floor function
logarithms
inequalities
algebra
polynomial
number theory
relatively prime

Sorry to ask probably a stupid (or very difficult) question: suppose the set of prime factors dividing at least one of the numbers $[a], [a^2], [a^3],...$ is finite. Does it follow that $a$ is integer?

Just to be sure:
You are asking whetever $| \{ p \in \mathbb{P} : p| \lfloor a^n \rfloor, \forall n \in N \} | < + \infty \Rightarrow a \in Z$ holds, right ? (by $\mathbb{P}$ I mean the set of all primes).
By the way, isn't this somehow related to : http://www.mathlinks.ro/Forum/viewtopic.php?t=59040 ;)

No, he askes for $| \{ p \in \mathbb{P} | \exists n \in \mathbb{N}: p| \lfloor a^n \rfloor\} | < \infty \Rightarrow a \in \mathbb{Z}$

Yes, the answer is positive, and this follows from... guess which theorem :) In fact, Schmidt's subspace theorem [or even, better, just Roth's theorem for several places over number fields for the first question; and Roth's theorem for several places [primes] over $\mathbb{Q}$ [a.k.a. Ridout's theorem] in the second question] implies the following claims, the second answering the question in this thread [and the first showing a much more general result, but only in the case that $a \in \overline{\mathbb{Q}}$ is assumed algebraic]:

[b] Unsubstantial claim. [/b] Roth's theorem implies the following: if $a \in \overline{\mathbb{Q}}$ is a real algebraic number $>1$,  $S$ a finite set of primes, and $c < a$ is arbitrary such that the interval $(a^{n}-c^{n}, a^{n}+c^{n})$ contains an $S$-unit for infinitely many $n$, THEN there exists an integer $d > 0$ such that $a^{d}\in \mathbb{Z}$. In particular, if $\lfloor a^{n}\rfloor$ is an $S$-unit for infinitely many $n$, then there exists an integer $d > 0$ such that $a^{d}\in \mathbb{Z}$.

[b] Another unsubstantial claim, which nevertheless answers the problem. [/b] Roth's theorem implies the following: if $a \in (1, \infty)$ is a real number and $S$ any finite set of primes, and $c < a$ is arbitrary, then: if there are infinitely many $n$ such that each of the intervals $(a^{n}-c^{n},a^{n}+c^{n}), (a^{n+1}-c^{n}, a^{n+1}+c^{n}), (a^{n+2}-c^{n}, a^{n+2}+c^{n})$ contains an $S$-unit, then $a \in \mathbb{Z}$.

In particular, if $a > 1$ is such that $\lfloor a^{n}\rfloor, \lfloor a^{n+1}\rfloor, \lfloor a^{n+2}\rfloor$ are all $S$-units for infinitely many $n$, then $a \in \mathbb{Z}$.

For a sketch of the proof of the first substantial claim, see the first hidden text.

[hide] For simplicity I assume $a = r/s$ is rational, with $r > s > 1$ relatively prime, but the general (algebraic) case is no more difficult (granted, of course, that we apply Roth's theorem for the corresponding number field, rather than just for $\mathbb{Q}$ :)). In this case, the result will follow from Ridout's theorem, which is Roth's theorem for several absolute values over $\mathbb{Q}$ and is formulated in post #13 from the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=140101 .  For the finite set $S$ we take the archimedean place $\infty$, the finitely many primes dividing $rs$, and, assuming the contrapositive of the UNSUBSTANTIAL CLAIM, finitely many other primes such that for infinitely many $n$ the interval $(a^{n}-c^{n},a^{n}+c^{n})$ contains an $S$-unit [where $c < a$ is fixed throughout]. For the infinitely many $n$ in question, there is an $S$-unit $s_{n}$ with $|a^{n}-s_{n}| < c^{n}$. Then $t_{n}: = s_{n}/a^{n}= s_{n}s^{n}r^{-n}$ is an $S$-unit, and it turns out that the rational number $\frac{t_{n}}{t_{n}+1}$, of denominator $< 4r^{n}$, is a sufficiently good archimedean approximation to the [algebraic] number $1/2$; and at the same time sufficiently good $p$-adic approximation, for each prime $p \mid r$, to the [algebraic] number $1$; and at the same time sufficiently good $p$-adic approximation, for each prime $p \mid s$, to the [algebraic] number $0$ for each of the remaining primes in $S$. Here, "sufficiently good" meant that any choice of $\varepsilon > 0$ with $a^{1-\varepsilon}> c$ suffices for Ridout's theorem, as formulated in the 13th post of the mentioned thread, to force a contradiction with our assumption on the infinitude of $n$ for which $(a^{n}-c^{n},a^{n}+c^{n})$ contains an $S$-unit. I leave the detail of the simple verification to anyone who is willing to check it :).
 [/hide]

For a sketch of the proof of the second unsubstantial claim, see the second hidden text.

[hide] This is rather different in spirit, even though in the end it also boils down to Roth's theorem. What we need is a rather strong, but sadly, ineffective at present theorem which is a stronger version of the $S$-unit equation, and which in the most basic case of two variables reads:

[b] Theorem following from Roth's. [/b] Let $S = \{p_{1},\ldots,p_{k}\}$ be a finite set of primes. Let $\varepsilon > 0$. Then there are only finitely many pairs of distinct $k$-tuples $(n_{1},\ldots,n_{k}), (m_{1},\ldots,m_{k})$ of nonnegative integers satisfying
\[| p_{1}^{n_{1}}\cdots p_{k}^{n_{k}}-p_{1}^{m_{1}}\cdots p_{k}^{m_{k}}| < \Big( p_{1}^{\max{(n_{1},m_{1})}}\cdots p_{k}^{\max{(n_{k},m_{k})}}\Big)^{1-\varepsilon}. \]
This is stronger than the bounds obtained from linear forms in logarithms of algebraic numbers, but the latter have the virtue of being effective, while ours doesn't give us any bound of $\max \{n_{i},m_{j}\}$; it can only give us a bound (quite explicit in terms of $\varepsilon$ and $|S|$) on the [i] number[/i] of such solutions, but nothing more than this!

Also, Schmidt's subspace theorem implies similar finiteness results for diophantine inequalities in any (fixed) number of summands of $S$-units, and over any number field; but this is besides the point.

To deduce our second UNSUBSTANTIAL CLAIM that nevertheless answered the problem, assume it's [the claim's] contrapositive. For infinitely many $n$ we may then choose a triple of $S$-units $s_{n}, s_{n+1},s_{n+2}$ with $|a^{n}-s_{n}| < c^{n}, |a^{n+1}-s_{n+1}| < c^{n}, |a^{n+2}-s^{n+2}| < c^{n}$. Using $\frac{s_{n}}{s_{n+1}}= \frac{a^{n}+O(c^{n})}{a^{n+1}+O(c^{n})}= a^{-1}\frac{1+O\big( (c/a)^{n}\big)}{1+O\big( (c/a)^{n}\big)}= a^{-1}+O\big( (c/a)^{n}\big)$ and similarly $\frac{s_{n+2}}{s_{n+1}}= a+O\big( (c/a)^{n}\big)$, we obtain $\frac{s_{n}s_{n+2}}{s_{n+1}^{2}}= 1+O\big( (c/a)^{n}\big)$, whence $|s_{n}s_{n+2}-s_{n+1}^{2}| = O\big( s_{n+1}^{2}(c/a)^{n}\big) = O\big( (ca)^{n}\big)$. This holds for infinitely many $n$, in contradiction with the above-formulated theorem for any $\varepsilon > 0$ satisfying $a^{1-\varepsilon}> c$. The contradiction proves the second UNSUBSTANTIAL CLAIM.   (I hope... :))

I leave to the potential reader the deduction from Roth's theorem [version of your choice :)] of the crucial claim herein stated.
[/hide]

:)

Also, the FIRST CLAIM follows immediately from the theorem, generalized to number fields, used in the proof of the SECOND CLAIM. The appropriate generalization of the mentioned result is: let $K$ be number field, $S \subset M_{K}$ a finite set of places, and $\varepsilon > 0$; then $0 < |u+v| < H(u)^{1-\varepsilon}$ has finitely many solutions in $S$-units $u,v \in \mathcal{O}_{K,S}$, where $H(u)$ is an appropriate measure for the [i] size[/i] of the algebraic number $u$, analogous to (and generalizing) the notion $\max \{|p|, |q|\}$ or $|p|+|q|$ for a rational number $p/q$ (it could be the absolute multiplicative height, it could be simply the maximal of the absolute values of the coefficients of the minimal reduced integer polynomial of $u$).

To apply this result: take $K : = \mathbb{Q}(a)$; under our hypothesis, there exists a finite set $S \subset M_{K}$ of places such that $a$ is an $S$-unit and, for infinitely many $n$, the interval $(a^{n}-c^{n},a^{n}+c^{n})$ contains an $S$-unit. For every such $n$ and such $S$-unit $v$, one has $|a^{n}-v| < c^{n}< a^{n(1-\varepsilon)}\ll_{n}H(a^{n})^{1-\varepsilon}$; where $\varepsilon$ has once again been chosen to satisfy $a^{1-\varepsilon}> c$. By the quoted result, for all but finitely many such $n$ this forces $a^{n}= v \in \mathbb{Z}$, whence $a^{d}\in \mathbb{Z}$ for some $d > 0$.


dunno if nyone has posted this ineq but i found it today only

\[ (a^2 + k)(b^2 + k)(c^2 + k)\geq (k+1)^2((ab+bc+ca) + k-2) \]

for all \[ k\geq1  \]
k belongs to naturals

and \[ a,b,c\geq0  \]

very similar to APMO 04 :)

i think this also holds

\[ (a^2 + k)(b^2 + k)(c^2 + k)\geq (k+1)(n+1)(ab+bc+ca+k-2) \]

for all \[ k\geq1 \]
and \[ a,b,c\geq0 \]

where \[ n = [k] \]
ie greatest int func

:lol:


What is
$ \frac{3\plus{}i\sqrt{3}}{2}^{8}\plus{}\frac{3\minus{}i\sqrt{3}}{2}^8$

I have no idea.

[quote="iamhe"]What is
$ \frac {3 \plus{} i\sqrt {3}}{2}^{8} \plus{} \frac {3 \minus{} i\sqrt {3}}{2}^8$

I have no idea.[/quote]

Split the fractions apart like so
\[ \frac{3}{2}\plus{}\frac{i\sqrt{3}^8}{2}\plus{}\frac{3}{2}\minus{}\frac{i\sqrt{3}^8}{2}
\]
and the second and fourth terms cancel so we are left with $ \boxed{3}$.

It was supposed to say
$ (\frac {3 \plus{} i\sqrt {3}}{2})^{8} \plus{} (\frac {3 \minus{} i\sqrt {3}}{2})^8$

My bad...

[b]Hint:[/b] The numbers are written in rectangular coordinates. What kind of coordinates are easiest to exponentiate with?

I have no clue.
I'm new to this. I'm learning this by myself.

I'm guessing polar? That's the only other coordinate I know  :lol:

[quote="iamhe"]I'm guessing polar? That's the only other coordinate I know  :lol:[/quote]

Correct  :) . Try http://www.artofproblemsolving.com/Wiki/index.php/De_Moivre%27s_Theorem (after converting your polar coordinates into the trigonometric representation).


Hello to all,
     I'm relatively new to this board, and I was wondering what would be good preparation for the AMC 12 and the AIME.  I have all the AoPS books and some of the past competitions, but I was wondering what else could help me with these tests.
                                    Thanks  :D

You could apply for the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#class1]AMC/AIME Problem Series[/url], as well as the rest of them if you have the time.

[quote="tetrahedr0n"]buy a series of past tests from http://www.unl.edu/amc[/quote]

He said he already had some past competitions.

I would suggest buying one of the Contest Problem Books, since they contain challenging problems from years past as well as solutions upon solutions.  My math teacher has one of them; I wish he had the others, too...

oops. i just saw amc 12 preparation and posted. mathfanatic which book do you have? ive been doing problems from the oldest book, and they are quite difficult compared to recent questions.

How to buy math books without any money:
a.) Ask the math department head to purchase it "as a classroom resource"
con: It has to be about the subject you're doing in school
b.) Ask the librarian to "add it to the collection"
con: It'll only be available at the start of next school year
c.) Ask the teacher because "the book(s) will help me understand the context more" or "can use it for extra help" (Use the second [b]after[/b] you complain to someone, preferrably someone above the teacher via your parents, [b]and[/b] when the teacher does [b]not[/b] give you any help; the teacher will be afraid he/she'll be reported again :twisted:)
con: use the second one with caution; only use when you're sure you'll get away with it with the second one
d.) Ask the Math Team/MATHCOUNTS/Etc. coach for it
con: Again, it'll be restricted to only materials about that specific competition

and finally,
e.) Start tutoring [b]as a volunteering project[/b] and after the first/second meeting suggest a "Book list" with all the books you want and say "It'll nourish/enrich the(se) student's study in mathematics" or something like that to [b]whoever is grateful[/b] for this project whether it'd be the parents, principal, whatever.

The only one I haven't done yet is b :)

Just in case people wanted books...

I also recommend asking for advice from previous AMC12/AIME takers, it'll really help. You can even ask them here, their wisdom will pass on, to [b]you[/b]:lol:

Those are good ideas. Unfortunately my school has "budget difficulties," so they have cut down purchases of books. Our math library is limited to two mathleague problem books, and whatever our teachers still have from competitions of years past. 100% of our budget is now going toward our award-winning football team.

[b]HA[/b] "budget difficulties," they make me laugh. Everyone and every school say that but in truth they're spending it on computers, golf clubs, lobsters, etc. (all of which I've witnessed here 8-) )
All you have to do is simply mention them, not to everyone, but to people with the budget :wink:

Either that or get the PTA/PTO to pressure the administration to screw whatever's using up the budget and spend more money on math or science or whatever, which is extremely powerful yet extremely dangerous...just don't mantion your name and you should be fine :)  8-)

Most schools are having budget difficulties [i]because[/i] they spend their money on golf clubs, lobsters, and computers...well, maybe not the lobsters.  That's a very isolated situation you have, Tare.  Most people don't have to worry about the little pinchy things...

Of course, if the book's only going to be used by [i]you[/i] (I mean, get [i]real[/i] here, no one's going to keep their head stuck in a book, let alone a math book...c'mon, this is the 21st century...books are [i]so[/i] 1994...) then maybe you should just buy it.  Or have a bake sale!  Or sell candy bars ("Hello, ma'am, would you like to buy an extra-crunchy, extra-fattening, extra-artery-clogging hershey's bar so that I can devote more of my [spare] time to my in-depth study of the Riemann Hypothesis with respect to p-adic numbers?"  -slam-  "Sigh.")

Oh, yeah, my school has the latest one, Volume VI, and none of the others.  From what I hear, the early problems were quite hard.

..."little pinchy things"? :lol:
Yeah, I guess it only applies to the Maine/Massachusetts people.

...maybe I should sell them online...no, that would just confuse more people...then again... :twisted:

Since I do in fact tutor I'll just use them in my lessons so others will have to read :twisted:
...don't worry, I just want past MATHCOUNTS problems and such, not the...er, "Riemann Hypothesis with respect to pediatric numbers" :lol:

p-adic, not pediatric.  Good try, though.

http://mathworld.wolfram.com/p-adicNumber.html

Score! Two people actually recognised my joke! Hurray! *champain bottle opens*

...oh wait, I'm still in middle school...oh well :twisted:


linear algebra
matrix
linear algebra theorems

[color=purple]
Prove that any square matrix $ A$ may be presented uniquely in the form $ A\equal{}B\plus{}C$ where $ B$ is a symmetric matrix and $ C$ is a skew symmetric matrix.
[/color]


factorial
induction
inequalities unsolved
inequalities

[quote="amparvardi"]Let $ n$ be a natural number. Prove that:
$ 2! \times 4! \times \ldots \times (2n)! \geq [(n \plus{} 1)!]^n$
[/quote]

Induction on $ n$ easily solves the problem!

For $ n\equal{}1$, we have equality, so let's assume that the statement holds for some $ n\in\mathbb{N}$. We want to prove
\[ \prod_{i\equal{}1}^{n\plus{}1} (2i)! \geqslant\left((n\plus{}2)!\right)^{n\plus{}1}
\Longleftrightarrow \prod_{i\equal{}1}^{n} (2i)! \cdot (2n\plus{}2)! \geqslant\left((n\plus{}1)!\right)^{n}\cdot (n\plus{}1)!\cdot(n\plus{}2)^{n\plus{}1}\mbox{.}\]
Thus, by the assumption, it suffices to prove
\[ \frac{(2n\plus{}2)!}{(n\plus{}1)!} \equal{} \prod_{i\equal{}0}^{n}(n\plus{}2\plus{}i)\geqslant(n\plus{}2)^{n\plus{}1}\mbox{,}\]
but this is obvious as every factor on the left-handed side is greater than or equal to $ n\plus{}2$.  :P

[quote="amparvardi"]Let $ n$ be a natural number.Prove that:

$ 2! \times 4! \times ... \times (2n)! \geq [(n \plus{} 1)!]^n$

[hide="Read this too"](sorry if is already posted,It was in a test today.)[/hide][/quote]

$ (n\plus{}1\minus{}k)!(n\plus{}1\plus{}k)!\geq [(n\plus{}1)!]^{2}$[list][/list]

[quote="Martin N."][quote="amparvardi"]Let $ n$ be a natural number. Prove that:
$ 2! \times 4! \times \ldots \times (2n)! \geq [(n \plus{} 1)!]^n$
[/quote]

Induction on $ n$ easily solves the problem!

For $ n \equal{} 1$, we have equality, so let's assume that the statement holds for some $ n\in\mathbb{N}$. We want to prove
\[ \prod_{i \equal{} 1}^{n \plus{} 1} (2i)! \geqslant\left((n \plus{} 2)!\right)^{n \plus{} 1} \Longleftrightarrow \prod_{i \equal{} 1}^{n} (2i)! \cdot (2n \plus{} 2)! \geqslant\left((n \plus{} 1)!\right)^{n}\cdot (n \plus{} 1)!\cdot(n \plus{} 2)^{n \plus{} 1}\mbox{.}\]
Thus, by the assumption, it suffices to prove
\[ \frac {(2n \plus{} 2)!}{(n \plus{} 1)!} \equal{} \prod_{i \equal{} 0}^{n}(n \plus{} 2 \plus{} i)\geqslant(n \plus{} 2)^{n \plus{} 1}\mbox{,}\]
but this is obvious as every factor on the left-handed side is greater than or equal to $ n \plus{} 2$.  :P[/quote]

Thank you,using induction is so beautiful,what about other ways of the proof?


[quote="Chocolate M&M"]Let $x,y,z$ are positive numbers such that $x+y+z=xyz$ . Calculate the expression : $\sum \limits_{cyclic} \frac{\sqrt{(1+x^2)(1+y^2)}-\sqrt{1+x^2}-\sqrt{1+y^2}}{xy}$ :)[/quote]

Put x=1 y=2 and z= 3 or in random fashion you will get the solution

so what is the answer??
by the way, how would plugging in numbers help, woldn't it only steer you to the answer?

We just need to solute
$\sum\frac{\frac{1}{\cos A\cos B}-\frac{1}{\cos A}-\frac{1}{\cos B}}{\tan A\tan B}\\ =\sum\frac{1-\cos A\cos B}{\sin A\sin B}\\ =\frac{\sum \sin C-\sum \sin C(\cos A+\cos B)}{\sin A\sin B\sin C}$
Since $\sum \sin C(\cos A+\cos B)=\sum (\sin A\cos B+\cos A\sin B)=\sum\sin (A+B)=\sum\sin C$
So the sum is 0.


Find the formula to express the number of $ n\minus{}$series of letters which contain an even number of vocals (A,I,U,E,O).


trigonometry
inequalities unsolved
inequalities

$ a,b,c$ are non-negative real numbers. Prove that
$ (a^2\plus{}b^2\plus{}c^2)^2 \geq 4(a\minus{}b)(b\minus{}c)(c\minus{}a)(a\plus{}b\plus{}c)$
When does equality occurs?

N.T.Tuan, you can lock this topic  :D It is one of problems in M&Y Magazine  :D

[quote="arqady"][quote="ricardokaka"]$ a,b,c$ are non-negative real numbers. Prove that
$ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 4(a \minus{} b)(b \minus{} c)(c \minus{} a)(a \plus{} b \plus{} c)$
When does equality occurs?[/quote]
Nice problem! :lol:[/quote]

But it is easy to prove  :D

It is problem 6.75 in my book. Equality occurs for $ a\equal{}0, b\equal{}\sin\frac{\pi}8$, $ c\equal{}\cos\frac{\pi}8$.


geometry
3D geometry
tetrahedron
geometry proposed

$ABCD$ is a tetrahedron.
Let $K$ be the center of the incircle of $CBD$.
Let $M$ be the center of the incircle of $ABD$.
Let $L$ be the gravycenter of $DAC$.
Let $N$ be the gravycenter of $BAC$.

Suppose $AK$, $BL$, $CM$, $DN$ have one common point.
Is $ABCD$ necessarily regular?

[quote=ythomashu][quote=jg123]What is the "gravycenter" of a triangle?[/quote]

gravity center or centroid[/quote]

lol gravity center sounds like some term engineers created

My attempt: We start by introducing a three dimensional barycentric coordinate system. Let $\left(x,y,z,w\right)=\frac{x\vec{A}+y\vec{B}+z\vec{C}+w\vec{D}}{x+y+z+w}$
We can use the results of barycentrics in two dimensions to get:
$$K=\left(0,\ CD,\ DB,\ BC\right)$$
$$M=\left(CD,\ 0,\ DA,\ AC\right)$$
$$L=\left(\frac{1}{3},\ \frac{1}{3},\ 0,\ \frac{1}{3}\right)$$
$$N=\left(\frac{1}{3},\ \frac{1}{3},\ \frac{1}{3},\ 0\right)$$
We can then find the equations of the lines:
$$AK=\ y:z:w=CD:DB:BC$$
$$BM=\ x:z:w=CD:DA:AC$$
$$CL=\ x:y:w=1:1:1$$
$$DN=\ x:y:z=1:1:1$$
These four lines are concurrent iff $CD=DB=BC=DA=AC$. However $AB$ is not necessarily equal to the other five. Consider having the equilateral triangle $ACD$, and rotating a copy around the line $CD$ to get a new equilateral triangle $BCD$, such that the length $AB$ is different from the lengths of the other sides. This is a counterexample and so the answer to the question is no.

$ABCD$ is NOT necessarily regular. However, triangles $ABD$ and $BCD$ must be equilateral.

At first, we'll show the following: Segments joining each vertex of the tetrahedron to the gravity center of the opposite triangle concur at the same point, which is the gravity center of the tetrahedron. Le $A_1, A_2, A_3$ and $A_4$ be the gravity centers of the triangles $BCD, ACD, ABD$ and $ABC$, respectively. Suppose $DA_1$ and $DA_2$ meet the edges $BC$ and $AC$, respectively, at $M$ and $N$. Notice that $DA_1:A_1M=DA_2:A_2N=2:1$. So, $A_1A_2$ // $MN$ // $AB$ and then $A_1A_2$ and $AB$ are coplanar. Also,

$$A_1A_2=\dfrac{2}{3}\cdot MN=\dfrac{2}{3}\cdot(\dfrac{1}{2}\cdot AB)=\dfrac{1}{3}\cdot AB.$$

Thus, $AA_1$ and $BB_1$ concur at a point $G$ such that $AG:GA_1=BG:GB_1=3:1$. Analogously, $CC_1$ and $DD_1$ concur at $G$ and satisfy $CG:GC_1=DG:GD_1=3:1$.

Now, let's solve the problem.

Let be $G\equiv BL\cap DN$. As  $AK$ and $CM$ concur with these segments, then $K$ and $M$ coincide with the gravity centers of $CBD$ and $ABD$. So, at these triangles, incenter coincide with barycenter, and then the triangles are equilateral. This condition is enough for the statement of the problem. For a counterexample, construct a tetrahedron such that $ABD$ and $BCD$ are equilateral sized $\ell$ and $AC\ne \ell$

[hide=is this correct?]Intersection point of $BL$ and $DN$ is just a tetrahedron's centroid, so $K$ and $M$ must be also centroids of their edges, so the problem statement is just equivalent to $\triangle{CBD}$ and $\triangle{ABD}$ are being equilateral, and it's possible for tetrahedron to be non-regular.[/hide]


function
limit
real analysis
real analysis unsolved

Given a function $ g(x)$, define $ m_{1}(x) \equal{} g(x)$ and

\[ m_{n\plus{}1}(x)\equal{}\displaystyle\min_{0\le t\le x}[m_{n}(t)\plus{}g(x\minus{}t)]\]

where $ n \equal{} 1, 2, \dots$. Find $ \displaystyle\lim_{n\rightarrow\infty}m_{n}(x)$ is $ g(x) \equal{} x^2$.

We assume $ x \geq 0$.

[b]Claim)[/b] $ m_n(x) \equal{} \frac{x^2}{n}$.

[b]proof)[/b] We use induction. If $ n \equal{} 1$, it is clear by definition.

Now assume the claim holds for $ n$. Then $ m_{n}(t) \plus{} g(x\minus{}t) \equal{} \frac{t^2}{n} \plus{} (x \minus{} t)^2$, whose global minimum is attained for $ t$ satisfying $ \frac{2t}{n} \plus{} 2(t\minus{}x) \equal{} 0$, or $ t \equal{} \frac{nx}{n\plus{}1}$.

Since this $ t$ lies on $ [0, x]$, we have $ m_{n\plus{}1}(x) \equal{} m_{n}\left( \frac{nx}{n\plus{}1} \right) \plus{} g\left( \frac{x}{n\plus{}1} \right) \equal{} \frac{x^2}{n\plus{}1}$, which proves the claim.

Then $ m_n (x) \to 0$ pointwise.

Generally, if we alter $ g(x) \equal{} x^2$ by $ g(x) \equal{} x^r$ ($ r \geq 1$), then $ m_n(x) \equal{} n^{1\minus{}r} x^n$.


Now that they're up: of the 19 people listed with 3-round scores of 75, three are current members of the Southern California ARML team and two others are former members who competed with us before teams were organized in their home territories.

Are you sure about that? I only see five California 75s, one of which is me, and I don't believe I fall in either of your categories...

Three people from Albuquerque, two of them middle schoolers at the time, joined us in 2004. There's a New Mexico team now. So I wasn't trying to count you, Lynnelle. (You're SFBA, right?)

Oh, I see. Sorry. Yes, I'm in the Bay Area.


limit
trigonometry
parameterization
algebra unsolved
algebra

$ u_{n\plus{}1} \equal{} 2u_{n}^{2}\minus{}1$; $ u_{0}$ is rational; $ u{i} \equal{} u_{j}$ for some $ i, j$ with $ i \neq j$.

Is it true to say that

if $ \minus{}1 < u_{0} < 1$ with $ u_{0} \neq \frac{\minus{}1}{2}, 0$ or $ \frac{1}{2}$, $ \lim_{n \to \infty} u_{n} \equal{} 1$

and so for any $ d\equal{}u_{i}\minus{}u_{j}$, $ d$ can be made as close to $ 0$ as we like?

we're gonna show that, if $ u_0\in ( \minus{} 1, 1)$, the denominators of the sequence of rationals numbers $ u_n$ is strictly increasing, unless $ u_0 \equal{} \pm\frac {1}{2}$. it's easy to see that if $ u_0\in ( \minus{} 1,1)$, then $ u_n\in ( \minus{} 1, 1)$ for all $ n$.

assume that $ u_k \equal{} \frac {m}{n}\in ( \minus{} 1,1)$, with $ \gcd(m,n) \equal{} 1$. then $ u_{k \plus{} 1} \equal{} \frac {2m^2 \minus{} n^2}{n^2}$. if $ n$ is odd, $ \frac {2m^2 \minus{} n^2}{n^2}$ is in lowest terms, and the denominator, $ n^2$, of $ u_{k \plus{} 1}$, is strictly greater than the denominator, $ n$, of $ u_k$. (since $ \frac {m}{n}\in ( \minus{} 1,1)$, $ n > 1$.)  if, on the other hand, $ n \equal{} 2^ra$, where $ a$ is odd and $ r\geq 1$, then $ u_{k \plus{} 1} \equal{} \frac {m^2 \minus{} 2^{2r \minus{} 1}a^2}{2^{2r \minus{} 1}a^2}$, which is in lowest terms. now note that $ 2^{2r \minus{} 1}a^2\not > 2^ra\Leftrightarrow a \equal{} r \equal{} 1 \Leftrightarrow u_k \equal{} \pm \frac {1}{2}$. in other words, the denominator will always increase from one term to the next, unless we reach one of the values $ \pm \frac {1}{2}$. but it's easy to see that this can only happen if $ u_0\equal{}\pm\frac {1}{2}$.

since $ u_i \equal{} u_j$ for some $ i\neq j$ -- i.e., the sequence is periodic -- we must therefore have $ u_0 \equal{} \pm\frac {1}{2}$. for either of these initial values, $ \lim_{n\to \infty}u_n \equal{} \minus{} \frac {1}{2}$.

Thanks pleurestique.

I can see that for $ \minus{}1<u_{n}<1$, with $ u_{0} \neq \frac{\minus{}1}{2},0,$ or $ \frac{1}{2}$, if $ u_{n}$ converges then $ \lim_{n \to \infty} u_{n} \equal{} 1$. But how can we prove that the sequence converges?

Because $ \minus{} 1 < u_0 < 1$ we could find such $ \phi_0$, that $ u_0 \equal{} \cos{\phi_0}$. Then:
\[ u_{n \plus{} 1} \equal{} 2u_{n}^{2} \minus{} 1 \equal{} 2\cos^2{\phi_n} \minus{} 1 \equal{} \cos{2\phi_n}\]
so for every n: $ u_n \equal{} \cos{2^n\phi_0}$.
UPD: It's a variant of [url=http://en.wikipedia.org/wiki/Logistic_map]Logistic map[/url] with parameter $ \lambda \equal{} 4$ and diverges for almost all initial values.
If you write that $ u_n \equal{} 1\minus{}2y_n$ then: $ y_{n \plus{} 1} \equal{} 4y_n(1 \minus{} y_n)$

[quote="AndrewTom"]Thanks pleurestique.

I can see that for $ \minus{} 1 < u_{n} < 1$, with $ u_{0} \neq \frac { \minus{} 1}{2},0,$ or $ \frac {1}{2}$, if $ u_{n}$ converges then $ \lim_{n \to \infty} u_{n} \equal{} 1$. But how can we prove that the sequence converges?[/quote]

you said that $ u_i \equal{} u_j$ for some $ i\neq j$. is this really a requirement? if this is a requirement, then the sequence necessarily converges to $ \minus{} \frac {1}{2}$, as i showed in my post above.

on the other hand, it looks like the sequence won't converge unless $ u_0 \equal{} \pm\frac {1}{2}$...

as greenvert points out, $ u_n \equal{} \cos 2^n\phi_0$, where $ u_0 \equal{} \cos \phi_0$. it's well-known that the only rational multiples of $ \pi$ whose cosines are rational are essentially $ \pi/2, \pi/3, 2\pi/3$. therefore, if $ \cos\phi_0\neq \pm\frac {1}{2}$, we may write $ \phi_0 \equal{} r\pi$, where $ r$ is irrational. note that $ \cos 2^nr\pi$ depends on $ \displaystyle \{\frac {2^nr\pi}{2\pi}\} \equal{} \{2^{n \minus{} 1}r\}$. ($ \{x\}$ denotes the fractional part of $ x$.) write $ r$ in base $ 2$. since $ r$ is irrational, there are both infinitely many 1's and infinitely many 0's after the decimal point. we claim that there must also be infinitely many occurrences of either 00 or 11. if there weren't, then from some point on there are no occurrences of either 00 or 11. but since there are infinitely many of both 0's and 1's, the sequence of digits must therefore eventually look like 010101..., i.e. periodic, contradicting the irrationality of $ r$. if there are infinitely many occurrences of 00, then there are infinitely many $ k$ for which $ 2^kr$ has fractional part in $ (0,\frac {1}{4})$. however, since there are infinitely many 1's, there are infinitely many $ k$ for which $ 2^kr$ has fractional part in $ (\frac {1}{2},1)$. this implies that the sequence of fractional parts of $ 2^kr$ doesn't converge and thus $ u_n$ doesn't converge. similarly, if there are infinitely many occurrences of 11, then there are infinitely many fractional parts that are in $ (\frac {3}{4},1)$. but since there are infinitely many 0's, there are infinitely many fractional parts in $ (0,\frac {1}{2})$. so once again the sequence of fractional parts cannot converge.

sorry, i just realized my "proof" is flawed. still, if someone can show that $ \{2^nr\}$ is dense in $ (0,1)$ for $ r$ irrational (which i think it is??), that would do the job.


Let a,b,c be positive real numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$.Prove that:
\[ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}d}\plus{}\frac{c^2}{d\plus{}a}\plus{}\frac{d^2}{a\plus{}b}\geq 1\]

it's too easy , it could have two solutions i think  :)

[quote="MrMinh"]Let a,b,c be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$.Prove that:
\[ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} d} \plus{} \frac {c^2}{d \plus{} a} \plus{} \frac {d^2}{a \plus{} b}\geq 1\]
[/quote]

By Using Cauchy-Shwraz inequality we have :

$ \sum_{cyc}\frac{a^2}{b\plus{}c}\sum_{cyc}a^2(b\plus{}c) \ge (a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2\equal{}1$

Thus , It suffices to show that :

$ 1 \ge \sum_{cyc}(a^2b\plus{}a^2c)$  and Acording to Am-Gm inequality we have :

$ \sum_{cyc}\left(\frac{a^2b}{2}\plus{}\frac{a^2c}{2}\right) \le \sum_{cyc}\left(\frac{\frac{a^2}{4}\plus{}a^2b^2}{2}\plus{}\frac{\frac{a^2}{4}\plus{}a^2c^2}{2}\right)\equal{}\frac{1}{4}\plus{}\frac{1}{2}\sum_{sym}a^2b^2$
$ \le \frac{1}{4}\plus{}\frac{(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2}{4}\equal{}\frac{1}{2}$

And we are done  :P

By the Cauchy-Shwraz inequality, we have :
$ \sum_{cyc}\frac{a^{2}}{b\plus{}c}\geq \frac{(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})^{2}}{A\plus{}B}$
where
\[ A\equal{}a^2b\plus{}a^2c\plus{}c^2d\plus{}c^2a\]
\[ B\equal{}b^2c\plus{}b^2d\plus{}d^2a\plus{}d^2b\]
Using Cauchy-Shwraz inequality again
\[ A\leq \sqrt{(a^2\plus{}b^2\plus{}c^2\plus{}d^2)(a^4\plus{}c^4\plus{}2a^2c^2)}\equal{}a^2\plus{}c^2\]
\[ B\leq \sqrt{(a^2\plus{}b^2\plus{}c^2\plus{}d^2)(b^4\plus{}d^4\plus{}2b^2d^2)}\equal{}b^2\plus{}d^2\]
Thus
\[ A\plus{}B\leq a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1\]
The proof is complete.

[quote="MrMinh"]By the Cauchy-Shwraz inequality, we have :
$ \sum_{cyc}\frac {a^{2}}{b \plus{} c}\geq \frac {(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2})^{2}}{A \plus{} B}$
where
\[ A \equal{} a^2b \plus{} a^2c \plus{} c^2d \plus{} c^2a\]

\[ B \equal{} b^2c \plus{} b^2d \plus{} d^2a \plus{} d^2b\]
Using Cauchy-Shwraz inequality again
\[ A\leq \sqrt {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)(a^4 \plus{} c^4 \plus{} 2a^2c^2)} \equal{} a^2 \plus{} c^2\]

\[ B\leq \sqrt {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)(b^4 \plus{} d^4 \plus{} 2b^2d^2)} \equal{} b^2 \plus{} d^2\]
Thus
\[ A \plus{} B\leq a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1\]
The proof is complete.[/quote]Nice idea!  :)
Written more suggestively
\[ A\leq \sqrt {(b^2 \plus{} a^2 \plus{} d^2 \plus{} c^2)(a^4 \plus{} a^2c^2\plus{}c^4 \plus{} a^2c^2)} \equal{} a^2 \plus{} c^2\]

\[ B\leq \sqrt {(c^2 \plus{} b^2 \plus{} a^2 \plus{} d^2)(b^4 \plus{} b^2d^2\plus{}d^4 \plus{} b^2d^2)} \equal{} b^2 \plus{} d^2\]


superior algebra
superior algebra unsolved

A ring $R$ is said to be Boolean if $a^{2}=a\ \forall a\in R$. Prove that if $R$ is Boolean then $2a=0\ \forall a\in R$. Prove also that all Boolean rings are commutative.

From Wikipedia:

$a+a = (a+a)^{2}= a^{2}+2a^{2}+a^{2}= a+2a+a = a+a+a+a$

Hence $a+a = 0 \Leftrightarrow a =-a$.

Then $a+b = (a+b)^{2}= a^{2}+ab+ba+b^{2}= a+ab+ba+b$

Hence $0 = ab+ba \Leftrightarrow ab =-ba = ba$.

Can somebody also post other properties for boolean rings?  Some days ago I saw some regarding cardinals that make a ring boolean....

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=boolean&t=45041]this topic[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=boolean&t=46877]this one[/url] for example.


Παιδιά χρόνια πολλά και από εμένα με μια ασκησούλα. Οι έμπειροι στον τομέα των ανισοτήτων θα την καθαρίσουν αμέσως φαντάζομαι. Αν είναι κάποιο γνωστό θέμα πείτε μου να το καταχωρήσω ...

Y.Σ.  Siluan αν τη λύσεις σε δευτερόλεπτα  :P  βάλε τη λύση hint.

Αν $ a,b,c\ge0$  δείξτε ότι  $ \dfrac{ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{a\plus{}b\plus{}c\plus{}1}\le\frac{\sqrt{3}}{2}$

Meta apo polu kairo mporesa na mpw k egw sto site, mias k exoume tis diakopes twn xristougennwn!
Gia na mh polulogw,
8etoume $ \sqrt {a} \equal{} x$, $ \sqrt {b} \equal{} y$ kai $ \sqrt {c} \equal{} z$ opote meta apo pra3eis katalhgoume sthn isodunamh
$ \sqrt {3}x^2 \minus{} 2x \plus{} \sqrt {3}(y^2 \plus{} z^2 \plus{} 1) \minus{} 2(y \plus{} z) > \equal{} 0$ auto to triwnumo exei diakrinousa (meta apo tis katalhles pra3eis) $ D \equal{} \minus{} 12(y \minus{} 1/\sqrt {3})^2 \plus{} \minus{} 12(z \minus{} 1/\sqrt {3})^2 < \equal{} 0$ opote h dosmenh isxuei.
Elpizw na mhn ekana kanena la8os stis pra3eis!
Hlias

Aπό την $ (a \plus{} b \plus{} c)^2\geq 3(a^2 \plus{} b^2 \plus{} c^2)$ έχουμε:

$ LHS\leq \frac {\sqrt {3(a \plus{} b \plus{} c)}}{a \plus{} b \plus{} c \plus{} 1}$

Αρα αρκεί αυτό $ \leq\frac {\sqrt {3}}{2}\Longleftrightarrow 2\sqrt {a \plus{} b \plus{} c}\leq (a \plus{} b \plus{} c) \plus{} 1$ το οποίο είναι φανερό απ' την απλή ανισότητα:

$ x \plus{} 1\geq 2\sqrt {x}$

Παιδιά κάτι τέτοιο είχα σκεφτεί όταν έκανα post το θέμα.  Αρχικά η ιδέα μου για την ανισότητα $ \sqrt {3}\sqrt {a \plus{} b \plus{} c}\ge \ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$... ήταν από τη jensens στη συνάρτηση $ \sqrt {x}$...αλλά μετά διαπίστωσα ότι βγαίναι και με Cauchy.  Για τη συνέχεια με καλύψατε .... :)  Δέν ήταν κάτι το εξαιρετικό ... :|

Και μια πιο ανεβασμένη ... :)

Αν a,b,c > 0  και  a+b+c=1  δείξτε ότι  $ \sum_{cyc}\frac {a \plus{} b \minus{} 2ab}{c \plus{} 1}\ge {1}$.

ΥΣ.  Παιδιά σόρυ αν έχω γίνει βαρετός με τις ανισότητες αλλά αυτόν το καιρό είμαι ενθουσιασμένος με τον χώρο αυτό.
Επίσης οι ανισότητες που κάνω post είναι από κάποιες ιδέες που μου έρχονται στην προσπάθεια να λύσω άλλα θέματα. Με ενδιαφέρει να βλέπω και άλους τρόπους επίλυσης , διαφορετικούς από την αρχική ιδέα. Αν ο τρόπος που προτείνετε είναι διαφορετικός, θα κάνω post τη λύση που στηρίζεται στην αρχική ιδέα.Επίσης άν κάποιο θέμα μοιάζει με κάποιο άλλο που έχετε δεί, πείτε μου να το καταχωρίσω :)
Ευχαριστώ.

Αυτα από εμένα και χρόνια πολλά καλά Χριστούγεννα :)

[quote="vangelismak"]
Αν a,b,c > 0  και  a+b+c=1  δείξτε ότι  $ \sum_{cyc}\frac {a \plus{} b \minus{} 2ab}{c \plus{} 1}\ge {1}$.[/quote]

Συμπαθητική είναι  :P

Λοιπόν έχουμε $ \frac{2a\plus{}2b\minus{}4ab}{c\plus{}1}\geq\frac{2a\plus{}2b\minus{}(a\plus{}b)^2}{c\plus{}1}\equal{}\frac{(a\plus{}b)(1\plus{}c)}{1\plus{}c}\equal{}a\plus{}b$

Άρα $ \frac {a \plus{} b \minus{} 2ab}{c \plus{} 1}\geq\frac{a\plus{}b}{2}$ . Προσθέτοντας ..., το ζητούμενο  :)

[quote="vangelismak"]
Αν a,b,c > 0  και  a+b+c=1  δείξτε ότι  $ \sum_{cyc}\frac {a \plus{} b \minus{} 2ab}{c \plus{} 1}\ge {1}$.

ΥΣ.  Παιδιά σόρυ αν έχω γίνει βαρετός με τις ανισότητες αλλά αυτόν το καιρό είμαι ενθουσιασμένος με τον χώρο αυτό.
[/quote]

Οι ανισότητες είναι πραγματικά πόρωση (ιδιαίτερα όταν εμφανίζονται σε συνδυασμό με άλλα παιδία άλγεβρας). Είναι μάλλον τα πιο θαυμάσια θέματα για λύση (χμμ..., ίσως θα μπορούσαμε να εξαιρέσουμε τις διεθνείς ολυμπιάδες  :P ).

Στο παραπάνω θέμα αρκεί να δείξουμε ότι $ \frac {a \plus{} b \minus{} 2ab}{c \plus{} 1} >\equal{} \frac {a \plus{} b}{2}$ (η ίδια σκέψη με τον Σιλουανό, απλά λίγο διαφορετική εκτέλεση)
$ < \equal{} > \frac {a \plus{} b \minus{} 2ab}{2 \minus{} a \minus{} b} >\equal{} \frac {a \plus{} b}{2}$
$ < \equal{} > 2a \plus{} 2b \minus{} 4ab >\equal{} 2a \plus{} 2b \minus{} (a \plus{} b)^2 < \equal{} > \minus{} (a \plus{} b)^2 <\equal{} \minus{} 4ab < \equal{} > (a \plus{} b)^2 >\equal{} 4ab$
$ < \equal{} > a^2 \plus{} b^2 \plus{} 2ab > \equal{} 4ab < \equal{} > a^2 \plus{} b^2 \minus{} 2ab >\equal{} 0 < \equal{} > (a \minus{} b)^2 >\equal{} 0$ που ισχύει
άρα ισχύει και ότι $ \frac {a \plus{} b \minus{} 2ab}{c \plus{} 1} >\equal{} \frac {a \plus{} b}{2}$
Προσθέτοντας τώρα κατά μέλη έχουμε:
$ \sum_{cyc}\frac {a \plus{} b \minus{} 2ab}{c \plus{} 1} >\equal{} \sum_{cyc}\frac {a \plus{} b}{2} \equal{} \frac {a \plus{} b \plus{} b \plus{} c \plus{} c \plus{} a}{2} \equal{} 1$

Πάντως είναι φοβερό απο πού ξεκινάει κανείς ( όταν του έρχεται η ιδέα να φτιάξει ένα θεματάκι )  και τελικά πόσες διαφορετικές λύσεις εμφανίζονται ...  Μόλις βάλω internet σπίτι ( το περιμένω σε λίγες μέρες ) θα κάνω post την όλη διαδικασία με την οποία μου προέκυψε αυτό το θεματάκι ... Είναι φοβερό !!! Εκεί που πειραματίζεσαι με κάτι ... λές "ρε συ δεν προσθέτω και αυτό ... δεν το μετατρέπω και λίγο ...κ.λ.π  και ΤΣΟΥΠ !!! βγήκε το θεματάκι!
Χρειάζεται βέβαια και λίγη προσοχή μην έχεις κάνει καμία πατάτα.
Ενοείται ότι κάποιες φορές εμφανίζονται εντελώς διαφορετικές ιδέες στις λύσεις που παρουσιάζονται.
Αυτό παιδιά ήταν πάντα κάτι το οποίο μου προξενούσε Θαυμασμό σε κάποιες ασκήσεις. Η ιδέα αυτού που κατασκεύασε την άσκηση. Βλέπεις κάποιες ασκήσεις που τελικά όταν φανερωθεί η ιδέα μένεις έκπληκτος από θαυμασμό! Είναι κάτι το οποίο είναι εξίσου όμορφο με μία κομψή και θεαματική λύση!


function
real analysis
real analysis unsolved

Let $f$ be a function in the plane wich satisfies $d(P,Q)=1 \Rightarrow d(f(P),f(Q))=1$ for all the points $P,Q$ in the plane. Show that $d(f(P),f(Q))=d(P,Q)$ for all the points $P,Q$ in the plane.

This one has already been posted by sam-n and solved by grobber. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17406[/url].


[quote="JBL"]Have you intentionally done away with post ratings following the update this afternoon, and are they coming back?  (Everyone still seems to have their personal ratings.)[/quote]What do you mean?

I mean that on my system (Windows XP, firefox 1.0.7 or IE6, AoPS skin) I can no longer see the little cubes which show the ratings of individual posts.

Edit: Using IE on the mathlinks skin, I see the hexagons.

I'm not seeing the hexagons either- both in Firefox and Safari.

I can see them in Firefox 2, with the MathLinks skin.


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_{ \minus{} 1}^ {a^2} \frac {1}{x^2 \plus{} a^2}\ dx\ (a > 0).$

You may not use $ \tan ^{ \minus{} 1} x$ or Complex Integral here.

[hide] $ \int \frac{1}{x^{2}\plus{}a^{2}}\ dx \equal{}{i \over {2a}} ( \int {{dx} \over{x\plus{}ia}} \minus{} \int {{dx}\over{x\minus{}ia}}) \equal{}   ln {({{x\plus{}ia}\over{x\minus{}ia}}) }^{i\over{2a}}$[/hide].
 The limits (mainly the upper one) and the condition a>0, are kind of strange...

that simplifies to $ \frac {i}{2a}\ln ( \minus{} 1)$ which.. DOES make sense i take it back

$ \minus{}1\equal{}e^{i\pi},$ so we get... $ \frac{\pi}{2a}$?

That explains the upper limit. But what about a>0 ?

$ \ln (\minus{}1) \equal{}i\pi$

[quote="kunny"][quote="Immanuel Bonfils"]$ \int \frac {1}{x^{2} \plus{} a^{2}}\ dx \equal{} {i \over {2a}} ( \int {{dx} \over{x \plus{} ia}} \minus{} \int {{dx}\over{x \minus{} ia}}) \equal{} ln {({{x \plus{} ia}\over{x \minus{} ia}}) }^{i\over{2a}}$.
 The limits (mainly the upper one) and the condition a>0, are kind of strange...[/quote]

That's all Greek to me![/quote]

Fortunately it's not Greek to everyone else! :rotfl:

i tried to do it by expanding into power series then integrating and plugging in, but at the end i got 1.. which didnt sound right at all.


ratio
geometry
angle bisector
perpendicular bisector
geometry proposed

In triangle non-isosceles ABC, let AB=c, BC=a, CA=b. The A-altitude, angle bisector, and median meet BC at D, E, M, respectively. Prove that $ \dfrac{DM}{ME}\equal{}\dfrac{(b\plus{}c)^{2}}{a^{2}}$.

Looking for a synthetic solution; it's pretty easy to just compute the lengths, but it's somewhat ugly and I think it requires the law of cosines. Also I know if a pretty contrived solution that involves taking MI and intersecting it with AD, but perhaps there is something simpler?

Perpendicular bisector of $BC$ and angle bisector $ AE$ meet at the midpoint $ N$ of the arc $BC$ and $ \frac {_{DM}}{^{ME}}=\frac {_{AN}}{^{EN}}.$ Power of $ E$ to $ (O)$ is $ EB \cdot EC \equal{} AE \cdot EN$  and since $ AN$ is self-isogonal WRT $\angle A,$ it follows that $ AN \cdot AE \equal{} AB \cdot AC $

$ \Longrightarrow \frac {DM}{ME} \equal{} \frac {AN}{NE} \equal{} \frac {AE \cdot AB \cdot AC}{BE \cdot EC \cdot AE} \equal{} \frac {AB \cdot AC}{BE \cdot EC}$

Using angle bisector theorem we get:

$ BE \equal{} \frac {ac}{b \plus{} c} \ , \ CE \equal{} \frac {ab}{b \plus{} c} \ \Longrightarrow \ \frac {DM}{ME} \equal{} \frac {bc(b \plus{} c)^2}{a^2bc} \equal{} \frac {(b \plus{} c)^2}{a^2}$


We know that a [b]reference angle [/b]is an acute angle formed by the terminal side and the x-axis.  As we can see in the picture, θ=120, so i think its reference angle is POQ which is equal to 180-120=60.  cosθ=x/r, right? cos POQ(60) also equals to x/r, so cosθ=cosPOQ(60). But why we say: cosθ=-cos60?  Why do we add a negative sign in it?  Is that because cos 60 here is in standard position,it is in the first quadrant. It is not the angle which is in that right triangle POQ. so we say:cosθ=-cos60? But i think we can't call it is the relationship between the trigonometric function values of any angle and its reference angle, because the angle 60 in the standard position isn't the angle θ's reference angle.

Hope you can tell me,thanks.

Because of the following two conventions:

1. Reference angle is always measured between the POSITIVE direction of x-axis and the terminal side, in THAT order;

2. Angles are considered positive if measured COUNTERCLOCKWISE, and negative in the opposite case.

Having those two in mind, the reference angle of your point P is +120 degrees or -240 degrees, depending on whether you measure it CW or CCW. (The important thing is that you must always begin from the positive direction of x-axis.) Of course, you can always add to it as many positive or negative multiples of 360 degrees as you want.


Why does the right angle mark not work: :maybe:
 [geogebra]2440c2ae5511e21fdf40291decde6eaf73d03094[/geogebra]

Seems like a site bug. I know that the additional tools worked when geogebra just appeared on AoPS but now they seem to be malfunctioning :(. Alas, only Valentin can answer why that happened. Check whether somebody else posted this problem in the support forum and, if not, post it there.

I'll investigate this, it seems that it is indeed a problem on the site.

[geogebra]ae70a7567f23f77f327a35001fd127940209d58a[/geogebra]

I think I've fixed the problem. What I don't understand is how they were working before ... (can't fix the old images.)

It might happen shortly after GeoGebra was added. All I know is that I tested everything right then but didn't even bother to post the test, just checked that it worked in the preview window (which might be a mistake) and forgot about it (I'm more of an Asymptote fan than that of GeoGebra). What puzzles me is why then nobody noticed it for so long time. :?

By the way, Valentin, as far as Asymptote is concerned, will you take a look at the current problem with 3D images or you are too busy nowadays?

The latest asymptote version is installed right now so the 3D thingies should work ... give them a try.

I'm trying, you see :( Still the same problem, it seems...
[asy]import three;
size(150);
currentprojection=orthographic(1,1,1);
draw((0,0,0)--(1,1,1));[/asy]

I tried that code and I obtain some sort of weird eps file (if it is an eps file) which has weird options and cannot be converted into a .png as far as I can see.

Did you try to add

access settings;
settings.render=0;

to the code and see if the result would be the normal .eps file (i.e., convertible to .png, etc.)? I cannot do it from the user end but you can easily do it from your end. If we are to believe the documentation, these 2 extra lines should just create a good old 2D projection of a 3D picture rather than the interactive "monster" (I do not think you want to support the interactive 3D pictures on the site anyway :)).

I feel in the dark.
So 3d is not working? (That could explain a couple things)

Your code fedja has a problem. A correct 3D code now works, as seen below:

[asy]import three;

size(101);

path3 g=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle;
draw(g);
draw(O--Z,red+dashed,Arrow3);
draw(((-1,-1,0)--(1,-1,0)--(1,1,0)--(-1,1,0)--cycle));
dot(g,red);[/asy]

or

[asy]import three;
import graph3;

size(0,100);
currentprojection=perspective(3,-2,2);

real V(real r) {return r^4-r^2;}
real V(pair pos) {return V(abs(pos));}

real R=1/sqrt(2);
real z=-0.2;

bool activebelow(pair pos) {return abs(pos) < R && V(pos) < z;}
bool activeabove(pair pos) {return abs(pos) < R && V(pos) >= z;}

pair a=(-1.5,-1);
pair b=(0.5,1);
real f=1.2;

draw(surface(plane(f*(b.x-a.x,0,z),(0,f*(b.y-a.y),z),(a.x,a.y,z))),
     lightgrey+opacity(0.5));
draw(surface(V,a,b,Spline,activebelow),lightblue,black);
draw(surface(V,a,b,Spline,activeabove),lightgreen,black);

xaxis3(Label("$\phi^\dagger\phi$",1),red,Arrow3);
zaxis3(Label("$V(\phi^\dagger\phi)$",1),0,0.3,red,Arrow3);[/asy]

Use import three; to add the settings.render =0 automatically.


What I want to ask people is whether there is a right way to things that are to us not known to be exact. I am refering to for example, is a political party, the democrats or republicans right, which one is correct? How about what is the idle profession? I think that most people would say that there is no right answer, and there will always be arguments in everything, in that case I ask, what is the point of arguing if there are different ways of looking at things and both can be viewed as correct?

Discourse allows ideas to be refined, and arguments to be strengthened. And people do get persuaded to switch to a different viewpoint.

Just because we do not know whether they are exact, does not mean they are not exact. Just because we don't have the facts at the moment, does not mean that it is an opinion. Admittedly some things are opinions and cannot be verified by facts, but you'd also need to define "facts" and "exact".

I am talking about things that are always debated and are never known to have an exact answer, no matter how much progress has been made, there is no consensus on which party is right in ideology the REpublicans or Democrats.

Both of them are right in some ways and both of them are wrong in some ways. Most things are like that :D

The Democrats and the Republicans are political parties with a views on a wide variety of things.  There are even separate factions within each party.  It is pretty much impossible to say that the Republicans are right or vice versa since doing so would mean that every single one of their views is correct.

Again, I maintain that just because we don't know the right answer doesn't mean there is no right answer. The cult of the Pythagoreans was held to be just that until someone proved its belief about rational numbers false.

Mathnerd, I agree with what you are saying. I believe that in a lot of cases there is a right answer, which would make a lot of a particular parties' ideology incorrect.

Since ideology is more or less a collection of opinions, I'm not sure that you can ever say that one set of beliefs is proven to be "correct" while another is "incorrect". In the US, most of us have been used to (and even taken for granted) many of our freedoms. And indeed, in our minds it would better the world if we spread democracy to all the corners of the world that we can influence. But who is to say that Communism or a Monarchy is not a better form of government? Perhaps statistics can show that democracy is a more effective form of government economically or psychologists can argue that people with more freedoms are happier, but in the end still no one has proven that what they believe is correct. More often than not, politicans simply provide quite convincing arguments on behalf of their beliefs. But every AoPSer should know that convincing certainly doesn't equal rigorous, and neither of them implies "right".

Four words: Refer to hello's signature.

Even in mathematics, though, proof can be argued against if you have a different set of axioms. Everything in math is based on a set of axioms, namely: equals and equals are equals, equals minus equals are equals, equals times equals are equals, two points determine a line, all right angles are equal, etc. Everything in science is based on uniformitarianism, that is, that all processes will stay at the same rate and in the same way as they have been. And so on.

mathnerd, I am not sure what you are refering to in all your equals and unalitarianism but even though we use certain axioms to build a theory, I want to ask, can we assume that for example 1+1 equals something other then 2 and then build a different mathematical theory? CAn the meaning of the axiom change? I dont see how this can be done.

I think that before even looking at the economic effects on the people, we should look at whether all actions give all people liberty. For example, I think welfare programs might make poorer people richer then they would be under free markets, but does that mean we should have welfare to decrease inequality? I dont think so. What I do think is that people say coming to the US from other nations with a degree (like from European or ASian countries etc) their degrees should count as American ones so they have a chance to apply for a decent job. I think that this is largely absent currently and people from other nations immigrating here have difficulties with this.

Equals and such: examples of axioms.

What do you mean by "changing the meaning of an axiom"? For example, in Euclidean geometry you can prove that the angles of a triangle sum to 180 degrees, but in Hyperbolic geometry you can only prove that they sum to LESS THAN 180 degrees, because of the different fifth postulates.

[quote="hello"]I think that before even looking at the economic effects on the people, we should look at whether all actions give all people liberty.[/quote]

Most in modern times would agree with you, but society can function without this. Consider Roman times - their society existed fairly well with a slavery system.

Debates are actually quite specious to people who have very little knowledge about the topic that is being debated.  Debates/discussions go on until both sides of the debate reach equilibrium relating the amount of knowledge about the topic; therefore, there will always be people who thing that something is wrong.  There usually is a correct side as long as the topic of the debate reaches its entirety regarding logic.

Masoud Zargar

[quote="banblink182"]Although I do not consider there to be a "right" way, I believe that given a specific goal there is a "better" way, a way that achieves more of what was intended.

This belief relies on the assumption that two things can never be equal.[/quote]

Define a "thing". 1 and 1 are equal.

[quote="mathnerd314"]Define a "thing". 1 and 1 are equal.[/quote]

No, one of the 1's is to the left of the other one. They are located in different "positions" in Cyberspace as well as in your brain.

I do not believe that what I have said is a pointless technicality in any way.

A thing is anything that in math would be considered itself. So 1 is a thing, because in math $1=1$.


find the smallest number that is not a multiple of 2 or 3, and is not prime.

[hide]smallest prime greater than 3 is 5. so 5^2 is the answer.[/hide]

well,

[hide]the actual answer would be
negative infinity,

but the positive answer is 25[/hide]

why not 1, according to what my math teachers say to me 1 is not a prime number, its not divisible by 2 or 3 so why wouldnt 1 work?

zolojetto
one wont work because some still think it is a prime number, and other reasons.  Also, you don't use it because its mutually known that you can't use one for these questions... check around

I guess the answer would actually not be 1, but -infinity as thacheezinator said.

If must be non-prime and natural #, then 1.

If must be non-prime and composite, then 25.

Why would it be negative infinity?  :maybe:

$ \minus{}\infty$ isn't an integer, and many weird things happen

for example, if it was an integer, is it odd or even?

I don't really know how to explain it....

[hide]25 or negative infinity[/hide]

All right; negative infinity is not really defined. Like Alan said, how do you know if it's not prime, divisible by 2 or 3?

God. Stop arguing about this -_-. Negative infinity has no value and it would definitely not be the answer on any such test.

Assuming that he means the smallest natural number that is not a multiple of 2 or 3 and is composite, then the answer is obviously 25.

[quote="unimpossible"]negative infinity is not a number

period.[/quote]

Exclamation point.

But there is no way that would be an answer to a question.  If this was on a mc test, I would put 1 and protest if it was wrong.


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
AMC 8
AMC 10

What kind of math do you like?

PURE=proofs, theory, no uses in the real world, just thinking,...
APPLIED=using math in the real world, using math to perform challenging tasks,...
COMPETITION=MathCounts, other competitions, practicing, countdown, trivia, speed,...

I think I like all three, but mostly applied. My parents and mathy people I know tell me to figure out what kind of math you like. I just wanted to see what you guys like.

PowerofPi, I really like this competition category because I like trivia and tricks to solve problems fast.

I like competition math, too, because that's what I usually do (these days  :D). I also like the other two kinds of math, but I believe that competition math is the most enjoying as you learn a lot of tricks, you must organize your time so that you can finish solving the problems very quickly, and because the problems are not very hard. I usually get bored, if I can't solve a problem. I can't stand in front of a problem more than two or maximum three days. The olympiads are some of the most beautiful reminiscences, and I will miss them, as the next year is the last year when I can participate.

Math, overall, is just an amazing subject. As far as what I like...I just looove Pure Mathematics, more specifically Number Theory. The proofs are so amazing...I mean they just fit together so well. I also like competition math. I like how they make you think outside the box.  :D

I like competition because it's more intense and exciting. You get to interact with the smartest kids in the world/country/state/city. It's so fun learning from other kids and meeting people just like you who love math.

I like competition math because its more nerve-racking, more dissapointing :( , but highly rewarding and satisfying :) . I also like it because I love the use of tricks and speed.

1) competition
2)pure
3)applied

although pure maybe number one. competition 4 now  :D

Hmm, this should be moved to round table/fun factory. Anyways, I like pure mathematics the best.

They are all amazing! I, and probably a lot of other people out there, spend the most time with competition, however.

I like applied math the most because it gives me something to do when I am spacing out in class. I figure things out like the percentages I need to get such and such grade, the percentage saved when buying this over that, and other things; however, I still like pure math the most because it really stimulates my imagination and challenges me (though I often fail).

In USA, there are a lot of math competitions, such as MathCounts, MOEMS, AMC 8, AMC 10, AMC 12, AIME, USAMO, Purple Comet, Math Kangaroo, and the list goes on. Most schools participate in some of them. I personally like Applied the best, but I like competition because it motivates you to study more things.

I think competitions are great to, but where I'm from, competitions gave way to a dark side and it became more about beating people than math (people would have arguments in class about what their results meant, who was better, slamming others). It was pretty creepy and kind of reminds me of the whole Poincare conjecture issue

Applied, definitely.

For me, real-world uses are so commonplace that if I ever was to teach any math (provided there are real-world uses for the math I would teach), I would focus on applications. I just can't avoid applications, otherwise it would be a waste.

Here is something I found [url=http://www.juliantrubin.com/mathjokes.html]here[/url]:

[quote]An engineer [applied mathematician] thinks that his equations are an approximation to reality. A physicist thinks reality is an approximation to his equations. A mathematician [pure mathematician] doesn't care.  [/quote]

Both pure and applied mathematicians are needed. Without the pure mathematicians, applied mathematicians wouldn't be able to apply anything. Without applied mathematicians, pure mathematicians' work would be useless.[/quote]

Also, I think that the needs of pure mathematicians would be best catered by someone other than me. (If I was to teach any math, of course)

I like competitive math the best, because it's a great way to show off your skills. Applied is my second favorite, because it would most likely benefit my future and my career.

[quote="PowerOfPi"]Then what do you like? Most people on AoPS like math.[/quote]

I am not one of these persons.

So then why did you sign up? This forum is full of math.

Come on man, can you not understand a joke?

Ohhh, wow. I get it. I don't get jokes from the beginning. To not make this spam:

What kind do you like? The poll says the majority of people like competition. I am guessing that is because it is fun and exciting. I like that too. What about for the future, between applied and pure? I like applied, pure is too deep. Applied uses more problem solving skills, which I like (art of PROBLEM SOLVING).

Science is applied? Then applied is best. Then competition. Then pure. Yea.

I loooove competition math. It's just so exciting, and I love finding out ways to figure out a math problem very fast. My second would be pure mathematics, because it's just so logical. :lol:

[quote="tennis123"]I like math where it's not just telling you to multiply or divide or factor, where you have to find out [i]how[/i] to do the problems, if you know what I mean. I don't really know how to phrase that. I guess that's applied math?[/quote]

really? when I really need to calm down I love doing math that tells u what 2 do..... It makes me feel so much better. BUt besides that I like math where its kinda like a puzzle, I'm pretty sure that applied  math but idk  :)

i think the original poster got what he/she wanted.
Mods, pls lock.


email
algorithm
number theory
prime factorization

RSA is a widely-used public-key cryptography. My question is, is RSA still a good choice among various methods, when there's no need to publicize the encoding key? For example, when a CIA agent tries to communicate with his headquarterz?

I don't think they would intentionally publicize the key but simply send it by email or something, and if it does get hacked, it wouldn't matter.

How do you send a key securely so you make sure no one can read it?

You dont need to, the RSA cryptosystem uses a public key which everyone can access, but only the intended receiver can use the public key combined with his/her private key to decode the message.

I think RSA or another public-key crypto system is good for some special applications, such as key distribution or information blinding. But for just simple data transmission over an open channel, it can be better to use a conventional cryptosystem like DES or IDEA. Symmetric cryptosystems use in most cases algorithms that are optimized for the hardware on which they are executed and therefore the encryption/decryption can be done faster and more efficient.

RSA is ok.  I read somewhere that if all the computers in the world were put together to solve for the prime factorization of the key,  It could take over 6 times the lifetime of the universe.  But if some guy found a way to factor really big numbers into primes, then RSA is dead.  If you use a 256 bit key it can be broken in hours with a PC.

[quote="L_Li"]has anyone else read The Code Book? i what to know what others think of using the quantum principles to decipher RSA or for mathematically completely secure cryptography (in terms of feasibility, cost, and when if ever it'll happen)[/quote]

A very primitive quantum computer has already been made by IBM.. I am optimistic but I say that within 15 years public key crypto based on RSA and Diffie Hellman will both be history (since both can be broken by quantum computers)


geometry
incenter
circumcircle
angle bisector
IMO Shortlist
geometry solved

Given a triangle $\triangle ABC$ , let $k$ be its circumcircle.  Let $AD$ be the bisector of the angle $\angle BAC$ (with $D\in BC$) and let $l$ be the circle which is internally tangent to $k$ and which is tangent to the segments $AD$ and $BD$ . Denote by $E$ the point of tangency of the circle $l$ and the line $AD$ . Prove that $E$ is the center of the circle inscribed in $\triangle ABC$.

This is not hard at all.
Let $AD\cap k=X,l\cap k=P,l\cap BC=Q$. Then P,Q,X are collinear. Note that since $\angle BPX=\angle CBX$, $BX^2=XQ\cdot XP$. On the other hand, since XE is tangent to k, $XE^2=XQ\cdot XP$. Thus, BX=XE. This implies E is the incenter of $\triangle ABC$.

I don't see collinearity of P,Q,X - [b]mecrazywrong[/b]  (or anybody else) can you enlarge upon it a little bit ?

Fact: If circle $C_{1}$ is internally tangent to circle $C_{2}$ at $X$ and a line $L$ tangent to $C_{1}$ at $Y$ cuts $C_{2}$ at $M,N$, then $\angle MXY=\angle NXY$.
Back to mecrazywong's solution. Let $PQ$ cuts the circumcircle at $X'$, by this fact arc $BX'$ equals arc $CX'$. But since AX is the angle bisector of $A$, so arc $BX$ = arc $CX$. Hence $X=X'$, i.e. $P,Q,X$ are collinear.

yptsoi, thanks for explaining this in details :)

An, in my opinion, much nicer problem is Lemma 1 from the paper

  [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, [i]Sawayama and Thébault's Theorem[/i], Forum Geometricorum 3 (2003) pp. 225-229[/url].

This lemma asserts that, for [i]any[/i] point D on the line BC (not necessarily such that AD is the angle bisector of the angle BAC), if the circle l touches the line BD at a point F, then the line EF passes through the incenter of triangle ABC. Of course, in your case, when the line AD is the angle bisector of the angle BAC, this yields that the point E is the incenter of triangle ABC (in fact, the point E lies on the line EF and on the angle bisector of the angle BAC, while the incenter of triangle ABC also lies on the line EF and on the angle bisector of the angle BAC; but two distinct lines can have only one common point, and thus the point E coincides with the incenter of triangle ABC). So your problem is merely a corollary of the above-mentioned lemma.

Another application of that lemma is IMO Shortlist 1992 problem 7 ( http://www.mathlinks.ro/Forum/viewtopic.php?t=1370 , http://www.mathlinks.ro/Forum/viewtopic.php?t=18743 , http://www.mathlinks.ro/Forum/viewtopic.php?t=24869 ).

  Darij

Consider the circle with center $AD$^$k$ passing through $B, C, I$ (I being the incenter). Inverting in this circle keeps $l$ fixed, so $E$ inverts to itself $\Rightarrow E = I$.

edit: sorry...I see darij has already massively generalized this problem ;)

If we talk about similar problems I've just today luckily come across this problem from Romania 1997:

Let $ABC$ be a triangle, $D$ a point on side $BC$ and $w$ the circumcircle
of $ABC$. Show that the circles tangent to  $w,AD, BD$ and  to  $w,AD, DC$, respectively, are tangent to each other if and only if
 $<BAD = <CAD$.

Problem above from Belgrade could be applied to solve it at least to prove it in one direction.


group theory
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Prove that $\mathbb{Z}$ is nor isomorphic with the direct sum with any two subgroups.

Lets show it for the additive group, it will follow for the ring structure then:

The subgroups are all of type $a\mathbb{Z}$ (it is a PID, and the same reasoning shows this). I think you don't want the case $a=0$.
All these subgroups are isomorphic to $\mathbb{Z}$ itself then.
Se we want to show that $\mathbb{Z}\not\cong \mathbb{Z}^{2}$. We know that $\mathbb{Z}$ is generated by $1$, and let the image under an isomorphism be $(a,b)$ with $a,b \in \mathbb{Z}$. Then it's not hard to check that not both $(1,0)$ and $(0,1)$ can be multiples of $(a,b)$, where multiple means to multiply with an integer.
But an isomorphism would make it necessary that both occure since it has to be surjective.


group theory
abstract algebra
superior algebra
superior algebra unsolved

For which commutative finite groups is the product of all elements equal to the unit element?

It's the groups whose $ 2$-Sylow subgroup is either trivial or not cyclic.  Here's a proof using the structure theorem, which I don't know how to avoid.

For groups of odd order we have $ g\neq g^{\minus{}1}$ for $ g\neq e$, so the product is $ e$.

Otherwise, write $ G$ using elementary divisors as $ G \cong H \oplus \bigoplus_{i\equal{}1}^k C_i$, where $ H$ has odd order, $ C_i\equal{}\langle g_i\rangle$ is cyclic of order $ 2^{a_i}$, and $ k\geq 1$.  If $ \pi$ is the product of all elements of $ G$, the $ H$-component of $ \pi$ is $ e^{|G| / |H|} \equal{} e$ by the above.

It is not hard to show that the product of elements in $ C_i$ is $ g_i^{a_i / 2}$. (this is the real meat of the argument)  Then the $ C_i$-component of $ \pi$ is $ (g_i^{a_i / 2})^{|G|/|C_i|}$, which is $ e$ exactly when $ |G| / |C_i|$ is even--in other words, when $ k > 1$.

I would better describe it the following way:
If in an abelian group $ G$ the equation $ x^2\equal{}e$ has exactly two solutions ($ e$ and $ a$), then the product of all elements is $ a$. Otherwise it's $ e$.


What is the pH of a 0.20 M HA solution ($ K_a\equal{}1.0 \times 10^{\minus{}6}$) that contains 0.40 M NaA?

A 0.1 M solution of which salt will have a pH less than 7?
(a) NaCl
(b) NH4Br
(c) KF
(D) NaO2CCH3

Thanks!

1. Use hendersen hasselbach

2. its the ammonium bromide, since ammonium is acidic.


number theory
greatest common divisor
number theory solved

I don't know if is true or not. Can you prove that if

am + bn = 1  implies that gcd (a, b) = 1

I mean if gcd (a, b) = 1 is easy to find m and n using the Euclidean Algorithm. But prove what is above I can't find it true. :lol:

gcd(a,b)|a and gcd(a,b)|b
Hence for any integer m and n
gcd(a,b) | (m*a+n*b)
But if m*a+n*b=1 then gcd(a,b)|1
Hence gcd(a,b)=1 (since gcd(a,b) cannot be -1).
P.S: This should go to intermediate or even before.

Sorry, I just thought I had a counter example without even proving that. But I just found I was wrong. Thx anyways.


AMC USA(J)MO USAMO inequalities geometry

Which would you go for if you got into both, RSI or MOP?

RSI (so as to leave more mosp spaces for the rest of us)

RSI is much closer to the kind of mathematical research that real mathematicians do. USAMO, albeit extremely difficult, is still a timed test, and is nowhere near that kind of serious mathematics. So for me, I would prefer RSI.

If you have done MOP before I would definitely consider RSI.

If you have done Siemens before and then MOP would probably be good and also if you want to really improve your Olympiad math a lot.

[quote="jb05"]RSI (so as to leave more mop spaces for the rest of us)[/quote]

haha, nice

you know, especially if you're a winner

you should definitely consider going to RSI

like I don't even care if you're a junior or not

just go

[quote="tjhance"]whats rsi?  :blush:[/quote]
Google says: [url=http://www.cee.org/rsi/]RSI[/url]

If I had been accepted, RSI.  But if I get into MOP, I'm very happy to get that.

RSI; I say this as an alumni :wink:. Not that I ever made MOP, lol.

I probably have a much better chance making RSI next year than MOP (TJ almost always gets like 3 people into RSI, as one of the AP physics teachers here directed RSI for a year). That being said, if I got into both, I'm not entirely sure which I'd choose.

I thought there was at most 2 people from one state that goes to RSI...0.o

I enjoyed RSI a lot more than MOP and learned a lot more and made a lot more friends... so I would vote RSI.  But I feel my MOP experience may have subpar, so take my advice with half a grain of salt.

I LOVED mop and would never miss it for anything else. I will at least apply for RSI next year, though.

By MOP do you mean black blue or red well i guess red isnt likely since you would be a fish but black or blue because it probably makes a difference in your decision.

I was faced with this choice 20 years ago (wow, I'm old), when in 1987 I got into both MOP and RSI.  They only overlapped by a week (the last week of MOP was the first week of RSI), and my dad personally called the directors of both programs to see if there was any way that I miss a few days of either program, but both programs wanted a full commitment (which I think is pretty reasonable).

I chose RSI, and unless you have a legitimate chance to make the IMO team, I would advocate for that choice today too.  There's basically two criteria -- academics and social -- and I think RSI "wins" on both counts.

Academically, you'll learn a lot of math at MOP, but a lot of it tends to be the narrow, Olympiad-style math that doesn't necessarily "help" beyond MOP (obscure inequalities and geometric formula and that sort of thing).  The biggest academic plus to MOP is that you'll get lots of practice doing really hard problem solving.  This is a very valuable skill.  On the other hand, at RSI you'll get exposure to doing research, which is something that's very hard to get at the high-school level, and is also a very valuable skill.  For most of the people reading this board, exposure to hard problem solving is already a given, but exposure to hard research may be a new experience.  Also, at RSI, the focus is not so narrow -- there will be biologists and chemists and physicists there too.

Socially, at both places you'll make great friends and have a great time.  But at MOP you'll meet a bunch of math geeks from the US.  At RSI you'll still meet a bunch of geeks, but they'll be geeks in many different and varied ways, plus they'll be from all over the world.  Subjectively, RSI was a lot more fun for me than MOP was the following year.  Also, RSI has a much wider and more active alumni network than does MOP.

If there's someone that went to both programs that has a different -- and more current! -- take on this than I do, I'd love to see an alternative viewpoint.

Gabe Carroll (to correct the spelling). And I think (also around 1999 or 2000) that Sasha Schwartz opted for RSI.

I believe that RSI acceptance letter asked you to respond before Mar, 31, 2007. If you have been accepted into RSI, should you have already made your decision (either commitment to RSI or declining their acceptance)?

Brian Lawrence and Richard McCutchen (both last year’s USAMO winners) went to RSI in 2006, both were among 40 finalists in Intel Science Competition in 2007. Yi Sun (2005 and 2006 USAMO winner) went to RSI in 2005 and was the second price winner of Intel Science Competition in 2006.

And if Brian Lawrence, Richard McCutchen, and Yi Sun jumped off a cliff, would you?

(Actually, I would.  For one, I'd never have to hear that stupid question again.  For two, it has to be a pretty sneaky cliff to fool those three.  It would probably get me, too.)

MOP is completely free; RSI you have to pay for transportation (for the sake of contradiction I will argue with jb05 that MOP > RSI)

[quote="Valentin Vornicu"][quote="DPatrick"]I chose RSI, and unless you have a legitimate chance to make the IMO team, I would advocate for that choice today too.[/quote]Tiankai Liu and Gabriell Carol (spelling correct?) both with 100% chances of making the IMO team (since they both had Gold in 9th grade at the IMO) choose RSI over MOP :) In the very recent history Zach Abel also did that so I guess this is not an compelling argument for MOP :)[/quote]

Hm, but both of those guys were already IMO gold medalists and moppers. Maybe they just wanted to try something new.

I had to forgo any possible shot I might've had at MOP this year when I accepted RSI.  My thinking was that the opportunity to experience a research environment at RSI would ultimately be more valuable than just Olympiad training.

And I was hearing that RSI would help more with college admissions on the side  :P

[quote="mysmartmouth"]MOP is probably more prestigious, but I think RSI would be the better choice to actually attend.  I would never really apply anything I learned at MOP, because I would never make the IMO team.[/quote]

When I talk to ppl, prestige is usually the word everyone associates with RSI. And don't choose MOP because of the cost of a plane ticket...Go to RSI so I get the chance to meet all you smart ppl!

Mosp is certainly more prestigious, but rsi is more well-known to the entire science community, and is widely known as prestigious. It's just that mosp is only known in the math community.

But that should have nothing to do with you choice.

[quote="noneoftheabove"]Is RSI really more prestigous than MOP?[/quote]To this question, I would say "why does it matter"?

Both programs are very prestigious.  Both programs are invitation-only and only invite a few dozen US students.  Both program are strictly merit-based in the sense that they don't charge tuition and admit students based on merit.

It's very hard to compare the two programs on some abstract term like "prestige".  One (MOP) is math only and admits students solely based on performance on exams.  The other (RSI) covers all sciences and is more college-like in its admission process (test scores, essays, recommendations).

[quote="jb05"]Mosp is certainly more prestigious[/quote]What are you basing this assertion on?  (This is not an accusation -- I am genuinely curious why this perception exists.)

[quote="DPatrick"][quote="jb05"]Mosp is certainly more prestigious[/quote]What are you basing this assertion on?  (This is not an accusation -- I am genuinely curious why this perception exists.)[/quote]

Simply because I get that impression from the math community at large, and that's really all that matters in defining prestige. Of course, it's different if you're talking about the general science community.

Some more concrete, but less provocative reasons: MOSP selects from a much larger pool (well, this is only a guess, but I doubt RSI has hundreds of thousands of applicants), and has half as many participants. I know that doesn't sound like a big difference, but think about the effect that doubling the size of MOSP would have on the level of students there. I think it would actually drastically change it.

As I said, that shouldn't have any impact on your choice, as RSI and MOSP are both wonderful programs (and both extremely prestigious). I think it all depends on who you want to be with - those who share a deep interest/understanding of math (which I found extremely refreshing coming from a math-dead area), or just amazing scientists and general intellectuals.

Be careful with this comparison - RSI admits about 60 (I would actually say fewer) domestic students, but no more than (about) 15 are math students.  On the other hand, MOP admits about 20 students in blue group and about 12 in black group - the two groups for which juniors are eligible.

Also, while the RSI applicants number only in the thousands, only juniors (well, some sophomores) apply, and the group that applies is self-selective. It's really hard to argue that one is harder to get into than the other, but I'm sure both are worthwhile. And if you're looking at poll results, don't forget that this forum is naturally biased towards math competitions. I'm not sure about the prestige; from purely anecdotal evidence I've gotten the impression that people usually choose RSI over MOP.

[quote="iPod Nano"]I'm sure that both programs, as many before me have noted, would be very worthwhile.  However, I would keep in mind that an opportunity such as RSI comes literally once in a lifetime, as it is available only for high school juniors.  Thus, if I were ever lucky enough to be facing the dilemma you are proposing, I would probably go for RSI.  After all, even if MOP is more "prestigious", which is debatable, one could still mention it on one's college app :)[/quote]

Similarly, assuming that the student hasn't already make it to MOP(because if he had RSI would, in my opinion, be the better choice). If they were accepted to both Junior Year, then it would most likely be their last chance to go to MOP unless they are a winner Senior Year, so it would be somewhat of a once in a lifetime chance.


Is it possible to number the $8$ vertices of a cube from $1$ to $8$ in such a way that the value of the sum on every edge is different?

[hide]

I keep getting an answer of no but alas my proofs are so inelegant (I wish I knew how to write them well) and don't even make much sense in certain areas so I will not bother to post.

[/hide]

really? the proof is pretty straightforward though.

anyway it's a challenge so... keep trying!! :D

Why don't you post your prove Lucky? It may be inelegant, but it's better than nothing! ;)

I have a solution that, I believe, is the best for this problem:

the lowest value of a edge is 3 and the higher is 15.
in the sum of all edges we have 3 times each edge, so the sum of alledges is [tex]( \frac{(1+8)8}{2} )3 = 108[/tex] now the higher 12 values between 3 and 15 are the number from 4 to 15 wich sum is 104.

[quote="Bajoukx"] now the higher 12 values between 3 and 15 are the number from 4 to 15 wich sum is 104.[/quote]
No. Their sum is 114 :D

Try again! :)

such a stupid mystake...

well, I try again...
between 3 and 15 we have 13 numbers, we need 12 diferent of them to the values of the sum in the edges.
and we need that the sum of these 12values = 108, so these 12 numbers are the numbers between 3 and 15 less the 9.
the only way to do
15= 8+7
14= 8+6
13= 8+5 since 6 and 7 can be in the the corners of the same edge
and now we see that we can't do the 12 since we just can use each number three times and we can't put the 5 and 7 in the corners of the same edge...

 :)  I hope now it's everything correct

[quote="Johan Gunardi"]I think it is still not perfect. You should prove that the 12 values include 12,13,14,15.

(It is easy though.) :D[/quote]

[quote="Bajoukx"]such a stupid mystake...

well, I try again...
between 3 and 15 we have 13 numbers, we need 12 diferent of them to the values of the sum in the edges.
and we need that the sum of these 12values = 108, [b]so these 12 numbers are the numbers between 3 and 15 less the 9.[/b]
the only way to do
15= 8+7
14= 8+6
13= 8+5 since 6 and 7 can be in the the corners of the same edge
and now we see that we can't do the 12 since we just can use each number three times and we can't put the 5 and 7 in the corners of the same edge...

 :)  I hope now it's everything correct[/quote]

he does prove it :P

No he didn't. Between 3 and 15, inclusive, there are 13 numbers. We only choose 12 of them. We have to prove that 12, 13, 14 and 15 are chosen.

What he said was that the sum of the 12 numbers must be 108.

We start out with the 13 numbers: 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. (Because each corner is labeled from 1-8, and no numbers are repeated, the minimum value for an edge is 3 (1+2), and the maximum, 15 (8+7).)

However, the sum of these 13 numbers equals 117, or 108 + 9.

Therefore, the number 9 must be excluded.

(Apart from grammar mistakes, all of this was stated.)

If you have any qualms, please ask them.

Oh, I see. I misunderstood him. He said that the numbers are 'less than 9'. So I thought he was referring to the values of the vertices. :blush:

I don't blame you. I had to reread that a few times to figure out what he meant, too... :P

We prove by contradiction. Suppose it is possible. Note that the minimum value of the sum of vertices on an edge is $1+2=3$, and the maximum is $7+8=15$. Hence there are $15-3+1=13$ possible values of the sum on an edge. Since there are $12$ edges on the cube, there is exactly one possible value of the sum on an edge which is not used. Observe that the sum of all edges is $3(1+2+\ldots+8)=108$ and the sum of all possible values is $3+4+\ldots+15=117$. Hence, there is no edge with value $117-108=9$. Therefore there exist edges with values $15,14,13,12$. Clearly, the edge with sum $15$ must have $7$ and $8$ on its vertices. Similarly, $14=6+8$. Therefore, $6$ and $7$ do not share an edge. Hence $13$ must be formed with $5$ and $8$. Now $12=4+8$ or $5+7$. Since $8$ already has $3$ edges connected to it, $12=4+8$ is immediately rejected. Also, since $5$ and $7$ do not share an edge, $12$ cannot be formed. Hence we have a contradiction, and it is impossible.

Among $1-8$, there are $13$ different sums that can be formed, all these sums give a total sum of $117$ and since each vertex is counted trice, we need a total sum of $108$ which means $9$ must be excluded from these sums. Now, we know $(1,2), (1,2), (1,4)$ must share an edge but then the sum of $6$ can't be formed. Thus, it is impossible.

mathmax12's solution (which he told me to post):
The sum of the numbers can be anywhere from $3-15$, adding the possible sums up we get $117$ so we need to exclued $9$, so it is not possible.


Hi guys,

Do any of you own TI-84 silvers? I just got one and I was wondering if any of you could let me know where to find out more on how to write programs for your calculator.

Thanks

http://www.google.com/search?rlz=1C1GGLS_en-USUS300US304&sourceid=chrome&ie=UTF-8&q=ti+basic

That could help.

easy, just play around with the functions

Please move to Computer Science and Informatics.

come and ask quantumaccountant or imaginary13 tomorrow! Also, just search the web a bit.


Will the [url=http://www.foxnews.com/story/0,2933,373302,00.html]Large Hadron Collider[/url] go kablooey?

No, it won't go kablooey... "kablooey" usually refers to an explosion... instead, the Earth will sort of implode on itself as it enters a human-created black hole.  Way to go scientists!!

Just kidding... nothing destructive will happen...  I hope. :maybe:


What would be the method to graph the following:

1) $ \lfloor x\rfloor^{2}\plus{}\lfloor y\rfloor^{2}\equal{}1$

2) $ \lfloor x\rfloor^{2}\minus{}\lfloor y\rfloor^{2}\equal{}3$

3) $ \lfloor x\plus{}y\rfloor^{2}\equal{}1$

4) $ \lfloor x\rfloor\plus{}\lfloor y\rfloor\equal{}1$

1.
$ 1\rightarrow$ either $ [x] \equal{} 0$ and $ [y] \equal{}\plus{}_{\minus{}}1$ or the other way round thus
first let $ y\geq 0$
$ 0\leq x < 1$ and $ 1\leq y < 2$
hence the graph is a region bounded by the right triangle $ \text{ABC}$ right angled at $ \text{B}$
where $ \text{A}(0,1) ,\text{B}(1,1),\text{C}(1,2)$
and the second case(when $ [x] \equal{} 1$ and $ y\geq 0$ but  would imply a right triangle otained by refletcing this triangle about it's base and shifting it right by $ 1$ unit.
now when $ [y] \equal{}\minus{}1$ then clearly the graph is a square with vertices $ (0,0) ,(1,0),(0,\minus{}1) ,(1,\minus{}1)$
similarly proceed when $ [x] \equal{}\minus{}1$

2.for this one
use $ a^{2}\minus{}b^{2}\equal{} (a\minus{}b)(a\plus{}b)$ hence factorize the expression and use the fact the $ 3$ is a prime and can only be $ 3\times 1$ or $ 1\times 3$

3.$ \rightarrow [x\plus{}y] \equal{}\plus{}_{\minus{}}1$
thus $ \minus{}1\leq x\plus{}y < 0$ or $ 1\leq x\plus{}y < 2$
let's solve $ \minus{}1\leq x\plus{}y < 0$
consider the lines $ x\plus{}y\plus{}1 \equal{} 0$ and $ x\plus{}y \equal{} 0$
use the fact that point $ (h,k)$ lies on one side of the line $ ax\plus{}by\plus{}c \equal{} 0$ in which origin lies  or the other as $ (ah\plus{}bk\plus{}c)c > 0$ or $ < 0$
solve for common region

4.same method as 1


Heather wants to make a bracelet with seven beads. Of the seven beads, one bead is unique and the other six are three distinct pairs of matching beads. If the bracelet's bead-pattern is symmetric when the bracelet is not closed, how many distinct bracelets can she make? One is shown here.

[asy]draw((0,0)..(10,-6)..(20,0));
draw(circle((-.4,.4),.5));
draw((19.5,.5)--(20.5,-.5));

draw(circle((10,-6),1.25));
fill(circle((12.5,-5.75),1.25),red);
fill(circle((7.5,-5.75),1.25),red);
fill(circle((14.75,-4.75),1.25),blue);
fill(circle((5.25,-4.75),1.25),blue);
fill(circle((17,-3.75),1.25),green);
fill(circle((3,-3.75),1.25),green);[/asy]

the right side will mimic the left, and the unique bead must be in the middle.
The number of ways to arrange the three colors of non-unique bead is 3! or $ \boxed{6}$


I need to solve a "game of darts" problem, where the dart isn't punctual, but had an area itself. Let's call A, B the areas the dart can reach (where A+B is the total area), and C the area of the dart. What is the probability of the dart reaching the B (or the A) area? Idealize the problem supposing that the dart can reach one area at a time, not a portion of one or another.

Thanks a lot... and sorry for my english... I'm working on that  :blush:

[quote="white_horse_king88"]Are areas A and B and C circular, or of no defined shape?[/quote]

It seems the shape and distribution of "the shapes" makes a big difference... I will try to explain the exact problem:

Imagine a big circular area (Ra radius) and a total of "N" little circles (Rb radius and N very big) INSIDE the big circle. The little circles are distributed uniformly into the big circle. Then you "throw" the dart (thats a circle too, with Rc radius). What is the probability of reaching some of the Ra radius circles? (If the dart reaches partially one of the little circles Rb, we assumed a success.)

Thanks again… I hope my english is good enough to be understood  :blush:


logarithms
function
analytic geometry
graphing lines
slope
calculus
algebra solved

yeah i guess i haven't said why..
to think about it, i don't actually know how to prove that $f(x)=3^x-2^x$ is a strictly increasing function for all $x$
 :D
oh maybe its because $3^x$ increases at a faster rate than $2^x$?

[quote="minsoens"]yeah i guess i haven't said why..
to think about it, i don't actually know how to prove that $f(x)=3^x-2^x$ is a strictly increasing function for all $x$
[/quote]

maybe we can differentiate $f(x)=3^x-2^x$, we'll get $3^xln3 - 2^xln2$, since $3^xln3 > 2^xln2$, the slope is always positive, so it's an increasing function?    ;)

hmm... thanks!  :)
actually i was thinking non-calculus methods because calculus is discouraged, but i think it is okay for olympiad level problems.

[quote="violetcraze"]i just thought of a non-calculus method:

$f(x)=3^x-2^x$

for any $x +1$

$f(x+1) = 3^{x+1} - 2^{x+1} = 3. 3^x - 2. 2^x > 3. 3^x - 3. 2^x = 3 (3^x - 2^x) = 3 f(x)$
obviously, $3f(x) > f(x)$

so for any $x + k > x, f(x+k) > f(x)$

 :)[/quote]

thats very nice  :) . Anyway , the condition $x> 0$ must be here so that $f(x)> 0$ can hold  :P . But that is easy to see from $3^x-2^x=1$ ,  we must have positive $x$

[quote="VANCHANH123"][quote="VANCHANH123"]put t=log_6 x
I HAVE :6^T+3^T=2^T[/quote]
$2^{x}+6^{x}+16^{x}=8^{x}+9^{x}+7^{x}$[/quote]
From: Phu Khanh
... contimchaymau


is it true that as of now only 14 amc 10s exist?

2000 amc 10
2001 amc 10
2002 amc 10A
2002 amc 10B
2003 amc 10A
2003 amc 10B
2004 amc 10A
2004 amc 10B
2005 amc 10A
2005 amc 10B
2006 amc 10A
2006 amc 10B
2007 amc 10A
2007 amc 10B

i just want to make sure i have them all

I believe so. Before that, the AMC 10 did not exist.

yes it is true. before, the AHSME was for high school. But, almost all qualifiers were juniors or seniors so they created the AMC 10 to give freshmen and sophomres a shot.


I have always had the curiosity to know how do you say "mathematics" in another languages other than spanish (and english). I'd like you post here how do you say "math" in you own language :oops: Let's begin :D

Spanish: Matemática
English: Mathematics
...

German: Mathematik

(I think it's very similar in all languages)

chinese: shu xue.

KIND OF like sushuay.

In cantonese, it's like soo-hawk. (I don't know any romanization systems for cantonese...) Taken literally, I means "counting study".

In Japanese we say 数学　su-u-ga-ku.
In French mathematique

[quote="Melissa"]In Afrikaans: Wiskunde :D[/quote]

 :rotfl:  "kunde" means bum in hindi..so is math wise+kunde(wise butt) :D No offence,i'm not humiliating math i hope...just having fun with the names.

Hashbrownie - actually thats more modern. Looking for a more classic term - 'cheshbon' seems to work. Though, 'mat-eh-mat-ih-ka' also works...

hey how do you insert foreign characters?[/code]

[quote="IntrepidMath"]Pig latin: mathematics==>athematicsay[/quote]
That's bad Pig Latin. mathematics -> athematicsmay.
Good one. :D

Azerbaijan: Riyaziyyat
Iceland: Stærðfræði

[quote="shadysaysurspammed"]Not in hindi(THe national language of india)

Hindi:Ganit[/quote]
Same in bangla.(and also in sanskrit)

Latin: MATHEMATICA
Greek: μαθηματικά (mah-thay-mah-tee-kah)
Italian: matematiche
Dutch: wiskunde
Russian: математика (mah-tyeh-mah-tee-kah)
Norweigian: matematikk

That's interesting that Afrikaans and Dutch use the same word!

[quote="kunny"]In French mathematique[/quote]

Well, in fact, it is : mathématiques :P

[quote="stupidkid"]/v\47l-l3/v\47|(5

In 1337.[/quote]
lol congrats, you win the grand prize! :laugh:

[quote="lightrhee"]Korean: 수학 (Soo hak)
Hmm... this seems to be far off from the real pronounciation. Can someone do better? :?[/quote]
Actually, I think that's how my Korean friends pronounce that word.

It is interesting that "soo hak" sounds very similar to "su gaku" ("math" in Japanese, as kunny mentioned).

[quote="ZennyK"]Italian: matematiche [/quote]no you're wrong...:P we say matematica if u say matematiche you have to say scienze matematiche

[quote="mathclass"]chinese: shu xue.

KIND OF like sushuay.[/quote] try 数学

Most languages have one term for mathematics in the formal sense (a science, a discipline, a body of knowledge, a school curriculum), and one or more terms that denote a more basic process such as counting, reckoning, arithmetic, figuring, computing, numeration.   In the European and Slavic languages, and probably in most places whose mathematical tradition derives from Euclid, the formal word is usually a variant of the Greek "matematika", and the words for geometry and trigonometry also come from the Greek.  The informal words are typically older, and native to the language.

Bah, I was beaten to both 1337 and Russian :(.

Hmmph! ZennyK got Latin/Greek.

Old English: $eor\eth cr\ae ft$ (ay,-yorth'-craft)
Also of interest: $circumwyrde$ (mathematician)


linear algebra
matrix
vector
induction
linear algebra unsolved

Given an $ n\times n$ matrix $ A$ with $ n$ different eigenvalues $ \lambda_1$, $ \lambda_2$, $ \ldots$, $ \lambda_n$ with corresponding $ n$ eigenvectors $ x_1$, $ x_2$, $ \ldots$, $ x_n$. Prove that the set of $ n$ vectors $ \{x_1, x_2, \ldots, x_n\}$ is linearly independent.

We prove it by induction on $ n$.
$ \bullet \ n = 1$ is true.

$ \bullet$ Let's suppose the statement is true for $ n - 1$. We prove it for $ n$.

$ \bullet$ Let's take $ c_1,c_2,\cdots ,c_n \in \mathbb{R}\ (*)$ such that $ c_1x_1 + c_2x_2 + \cdots + c_nx_n = 0$
Then multiply by $ A$ both sides: $ A(c_1x_1 + c_2x_2 + \cdots + c_nx_n) = A\cdot 0$
One gets $ Ac_1x_1 + Ac_2x_2 + \cdots + Ac_nx_n = 0\Rightarrow c_1\lambda _1x_1 + c_2\lambda _2x_2 + \cdots + c_n\lambda _nx_n = 0 \ (**)$

Now, multiply both sides of $ (*)$ by $ \lambda _n$ and subtract it from  $ (**)$. We get:
\[ c_1x_1(\lambda _1 - \lambda _n) + c_2x_2(\lambda _2 - \lambda _n) + \cdots + c_{n - 1}x_{n - 1}(\lambda _{n - 1} - \lambda _{n}) = 0
\]
By induction hypotesis  we have that $ x_1,x_2,\dots x_{n - 1}$ are linearly independent so it must be:
\[{ c_1(\lambda _{1} - \lambda _{n}) = c_2(\lambda _2 - \lambda _n) = \cdots = c_{n - 1}(\lambda _{n - 1} - \lambda_{n}) = 0}
\]
Now because of the hypotesis on the eigenvalues being different: $ c_1 = c_2 = \cdots = c_{n - 1} = 0$.

Substituting the result in $ (**)$ and considering that $ x_n$ is an eigenvector it must be also $ c_n = 0$.

If A has n different eigenvalues $ \lambda _1, \lambda _2, ..., \lambda _n$, then the Vandermond determinant of $ \lambda _1, \lambda _2, ..., \lambda _n$ is not zero.

this is a problem in the contest of my university today :D
my idea solution is nearly like EUCLA but i don't use induction , i suppose $ X_1,..X_n$ are not independent and refer $ X_1,...,X_{n\minus{}1}$are not independent too. so repeat it n-2 times we have $ X_1,X_2$ are dependent and it is easy to get a contradiction.

Thank you for your solution, [b]EUCLA[/b]! My solution is similar to yours.

Assume that $ \text{rank} \{X_1, X_2, \ldots, X_n\}=r$. And WLOG, assume the set of $ \{X_1, X_2, \ldots, X_r\}$ is linearly independent.

If $ r<n$, we have:
\[ X_{r+1}=c_1X_1+c_2X_2+\cdots+c_rX_r\qquad (*)
\]
where $ c_1, c_2, \ldots, c_r\in\mathbb{R}$ and there exists at least one real number $ c_i\neq 0,\quad i=\overline{1, r}$.

Then multiply by $ A$ to both sides of $ (*)$, we conclude:
\begin{eqnarray*} AX_{r+1}&=&c_1AX_1+c_2AX_2+\cdots+c_rAX_r\\ \Rightarrow \lambda_{r+1}X_{r+1}&=&c_1\lambda_1X_1+c_2\lambda_2X_2+\cdots+c_r\lambda_rX_r\quad (1)\end{eqnarray*}
On the other hand, multiply both sides of $ (*)$ by $ \lambda_{r+1}$, we have
\[ \lambda_{r+1}X_{r+1}=c_1\lambda_{r+1}X_1+c_2\lambda_{r+1}X_2+\cdots+c_r\lambda_{r+1}X_r\quad (2)
\]
Combining $ (1)$ and $ (2)$, we get:
\[ c_1\left(\lambda_1-\lambda_{r+1}\right)X_1+c_2\left(\lambda_2-\lambda_{r+1}\right)X_2+\cdots+c_r\left(\lambda_r-\lambda_{r+1}\right)X_r=0
\]
Because the set of $ \{X_1, X_2, \ldots, X_r\}$ is linearly independent and the eigenvalues are different, so it implies that $ c_1=c_2=\cdots=c_r=0$ - Contradiction!

Hence, $ \text{rank}\{X_1, X_2, \ldots, X_n\}=n$, this completes our solution.


induction
calculus
integration
Pascal\u0027s Triangle

Prove that $\binom{n}{k}$ is an integer for $0\le k\le\ n$.

Let us consider $n$ objects in a row. We want to pick $k$ of these objects. How many ways can this be done?
There $n$ different orders that the objects can be put into. Of each of these, we can pick the first $k$. But the order of the first $k$ objects doesn't matter, so we divide by $k!$ (the number of orders the first $k$ can be in). Also, it doesn't matter what order the last $(n-k)$ are in, so we divide by $(n-k)!$. Hence the number of ways of choosing $k$ objects from a set of $n$ objects $=\frac{n!}{k!(n-k)!}=\binom{n}{k}$. By its interpretation, this must be an integer.

I think I once saw an algebraic proof to this problem..I wasn't really sure how it worked  :|

Thanks though :P

[hide="nice algebreic/combinatorical proof"]
We know from Pascal's Identity that

${n\choose r}+{n\choose{r+1}}={{n+1}\choose{r+1}}$
(you can prove this with algebra or with a combinatorical argument)

Using induction, we will show that the $n$th row of Pascal's triangle is integral

Base Case: n=1 ${1 \choose 0}= 1$, ${1 \choose 1}= 1$

Assume that the statement holds for $\forall n < k$. We know that ${k \choose 0}={k \choose k}= 1$.  For all other values of ${k \choose r}={{k-1}\choose r}+{{k-1}\choose{r+1}}$ since we assumed that the previous row is integral, we know that ${{k-1}\choose r}+{{k-1}\choose{r+1}}$ is integral, so the $k$th row is integral.[/hide]


calculus
trigonometry
logarithms
inequalities
algebra unsolved
algebra

prove that
$ 2^{\mid sin{x}\mid}+2^{\mid cos{x}\mid} \geq3 $ ($x \in R)$

for $x \in [0,\frac{\pi}{2}]$ it's true that $|sinx| \geq \frac{2}{\pi}x$ and also
$|cosx| \geq 1-\frac{2}{\pi}x$ so $2^{|sinx|}+2^{|cosx|} \geq 2^{\frac{2}{\pi}x}+2^{1-\frac{2}{\pi}x}=
t+\frac{2}{t}$ after putting $t=2^{\frac{2}{\pi}x}$, but of course $t+\frac{2}{t} \geq 3$ for $t \geq 1$ - what we obtain using - for instance - derivatives.
We can generalise this method for $[\frac{\pi}{2},\pi],[\pi,\frac{3\pi}{2}],[\frac{3\pi}{2},2\pi]$, replacing
$|cosx|$ and $|sinx|$ so we are done for $x \in R$.

[quote="Rafal"]...but of course $t+\frac{2}{t} \geq 3$ for $t \geq 1$ - what we obtain using - for instance - derivatives.[/quote]

But when $t = \sqrt{2} > 1$,
\[t + \frac{2}{t} = \sqrt{2} + \frac{2}{\sqrt{2}} = 2 \sqrt{2} < 3.\]

I was too lazy to check what I've written...shame on me :)

Here's a solution (yes, I'm using calculus, so sue me).

Let $f(x) = 2^{|\sin x|} + 2^{|\cos x|}$.  Then $f$ has period $\pi/2$, and furthermore, $f(x) = f(\pi/2 - x)$, so it suffices to prove that $f(x) \ge 3$ for $0 \le x \le \pi/4$.  On this interval, $f(x) = 2^{\sin x} + 2^{\cos x}$.

We claim that $f$ is increasing on $[0, \pi/4]$.  Note that
\[f'(x) = \log 2 \cdot 2^{\sin x} \cdot \cos x - \log 2 \cdot 2^{\cos x} \cdot \sin x.\]
So then, it suffices to prove that
\[\frac{2^{\sin x}}{\sin x} \ge \frac{2^{\cos x}}{\cos x}\]
for all $x \in (0, \pi/4]$ (we see that $f'(0) = \log 2 > 0$.)

Note that $\sin x$ is increasing on $(0, \pi/4]$ and $\cos x$ is decreasing on $(0, \pi/4]$, and furthermore $\sin \pi/4 = \cos \pi/4 = 1/\sqrt{2}$.  Thus, it suffices to prove that
\[g(x) = \frac{2^x}{x}\]
is decreasing on $(0, 1/\sqrt{2}]$.

We note that
\[g'(x) = \frac{2^x (x \log 2 - 1)}{x^2},\]
which is negative for $x < 1/\log 2$, and $1/\log 2 > 1 > 1/\sqrt{2}$.

I planned to put a similar solution in the morning :), however I thought that maybe there is a non-strictly analytic solution...but seems that there is no such one.

I have proved the following inequality: $2^{sinx}+2^{cosx}\ge\ 2^{\frac{2-\sqrt{2}}{2}} $  for any $x\in\mathbb{R}$ ;)

It is not difficult to show that $2^x\ge x^2+1$, for all $x\in [0,1]$. Therefore
\[2^{|sinx|}+2^{|cosx|}\ge sin^2x+cos^2x+2=3
\].

harazi posted an elementary soln here : http://www.mathlinks.ro/Forum/viewtopic.php?t=28960&highlight=

(it is one of the first problems i posted on ML)

And what about $2^x \geq x^2 + 1$? How do you prove it? (without calculus that is)


A straight line joins the points $ (\minus{}1,1)$ and $ (3,9)$. Its $ x$-intercept is:

$ \textbf{(A)}\ \minus{}\frac{3}{2}\qquad
\textbf{(B)}\ \minus{}\frac{2}{3}\qquad
\textbf{(C)}\ \frac{2}{5}\qquad
\textbf{(D)}\ 2\qquad
\textbf{(E)}\ 3$

the equation is $ y\equal{}2x\plus{}3$. setting this equal to $ 0$, we get $ 2x\equal{}\minus{}3$, and $ \boxed{x\equal{}\frac{\minus{}3}2}$

Alternatively realize that the y intercept where y=0 is 9/8 past (3,9) than (-1,1). So 4*9/8=9/2 and 3-9/2=-3/2


[quote="forevercontinue"]Let $ a,b,c \geq 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that :
\[ \frac {a}{\sqrt {8041b \minus{} 1683c^2 \plus{} 15201}} \plus{} \frac {b}{\sqrt {8041c \minus{} 1683a^2 \plus{} 15201}} \plus{} \frac {c}{\sqrt {8041a \minus{} 1683b^2 \plus{} 15201}} \geq \frac {3}{\sqrt {21559}}
\]
:([/quote]
This inequality not nice and easy :oops:
By Holder inequality,we have:
$ LHS^2(\sum \ a(8041b\minus{}1683c^2\plus{}15201)) \ge (a\plus{}b\plus{}c)^3$
$ (a^2b\plus{}b^2c\plus{}c^2a) \ge \frac{(ab\plus{}bc\plus{}ca)^2}{3} \ge 2(ab\plus{}bc\plus{}ca)\minus{}3$
So$ \sum \ a(8041b\minus{}1683c^2\plus{}15201) \le 4675(ab\plus{}bc\plus{}ca)\plus{}50278 \le 64303$
 Then, we have Q.E.D :wink: