it yourself- $dz$ is multiplied by $e^{i\theta}$ and $\frac1z$ by $\frac1{e^{i\theta}}$.

Actually, it's invariant under multiplication by any constant.

Yes... of course. I'm dumb. I forgot that constants go to $dz$ as well   :blush:


The integer $9$ can be written as a sum of two consecutive integers: 9=4+5. Moreover it can be written as a sum of (more than one) consecutive positive integers in exactly two ways, namely 9=4+5= 2+3+4. Is there an integer which can be written as a sum of $1990$  consecutive integers and which can be written as a sum of (more than one) consecutive positive integers in exactly $1990$ ways?

Define $n$ such a integer, if it exists; then by assumption there exists a integer $m$ such that $n=\sum_{i=0}^{1989}{m+i}=5\cdot 199 (2m+1989)$, and in particular $n$ is odd. The other assumption says that $2n$ can be expressed as $(k+1)(2n+k)$ in exactly $1990$ ways: but if $2n=ab$ for some integers $1<a<b$ then the solution is unique $(n,k)=(\frac{1}{2}(b+1-a),a-1)$, i.e. there are exactly $\frac{1}{2}\sigma_0(2n)-1$ such factorization. It implied that $\sigma_0(2n)=3982$. But $2n=2\cdot 5^{e_2}\cdot 199^{e_3} \cdot \prod_{i=4}^{\omega}{p_i^{e_i}}$ with the usual notation, and $\sigma_0(2n)=2\prod_{i=1}^\omega{(e_i+1)}=11\cdot 181$. By force $\{e_1,e_2\}=\{10,180\}$, so that we have exactly two solutions: $n_1=5^{10}\cdot 199^{180}$ and $n_2=5^{180}\cdot 199^{10}$. []


Last time I checked, Mr. Crawford won the 1990 MATHCOUNTS national competition, while on the website with historical winners says that Mathew Crawford never won. Whats going on?

This thread also open for discussion of [url=http://www.mathcounts.org/webarticles/articlefiles/Historical_Winners.pdf]www.mathcounts.org/webarticles/articlefiles/Historical_Winners.pdf[/url], the pdf of the winners.

Did he win written, not countdown?

Maybe.

We need MCrawford to make sure.

Tong's All Time Greatest MAthcounts problems book says the 1990 champion was Brian Jenkings

He could've lost countdown both times and finished 2nd.

exactly. note that under about us on AoPS it says "1990 national TEST champion"


Use the identity
\[\displaystyle \tan^{-1}{\frac{1}{2r^2}}=\tan^{-1}{(2r+1)}-\tan^{-1}{(2r-1)}\]
Then the given sum telescopes to
\[\displaystyle \tan^{-1}{\infty}-\tan^{-1}{1}=\frac{\pi}{4}\]


I went to the North Springs tournament with my school today and I would like to strongly urge anyone who reads this to never attend one of their tournaments.

My list of problems with the tournament:

1. They never provided us with answers to the written test questions after the test - in fact, they didn't even know the answers. When we got to the tournament, one of the sponsors asked our coach to help them write the answer key because they hadn't finished it yet.

2. I spotted at least 10 errors in the 45 written and ciphering questions. The mistakes were so numerous that by the time I reached the end of the test, I didn't even expect my answer to be in the answer choices. The women in charge of the tournament was obviously responsible for most of them:
She thought (and could hardly be convinced otherwise) that $ \minus{}6<x<6$ and $ \minus{}6<y<6$ was the locus for a circle. She counted us wrong for writing $ \binom{10}{7}$ in place of $ \binom{10}{3}$, and when our coach explained to her why the two expressions were equivalent, she said that she would "look into it."
A ciphering problem asked us to find an equation with four roots with which we were provided. The correct answer to the question was $ y\equal{}25x^4\minus{}6x^2\plus{}2$. The answer that she called out was "zero." Now I may be wrong, but I don't think that "zero" is an equation. :rotfl:

3. There were about ten statistics question on the 35 question written test. A couple of statistics might be okay, but ten of them put students who have not taken statistics at a major disadvantage. There was not one question during the entire test that I found interesting. The questions were either way too easy or used statistics terms that I haven't learnt yet.

4. There are plenty more problems with the tournament that my team laughed at over lunch that don't come to mind right now.

I know that it is hard work writing a tournament, but I don't see why someone who doesn't know anything about math would write one. Just to try to lessen the harshness of my comment, I will comment on the two things that did please me: the snacks were numerous and were all ready for us between the ciphering and written test and the trophies and medals were very high quality (I got a fifth place medal and my team got a 3rd place trophy).

At least the UGA tournament next week should make up for this one.

Robert Grosse won individual and Chamblee won team.

[quote="Math Geek"]I agree with Twiz.  They graded like half of my questions wrong, and I should have placed at least six places higher than I did if good test writers made the questions.  Sure, they had free food, but they obviously didn't know what they were doing.
Also, I almost got completely owned on the stat questions.  I've never taken stat, and they were like the only reason I got actual questions wrong.[/quote]

Fortunately, I don't think many people had taken stat, so I don't think many people got too much of an unfair advantage. Another great ciphering question that I forgot to mention (wording may vary slightly):
"Three yellow marbles and nine other marbles are placed in a bag. If three marbles are drawn at random, what is the probability that all three of them are blue?" Um... I don't think that they mentioned anything about blue marbles...
Also, I don't think you should worry about how you placed. As far as I'm concerned the results of that tournament are meaningless.


Multiplying Probabilities Part 1

We find the probability that a series of independent events occurs by multiplying the probabilities that each event occurs.

He says...

"How do you get it wrong? You get it wrong, every time, just like congress."

lol

[hide="My epic guessing skills"]Wow I guessed:
Heart
Spade
Diamond
Club

The answers were:
Diamond
Spade
Diamond
Club

$\frac{3}{4}$ right![/hide]
[hide="Probability you get x right"]
[b]The probability you get 0 cards out of 4 right[/b], is $\frac{3}{4}*\frac{3}{4}*\frac{3}{4}*\frac{3}{4}*\binom{4}{0}\implies \frac{3}{4}^4*1\implies \frac{81}{256}$
[b]The probability you get 1 card right:[/b]  $\frac{3}{4}*\frac{3}{4}*\frac{3}{4}*\frac{1}{4}*\binom{4}{1}$ because their is $\binom{4}{1}$ ways for which card you get to be correct.
$\frac{3}{4}^3*\frac{1}{4}*\binom{4}{1}\implies \frac{3}{4}^3*\frac{1}{4}*4\implies \frac{3}{4}^3*\cancel{\frac{1}{4}*\binom{4}{1}}\implies \frac{27}{64}$
[b] The probability you get 2 cards right: [/b]  $\frac{2}{3}*\frac{2}{3}*\frac{1}{3}*\frac{1}{3}*\binom{4}{2}\implies \frac{4}{81}\frac{4*3}{2}\implies \frac{4}{81}*6\implies \frac{24}{81}\implies \frac{8}{27}$
[b] The probability that you get 3 cards right: [/b] $\frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\frac{2}{3}*\binom{4}{3}\implies \frac{2}{81}*\binom{4}{3}\implies \frac{2}{81}*4\implies \frac{8}{81}$
[b] The probability that you got all 4 cards correct: [/b] $\frac{1}{4}*\frac{1}{4}*\frac{1}{4}*\frac{1}{4}*\binom{4}{4}\implies \frac{1}{4}^4*1\implies \frac{1}{256}$ [/hide]


$tan{\alpha}=2tan{\beta}$
what can you say about the connection between $\alpha$ and $\beta$ ?

...
Why is this in Pre-Olmypiad?
Can't you just take out t and g if neither is 0, and then you have an answer?

I don't think t and g are variables here, i think he means $\tan{\alpha}$

say we have a right triangle, and $\alpha$ is an acute angle, then double the opposite side, connect this to opposite vertex, this forms $\beta$...this is too general, also i suppose you can say that $|\beta| \ge |\alpha|$

[quote="Kurt Gödel"]I'm not sure of what you really want, but euhm...what about this observation?

[hide]
As $\tan\alpha=2\tan\beta$, we know that $\frac{\sin\alpha}{\cos\alpha}=\frac{2\sin\beta}{\cos\beta}$. Therefore, $\sin\alpha\cos\beta = 2\sin\beta\cos\alpha$. Thanks to Simpson, we now know that $\frac{1}{2}\cdot\left(\sin(\alpha+\beta)+\sin(\alpha-\beta)\right) = \sin(\alpha+\beta) + \sin(\beta-\alpha) = \sin(\alpha+\beta) - \sin(\alpha-\beta)$.
Finally, this results is $3\sin(\alpha-\beta) = \sin(\alpha+\beta)$...  :huh:
[/hide][/quote]

I have doubts about this.  From $\sin \alpha \cos \beta = 2 \sin \beta \cos \alpha$ we can add $\sin \beta \cos \alpha$ to both sides and get $\sin (\alpha + \beta) = 3 \sin \beta \cos \alpha$  :huh:

I don't think a general relationship of such kind exists between $\alpha$ and $\beta$.

[quote="pavel25"]sorry if it was not clear! I meant if $\alpha=k\beta$ so find $k$ ![/quote]

So $\tan k \beta = 2 \tan \beta$?

This seems to be true only for particular pairs of numbers $(k, \beta)$, I do not think there exists such a $k$ independent of $\beta$  :huh:

[quote="pavel25"]$tg\alpha=2tg\beta$
[/quote]
This has got something to do with physics: The projectile motion.

Maybe there isn't relationship between $\alpha$ and $\beta$ , but why ?
I just gave an example , maybe the relationship isn't linear , if someone has an idea, please tell me!


How to prove the theorem, that if
$n\in\mathbb{Z}$ and $p$ is a prime number then
$p$ is a divisor of $n^p-n$

[url]http://www.om.edu.pl/zadania/om/om51_calosc_r.pdf[/url] page 80 (it's in Polish).

first we proof that $\rm p|{n \choose k}$  for all $1\leq k \leq {p-1}$
and deduce that $\rm p|[(n+1)^p - (n^p +1)]$ for all $n\in N$
and we proof that $n^p = n[p]$


MATHCOUNTS
geometry
analytic geometry
function
inequalities
probability
power of a point
\/closed

Hey,
I'm preparing for mathcounts, and was wondering,
I have the AOPS original Basic book,
this one here:
http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1)

and as I was looking through, I thought not all subjects were needed to prepare for mathcounts. I am planning to finish the book evantually, but I was wondering out of these topics, which ones are totally neccessary and which ones are optional if I'm studying mathcounts material?

Chapters:

Polygons,
Angle Chasing,
Areas,
The Power of Coordinates,
Power of a Point,
Three Dimentional Geometry,
Shifts, turns, stretches, and squeezes,
A potpouri of geometry,
Functions,
Inequalities,
Operations and Relations,
Sequences and Series,
Learning to COunt,
Statistics and Probability,
Sets,
Prove it,
Parting Shots.

and the ones before that, I already did so...
Thanks,  :)

-Kevin

Those topics all appear in MathCounts, more or less.  However, they go light on functions, inequalities, sets, and a few other things.  However, one my of my greatest regrets from MathCounts is not studying the entire Volume 1 book, as I can go back and point out problems on the State and National test that I could have done, had I worked the AoPS book extensively.  So I would recommend doing the whole book, and it will definitely come in handy in the near future, if not for MathCounts.  But that's just my opinion...

[quote="MCrawford"]If you really want to do well at MATHCOUNTS, study it all.  The more you understand about math, the more you'll understand subtle connections between MATHCOUNTS problems and higher level topics, even though higher level math isn't necessary to solve the problems.

I have an anecdote about how coordinates got me out of a jam on a problem.  It was at the state level during my first year.  I couldn't figure out how to solve a target round problem without coordinates, but since I was able to solve the other of the pair of problems quickly, I had plenty of time to use coordinates.  I can't say whether or not that problem made a difference in rankings, but I doubt it's the only time that knowing a broader array of problem solving tools helped me when I didn't quickly see the nice solution.[/quote]
What was the problem?

[quote="bpms"]
What was the problem?[/quote]

That was over 16 years ago.  I remember very few of the actual problems from back then without seeing them again.  I just always remembered the fact that coordinates saved me a couple of times, but then choosing to try using them on a problem at nationals might have cost me a chance to solve that problem.

[quote="mathgeniuse^ln(x)"]Yes, AOPS Books will definitely help you solve problems.

also you can order past MathCounts Books from http://www.mathcounts.org

AOPS volume 1 helps extensively for middle school, especially National MathCounts.

AOPS volume 2 helps extensively for High School, especially AMC and AIME.

All of this is my opinion.[/quote]

I'll add that AoPSers think of AoPS Volume 2 as the book most helpful on the AMC exams, but that's because most AoPSers are working on the last 10 problems.  Most of the material on the first 12-15 problems on the AMC 12 and 20 or so problems on the AMC 10 is covered in AoPS Volume 1.


geometry
circumcircle
geometric transformation
reflection
parallelogram
trigonometry
ratio

Let in triangle ABC, D be the midpoint of BC and P be a point on AD such that
$ \angle BAC \plus{} \angle BPC \equal{} 180^0$
Prove that $ AC.BP\equal{}AB.CP$

Let $ Q$ be the intersection of $ AD$ and the circumcircle of $ ABC$. Then $ PCQB$ is a parallelogram. The conclusion is equivalent to $ \text{Area}(ABQ) \equal{} \text{Area}(ACQ)$.

[quote="Ahiles"][img]http://pic.ipicture.ru/uploads/081022/56DCC3KFtF.png[/img]
[color=darkblue]
Let $ A'$ be reflection of $ A$ w.r.t. $ D$, Then quadrilateral $ ABA'C$ is a paralelogram, so

$ A'B = AC$     (1)
$ A'C = AB$     (2)
$ \angle{BA'C}\equiv\angle{A}$

But from the condition $ m(\angle{A) + m(\angle{BPC}) = 180^\circ}$ we get

$ m(\angle{BA'C) + m(\angle{BPC}) = 180^\circ}$

Therefore quadrilateral $ PBA'C$ is cyclic, so $ m(\angle{PBA') + m(\angle{BCA'}) = 180^\circ}$ and

$ \sin{PBA'} = \sin{PCA'}$     (3)

Becuase $ D$ is midpoint of the segment $ BC$ we get

$ [PBD] = [PCD]$
$ [BDA'] = [CDA']$

By adding them we get

$ [PBA'] = [PCA']$

or

$ \frac {1}{2}BP\cdot BA' \sin{PBA'} = \frac {1}{2}PC\cdot CA'\sin{PCA'}$

Also from relations (1), (2), (3) we get

$ AC\cdot BP = AB \cdot CP$[/color][/quote]

Thanks so much!  :)

it can be done by this
$ (ABD)\equal{}(ADC)$ and $ (BPD)\equal{}(CPD)$.....now find the areas of these triangles and take the ratio and then use sine rule for these 4 triangles ..you will get your desired result


induction
LaTeX
number theory proposed
number theory

[quote="fermat3"]but you think that you ara proved this:[ 1) n has 2007 divisors} :blush:[/quote]
Just think a little about Rust's post :wink: .

[quote="fermat3"]hy ! my problem is: prove that they are infinitly n natural number:1)n has 2007 deferrents divisor. 2) n/ 2006^n -1 .[/quote]

for the first part, i only have $n=p_{1}^{8}p_{2}^{222}, p_{1}^{2}p_{2}^{2}p_{3}^{222}$ or $p_{1}^{8}p_{2}^{668}$. with prime $p_{i}$. :P

you can use the lemma: if p/a-b then p^k+1 /a^x -b^x where x=p^k.by the way there is a generalistion for this problem who was a problem in imo 2002 where a=2 and b=1 :ninja:

do you mean if $p|(a-b)$ then $p^{k+1}|(a^{x}-b^{x})$? if you did i understand it. i thought that you meant $p^{k}+1|(a^{x}-b^{x})$ :)

no.let prove this lemma using induction.we have p/a-b; assume that p^(k+1)/a^x-b^x and lets try  to prove that p^(k+2)/a^y-b^y (y=p^k+1) .pose a^x=m and n=b^x => a^y-b^y=m^p-n^p=(m-n)A.and we can easly prove that p/A. :roll:

fermat3, it's time to learn $\LaTeX$...
By the way just a special case of Hensels Lemma.

by the way  i will post my last post without latex.and this problem is didicated to you: find all natural number such that  a^m +1/(1+b^k)^n :lol:

All pentupels $(a,b,k,m,n)$ with $a^{m}+1 | (1+b^{k})^{n}$ or with $a^{m+1}| (1+b^{k})^{n}$¿ This is the reason why you should use $\LaTeX$...
For the first, use Zsygmoni's theorem, the second is only factor-chasing.

this problem is the generalistion for a short listed problem 2000. you can tell me zsygmoni theorem :maybe:


LaTeX
parameterization
combinatorics unsolved
combinatorics

Hi, all when I was a student I was more intelligent than now... I've invented the following problem I knew I solved it by some use of calculus. But i think there is an easier solution. The problem is the following:

///Prove that for all k in Z: (0<=k<n-i) and n in N is true:
     n
   SUM((-1)^i.C(i,n).V(k,n-i))=0
    i=0
/// it were wrong.
Corrected version of the problem is:

Prove that for all k in Z: (0<=k<n) and n in N is true:
    n-k
   SUM((-1)^i.C(i,n).V(k,n-i))=0
    i=0

What is $ V(k, n\minus{}i)$, and could you use LaTeX, please?

V(k,n) is n!/(n-k)!
C(k,n) is n!/(k!.(n-k)!)

I've never used Latex in my life, but i think it is not difficult - tomorrow I'll spend some time of learning Latex. And I'll use it - I promise.

[quote="borislav_mirchev"]Prove that for all $ k\in\bf Z$ and $ n\in\bf N$,  $ 0\leq k < n\minus{}i$, we have
\[ \sum_{i\equal{}0}^{n}(\minus{}1)^{i}\binom{n}{i}\frac{(n\minus{}i)!}{(n\minus{}i\minus{}k)!}\equal{}0\]
[/quote]

But actually some things don't make sense.  $ i$ is not some fixed value but rather a parameter over which we sum; what does it mean to say that $ k < n\minus{}i$?  Perhaps you simply mean the upper limit of the summation to be $ n\minus{}k$?

==Edit==
Corrected version (see following post):
[quote="borislav_mirchev"]Prove that for all $ k\in\bf Z$ and $ n\in\bf N$,  $ 0\leq k<n$, we have
\[ \sum_{i\equal{}0}^{n\minus{}k}(\minus{}1)^{i}\binom{n}{i}\frac{(n\minus{}i)!}{(n\minus{}i\minus{}k)!}\equal{}0\]
[/quote]

Thank you for the question. The correct version of the problem should be:

Prove that for all k in Z: (0<=k<n) and n in N is true:
n-k
SUM((-1)^i.C(i,n).V(k,n-i))=0
i=0

P.S. Please if you know Latex rewrite the problem in Latex.

I have added the corrected version above.

[hide="Solution"]\[ (1\plus{}x\plus{}y)^{n}\equal{}\sum_{\substack{i\plus{}k\leq n\\ 0\leq i, k}}\frac{n! }{i!k!(n\minus{}i\minus{}k)!}x^{i}y^{k}.\]  Thus \[ y^{n}\equal{}\sum_{\substack{i\plus{}k\leq n\\ 0\leq i, k}}(\minus{}1)^{i}\frac{n! }{i!k!(n\minus{}i\minus{}k)!}y^{k}.\] Comparing coefficients on both sides, we have that for $ 0\leq k\leq n\minus{}1$, \[ \sum_{0\leq i\leq n\minus{}k}(\minus{}1)^{i}\frac{n!}{i!k!(n\minus{}i\minus{}k)!}\equal{} 0.\]  Multiplying by $ k!$ gives the desired result.[/hide]

I've a question. Have you ever seen this problem before? What do you think about it? Is it difficult or is it easy? May it be given at Mathematical Olympiad?

I don't recall seeing this particular problem before, but I suspect someone somewhere has noticed it sometime in the past.  This problem is easy from my current point of view, which is that of someone just beginning graduate studies in algebraic combinatorics -- it only took me a few (maybe 10?) minutes to figure out how to manipulate it to spit out something nice.  I would have found it much less easy (although perhaps still tractable) when I was in high school.  For the last question, I think it would be better to wait for an answer from someone who knows better than I the appropriate difficulty of olympiad questions -- I'm rather out of practice in that sense.

I think this wouldn't be a bad olympiad problem - not too hard and not too easy - but of course, if you want to use or propose it as an olympiad problem, you shouldn't post it here on Mathlinks.

Thanks for your opinion. I've lots of things to learn - combinatorics definitely isn't my strongest side :) And sometimes I reinvent some well known facts. (For example I've found common formulae for sin(A+B) before to see it in the textbook.

Arne, I know lots of people know mathlinks.ro but do you believe they know all the problems posted here? I may give you example from lots of problems that were given at national level math olympiads.
1. For example: 2^m+3^n=t^2 to be solved in integers is given in bulgaria and UK... I'm not sure for india.
2. The next one is given in poland 1998-1999, i think, and austria 1989:
If u,v,w>0 u+v+w=1 prove that: sqrt(12uvw)+u^2+v^2+w^2<=1.
I may continue the list to infinity, especially, for the problems from the indian math olympiad you may see lots of russian problems. Not used problems for IMO are used in many countries: Bulgaria, Iran, Poland, ....

Here's a more obvious solution:[hide]\begin{eqnarray*}\sum_{i=0}^{n-k}(-1)^{i}\binom{n}{i}\frac{(n-i)!}{(n-i-k)!}&=&\sum_{i=0}^{n-k}(-1)^{i}\frac{n!}{i!(n-i-k)!}\\ &=&\frac{n!}{(n-k)!}\sum_{i=0}^{n-k}(-1)^{i}\frac{(n-k)!}{i!(n-i-k)!}\\ &=&\frac{n!}{(n-k)!}\sum_{i=0}^{n-k}(-1)^{i}\binom{n-k}{i}\end{eqnarray*}
and this last summation is well-known to be 0 if $ n-k > 0$.[/hide]

If it is your solution - I think it is more than excellent! I may only dream - when I grow up I'll be smart like you ;)


algebra
polynomial
function
complex analysis
complex analysis unsolved
genus

If  f has finite genus $h$, what is the possible range for the genus of $f(z^{2})$?

Let k be the genus of the canonical product associated to the zeros of $f(z)$. Clearly, the genus of the canonical product associated to the zeros of $f(z^{2})$ is $2k$ or $2k+1$, but how do I determine the degree of the polynomial $g$ in the factor $e^{g}$, or even show that $g$ is a polynomial?


Find $ x \in R$ of this equation: $ 5x^4 \minus{} 10x^3 \plus{} 10x^2 \minus{} 5x \plus{} 1 \equal{} 0$.

Use Maple I have a solution:

Equation $ \Longleftrightarrow \left ( x^2 \minus{} x \plus{} \frac {5 \minus{} \sqrt {5}}{10} \right )$$ \left ( x^2 \minus{} x \plus{} \frac {5 \plus{} \sqrt {5}}{10} \right ) \equal{} 0$

$ \Longrightarrow$ Equation have'nt get any root $ x \in R$.

Can you give me another solution?

hello, this may be helpfull
http://arxiv.org/abs/math.HO/0310449
Sonnhard.

It's easy

$ (1\minus{}x)^5\equal{}\ldots$

hello, but $ (1\minus{}x)^5\equal{}1\minus{}5x\plus{}10x^2\minus{}10x^3\plus{}5x^4\minus{}x^5$, this will not help.
Sonnhard.

Of course it helps!  If $ (1 \minus{} x)^5 \equal{} (\minus{}x)^5$ then what can you say about $ \frac{x \minus{} 1}{x}$?

Adding to the both sides of this equation $ \minus{}x^5$ we get $ (1\minus{}x)^5\equal{}\minus{}x^5 \Rightarrow 1\equal{}0$ contradict. So there is no real solution.

[quote="micf"]Adding to the both sides of this equation $ \minus{} x^5$ we get $ (1 \minus{} x)^5 \equal{} \minus{} x^5 \Rightarrow 1 \equal{} 0$ contradict. So there is no real solution.[/quote]

It's a nice solution. Thank [u][i][b]micf[/b][/i][/u] very much.


abstract algebra
superior algebra
superior algebra unsolved

What is the decomposition of $\mathbb{Z}_{18}\oplus \mathbb{Z}_{24}\oplus \mathbb{Z}_{30}\oplus \mathbb{Z}_{45}$ into p-primary modules?

Hi.

The task is to find the Sylowgroups, since the given group is finite. So the 2-Sylow (=2-primary) is:
$9\mathbb{Z}/18\mathbb{Z}\oplus 3\mathbb{Z}/24\mathbb{Z}\oplus 15\mathbb{Z}/30\mathbb{Z}\oplus 0/45\mathbb{Z}$

The 3-Sylow is:
$2\mathbb{Z}/18\mathbb{Z}\oplus 8\mathbb{Z}/24\mathbb{Z}\oplus 10\mathbb{Z}/30\mathbb{Z}\oplus 5/45\mathbb{Z}$

The 5-Sylow is:
$0/18\mathbb{Z}\oplus 0\mathbb{Z}/24\mathbb{Z}\oplus 6\mathbb{Z}/30\mathbb{Z}\oplus 9/45\mathbb{Z}$

All other Sylow-groups are trivial.


induction
modular arithmetic
number theory proposed
number theory

letp be a prime number and $a_1,a_2,...,a_p+1$be real number. The set ${a_1,a_2,...,a_p+1}$ has the property yhat no matter what elemen of the set we leave out,the remaining p elemens can be partitioned into several disjoint subset such that the mean of elemens in all these subset the same. Prove that $a_1=a_2=...=a_n$.

First, a lemma: if two disjoint sets of numbers have the same mean, then their union also has that mean. Proof: let $x_1+\ldots+x_a=am, y_1+\ldots+y_b=bm$. Then their sum is $am+bm=(a+b)m$. Dividing by $a+b$ we get the mean $m$. By induction this holds for the union of any number of sets.

This implies that the given condition is equivalent to the condition that, for any $p$ numbers, there's a non-empty subset of them with the same mean as the set of $p$: indeed, if there's a partition $A_1, \ldots, A_k$, then their union (the set of $p$) has the same mean as any of them, say $A_1$. On the other hand, if there's a subset $A$ with the same mean as the set of $p$ $B$, then $B-A$ has that same mean and we have a partition.

Another useful observation is that, if a set of numbers works, we can multiply them all by a constant, or add a constant to all of them, and they should keep working.

Now we'll prove the result for $a_1, \ldots, a_{p+1}$ integers. Let $a_1+\ldots+a_{p+1}=S$. Then $S$ is an integer. Now, take $a_1, \ldots, a_p$. The sum of these numbers is $S-a_{p+1}$. If there's a non-empty subset with the same mean, with $k$ elements for some $0<k<p$, then the sum of the elements of this set is $\frac{k(S-a_{p+1})}{p}$ which must be an integer, so $p \mid S-a_{p+1}$. Conversely $p \mid S-a_i$ for all $i$, so $S \equiv a_1 \equiv \ldots \equiv a_{p+1} \pmod{p}$. Let this common residue be $r$. Then take the integers $\frac{a_i-r}{p}$, which also work. Let $D$ be the difference between the largest and the smallest numbers in the set. Then the new set has $D_2=\frac{D}{p}$. Eventually we must come to a $-p<D_i<p$ so that $D_i=0$ and then $D=0$. So all the numbers are equal, and we're done.

Generalizing it to rationals is immediate (just multiply by the product of the denominators). To generalize it to reals, you can use Kronecker's lemma, but that is a rather cumbersome argument that I won't type out here.


algebra
polynomial
search
limit
real analysis
real analysis unsolved

$\alpha \in \mathbb{R}$ and $\alpha_{2},\alpha_{3}, .., \alpha_{r}$ its algebraic conjugates (i.e. the other
roots of its minimal polynomial). $\alpha$ is called a Pisot number if:
$\mid \alpha\mid ,\mid \alpha_{2}\mid ,..,\mid \alpha_{r}\mid <1$
Is the set of Pisot number closed.

maybe i am not thinking properly,but what if one considers the sequence ${1-\frac{1}{n}}$,as they r rational they do'nt have any conjugates other than themselves and $\mid..\mid < 1$,but they tend to $1$ which is not a Pisot number  :maybe:

:blush: Can somebody tell what is the defination of algebraic conjugates?

[quote="Enigma1984"]usually if $\alpha \in \mathbb{C}$,then if it satisfies a polynomial over $\mathbb{Q}$ we call it algebraic and then there is something called a minimal polynomial for $\alpha$ which is the polynomial with minimum degree in rationals which satisfies $\alpha$ and all its roots are called conjugates of $\alpha$.[/quote]
Thank you very much.But it may too hard to me.
Anyway I think your counterexample is right.  :)

Actually, looking up the definition of a Pisot number with a search engine will reveal that they are actually required to be algebraic [i]integers[/i], so your example does not work, Enigma.

<just_to_revive_this_topic>

In other words, what we want to prove is :
Let $(a_{n})_{n \geq 0}$ be a converging sequence of Pisot numbers and let $l = \lim_{n \to \infty}a_{n}$, then $l$ is a Pisot number.

</just_to_revive_this_topic>

According to MathWorld, Salem (1944) proved that the set of Pisot numbers is closed.  See the end of
[url]http://mathworld.wolfram.com/PisotNumber.html[/url].

Thank you for this reference [b]Ravi B[/b].
Now let's try to reconstruct the whole proof..

Also, another related question I have is :
let's denote by $P$ the set of Pisot numbers and by $S$ the set of Salem numbers, is it true that $P \cup S$ is closed?

[quote="mathmanman"]Thank you for this reference [b]Ravi B[/b].
Now let's try to reconstruct the whole proof.. [/quote]

The proof is really *not* obvious, it's easier to look it up ...

[quote]Also, another related question I have is :
let's denote by $P$ the set of Pisot numbers and by $S$ the set of Salem numbers, is it true that $P \cup S$ is closed?[/quote]

This is an open question.

[quote="{x}"]
The proof is really *not* obvious, it's easier to look it up ...
[/quote]
Ok.. do you know if I can find it online?

[quote="{x}"]
This is an open question.[/quote]
Ah.. thanks anyway, I didn't know this.  :oops:


A well-known method to memorize things is to create a sentence using the first letter of each word. For example, say I want to remember the four things to write in the opening paragraph of a typical essay: Hook, Information, Question, Thesis statement, I create a sentence Hank Is Quite Talented.

What's the name of this method? How does it work for you? And is there any computer program/website that can generate sentences so that you don't have to come up with one by yourself?

in my opinion , this memorising method is very useful iff you create the sentences yourself . I used to use this method in school so I feel that I memorize better if I create the sentences myself .Of course , if you can create a very " silly " or " funny " sentence which maybe making no sense , then it would be more effective .

So maybe for example the

Hook , Infomation , Question ,Thesis

==> Q HInT  ( sounds like Cute Hint  :P )

SOHCAHTOA: Some Old Heavy Cows Are Heavier Than Others Are

Worked for me.

whoa

isn't sohcahtoa itself an acronym for the trig functinos?

I learned sincostan as Oscar Had A Hold On Arthur.

[quote="shyong"]in my opinion , this memorising method is very useful iff you create the sentences yourself . [/quote]

That's an interesting point!  :)

Anyway, I would still be interested in such a program/website.

Moreover , i think that the sentences do not nessecary in ALL english . I usually mix with other language when trying to create funny sentence . i.e in my country we have english , chinese , malay , hokkien .... so using many different type of languages also create uniqueness so that you wont "lack of idea" :)


See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url] 10


define

$ \textbf S(x,y)\;\equal{}\;\sum_{k\equal{}1}^\infty\;\frac{1}{k^x\cdot(e\,^{y\,k}\minus{}1)}$

show

$ \text{(i)}\qquad 2\,\textbf S(3,2\pi)\;\equal{}\;\frac{7\pi^3}{180}\minus{}\zeta(3)$

and then show

$ \text{(ii)}\qquad 4\,\textbf S(3,\pi)\minus{}5\,\textbf S(3,2\pi)\plus{}\textbf S(3,4\pi)\;\equal{}\;\frac{\pi^3}{180}$

Use this: $ \frac {2}{k^x\cdot(e\,^{y\,k} \minus{} 1)} \plus{} \frac {1}{k^x} \equal{} \frac {1}{k^x}\cdot \frac {e^{yk} \plus{} 1}{e^{yk} \minus{} 1} \equal{} \frac {1}{k^x}\coth \left(\frac {yk}{2}\right)$
So: $ \textbf S(x,y) \equal{} \frac {1}{2}\sum_{k \equal{} 1}^{\infty}{\frac {1}{k^x}\coth \left(\frac {yk}{2}\right)} \minus{} \frac {\zeta (x)}{2}$
Now look at this [url]http://www.mathlinks.ro/viewtopic.php?t=183059[/url]
So, this problems has the same method of solution.
Note that $ \coth (u) \equal{} \frac {1}{u} \plus{} 2u\sum_{k \equal{} 1}^{\infty}\frac {1}{k^2 \pi^2 \plus{} u^2}.$ Maybe we should use this.

To your condition $ (ii):$ using the same method we can get that $ \textbf{S}(3, \pi) \equal{} \frac {\pi^3}{45} \minus{} \frac {\zeta (3)}{2}$ and $ \textbf{S}(3, 4\pi) \equal{} \frac {\pi^3}{72} \minus{} \frac {\zeta (3)}{2}$. Using this we immidiately get your result.


linear algebra
matrix
linear algebra theorems

Everyone knows the inverse of the partitioned matrix

$ \left[\begin{array}{cc} A & B \\
C & D\end{array}\right]^{ \minus{} 1} \equal{} \left[\begin{array}{cc} A^{ \minus{} 1} \plus{} A^{ \minus{} 1}BNCA^{ \minus{} 1} & \minus{} A^{ \minus{} 1}BN \\
\minus{} NCA^{ \minus{} 1} & N\end{array}\right]$
where $ N \equal{} (D \minus{} CA^{ \minus{} 1}B)^{ \minus{} 1}$ and $ A\in{\mathbb R}^{m\times m}, B\in {\mathbb R}^{n\times n}$. However do we have a formula if $ A,D$ are not square matrices but only $ B\in {\mathbb R}^{m\times m}$ and $ C\in {\mathbb R}^{n\times n}$ are square matrices   ??  :oops:

To my shame, I did not know that formula. But I know that multiplication on the right by the block matrix $ \begin{pmatrix}0&1\\1&0\end{pmatrix}$ reduces one problem to the other.