function
Functional Analysis
absolute value
complex numbers
advanced fields
advanced fields unsolved

Let $ T$ be a unitary operator on a Hilbert space $ H$ such that its spectrum $ \sigma(T)$ is a countable set. Prove that there exists a sequence $ \{ n_{j}\}\in\mathbb{N}$ such that $ T^{n_{j}}\to I$ (the identity operator) in the strong operator norm topology.

It suffices to solve it for separable Hilbert spaces, since every finite subset of $ H$ is contained ina separable reducing subspace of $ T$. On a separable space, the problem is reduced, via the Spectral Theorem, to the following:

Let $ (X,\mu)$ be a finite measure space, and $ \varphi: X\to\mathbb C$ a measurable function with countable range and absolute value $ 1$ everywhere. Also, let $ T\equal{}M_{\varphi}$ be the multiplication operator on $ \mathcal L^{2}(\mu)$ induced by $ \varphi$ (defined by $ f\mapsto\varphi f,\ \forall f\in\mathcal L^{2}(\mu)$). Then we can find a sequence $ (n_{j})_{j}$ of positive integers such that $ T^{n_{j}}\to 1$ in the strong operator topology.

For every finite set $ \{z_{1},\ldots,z_{k}\}$ of complex numbers of absolute value $ 1$, we can find a positive integer $ n$ such that $ z_{i}^{n}$ are all arbitrarily close to $ 1$. Using this and the fact that $ \varphi$ has countable range, we can choose our sequence $ n_{j}$ so as to make $ \varphi^{n_{j}}$ converge pointwise to $ 1$. The fact that such a sequence $ n_{j}$ works is a consequence of Egorov's Theorem, stating that under the present circumstances the sequence $ \varphi^{n_{j}}$ converges uniformly to $ 1$ outside subsets of arbitrarily small measure.


i know for AIME and USAMO and etc problems you can go to kalva and get the whole test from each year

but i was wondering if they had a site like that for AMC 10A, 10B, 12A or 12B problems

anyone kno of ne good ones?

So far I have not discovered such a site. There are a few tests here and there but not many. You can order them for a relatively low price.

The Harold REiter page is the only one I know of. Google his name u will get to his website.

http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm

has some old AMC12/AHSME.  However, the old ones are fuzzy and bad quality, and the collection is not complete.

I suggest you either order the copies you want using the forms at:

http://www.unl.edu/amc/d-publication/publication.html

or get Problem Books I - VI from the online bookstore at:
http://www.maa.org

You get not only the  problems but the well-edited solutions as well.

Steve Dunbar
AMC Director

Yeah the solutions are awesome, and if you want to compare them to kalva, get ready to be blown out of the water :P And, as mentioned a few posts up, they are pretty inexpensive.


[hide]
well absolute magnitudes can't be negative, so both |a+b| and |a-b| equal 0, wherein a=b=0.  Thus the given exprsesion also simplifiefs to zero.

Absolute value cannot be negative, so they are both zero, aalong with a and b, therefore it simplifies to 0

The solutions are both 0 because absolute value can't be negative.

[hide]Since the expressions are in absolute value, the values can't be negative, therefore they must be equal to 0. So a and b are also both equal to 0. Therefore the equation must be equal to 0[/hide]

[hide]a and b are both zero, meaning that $a^{1999}\text{ is zero and } b^{1999}=0$[/hide]

0,[hide] because the |b|+|-b| =0 => b=0, a=0[/hide]

[hide]

since $  \mid a+b \mid + \mid a-b \mid = 0  $, then you can also simplify
$  \mid a^{1999}+ b^{1999} \mid + \mid a^{1999}- b^{1999} \mid = 0  $ as well because the rations are proportional

So the answer is $  0  $[/hide]


solve equation:
$7x^{2}+7x= \sqrt{\frac{4x+9}{28}}$


algebra
polynomial
superior algebra
superior algebra solved

Yes!!!
from the fact that
AC+BC=(A+B)C=ABC=A(B+C)=AB+AC=>BC=AB=>B+C=A+B=>C=A
doing the same for BC+CA you get that A=B so A=B=C
so we get that A^2=2A and so: A^3=4A so:
A+B+C=3A=3/4A^2=3/4ABC so m=3/4 :D

cheers!

AB-A-B=0
(A-I)(B-I)=I so A-I , B-I are invertible they commute
(B-I)(A-I)=I
BA-A-B=0
We find AB=BA, by symetry A,B,C commute by pair
Now
(AB)C=AC+BC=A+B+2C
A(BC)=AB+AC=2A+B+C this gives A=C by symetry A=B=C
now we get A^3=4A and continu like Lagrangia did.
Remark: A^3=4A, A vanishe the polynomial x^3-x=x(x-2)(x+2) split and simple roots, so A is diagonalisable in Mn(C), even the matrices B,C.
Since they commute by pair, A,B,C are diagonalisable in a same basis.


I'm not sure if this is the right level, as I'm not sure if it can be done with Intermediate level methods only. Please move if it is in the wrong forum.

$\prod_{k=2}^{\infty}(1-\frac{1}{k^{2}})$

[hide="what it seems to be"]I went out to the thirtieth term I believe, and it was really close to 1/2. [/hide]

[hide="Sure it can"]
Express $1-\frac{1}{k^{2}}$ as $\frac{(k-1)(k+1)}{k^{2}}$ and you'll find it telescopes pretty nicely.
[/hide]

I would say that it tends to $\frac{1}{2}$.

My logic:

$\frac{3}{4}\cdot \frac{8}{9}\cdot \frac{15}{16}\cdot \frac{24}{25}\cdot \frac{35}{36}\cdot \frac{48}{49}...$

So the products of the terms, factored in one at a time are:

$\frac{3}{4}, \frac{2}{3}, \frac{5}{8}, \frac{3}{5}$.

Clearly we see a pattern.  The sum up to the will always be of the form $\frac{1}{2}+\frac{1}{x}$.  As $x$ gets larger and larger, we see that the 1/x will become smaller and smaller.  Around infinity, we can just ignore it.  Thus the sum becomes $\frac{1}{2}$.

EDIT: Or you could listen to treething and it works out much more nicely.

[hide="One line"] $\prod_{k=2}^{n}\left( \frac{k-1}{k}\cdot \frac{k+1}{k}\right) = \frac{1}{2}\cdot \frac{n+1}{n}$ [/hide]

[quote="pianoforte"]$(1+1/2)(1+1/3)...=(3/2)(4/3)...=1/2$, telescopes.
$(1-1/2)(1-1/3)...=(1/2)(2/3)...=1$, telescopes.[/quote]

Absolutely insensible.  Clearly every term of the first product is greater than $1$ and clearly every term of the second is less than $1$, so neither result can be correct.

In fact, the first product diverges to infinity while the second converges to $0$, so this is not correct at all.

[hide="The relevant lemma"] $\prod_{k=1}^{\infty}(1+a_{k})$ converges [b]iff[/b] $\sum_{k=1}^{\infty}a_{k}$ converges. [/hide]

Yes wouldn't the top end up being $\lim_{n\rightarrow\infty}\frac{1}{n}$ and the bottom end up being $\lim_{n\rightarrow\infty}\frac{n}{2}$?

[quote="t0rajir0u"]
In fact, the first product diverges to infinity while the second converges to $0$, so this is not correct at all.

[hide="The relevant lemma"] $\prod_{k=1}^{\infty}(1+a_{k})$ converges [b]iff[/b] $\sum_{k=1}^{\infty}a_{k}$ converges. [/hide][/quote]
This is not correct by your definition of convergence. It is correct if we follow the usual conventions and say that a product diverges to 0 if it has partial limits converging to 0. Actually it is even more complicated than that. A product is said to converge if it has only finitely many zero factors and the partial products obtained when removing these factors converge to a non-zero number. The reason for this convention is that we want to be able to say things about convergent products, while this is not possible for products converging to 0 (basically anything can happen).

[quote="pianoforte"]$(1+1/2)(1+1/3)...=(3/2)(4/3)...=1/2$, telescopes.
$(1-1/2)(1-1/3)...=(1/2)(2/3)...=1$, telescopes.[/quote]

nope.

[hide="reason #1:"]The values are always greater than 1, and 3/2 times 4/3 is 2. That times 5/4 is 5/2. That tims 6/5 is 3. so on up to infinite.[/hide]

[hide="reason #2:"]The values are LESS than 1, and 1/2*2/3 is 1/3. That times 3/4 is one fourth. so on to 0.[/hide]

[hide="my reasoning:"] The denominator gets to 0 and the numbeator gets to infinite. So I say $\infty$.[/hide]


Will tthe new mathcounts book help me for nats?

no. the mathcounts book, written for mathcounts, will not help you with...mathcounts. sorry for the caustic sarcasm, but i feel that your question is rather...self-answerable (if that's a word) and that [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=270066]this[/url] would be a better place to post it.

Hey nike..r u buyin it?
Who is buyin it?
pm me ur name..

[quote="USAMO13"]Hey! Thats mean!!
You called me annoying, and purposefully put the color in white!!!![/quote]

purposefully is something done with an intention, or a purpose...i think you mean intentionally?

Btw, the tests in the book are SUPPOSED to help you prepare for states(as mentioned in the announcement). So, they're not meant for helping you prepare for nats.

I bought the book and it was about states level. Some of them were between chapter/state and some were slightly harder than states (in my opinion).


On the $ 8 \time 8$ grid shown, the smallest square regions are $ 1\frac12$ cm on a side. What fractional part of the grid is shaded? Express your answer as a common fraction.
[asy]for(int i = 0; i < 9; ++i){
draw((i,0)--(i,8)); draw((0,i)--(8,i));
}
path p = (0,0)--(0,2)--(2,2)--(2,0)--cycle;
fill(shift((0,1))*p,grey); fill(shift((3,0))*p,grey);
fill(shift((6,0))*p,grey); fill(shift((3,3))*p,grey);
fill(shift((0,6))*p,grey); fill(shift((3,6))*p,grey);
fill(shift((6,5))*p,grey);[/asy]

The size of each square is irrelevant, as long as they all have the same side lengths. Assume they are unit squares. Thus the total area is 64, and the shaded is 7 squares each of area 4 to total 28.

$ \frac{28}{64} \equal{} \frac{7}{16}$


Find the value of x in the interval 0 degrees < or = to x < 360 degrees that satisfies the equation

sqrt{4cos^2(x) - 1} - 1 = 0

Are you sure $ \cos(x) \equal{} \sqrt{\frac{1}{2}} \equal{} \pm\frac{1}{4}$ ?

The square root of 1/2 is $ \pm\1/\sqrt2$, or $ \pm\sqrt2/2$.

 Thus, $ \cos(x) = \pm\sqrt2/2$

$ x = 45, 135$ degrees

AHHHH!
Epic fail!
*face palm*
Slaps self.

Then edits post. To any who are confused, the post above has been edited to correct a silly error.

$ \sqrt {4\cos^2x \minus{} 1} \minus{} 1 \equal{} 0$

$ \Longleftrightarrow$ $ \sqrt {4\cos^2x \minus{} 1} \equal{} 1$

$ \Longleftrightarrow$ $ 2\cos^2x \equal{} 1$

$ \Longleftrightarrow$ $ 2.\frac {1 \plus{} \cos{2x}}{2} \equal{} 1$

$ \Longleftrightarrow$ $ \cos{2x} \equal{} 0$

$ \Longleftrightarrow$ $ 2x \equal{} k\frac {\pi}{2}, k\in Z$

$ \Longleftrightarrow$ $ x \equal{} k\frac {\pi}{4}, k\in Z$

But $ 0\le x < 2\pi$ $ \Longrightarrow k \equal{} 0, k \equal{} 1, k \equal{} 2, k \equal{} 3, k \equal{} 4, k \equal{} 6, k \equal{} 7$ then OK! :P

Actually, no: $ x \equal{} k \frac {\pi}{4}$ for $ k \in \{1, 3, 5, 7\}$, since $ k \in \{0, 2, 4, 6\} \Longrightarrow x \in \left\{0, \frac {\pi}{2}, \pi, \frac {3\pi}{2}\right\} \Longrightarrow 4 \cos^{2} x \in \{4, 0\}$.

Note that neither $ (\sqrt {3} \minus{} 1)$ nor $ ( \minus{} 1 \plus{} i)$ equal zero.

There are actually $ 4$ values that satisfy the restrictions of $ x$: $ 45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}$.

[quote="metafor"]
There are actually $ 4$ values that satisfy the restrictions of $ x$: $ 45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}$.[/quote]

Ack, I got lazy at the end. Wasn't thinking properly.

Meant to say $ (\frac {\pi}{4}, \frac{3\pi}{4}) \plus{}\frac{\pi}{2}$


geometry
analytic geometry
graphing lines
slope
quadratics
trigonometry
perpendicular bisector

given that $(0,0)$ and $(-4,3)$ are the coordinates of two points of a triangle, find a third point that makes this an equilateral triangle.

[quote="t0rajir0u"][hide="Messy solution 1"] The third point must lie on the perpendicular bisector of the line segment connecting those initial two points.  This bisector has slope $\frac{4}{3}$ and passes through the point $(-2, \frac{3}{2})$, so it is given by

$y-\frac{3}{2}= \frac{4}{3}(x+2) \Leftrightarrow$
$y = \frac{4}{3}x+\frac{7}{6}$

The third point must have a distance of $\sqrt{3^{2}+4^{2}}= 5$ from the origin, so we solve the quadratic equation

$x^{2}+\left( \frac{4}{3}x+\frac{7}{6}\right)^{2}= 25$

To find our messy answer. [/hide]
[hide="Messy solution 2"] The side length of the triangle is $5$, so we know the third point must be at some $(5 \cos \theta, 5 \sin \theta)$.

Moreover, letting $\phi = \arctan \frac{3}{-4}$ (so that $\sin \phi = \frac{3}{5}, \cos \phi = \frac{-4}{5}$) we have $\theta = \phi-60^{\circ}$, so

$x = 5 \cos \theta = 5 \cos (\phi-60^{\circ}) = 5 \cos \phi \cos 60^{\circ}+5 \sin \phi \sin 60^{\circ}=-4 \cdot \frac{1}{2}+3 \cdot \frac{ \sqrt{3}}{2}$

And

$y = 5 \sin \theta = 5 \sin (\phi-60^{\circ}) = 5 \sin \phi \cos 60^{\circ}-5 \cos \phi \sin 60^{\circ}= 3 \cdot \frac{1}{2}-4 \cdot \frac{ \sqrt{3}}{2}$ [/hide][/quote]
I like the first messy way better because there is no trig involved. :)
I solved your way for the first way because I like solving these kinds of equations. I found x to be
[hide]$x=\frac{\sqrt{739}}{10}$.[/hide]
I'm sure theres some mistake but I'm too tired.


[b]0. Introducción:[/b] 0.1 Conjuntos 0.2 Conceptos primitivos: punto, recta y plano 0.3 Numeros Reales 0.4 Desigualdades 0.5 La recta Real

[b]1. Rectas, planos, separacion:[/b] 1.1 Definiciones 1.2 Conjuntos Convexos

[b]2. Angulos y triangulos:[/b] 2.1 Definiciones Fundamentales. 2.2 Medida Angular. 2.3 Ángulos Rectos, Perpendicularidad. 2.4 Isometrias y Congruencias. 2.5 Clasificación de Triángulos. 2.6 Líneas Notables de un Triangulo. 2.7 Perpendiculares. 2.8 Desigualdades Geométricas. 2.9 La Distancia Entre Una Recta y Un Punto. La Desigualdad Del Triangulo. 2.10 Alturas de Triángulos.2.11 Traslaciones. 2.12 Ángulos Correspondientes, suma de ángulos de un triangulo. 2.13 Polígonos. 2.14 Cuadriláteros. 2.15 Secantes a Varias Rectas Paralelas

[b]3. Circunferencia:[/b] 3.1 Definiciones. 3.2 Arcos de Circunferencia. 4.3 Posiciones relativas de dos circunferencias. 4.4 Posiciones Relativas de un Triangulo y una Circunferencia. 4.5 Cuadriláteros circunscriptibles e inscriptibles. 4.6  Los Teoremas de Miquel.

[b]4. Geometría del triangulo I:[/b] 4.1 Introducción. 4.2 Medianas, Baricentro 4.3 Alturas, ortocentro. 4.4 Mediatrices y Circuncentro 4.5 Bisectrices y centros tritangentes. 4.6 Recta de Euler. 4.7 Circunferencia de Euler. 4.8 La recta de Simson. 4.9 Punto de Fermat y Triángulos de Napoleón 4.10 Geometria del Cuadrilátero.

[b]5. Semejanza y Proporcionalidad:[/b] 5.1 Conceptos y Homotecia. 5.2 Semejanza de Triángulos 5.3 Semejanza de Circunferencias, Centros de Similitud 5.4 La Circunferencia de Apolonio. 5.5. Razones Trigonometricas 5.7 Teoremas de Semejanza 5.8 Proyecciones, Proyeccion Ortogonal 5.9 Relaciones Métricas en los Triángulos. 5.10 Relaciones en triángulos especiales.  5.11 Polígonos Regulares, división de un segmento en media y extrema razón. 5.12 Teoremas de semejanza en la circunferencia. 5.13 Relaciones en Circunferencias

[b]6. Colinealidad y Concurrencia: [/b]6.1 Teoremas de menelao Y Ceva 6.2 Concurrencia y colinealidad en Circunferencias. 6.3 division Armonica 6.4 Curvas ortogonales. 6.6 n puntos y n linea completo.

[b]7. Geometría Analítica:[/b] 7.1 Plano Coordenado, Lugares Geometricos 7.2 Razones trigonometricas, C.T. 7.3 Relaciones métricas en los cuadriláteros. 7.4 Regiones Poligonales 7.5 Áreas de Triángulos y Cuadriláteros 7.6 Teoremas de áreas. 7.7 Razones de Áreas de figuras semejantes 7.8 Longitud de la circunferencia y area del circulo

[b]7. Geometria del triangulo II:[/b] 9.1 Coordenadas Baricentricas y Trilineales, Teorema de Routh 9.2 Puntos de Nagel, Gergonne y Circunferencia de Spieker 9.3 Polos y Polares con respecto al triangulo 9.4 Transformaciones Isogonales e Isotomicas, Producto de dos puntos 9.5 Las circunferencias de Lemoine 9.6 Circunferencias de Apolonio. 9.7 Puntos de Brocard, Circunferencia de Brocard, Triangulos de Brocard. 9.8 Circunferencias de Tucker: Circunferencia de Apolonio, Circunferencia Circunscrita, Circunferencia Coseno, Primera Circunferencia de Lemoine, Ccunferencia de Gallaty, Circunferencia de Kenmotu, Circunferencia de Taylor. Circunferencia de Adams 9.9 Teorema de Feuerbach. 9.10 Triangulos de Morley

[b]9. Circunferencias II:[/b] 9.3 Relaciones en Cuadriláteros Inscritos. 9.4 Relaciones en Cuadriláteros Circunscritos 9.5 Triangulos Mixtilineos 9.6 Potencia, Eje radical, Circunferencias Coaxiales, Centro Radical. 9.7 El problema de Apolinio.

[b]10. Homografias:[/b] 10.1 Coordenadas Homogeneas 10.2 Homografias 10.1 Perspectividades 10.2 Razon doble 10.3 Los seis valores de la razón doble 10.4 Teorema de Pappus 10.5 Puntos Autocorrespondientes. 10.6 Homografias entre rectas 10.7 Homografias entre conicas 10.8 Homografias entre rectas y 10.9 Involuciones 10.10 Relaciones de razon cruzada de los seis puntos de la involución. 10.11 Haz de involución con vértice en una circunferencia. 10.12 Lineas conjugadas en ángulos rectos 10.13 Involución Ortogonal 10.14 Teoremas de Desargues

[b]11. Inversion:[/b] 11.1 Definiciones. 10.2 Invercion de rectas y circunferencias. 10.3 Inversion de un teorema. 10.4 Circunferencia de Asimilitud 10.5 Inversiones Especiales (Circunferencias en Circunferencias iguales, en si mismas, en dos que se intersequen en un angulo dado) 10.6 Celda de Paucellier 10.7 Porisma de Steiner

[b]12. Las Conicas:[/b] 12.1 Polos y Polares 12.2 Definiciones 12.3 Relaciones Armónicas 12.4 Triángulo Polar y Autopolar 12.5 Relaciones Metricas de las Conicas.

[b]13. Geometria del triangulo III: [/b]14.1 Teoremas Varios  14.2 Triangulos isologicos, paralelogicos, Ortologicos. 14.3 Isopolos, Ortopolo 14.4 Ortocorrespondencia 14.5 La teoria de los Poligonos Armonicos 14.6 La Teoria de las figuras adjuntas 14.7 Las cónicas en el triangulo. 14.8 Porisma de Poncelet.

[b]14. Temas Selectos: [/b]15.1 El Arbelos 15.2 Problemas Selectos.

Interesante contenido ..
esto es el indice del libro del que hablas?? (A sequel to Euclid de Casey)

donde podes conseguir una copia de este?? ...

esta info seria de gran ayuda par los miembros de la comunidad ......
en especial para la gente apasionada con Geo. ;)

Gracias.

Carlos Bravo ;)
Lima Peru

No, el libro de casey recien estoy leyendo. Este indice es propuesto por mi  :D. La copia me la paso alguien, si quieres puedo pasarla, pero no se de que website lo sacaron

José Carlos

mmmmmmmmmm ok

aunque aki existe un prob de attachment (max 2.00 Mb)

Ya veremos como colgar ebooks y cosas por el estilo

Hola el libro me parece realmente interesante alguien por favor pordria subirlo para pordelo descargar


A little point mass is released from the periphery of a vertical circle and slides without friction
along a chord to the lowest point on the circle. The chord makes an angle $\theta$ as in figure.
Which value of $\theta$ minimizes the time it takes to slide down?

${0\leq \theta \leq \frac{\pi }{2}}$

$BC = 2 r \sin(\theta )$

$a = g \sin(\theta )$

${\Delta s = \frac{1}{2}a t^{2}}$

${2 r \sin(\theta ) = \frac{1}{2}g \sin(\theta )t^{2}\Rightarrow t = \sqrt{\frac{4 r }{g}}}$

[quote="yagaron"]
$a = g \cos(\theta )$
[/quote]
now the nagle the $\vec g$ makes with the line towards the end point is $\frac{\pi}{2}-\theta$ so the component along the movememnt of the particle is not $g\cos \theta$ but rather $g\ sin \theta$ .
this gives $2r \sin \theta$ $=$ $\frac{1}{2}g \sin \theta t^{2}$

so we get $t$ $=$ $\sqrt{\frac{4r}{g}}$
so the time taken is independent of $\theta$

As the point mass slides without friction, we can use conservation of mechanical energy (where h is the distance between point C and the horizontal plan in the diagram of Yagaron):

$mgh = \frac{1}{2}mv_{B}^{2}\Rightarrow  v_{B}= \sqrt{2gh}$.

The acceleration of the ball, for a given angle $\theta$, is $a = g\sin \theta$, so the acceleration is constant and so we have $v_{B}= v_{0}+at$. But $v_{0}= 0$, and so

$t = \frac{v_{B}}{a}= \sqrt{\frac{2h}{g}}\cdot \frac{1}{\sin \theta}$.

As $h = 2r\sin^{2}\theta$, then we have $t = \sqrt{\frac{4r}{g}}$.


induction
real analysis
real analysis unsolved

There exists a set $ B$ such that $ B$ and $ \mathbb{R} \setminus B$ contains no perfect set?

In other words, is it possible that $ B$ and $ \mathbb R\setminus B$ contain no closed sets of cardinality continuum? The answer is "yes". More generally, if $ X$ is a set and $ \mathcal F$ is a family of its subsets of cardinality $ \alpha$ such that each $ F\in \mathcal F$ has cardinality $ \alpha$, then there exists $ B\subset X$ such that both $ B$ and $ X\setminus B$ intersect any $ F\in\mathbb F$. For $ \alpha\equal{}\aleph_0$ this is fairly obvious: just choose points in $ B$ and $ X\setminus B$ one by one serving one $ F$ a time. In general, the proof is the same only the usual induction should be replaced with transfinite one. AC is inavoidable: without it, every set on the line can be Lebesgue measurable and, thereby, the answer may be "no" in some models.

By the way, csabafarkas, there is no need to push :).


I heard from many others that we get TWO free classes....

It is not that I desperately need two, but that I cannot decide on whether to take Olympiad Geometry/problem Solving/Inequality, or Intermediate Trig/Complex Number.

It used to be two free classes, but that proved to be way too much work for the students, so we scaled back the Independent Study to one free class.  You can sign up for a second the regular way if you want to, but be forewarned, it will be a lot of work!

I agree completely with this statement. I was taking Intermediate Trig and Intermediate Counting at the same time during the schoolyear and I felt quite overwhelmed although the classes were VERY beneficial!


let {h_{a}} as distance from Point A to side BC
and {m_{a}} as distance from Point A to M which middle point in BC

then, prove inequaility that [code]{h  _{a}} over {m  _{a}} + {h  _{b}} over {m  _{b}}  LEQ  {4} over {sqrt {3}}[/code]

hello, do you mean the inequality
$ \frac{h_a}{m_a}\plus{}\frac{h_b}{m_b} \le \frac{4}{\sqrt{3}}$?
Sonnhard.

Isn't it fairly obvious that

$ \frac{h_a}{m_a}\plus{}\frac{h_b}{m_b}\le 2<\frac{4}{\sqrt{3}}$

??


compute the ratio  :D

$ \dfrac{1\plus{}\dfrac{1}{1\cdot 3}\plus{}\dfrac{1}{1\cdot 3\cdot 5}\plus{}\dfrac{1}{1\cdot 3\cdot 5\cdot 7} \plus{}\dfrac{1}{1\cdot 3\cdot 5\cdot 7\cdot 9}\plus{}\dfrac{1}{1\cdot 3\cdot  5\cdot 7\cdot 9\cdot 11 }\plus{}\ldots\;\plus{}\;\ldots}{ 1\minus{} \dfrac{1}{2\cdot 3}\plus{}\dfrac{1}{2^2\cdot 5\cdot 2!}\minus{}\dfrac{1}{2^3\cdot 7\cdot 3!}\plus{}\dfrac{1}{2^4\cdot 9\cdot 4!}\minus{}\dfrac{1}{2^5\cdot 11\cdot 5!}\plus{}\ldots \;\plus{}\;\minus{}\ldots}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193943]This[/url] unproved result might help :D

Making the answer ofc $ \sqrt {e}$   :D


trigonometry
limit
integration
function
calculus
derivative
calculus computations

Let sequence {$ {x_n}$}| $ x_1 \equal{} 2008$ and $ x_{n \plus{} 1} \equal{} \cos{x_n}$. Prove that there is exist $ \lim_{n\to\infty} x_n$

suppose $ x_{1} \equal{} c$
then easly seen that :
$ |x_{n}| < 1$ for $ n \geq 2$
thus I can say that :
$ \cos x_{n} \approx 1 \minus{} \frac {x_{n}^{2}}{2} \plus{} o(x_{n}^{2})$
such formula $ x_{n \plus{} 1} \approx 1 \minus{} \frac {x_{n}^{2}}{2} \plus{} o(x_{n}^{2})$ change value of $ x_{n}$ very slovely hence rougly speaking
we can wrote : $ x_{n \plus{} 1} \minus{} x_{n} \approx x'_{n}$
therefore we have :
\[ \int_{2}^{N} \frac {dx_{n}}{1 \minus{} x_{n}^{2}/2 \minus{} x_{n}} \approx const \plus{} N
\]
And
\[ x_{n} \approx const_{3}\tanh\left( const_{1}(N \plus{} const_{2}) \plus{} const_{5}\right)
\]
then it's obviusly that $ \lim_{n \to \infty}x_{n}$  exist.

[quote="Extremal"]
such formula $ x_{n \plus{} 1} \approx 1 \minus{} \frac {x_{n}^{2}}{2} \plus{} o(x_{n}^{2})$ change value of $ x_{n}$ very slovely hence rougly speaking ...
[/quote]
Sorry, I don't understand your solution.
What means "change value very slovely"?
We don't know anything about function $ r(x_{n}) \equal{} o(x_{n}^2)$.
We only know, that $ \lim_{x\to 0} \frac {r(x)}{x^2} \equal{} 0$.

But we can see that for all $ n$ $ x_n$ not near 0.

Here is a solution for Japanese High School students.

Let $ g(x) \equal{} \cos x \minus{} x$, we have $ g'(x) \equal{} \minus{} \sin x \minus{} 1\leq 0$ and $ g(0) \equal{} 1 > 0\ ,\ g\left(\frac {\pi}{2}\right) < 0$, thus $ g(x)$ is a decreasing function and by Intermidate value theorem, $ f(x) \equal{} 0$ has only one solution in $ \left(0,\ \frac {\pi}{2}\right)$, denote the solution by $ \alpha$. Let $ f(x)\equal{}\cos x$,
Now $ x_{n \plus{} 1} \equal{} f(x_n)$ and $ \alpha \equal{} f(\alpha)$, by Mean value theorem

we have $ |x_{n \plus{} 1} \minus{} \alpha| \equal{} |f(x_n) \minus{} f(\alpha)|$

$ \equal{} \left|\frac {f(x_n) \minus{} f(\alpha)}{x_n \minus{} \alpha}\right \parallel{}x_n \minus{} \alpha| \equal{} f'(c_n)|x_n\minus{}\alpha|\ (c_n\ is\ between\ x_n\ and\ \alpha )$.

$ \equal{} |\sin c_n\parallel{}x_n \minus{} \alpha|$. Let $ r \equal{} |\sin c_n|\ (0 < r < 1)$, we have $ 0\leq |x_n \minus{} \alpha| < r|x_{n \minus{} 1} \minus{} \alpha| < r^2|x_{n \minus{} 2} \minus{} \alpha| < \cdots < r^{n \minus{} 1}|x_1 \minus{} \alpha|$.

Since $ \lim_{n\to\infty} r^{n \minus{} 1} |x_1 \minus{} \alpha| \equal{} 0$, yielding $ \lim_{n\to\infty} |x_n \minus{} \alpha| \equal{} 0\Longleftrightarrow \lim_{n\to\infty} x_n \equal{} \alpha$.

[quote="kunny"]
$ \equal{} |\sin c\parallel{}x_n \minus{} \alpha|$. Let $ r \equal{} |\sin c|\ (0 < r < 1)$, we have $ 0\leq |x_n \minus{} \alpha| < r|x_{n \minus{} 1} \minus{} \alpha| < r^2|x_{n \minus{} 2} \minus{} \alpha| < \cdots < r^{n \minus{} 1}|x_1 \minus{} \alpha|$.
[/quote]
I think, you can't write $ r^2$, $ r^3$, ..., $ r^{n \minus{} 1}$, because $ r$ is different for different $ x_n$.

You wrote $ r \equal{} f'(c)$, but it's only for one value $ n$. It must be $ r_{n} \equal{} f'(c_n)$.

I mean:

$ 0\leq |x_n \minus{} \alpha| \equal{} r_{n \minus{} 1} |x_{n \minus{} 1} \minus{} \alpha| \equal{} r_{n \minus{} 1} r_{n \minus{} 2} |x_{n \minus{} 2} \minus{} \alpha| \equal{} ... \equal{} r_1 r_2 r_3 ... r_{n \minus{} 1} |x_1 \minus{} \alpha|$, where $ r_i$ between $ x_i$ and $ \alpha$.

But it must be all $ r_i$ from $ [0, 1 \minus{} \epsilon]$ - then $ lim_{n\to\infty} r_1 r_2 ... r_n |x_1 \minus{} \alpha| \equal{} 0$.

kunny, how can you prove, that exist $ \epsilon > 0$ that all $ r_i$ from $ [0, 1 \minus{} \epsilon]$ ?

$ x_2$ is between $ \minus{} 1$ and $ 1$. On $ [ \minus{} 1,1]$, the derivative $ \sin x$ is between $ \minus{} \sin 1$ and $ \sin 1$, which are strictly less than $ 1$ in absolute value. Apply the Banach contraction principle; the iteration converges to the unique fixed point.

kunny's argument is essentially this, proving instead of quoting the theorem usually seen in more advanced analysis classes. To make it properly rigorous, use $ r\equal{}\sin 1$ and $ |f(x_n)\minus{}f(a)|\equal{}|\sin c|\cdot |x_n\minus{}a|<r|x_n\minus{}a|$. That gives the desired exponential decay of error.

I'm not sure about error in my first post , I need to check such "roghly" steps , but now I have no time to do this. it must be wrong since $ x_{n}$ don't tends to $ 0$.  Anyway , when I read problems in much times I do not spend a lot of time on the problem,  (it's not good) and hence my post looks as a spam  :lol:  .
Now I try to think a bit and then post some idea.
in such problems in much times works kunny's argument , and also geometrical interpretations.
let's try to consider my picture.
I consider two functions : $ \cos$ and $ cos^{ \minus{} 1}$ on the interval $ [0, \pi/2]$
let's denote intersection of this two graphs by $ F$ (it will be limit or his symetric on $ OY$).  this point also satisfies equation:
$ \cos(x) \equal{} cos^{ \minus{} 1}x$ on $ 0,\pi/2$
suppose $ a$ is any real number . and consider $ \cos a$
suppose $ \cos a \in [0,1]$ (case $ \cos a \in [ \minus{} 1, 0)$ considered similary)
also suppose that $ \cos a \in [0,M)$  if $ \cos a \equal{} M$ then everything is done . and case $ \cos a \in (M,1)$ can solve similary any mathlinker's from this forum  :) by similary first.
Ok.
Now let's denote $ \cos a \equal{} A(0)$ as on picture. and let's it will be first term of our sequance say $ x_{1}$ easly seen as on the picture that  $ x_{2}$ will be equal  to the proection on $ OX$ of  $ A(1)$ , similary for $ x_{3} \to A(2)$, also $ x_{4} \to A(3)$ and e.t.c.
finnaly the point tends to $ F$,. you can find many way how to explain: why it tends to   $ F$.
 :roll:


vector
linear algebra
matrix
linear algebra unsolved

Prove the following two theorems:

1. The column vectors of a matrix form a linearly dependent set if and only if there is a column having no pivot.

2. The column vectors of a matrix form a linearly independent set if and only if there is a pivot position in each column of the matrix.

Do you think the theorems are different? :lol:


function
real analysis
calculus
derivative
topology
real analysis unsolved

This must have been posted, but I can't find it...

Given a closed set $ E$, is there a $ n$ times (or even infinitely) differentiable function $ f$ such that $ f^{\minus{}1} (0) \equal{} E$?

The name is spelled Walter [u]Rudin[/u].

Are we working on $ \mathbb{R}$ or $ \mathbb{R}^n?$

I'll assume the former, which is easier.

Consider $ \varphi(x)\equal{}\begin{cases}e^{\minus{}\frac1{x(1\minus{}x)}}&\text{if }0<x<1\\0&\text{otherwise}\end{cases}.$

Then $ \varphi\in C^{\infty}$ and $ \varphi(x)\ne0$ precisely when $ x\in(0,1).$

Suppose $ E\subset \mathbb{R}$ is closed. Then $ U\equal{}\mathbb{R}\setminus E$ is open. We have a theorem that states that $ U$ can be written uniquely as a (finite or) countable union of disjoint open intervals. That is, $ U\equal{}\bigcup_kI_k$ where all but at most two of the $ I_k$ are bounded open intervals.

If $ I_k\equal{}(a_k,b_k),$ then write $ \psi_k(x)\equal{}e^{\minus{}\frac1{b_k\minus{}a_k}}\varphi\left(\frac{x\minus{}a_k}{b_k\minus{}a_k}\right).$

If $ I_k\equal{}(a_k,\infty),$ then write $ \psi_k(x)\equal{}\begin{cases}e^{\minus{}\frac1{x\minus{}a_k}}&\text{if }x>a_k\\0&\text{otherwise}\end{cases}.$

If $ I_k\equal{}(\minus{}\infty,b_k),$ then write $ \psi_k(x)\equal{}\begin{cases}e^{\minus{}\frac1{b_k\minus{}x}}&\text{if }x<b_k\\0&\text{otherwise}\end{cases}.$

Now write $ f(x)\equal{}\sum_k\psi_k(x).$

My reason for that scaling factor of $ e^{\minus{}\frac1{b_k\minus{}a_k}}$ in the definition of $ \psi_k$ was so that the sum for the $ n$th derivative of $ f$ converges uniformly for every $ n,$ and hence $ f\in C^{\infty}.$ And we have arranged to have $ f(x)\equal{}0$ precisely when $ x\not\in U,$ in other words, $ x\in E.$


Prove or disprove this

$\sum^m_{i=1}\binom{m}{m-i+1}\binom{n-1}{n-m+i-1}=\binom{n+m-1}{n}$ for all $n,m\in \math{N}$

I think the identity is true with additional condition $n\geq m$. Suppose we have it, then interpret combinatorially both sides.

Suppose we have two piles of coins - one with $m$ and one with $n-1$ coins - we want to choose $n$ coins. Right side is obvious - we take normally randomily $n$ coins.

Try to interpret left side: call $T_i$ such a choice when we take $m-i+1$ coins from the pile with $m$ coins and the rest from $n-1$ pile (this is the moment in which I use $n\geq m$). Clearly number of choices of $T_i$ is $\binom{m}{m-i+1}\binom{n-1}{n-m+i-1}$. As $\sum _{i=1}^m T_i$ goes through all the possible choices of $n$ coins we established the result

so generally speaking it's not true for all $n,m$ - take some small $n$ and large enough $m$ to get a contradiction

Ya , i agree with you Megus that $n\geq m$ must hold .
Since we have $n-m+i-1\geq n-m\geq 0$ is necessary .

The combinatorial prove seems ok to me but im wondering is there a way to prove it algebrically ? :?


Assume that $ f(x),g(x)$ are continuous function on $ \mathbb R$ and $ f(g(x))=g(f(x)),\ \forall x\in \mathbb R$. Prove that if the equation $ f(x)\equal{}g(x)$ doesn't have real roots, then the equation $ f(f(x))\equal{}g(g(x))$ doesn't have real roots either.

[quote="tdl"]Assume that $ f(x),g(x)$ are continue function on $ R$ and $ f(g(x)) \equal{} g(f(x))\ \forall x\in R$. Prove that: If equation $ f(x) \equal{} g(x)$ hasn't real root then equation $ f(f(x)) \equal{} g(g(x))$ hasn't real root too.[/quote]

Let $ h(x) \equal{} f(x) \minus{} g(x)$. $ h(x)$ is continuous.

Suppose $ \exists a\in \mathbb{R}$ such that $ f(f(a)) \equal{} g(g(a))$

$ h(f(a)) \equal{} f(f(a)) \minus{} g(f(a)) \equal{} g(g(a)) \minus{} g(f(a))$ $ \equal{} \minus{} (g(f(a)) \minus{} g(g(a)))$ $ \equal{} \minus{} (f(g(a)) \minus{} g(g(a)))$ $ \equal{} \minus{} h(g(a))$

And so, since $ h(x)$ is continous, $ h(x)$ has a root $ b\in [f(a),g(a)]$ and so $ f(b) \equal{} g(b)$

Hence the result : if equation $ f(x) \equal{} g(x)$ has no real root then equation $ f(f(x)) \equal{} g(g(x))$ has no real root too


function
calculus
derivative
calculus computations

If $f$ is $C^{2}$, prove that $f$ cannot have a local max or min at an inflection point.

[quote="§outh§tar"]What is an inflection point?[/quote]

an inflection point is  a point where the second derivated function vanish
f''(x)=0 and f' dont changes the sign around this point

sorry for my bad english

[quote="countdownmath"]as far as i know, at an inflection point,
both $f'(x)=f''(x)=0$.[/quote]
by the Extremum theorem. because after seccond derivetiv is third derivetiv.
Sulution
Suppose that the function has local extremum in point $a$. if we write this function by Taylor  $f(t)-f(a)-f'(a)(t-a)-\frac{f''(a)(t-a)^{2}}{2}=O((x-a)^{3})$ $\Rightarrow $ $f(t)-f(a)=O((x-a)^{3})$ wich change his sign when $x>a$ and $x<a$ because in $O$ we have third power. we have wrong suppose therefore dano. Sorry My english

[quote="jmerry"]Actually, an inflection point is a point where the second derivative changes sign, not merely a zero of the second derivative. The first derivative can be anything.[/quote]
ok then $o(x-a)^{2}$  ( it is small number you can easly seen) has a same sign wich have $f''$ therefore we can write Tayler formula by next: $f(x)-f(a)=(x-a)^{2}(f''(x)+\alpha)$ this $f(x)-f(a)$ change his sign because$f''(x)-\alpha$  have same signt wich have $f''(x)$. $\alpha$ is smaller than $f''$ done. :wink:


vector
linear algebra
matrix
abstract algebra
superior algebra
superior algebra unsolved

Find an operation on real number which statisfies associate and non-commutative.

You want to make $\mathbb{R}$ into a noncommutative pseudoring, with the usual addition. This does not sound easy to me, especially if you want to explicit such an operation ... [well, I won't try but wait and see what other people propose; if you do not impose the usual addition, then the existence is easy, since you only need a bijection between $\mathbb{R}$ and $M_{2}(\mathbb{R})$].

[quote="Beginner"]Never mind i've got it. $a * b = b$
thanks

Now let make it harder:

Find an operation on real number which statisfies associate, non-commutative and distributive (right and left).[/quote]
How can we consider distributive?
Because we have only one operation :?:

we must define it or use that already defined "$+$" dan "$\times$". My instructor said that there are some of them.

Can anybody find another example other than $a*b=b$ for the first question (almost every my classmates have the same example) ?

[quote="Beginner"]we must define it[/quote]

If you mean that you can define your addition then it's easy, by what I said before. Otherwise, if you keep the regular addition on $\mathbb{R}$, well, I don't see how to do (but nothing tells me there's not a solution).

One can give the additive group of IR a scalar multiplication that turns it into an IR vector space that is isomorphic to IR^4 (use that IR and IR^4 are isomorphic as IQ vector spaces). This space can be regarded as the matrix space M_2(IR), which can be given the usual matrix multiplication, which fits your request, Beginner.
The only problem with this example is, that the scalar multiplication can probably not be given explicitly.

LiMa

Yes Julien_santini, we can give an example by making a bijection from $\mathbb{R}$ to $M_{2}(\mathbb{R})$.

I have another problem that is my self problem and i don't have any answer yet, that is:

Is there any such operator if we add the closer condition?

that is the operator $\mathbf{R}\times \mathbf{R}\rightarrow \mathbf{R}$.

thanks for all.

We have already considered maps R x R -> R. Could you specify your question?

who gave that? and i didn't see it here, perhaps i'm missing something.  :(

@ Lima: I probably misunderstood what you said: are you asserting that $\mathbb{R}$ and $\mathbb{R}^{4}$ are isomorphic as $\mathbb{R}$-vector spaces ?

@julien_santini: First, they are isomorphic as usual $Q$-vector spaces, and hence als abelian groups. This isomorphism carries the usual scalar multiplication on $R^{4}$ to an unusual scalar multiplication on $R$, s.t. $R^{4}$ and this modified $R$-vector space $R$ are isomorphic.

I meant it exactly as -oo- said. I equip the additive group of IR with an "unusual scalar multiplication" (induced by a IQ vector space isomorphism between IR and IR^4), which turns it into a 4-dimensional IR vector space.

Precisely:
Let $f: \mathbb R \to  \mathbb R^{4}$ be a IQ vector space isomorphism (both spaces have dimension card(IR)).
For a scalar r in IR and a vector v in IR let
$r.v : = f^{-1}(r f(v))$
An easy calculation shows that (IR, +) with this scalar multiplication is an IR vector space, and that f is a IR vector space isomorphism.

This "unusual IR^4" gives rise to an "unusual" multiplication (coming from the matrix multiplication of IR^4 regarded as M_2(IR)) that is distributive, associative, even has a unit, and is noncommutative.

LiMa

I got it now, thank you both for the precisions  :)


Hello everyone.
I am in desperate need of help with my statistics problem solving homework..  I have to pass this class in order to get my degree.  I have exausted myself trying to figure out how to do these problems.  Here is one:

A certain lottery is won by selecting the correct five numbers from 1, 2,------32.  The probabilty of winning the game is 1/201,376.  What is the probability of winning if the rules are changed so that in addition to selecting the correct five numbers, you must now select them in the same order as they are drawn?

Can someone PLEASE help me understand these problems.
Thank You
Felicia

For the first part, we count the number of ways of selecting $5$ tickets (order doesn't matter here) out of $32$ tickets. This is just $^{32}\text{C}_{5}= 201,376.$ But only one of the ways favors a win; hence the probability of winning is $\frac{1}{201376}.$

For the second part, we count the number of ways of selecting $5$ tickets out of $32$ tickets such that order is important. This number is just $^{32}\text{P}_{5}= 24,165,120.$ But only one of the ways favors a win. Hence the probability of winning the game is $\frac{1}{24165120}.$

Here's hoping you pass the course with flying colors.  :wink:

remember,
combinations -> order does not matter
permutations -> order does matter

Easy way to remember it is that a combination lock should in fact be called a permutation lock


Leonardo's Limo Service charges a $ \$7.50$ flat rate plus $ \$1.50$ per mile traveled. Jack and Jill rented the limo for a trip up the hill and it cost them $ \$19.50$. How many miles did they travel up the hill?

If the total cost was 19.50, and the initial cost was 7.50, the cost of just the miles would be 19.5-7.5 = 12.
12/1.5 (the rate per mile) = 8. So, they traveled [b]8[/b] miles.

$ \text{travel}\equal{}\text{shop}\to\text{hill}\to\text{shop}?$
or
$ \text{travel}\equal{}\text{shop}\to\text{hill}?$

If first, $ 8$ miles.
But second, $ 4$ miles.

[quote="myyellowducky82"][quote="Kouichi Nakagawa"]$ \text{travel} \equal{} \text{shop}\to\text{hill}\to\text{shop}?$
or
$ \text{travel} \equal{} \text{shop}\to\text{hill}?$

If first, $ 8$ miles.
But second, $ 4$ miles.[/quote]
I don't really understand what you mean by the
$ \text{travel} \equal{} \text{shop}\to\text{hill}$
and $ \text{travel} \equal{} \text{shop}\to\text{hill}\to\text{shop}?$[/quote]

travel is
from a rent-a-car shop to the hill?
or
from a rent-a-car shop to the hill and return to the rent-a-car shop?

dumb me >.<
I believe they just went from the shop to the hill and that's it. The problem never said they went back.


trigonometry
quadratics
algebra proposed
algebra

Let $ a$, $ b$, $ c$ be real numbers such that for every two of the equations
\[ x^2\plus{}ax\plus{}b\equal{}0, \quad x^2\plus{}bx\plus{}c\equal{}0, \quad x^2\plus{}cx\plus{}a\equal{}0\]
there is exactly one real number satisfying both of them. Determine all possible values of $ a^2\plus{}b^2\plus{}c^2$.

[quote="FelixD"]Let $ a$, $ b$, $ c$ be real numbers such that for every two of the equations
\[ x^2 \plus{} ax \plus{} b \equal{} 0, \quad x^2 \plus{} bx \plus{} c \equal{} 0, \quad x^2 \plus{} cx \plus{} a \equal{} 0\]
there is exactly one real number satisfying both of them. Determine all possible values of $ a^2 \plus{} b^2 \plus{} c^2$.[/quote]
If I'm not mistaken,the possible values of $ a^2 \plus{} b^2 \plus{} c^2$ are $ 0$ and $ 6$.If my answer is correct ( however,I'm not sure about it,because I'm a bit tired now and I did all the work in mind  ),I may  post my solution tomorrow, since It's already time to sleep now  :P

[quote="FelixD"]Let $ a$, $ b$, $ c$ be real numbers such that for every two of the equations
\[ x^2 \plus{} ax \plus{} b \equal{} 0, \quad x^2 \plus{} bx \plus{} c \equal{} 0, \quad x^2 \plus{} cx \plus{} a \equal{} 0\]
there is exactly one real number satisfying both of them. Determine all possible values of $ a^2 \plus{} b^2 \plus{} c^2$.[/quote]

We have two possibilities :

1) All three equations share a common root and so :
$ x^2\plus{}ax\plus{}b\equal{}(x\minus{}u)(x\minus{}v)$
$ x^2\plus{}bx\plus{}c\equal{}(x\minus{}u)(x\minus{}w)$
$ x^2\plus{}cx\plus{}a\equal{}(x\minus{}u)(x\minus{}t)$

Solving this implies $ v\equal{}w\equal{}t\equal{}\frac{\minus{}u^3\plus{}u^2\minus{}u}{u^3\plus{}1}$ ad so no solution because we want exactly one common root between any two equations and not two).

2) All three equation share three roots in a circular manner and so :
$ x^2\plus{}ax\plus{}b\equal{}(x\minus{}u)(x\minus{}v)$
$ x^2\plus{}bx\plus{}c\equal{}(x\minus{}v)(x\minus{}w)$
$ x^2\plus{}cx\plus{}a\equal{}(x\minus{}w)(x\minus{}u)$

Solving this gives $ u^4\plus{}u^3\minus{}4u^2\minus{}3u\plus{}2\equal{}0$ $ \iff$ $ (u\plus{}2)(u^3\minus{}u^2\minus{}2u\plus{}1)\equal{}0$

$ u\equal{}\minus{}2$ implies $ u\equal{}v\equal{}w\equal{}\minus{}2$ and does not fit the requirement "$ u,v,w$ pairwise different". And so $ u,v,w$ are the three roots of $ u^3\minus{}u^2\minus{}2u\plus{}1\equal{}0$

So $ u\plus{}v\plus{}w\equal{}1$ and $ uv\plus{}vw\plus{}wu\equal{}\minus{}2$

Then $ a^2\plus{}b^2\plus{}c^2\equal{}(u\plus{}v)^2\plus{}(v\plus{}w)^2\plus{}(w\plus{}u)^2$ $ \equal{}2(u\plus{}v\plus{}w)^2\minus{}2(uv\plus{}vw\plus{}wu)$ $ \equal{}6$

Hence the result : If such $ a,b,c$ exist, then $ \boxed{a^2\plus{}b^2\plus{}c^2\equal{}6}$

(and, indeed, they exist : choose $ (u,v,w)\equal{}(2\sin(\frac{\pi}{14}),2\sin(\frac{5\pi}{14}),\minus{}2\sin(\frac{3\pi}{14}))$)
Notice also that $ a^2\plus{}b^2\plus{}c^2\equal{}0$ cant be an answer since then $ a\equal{}b\equal{}c\equal{}0$ and all equation have two roots in common ($ 0$ and $ 0$) and not exactly one

[quote="pco"]Notice also that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$ cant be an answer since then $ a \equal{} b \equal{} c \equal{} 0$ and all equation have two roots in common ($ 0$ and $ 0$) and not exactly one[/quote]
My solution is very similar to yours pco ( a bit simpler,I must say  :P  ).However,I still can't understand this last point.If $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$ ( $ \Rightarrow a \equal{} b \equal{} c \equal{} 0$ ),then for every two of the equations : $ x^2 \equal{} 0,x^2 \equal{} 0$ and $ x^2 \equal{} 0$ there is exactly one real number which satisfies both of them,namely $ 0$  :roll:

Then $ a\equal{}b\equal{}c\equal{}4$ works also, which means $ a^2\plus{}b^2\plus{}c^2\equal{}48$ is also a solution

[quote="Rafikafi"]Then $ a \equal{} b \equal{} c \equal{} 4$ works also, which means $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 48$ is also a solution[/quote]

Quite good response, thanks :)

The problem is always the same : does $ x^2 \equal{} 0$ have one root or two roots (one double).

If you agree that $ x^2 \equal{} 0$ has two roots (as all quadratic, if we consider them in $ \mathbb C$), then all pair of equations have both roots equal.

[quote="Rafikafi"]Then $ a \equal{} b \equal{} c \equal{} 4$ works also, which means $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 48$ is also a solution[/quote]
You're right,thank you ( BTW,$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 48$ is the only left solution ) ! Indeed,in my solution I used uncarefully $ a \equal{} b \equal{} c \Rightarrow a \equal{} b \equal{} c \equal{} 0$ which is not true ( Note that I did all the work in mind and I was really tired as noted above :D  ),because $ a \equal{} b \equal{} c \Rightarrow$ The equation $ x^2 \plus{} ax \plus{} a \equal{} 0$ has only one real  solution $ \Rightarrow$ The discriminant $ \equal{} a^2 \minus{} 4a \equal{} 0 \Rightarrow a \equal{} 0$ or $ a \equal{} 4 \Rightarrow a \equal{} b \equal{} c \equal{} 0$ or $ a \equal{} b \equal{} c \equal{} 4$.

In conclusion,the only possible values of $ a^2 \plus{} b^2 \plus{} c^2$ are $ 0,6$ and $ 48$ ( I consider that the equation $ (x \minus{} a)^2 \equal{} 0$ has only one solution $ a$ )


[quote="conceive"][quote="myc"]How do you make a PDF file in crimson editor?  In detailed please.

Thanks
myc[/quote]

Note that this is a wrong forum to ask this question!! I would no wonder if some moderator will delete it..[/quote]

Very few topics are deleted except for those that are extreme spam, inappropriate, advertising, etc.  Additionally, I don't think this is really the wrong forum.  It might also fit in the computer science forum, but I think Round Table is fine also.

Also, if you believe a thread is misplaced, don't shout about it (I take the double exclamation points to be equivalent to shouting).  If noting that something is in the wrong forum is the full extent of your post, you should just send a PM to a moderator of the forum that the post is in.

[b]Moderators [/b]please delete posts number 2-8 and move it to LaTeX forum. Thanks.

Sorry about the posting in the wrong forum, I didn't know there was a Latex forum.  Thanks for moving the post.

"To make a PDF document from your LaTeX source file, locate the drop-down box that says LaTeX=>DVI in TeXnicCenter "

Does anyone know how to do this in Crimson Editor?  Is there any other way to turn tex files into PDFs without having any background in computer coding.

Thank you

I'd work on my temper if I were you conceive.

[b]Moderators[/b] please delete posts number 2-9. There are here for nothing.

[quote="myc"]Sorry about the posting in the wrong forum, I didn't know there was a Latex forum.  Thanks for moving the post.

"To make a PDF document from your LaTeX source file, locate the drop-down box that says LaTeX=>DVI in TeXnicCenter "

Does anyone know how to do this in Crimson Editor?  Is there any other way to turn tex files into PDFs without having any background in computer coding.

Thank you
[/quote]

Well I thought about it. As far as I know you cannot create a pdf file in Crimson editor. There can be only div files created. You will have to install some other software, eg TeXnicCenter. There you can make a pdf.

nice day
conceive

[quote="conceive"]Well I thought about it. As far as I know you cannot create a pdf file in Crimson editor. There can be only div files created. You will have to install some other software, eg TeXnicCenter. There you can make a pdf.[/quote]Not true -- I use Crimson Editor to compile LaTeX files to PDF all the time!

Choose "Configure User Tools..." from the Tools menu.

Pick one that's not yet in use.

Command: C:\texmf\miktex\bin\pdflatex.exe  (or wherever your pdflatex compiler is located; this is the default location for MikTeX installations in Windows)
Argument: $\$$(FileName)
Initial dir: $\$$(FileDir)

Check the "Capture output" and "Save before execute" boxes (though this is not required)

The main advantage of using Crimson Editor for LaTeX is that it recognizes LaTeX syntax (surprisingly, it's better than TeXnic Center in this regard).

The main disadvantage is that you have to close the PDF file in Adobe Reader before you recompile it.  (TeXnic Center can send a DDE command to do this automatically, but I don't know how to make Crimson do this.)


[quote="quantum leap"]Are we allowed to refer to books for these arml problems?[/quote]

Since this isn't an official contest or anything and nothing's at stake (unless you count the points you have on the scoreboard :P), you're free to use whatever you want to help you answer the question. Ideally, however, you should first try it on your own to see if you can solve it by yourself.

Hmm... I haven't been keeping up with the more recent questions :blush:

[quote="quantum leap"]Are we allowed to refer to books for these arml problems?[/quote]

Well, after giving it a little bit of thought, I am allowing the use of books.  Exceptions (i.e. books you cannot look in) are the ARML books.  Additionally, you shouldn't look anywhere that has the same exact problem and answer.  You all should have a good enough judgment of right and wrong.

That being said, I encourage you to try the problems on your own and as quickly as possibly.

Of course, you may NOT discuss the problems with anyone else until I have posted the solution.

The goal of the practice series is for you to learn.


Euler
function
number theory
greatest common divisor
modular arithmetic
AMC
AIME

hello, we have $ 3^5 \equiv 43\mod 100$,squaring we get $ 3^{10} \equiv 43^2\equiv 49 \mod 100$, squaring again gives the desired result: $ 3^{20} \equiv 1 \mod 100$.
Sonnhard.

Hi, thanks for the reply.

I have another question.

The solutions seems good enough for me. However, just curious, is that possible to have another solutions for this?

I still think whether we have other option to prove the above.

Many thanks to Dr. Sonnhard for his solution.

Wait, $ x^{\phi(n)}\equiv 1 \pmod{n}$ when $ n$ and $ x$ are relatively prime, right?

$ \phi(100)\equal{}100\left (\frac{1}{2} \right ) \left (\frac{4}{5} \right )\equal{}40$, so I think we can just show that $ 3^{40} \equiv 1 \pmod{100}$, but I might be wrong, as usual...

Eh AIME,

I also got the same results as you when using Euler totient function. I'm confused now as to whether prove it or disprove it.

 :(

[quote="chanming.isaac"]Show that 3^20 Ξ 1 (mod 100).

I'm not sure whether the Euler totient function works for this question. I just thought that it fits the requirement. Since 3 and 100 are relatively prime (gcd (3,100)=1 ).

Is it right?

I've been trying to show that using Euler totient function, but it comes out not really that right.

Anyone can show me the right way?

Thanksz... [/code][/quote]

First, it isn't necessary to show that 3 and 100 are relatively prime; 3 is prime and it follows that it is "relatively prime" to every other number.

Maybe someone else knows some trick to exploit the totient. I don't.

$ 3^{20} \equal{} 3^{q\phi(100) \plus{} r}$, where $ r\in S,\, S\equal{} \{0,1,2,3,\ldots,\phi(100)\minus{}1\}$

$ \implies 3^{20} \equal{} (3^{\phi(100)})^q\cdot 3^r \\ \\ \implies 3^{20}\equiv (1)^q\cdot 3^r \pmod{100} \\ \\ \implies 3^{20}\equiv 3^r \pmod{100}$.

Unfortunately, $ 20 \in S$ so you're stuck having to use some hammer and tongs method.

$ 100$ is a composite number with more than one prime power. As a general hint: avoid working with such numbers directly, if you can.

Let's look at $ 3^{20}\pmod{4}$ and $ 3^{20}\pmod{25}.$

If we're working mod $ 4,$ we see that $ 3\equiv\minus{}1$ so $ 3^2\equiv 1,$ which means that $ 3^{20}\equal{}(3^2)^{10}\equiv 1\pmod{4}.$

As for mod $ 25:$ it's here that Euler's theorem shows its usefulness, as $ \phi(25)\equal{}20.$ That means that $ 3^{20}\equiv 1\pmod{25}.$

So $ 3^{20}\equiv1\pmod{4}$ and $ 3^{20}\equiv 1\pmod{25}.$ Putting these together is a matter of the Chinese Remainder Theorem - but in this case, a CRT problem with a solution that is obvious on inspection.

$ 3^{20}\equiv 1\pmod{100}.$

Using CRT instead of using the totient function directly is equivalent to replacing the totient function with the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url].  And in fact $ \lambda(100) \equal{} 20$.  The improvement over the totient function is even better for $ 1000$.

In group-theoretic terms, all this means is that a product of cyclic groups can't have an element of order the size of the group or else it would itself be cyclic.

[quote="Dr. No"]First, it isn't necessary to show that 3 and 100 are relatively prime; 3 is prime and it follows that it is "relatively prime" to every other number.[/quote]
That's not divisible by $ 3$.

[quote="t0rajir0u"]
[quote="Dr. No"]First, it isn't necessary to show that 3 and 100 are relatively prime; 3 is prime and it follows that it is "relatively prime" to every other number.[/quote]
That's not divisible by $ 3$.[/quote] :lol:

From the movie [i]Little Man Tate[/i],

Teacher (pointing to a number line ringing the periphery of the classroom): "Tate, how many of these numbers are divisible by two?"

Tate (roused from preoccupation with an art project of his own making): "All of them."


Pls help solve this with a Markovian chain:
Census studies from 1960 reveal that in France 70% of the daughters of working women also work, and 50% of the daughters of nonworking women work. Assume that this trend remains unchanged from one generation to the next,
Set up a Markov chain by drawing the transition diagram and the transition matrix.
That time 40% of French women worked. Determine the percentage of working women in the next two generations.
In the long run what percentage of French women will work?

I don't know what a Markov chain is, but I think I can answer the question without it...

First, let $ W_i$ be the percentage of working women in france in generation $ i$, where $ i \equal{} 0$ corresponds to the generation in 1960.  Then we have the recursion $ W_{i \plus{} 1} \equal{} .7W_i \plus{} .5(1 \minus{} W_i) \equal{} .5 \plus{} .2W_i$, with $ W_0 \equal{} .4$.

Then $ W_1 \equal{} .5 \plus{} .2(.4) \equal{} .58$ and $ W_2 \equal{} .5 \plus{} .2(.58) \equal{} .616$

Note that $ \lim_{n \to \infty} W_{n \plus{} 1} \equal{} \lim_{n \to \infty} W_n$ if the sequence converges, so the limit ($ W$) must satisfy $ W \equal{} .5 \plus{} .2W$, giving that eventually $ 62.5\%$ of french women will work.


Note that $ D_{b}$, $ D_{c}$ are the midpoints of the arcs $ CA$, $ AB$ not containing the vertices $ B$, $ C$. Draw the circle $ \Gamma_{1}$ centered at $ D_{b}$ with radius $ D_{b}A$ and the circle $ \Gamma_{2}$ centered at $ D_{c}$ with radius $ D_{c}A$. Their second intersection is the incenter $ I$ of triangle $ ABC$. In this case, $ B$ is the second intersection of $ D_{b}I$ with $ \Gamma_{2}$ and $ C$ is the second intersection of $ D_{c}I$ and $ \Gamma_{1}$.


Show that the roots of the equation z^5-(z-i)^5=0, where z is not equal to i, are 1/2(cot(360k/5)+i) where k=1,2,3,4. (i is sqrt-1)

This problem belongs to Intermediate Topics, not HSB. Please read guidelines before posting.

[hide="Hint"]Take the 'usual' complex 5th root of $ z^5\equal{}(z\minus{}i)^5$[/hide]


Give a recursive definition of $S_m (n)$, the sum of positive integer $m$ and nonnegative integer $n$.

I consider this problem isn't  hard. but My English so Worst that i can't understand what the problem want :? .

$S_m(n+1)=S_m(n)+1,S_{m+1}(n)=S_m(n)+1,S_m(m)=2m,S_0(1)=1$ I don`t really understand the question


Prove

$ \frac {1*3*5*\ldots*99}{2*4*6*\ldots*100} < \frac {1}{10}$

Try to do it without using 'advanced' things like logarithems, limits, complex inequalities...

See the discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=230796]here[/url].  For simplest results you'll want to look at Ravi's first comment.


[quote="nguoivn"]Given $ a, b, c \geq 0$ and $ a \plus{} b \plus{} c \equal{} 8$. Prove that: $ 268 \plus{} 12a^2b^2c^2 \geq\ ab(a \minus{} b)^2 \plus{} bc(b \minus{} c)^2 \plus{} ca(c \minus{} a)^2$
When does equality occurs ?  :)[/quote]

By "pqr"  $ \Leftrightarrow 12(r\minus{}1)^2\plus{}\frac{4}{9}(q^2\minus{}40)^2 \ge 0$ :)


calculus
derivative
integration
function
absolute value
calculus computations

Hello all,

(First, I want to say that I am not sure if I'm posting in the right forum and the right section. Please feel free to move this thread around if you think there is a better section to fit, but here I go.)

This just happened today at school, and I thought to bring it here and see if you can tell me what is going on. It's very strange. Long story short, I went back at the last school hour to discuss with my math teacher about the quiz today. One of the problems was:

$\text{Differentiate}$ $\arcsin{\sqrt{1-4x^{2}}}$ $\text{with respect to x.}$

[b]"Yes, now tell me what do you want to do with it on the board," [/b]my teacher said.

[b]"On the board? Ok, well . . ."[/b] I started to solve the problem. It wasn't because that I didn't know how to do it, but something else. I started writing.
\[\frac{d}{dx}\left[\arcsin{\sqrt{1-4x^{2}}}\right] \]

\[\text{Let }\emph{u }= \sqrt{1-4x^{2}}\text{,}\]

\[\Rightarrow \emph{du}= \frac{1}{2}\left(1-4x^{2}\right)^{-\frac{1}{2}}\left(-8x\right) \emph{dx}\]

\[\text{Refer to the differentiation formula, }\frac{d}{du}\arcsin{\emph{u }}= \frac{1}{\sqrt{1-\emph{u}^{2}}}\emph{du}\text{ , we get}\]

\[\frac{1}{\sqrt{1-\left(1-4x^{2}\right)}}\cdot\frac{1}{2}\left(1-4x^{2}\right)^{-\frac{1}{2}}\left(-8x\right) \]

\[= \frac{1}{\sqrt{4x^{2}}}\left(1-4x^{2}\right)^{-\frac{1}{2}}\left(-4x\right) \]

\[= \frac{\emph{x}}{\mid\emph{x}\mid2x}\left(1-4x^{2}\right)^{-\frac{1}{2}}\left(-4x\right) \]

\[= \frac{-2\cdot\emph{x}}{\mid\emph{x}\mid\sqrt{1-4x^{2}}}\]

\[\text{In conclusion, }\frac{d}{dx}\left[\arcsin{\sqrt{1-4x^{2}}}\right] = \]

\[\frac{-2}{\sqrt{1-4x^{2}}}, x > 0 \]

\[\frac{2}{\sqrt{1-4x^{2}}}, x < 0 \]
[b]"But that doesn't work,"[/b] I said quickly.

[b]"How come that doesn't work?"[/b] My teacher said.

[b]"Well, if we integrate this answer, we should get the original problem, right?"[/b]
[b]
"Right."[/b]

[b]"Well,"[/b] I started to integrate. (Refer to the previous formula)
\[\text{if }x > 0 , \]

\[\int\frac{-2}{\sqrt{1-4x^{2}}}=-\int\frac{2}{\sqrt{1-(2x)^{2}}}=-\arcsin{2x}+C ; \]

\[\text{if }x < 0 , \]

\[\int\frac{2}{\sqrt{1-4x^{2}}}= \int\frac{2}{\sqrt{1-(2x)^{2}}}= \arcsin{2x}+C \]
I then said, [b]"But the two functions are not equal. . . I don't know. I don't think so anyway."[/b]
\[\pm\arcsin{2x}+C = \arcsin{\sqrt{1-4x^{2}}}\]
We spent the entire 8th period discussing about this, but we didn't come to an answer yet. What do you think? Is the above statement true?

(Note that if you use $\frac{\pi}{2}$ for $C$, the graphs match - or so it looked that they did.)

Any comments and thoughts welcome! Please let me know if I had any mistakes in typos or logics.

I'm not sure what you mean.

Why are you saying that the integrator is in one case positive and in another negative? i don't think that makes sense.

if you want to use the fundamental theorem of calculus then you have to include the absolute value sign and carry out the integration keeping in mind the interval of integration. What you have done is something like integrating over two different intervals where absolute value comes into play and then expecting the answers to be similar.

Indefinite integrals can be misleading like that. Try starting out with an interval to integrate over. If you do the intergral correctly (pay attention to signs), things come out fine.

For example, if I set out to find the indefinite Riemann integral of |x| using your approach, I would find that my answer is "either -(x^2)/2 + C or (x^2)/2 + C". But that's not how it works obviously, just by looking at the graph of |x|. Say I integrated over [-a,a].  Then that approach would leave me with a result of 0 regardless of what C is. You always have to pay attention to the functions behavior on a given interval.

Another way of seeing that your method isn't viable is by checking that it doesn't work if you want to find the indefinite integral of [x] because its value depends on the interval.

Maybe someone will come by and explain it better. :)

Oh, and I'm very afraid of your calc teacher. Heh.

I appreciate your reply, §outh§tar,

But pardon me. I'm just a highschooler who is currently taking my first year of Calculus.

Can you write out how you do it? I would be really grateful.

jmerry, thank you so much! I will look over that again with my teacher.  :wink:

Edited again:

Is there a way to mathematically prove those identities, by the way? We are just short one step.

[quote="§outh§tar"]It's always helpful to look at the graph too.[/quote]

I looked at it, but graphs can't convince me enough. It may on a test, but not now :p. Thanks anyways.

Anyway to algebraically prove those identities? Sorry for my ignorance and excessive request, but I really have no idea.

[quote="jmerry"]For $0<t<1$, $\arcsin\sqrt{1-t^{2}}=\arccos(t)=\frac{\pi}2-\arcsin(t)$. For $-1<t<0$, $\arcsin\sqrt{1-t^{2}}=\arccos(-t)=\frac{\pi}2-\arcsin(-t)= \frac{\pi}2+\arcsin(t)$. There's no contradiction here- it's just that you have to watch out for identities. The arcsin doesn't cross over the bumps.[/quote]
jmerry's post. Sorry for double copying.


[quote="ashpotter"]I know there is different rounds. What are the 4 rounds and what do you do in them?[/quote]
There are the sprint round. The target round,the team round, and countdown. In the sprint round there is 30 questions and you have I think 30 or 40 minutes to do them. The target is when you have 8 questions and you do 2 at a time. You also have 6 minutes to do 2 problems. Then we have team round. In team round you and three other people have a set of 10 problems to do in 20 minutes. Your individual score is determined by how many sprint you get right. Each one is worth one point. And how many target questions you get right. Each one is worth 2 points. And the top ten will compete in countdown. In couuntdown the 10th place will versus the 9th place in best 2 out of three. They give you a question and whoever answers first gets the point. Then the victor of the 9th and 10th place faces the 8th place. This happens until you get to the top three. Now in the top three, who ever reaches three points first is declared the winner.

There are four rounds Sprint, Target, Team, and Countdown. The Sprint and Target rounds affect your individual score. Each Problem you get right out of 30 in the sprint round gives you 1 point. Each problem you get right from the Target round out of 8 gets you 2 points. Therefore a perfect score for an individual is 46. The team round is a round where you work with 3 other members. It does not affect you individual score. The Countdown round is a round where a person puts a problem on the board and the first person to get 2 questions right wins the wins the countdown round. It also does not affect your individual score. Hope that helps! :wink:


algorithm
search
graph theory
combinatorics unsolved
combinatorics

In the parliament of Sikinia there are $ 51$ senators. Each one of them hates exactly $ 3$ other senators. The "hating" relation is not necessarily symmetric (i. e. if $ A$ hates $ B$, it doesn't imply that $ B$ hates $ A$ as well). Determine the minimal integer $ k$ such that we can always group the senators into $ k$ houses such that each senator belongs to exactly one house and that, for any $ 2$ senators $ A$ and $ B$ which are in the same house, $ A$ doesn't hate $ B$ and vice versa.

I just want to take the extreme case...

isn't it trivial that we just need 2 houses?

*or I misunderstand your problem?*

No, we need at least 7 houses. (Because the relations CAN be symmetric). Id est, If A hates B, then B can either hate A or do not hate him. It's a matter of choice.

I think you confused Raja by your use of "minimal". Of course it [i]can [/i]happen that 2 houses could suffice, so the "minimal" number appears to be 2. What you meant was, determine the minimal number of houses that are [i]guaranteed to suffice in all cases[/i].

OK, now it should be clear.

Please, I need a solution - this problem is driving me crazy. Is there any elegant approach?

Let's rephrase the problem in the language of graph theory:

Given integers $ n$ and $ d$, let $ \overrightarrow{G}$ be a directed graph on $ n$ vertices such that each vertex has out-degree $ d$. Let $ G$ be the corresponding undirected graph, obtained by simply ignoring the orientation of the edges. Consider the chromatic number $ \chi(G)$, namely the least number of colors required to color the vertices of $ G$ in such a way that adjacent vertices have different colors.

The question is, as $ \overrightarrow{G}$ varies over all possible choices (of graphs with $ n$ vertices and constant out-degree $ d$), what is the worst (i.e. largest) value of $ \chi(G)$?

First, notice that it's easy to force $ 2d\plus{}1$ colors, simply by forcing a $ (2d\plus{}1)$-clique in $ G$ (a $ k$-clique is $ k$ vertices such that every two are adjacent). Just take $ 2d\plus{}1$ "senators", seat them around a table, and make everyone hate the $ d$ people to his right (that is, put a directed edge from each vertex $ i$ to vertices $ i\plus{}1, \ldots, i\plus{}d$ where addition is modulo $ 2d\plus{}1$). The corresponding undirected graph is just the complete graph on $ 2d\plus{}1$ vertices, since each vertex points to $ d$ others and is pointed at by $ d$ others.

So in the original problem, the number of "houses" may need to be as high as 7.

For the upper bound, let's investigate the degree properties of subgraphs of $ G$. Let $ \overrightarrow{H}$ be some subgraph of $ \overrightarrow{G}$. Every vertex in $ \overrightarrow{H}$ has out-degree at most d. Thus if $ \overrightarrow{H}$ has $ m$ vertices, it has at most $ dm$ edges, so the same holds true for the undirected version $ H$ as well: every subgraph $ H$ of $ G$ with $ m$ vertices contains at most $ dm$ edges. Therefore $ H$ must contain a vertex with degree at most $ 2d$. This is because the number of edges of an undirected graph is obviously at least half the number of vertices times the minimum degree.

In other words, $ \delta(H) \leq 2d$ for every subgraph $ H$ of $ G$. (Here $ \delta(H)$ is the smallest degree of any vertex in $ H$).

To translate this back into political language: note that every senator hates three people but there's no limit on how many people can hate a senator. It is perfectly possible for someone to be hated by all 50 other senators. However, if we take any group of senators and focus on the hating relationships between them, at least one of them is hating/hated by at most 6 others of the group. This is the crux of the problem.

We now appeal to a simple result in graph theory:

$ \chi(G) \leq \max \{ \delta(H) | H \subseteq G \} \plus{} 1$

(This is very easy to prove by the "greedy" coloring algorithm applied to a proper ordering of the vertices. See e.g. Diestel, Proposition 5.2.2. Use [url=http://books.google.com/books?id=04YbQF8oscQC&pg=PA112&dq=diestel+graph+theory&source=gbs_toc_r&cad=0_0&sig=ACfU3U3oGTVM5OyMoN351lOZbQHrMYxRyw#PPA115,M1]Google Books[/url] and just search for "Proposition 5.2.2", you'll find it on page 115, and the proof is on the previous page).

So, $ \chi(G) \leq 2d\plus{}1$ regardless of how "bad" $ \overrightarrow{G}$ is. In other words 7 colors or "houses" always suffice to ensure no hateful partners are in the same house. The number $ n$ of senators is quite irrelevant, except of course that the lower bound requires at least 7 of them.

Note, by the way, that if the hating relationship was symmetric, $ G$ would simply be a cubic graph, so 4 colors would always suffice - in fact, 3 would suffice unless $ G$ happened to be the complete graph on 4 vertices. See [url=http://mathworld.wolfram.com/BrooksTheorem.html]Brooks Theorem[/url].

Just wanted to add that you need no google books for downloading Diestel's graph theory; just use

http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/GraphTheoryIII.pdf
(see also http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/ ).

I have edited the problem statement for clarity.

[b]EDIT:[/b] This problem has also been discussed at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61955
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6165
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=248902

  darij


[color=darkblue]Let $ \frac {x^2 \minus{} yz}{a} \equal{} \frac {y^2 \minus{} zx}{b} \equal{}\frac {z^2 \minus{} xy}{c}$. Prove that $ \boxed{\frac {a^2 \minus{} bc}{x} \equal{} \frac {b^2 \minus{} ca}{y} \equal{} \frac {c^2 \minus{} ab}{z}}$[/color]

let x^2-yz/a=y^2-zx/b=z^2-xy/c=1/m, obviously m cannot be 0 here

then

a=m(x^2-yz)
b=m(y^2-zx)
c=m(z^2-xy)

sub them in and brute force (which works 4 me faster than finding an elegant way to solve it)

btw, what is the code for the "click to reveal hidden content" thingy?

[code][hide=Proof]fdnbvfgmbgfm[/hide][/code]

Gives us

[hide="Proof"]fdnbvfgmbgfm[/hide]

Alternatively, highlight the text you wish to hide, and press the "hide" button above the text box in the reply page.

The equality doesn't hold for $ x \equal{} y \equal{} z$.


Suppose you have a lot of pictures on a folder. Everything unsorted, with random names etc.
You may want to sort it on numerical order. You can chose to sort by size, by date. But you want to sort it by your personal taste. The best picture has the number 001, the second best 002 etc.

We can make a GUI that show 2 pictures and we make a manual comparison with the 2 (you click on one or the other). If we chose the algorithm to be Quick Sort, on average we will need to make $ n \log n$ manual comparisons, (not exactly this, but something near it). In worst case we will make $ n^2$!!

If we have 300 pictures and each comparison takes 3 seconds we will spend 3 days in front the pc sorting pictures!

I were thinking of a algorithm that Almost Sort it. Like this.

Import all pictures and make a dictionary. All pictures have value 0. Pictures with same value are in the same Deck.

We make comparisons with pictures in the same deck. If You chose a picture its value is increased 1, the other decreased 1.

We make a list so we don't test the 2 same pictures twice. We can also build a binary tree, se we don't need to test A and C if A>B and B>C.

In the end we will have something almost sorted (don't know if it is really almost or is more random). So we open nautilus or explorer. We can see the pictures, so we rearrange it in order. Etc. Since we can see 30 pictures at same time in this, we can make faster comparisons.

The questions are:
How much comparisons will be needed to be made in the preliminar process?
There is a better algorithm for sorting images (google didn't help on this)

NOTE: This is just a schetch. A Better program would've be make a GUI that open 10,15 images at same time and we click the images in the order we like it and we store this in a list and open more 5. Etc and them make the test of the best images in each group etc. But this approach is more complicated (Don't know how to do)

If you just want a technical answer to the question "What comparison sort will use the fewer number of comparisons?" then you probably want to try either binary insertion sort (which is just insertion sort with a binary search inside) or heapsort (particularly with linear-time construction).

However, here is what I feel would be a better solution to your task:

First, ask the user for a 'rating' (say, from 1 to 100) for each picture. After sorting the pictures by this rating, we should have a list that is mostly in order, except for minor changes we might want to make. In order to do this, you could display multiple pictures at a time and provide some sort of functionality for the user to exchange adjacent ones.

If you have difficulties implementing the second part, it would probably suffice to just use the ratings for each picture, since it's probably too difficult to compare very close pictures anyway. (You may want to let the user re-rate some pictures though, especially after he/she has a chance to look at all of them.)

I think that the best way to make this faster is to have the user compare more than two pictures at the same time as you suggested. For example eight images at a time and ask the user to either sort them or just divide into two halves (good ones and not so good ones).

To answer your question, I think you'll end up using $ n\log n$ comparisons anyway. But that's actually not too bad, 2 hours according to your estimates. If you want to be sure that you'll do at most $ n\log n$ comparisons even in the worst case, use merge sort or insertion sort.

Rating! Why didn't I thought that!

This is the best idea. And is in linear time. So it's the best case possible.

The minor chances you said don't need to be implemented, we cannot distinguish between the 32th image and the 33th image. So we can manualy change the 7th image with the 8th, and a program to do this is unnecessary.

Thanks a lot. You were very helpfull.


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

$n$ points are given on the surface of a sphere. Show that the surface can be divided into $n$ congruent regions such that each of them contains exactly one of the given points.


[color=darkblue]Given are the circle $w=C(O)$, two fixed points $A,B$ which belong to the circle $w$ and a line $d$. For two mobile points $M\in AA$ and $N\in BB$ so that $MN\parallel d$ define the point $L$ for which $LA\perp OM$ and $LB\perp ON$. Ascertain the geometrical locus of the point $L$.

[b]A particular case.[/b] Given are three fixed lines $d$, $d_{1}$, $d_{2}$ so that $d_{1}\parallel d_{2}$, two fixed points $A\in d_{1}$, $B\in d_{2}$ so that $AB\perp d_{1}$. Denote the middlepoint $O$ of the segment $[AB]$. For two mobile points $M\in d_{1}$, $N\in d_{2}$ so that $MN\parallel d$ define the point $L$ for which $LA\perp OM$ and $LB\perp ON$. Ascertain the geometrical locus of the point $L$.

[b]Remark.[/b] Define $XX$ - the tangent in the point $X\in w$ to the circle $w$.[/color]

The locus is the perpendicular line from $O$ to $d$.  :)


Show that the equation:

$ x^{x}\equal{}y^3\plus{}z^3$

has got infinite solutions (where $ x, y, z$ are positive integers)

For any $ a$, we have the solution
\[ x\equal{}27a^3\plus{}1,y\equal{}3a(27a^3\plus{}1)^{9a^3},z\equal{}(27a^3\plus{}1)^{9a^3}\]

Beautiful!  I considered adding an extra condition, but it didn't occur to me that I should specify that $ z | y$.  Care to discuss a motivation?

It's even easier. Just take $ x\equal{}2^{6k\plus{}4}$ and $ y\equal{}z\equal{}2^{\frac{(6k\plus{}4)2^{6k\plus{}4} \minus{} 1}{3}}$. The motivation is clear.

Yep, but how to get this solutions? This is the problem


You might consider my topic spam but this is the truth:

If you are new, please read GUIDELINES before posting RANDOMLY about a question. You are more likely to have placed your question in the wrong forum section.

There have been numerous topics already in Classroom Math that contains topics and/or difficulty WAY above the intended level of this forum. [color=red]PLEASE READ GUIDELINES BEFORE POSTING[/color].

Thank you and have a good day.

Sadly, this is true for all forums, especially High School Basics.

Agreed.  Please read the guidelines.

[quote="10000th User"]You are more likely to have placed your question in the wrong forum section.[/quote]

It's because new users see Classroom Math as the first forum so they generically put all [homework] problems there. Had they spend time perusing this site, they would have discovered that the level of difficulty is branched out.


find $ f(x): R \minus{} > R$ such that:
$ 1.f(1) \equal{} 1$
$ 2.f(\frac {1}{x \plus{} y}) \equal{} f(\frac {1}{x}) \plus{} f(\frac {1}{y})$
$ 3.f(x \plus{} y)(x \plus{} y) \equal{} xyf(x)f(y)$
for all x,y#0

$ y\equal{}0$. Hence $ f(x)\equal{}1/x$.

What about $ x\equal{}0$? Remember that $ 0$ is in the domain.

But that's still easy.

Set $ x\equal{}y$ into (2) and (3) to get
(i) $ 2f(x)\equal{}f(x/2)$ and
(ii) $ 2xf(2x)\equal{}x^2f(x)^2$
and hence  $ f(x)\equal{}1/x$, if $ x \ne 0$. If $ x\equal{}0$, (2) is not defined and (3) not allowed, so we can't say anything about $ f(0)$.

You might wanna include $ x\plus{}y \neq 0$. Else try $ x\equal{}1,y\equal{}\minus{}1$ in $ 3$: $ f(\minus{}1)\equal{}0$. Then plug $ x\equal{}2,y\equal{}\minus{}1$: $ f(1) \equal{} 0$ contradiction.


topology
advanced fields
advanced fields unsolved

Let X be a topological space,and let A,B,andC be nonempty subsets of X such that A is a proper subset of B,and B is a proper subset of C.Which of the following statements is/are true?
[size=150]1.If A and C are connected,then B is connected.
2.If A and C are compact,then B is compact.
3.If A and C are hausdorff,then B is hausdorff.[/size]

1. and 2. are nonsense (consider intervalls of the real line, e.g.) and 3. is trivially true. subspaces of hausdorff spaces are also hausdorff.

Specific counterexamples for 1 and 2, following -oo-:
$ A\equal{}[0,1], B\equal{}[0,1]\cup [2,3], C\equal{}[0,3]$.
$ A\equal{}[1,2], B\equal{}(0,3), C\equal{}[0,3]$.


In topic [url=http://www.mathlinks.ro/Forum/topic-88692.html]"[u]Practical problems[/u]"[/url],[b]hungkhtn[/b] introduced a nice ineq:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} +\frac{abc}{2(a^3+b^3+c^3)} \ge \frac{5}{3}$  $\forall a,b,c>0$
  I will point out a new result:
  [b][u]New result[/u]:[/b]
a)Prove that the equation $4t^4+6t^3+6t^2-4t-6=0$ has only one positive root which is $t_0$(Furthermore,$t_0\in (0;1)$).
b)Let $K=\frac{3(t_0^3+2)}{2(t_0+1)(t_0+2)}$.Prove that $\forall a,b,c>0$ we have
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{Kabc}{a^3+b^3+c^3}\ge \frac{3}{2}+\frac{K}{3}$ ;)
And $K$ is the best constant. ;)
[u][b]Comment[/b]:[/u]
+)$t_0=0.88789...$(5 exactly digits after the dot)
+)$0.74<K<0.75$
+)With $k=\frac{3}{5}$ we have a nice form.

[quote="hungkhtn"][quote="nthd"][b]The answer for hungkhtn[/b]:
a)Prove that the equation $2t^3+6t^2+t-6=0$ has only one positive root which is $t_0$(Furthermore $t_0\in (0;1)$)
b)Let $K=\frac{(4t_0+5)(t_0^2+2)}{8(t_0+1)^2}$
Prove that $\forall a,b,c>0$ we have :
$\frac{a^3+b^3+c^3}{(a+b)(b+c)(c+a)}+\frac{K(ab+bc+ca)}{a^2+b^2+c^2}\ge \frac{3}{8}+K$
And $K$ is the best constant. ;)  ;)[/quote]

Nice. But I intrduce the problem finding the best value of $k$ because we can find exactly this value.[/quote]
So Iposted this pro in my topic.I did't post it in your topic. ;)

Yes. But problem which is quite the same as yours, I often choose a details and approcimate value of $k$. I don't like to solve an equation which its degree is more than $2$. :D.

[quote="nthd"][b]The answer for hungkhtn[/b]:
a)Prove that the equation $2t^3+6t^2+t-6=0$ has only one positive root which is $t_0$(Furthermore $t_0\in (0;1)$)
b)Let $K=\frac{(4t_0+5)(t_0^2+2)}{8(t_0+1)^2}$
Prove that $\forall a,b,c>0$ we have :
$\frac{a^3+b^3+c^3}{(a+b)(b+c)(c+a)}+\frac{K(ab+bc+ca)}{a^2+b^2+c^2}\ge \frac{3}{8}+K$
And $K$ is the best constant. ;)  ;)[/quote]
Yesterday,I realized that $t_0=\frac{\sqrt{7}-1}{2}\rightarrow K=\frac{14\sqrt{7}-17}{24}$ ;)


geometry
trapezoid
symmetry
trigonometry
number theory
relatively prime
trig identities

(Last problem on the test, so i figured it belonged in the hard forum)

30. A hexagon inscribed in a circle has three consecutive sides of length 3 and three consecutive sides of length 5.  The chord of the circle that divides the hexagon into 2 trapezoids (one with 3 sides of length 3 and one with 3 sides of length 5) has length equal to m/n, where m and n are relatively prime integers. Find m+n.

a) 309 b) 349 c) 369 d) 389 e) 409

How do you solve this, and what does "relatively prime" mean?

Relatively prime means they are prime to each other, ie. they can't be simplified.  It wants the fraction in lowest terms.  Another term they use for radicals is "not divisible by the square of any prime" ie. the radical is in simplest form.

Oh, so the numbers by themselves don't have to be prime? like 49/2 is relatively prime?

Basically they dont share any common factors.

Outline of a solution:
[hide]First, determine the radius of the circle. This can be done by noting that a hexagon ABCDEF of consecutive sides 3-5-3-5-3-5 can be inscribed in the same circle. By symmetry, mAngleAOC = mAngleABC = 120 degrees. Use the Law of cosines twice from these two angles to get the radius of the circle, and then the Law of Sines on the original hexagon to get the sine of the acute angle which intercepts a side of 3. Finally, use the sine triple angle formula and the Law of Sines again to get m/n.[/hide]

Let the hexagon be ABCDEF with AB = BC = CD = 3 and DE = EF = FA = 5.  Angle BAF is 120, so by the Law of Cosines,

$BF^2 = AB^2 + FA^2 - 2AB\cdot AF \cos 120 = 3^2 + 5^2 + 3\cdot 5 = 49 \implies BF = 7$

Let the distance between BC and EF be x.  Then by the Pythagorean Theorem,

$x^2 = BF^2 - 1^2 = 48$

Using the same theorem, we find that

$CF^2 = 4^2 + x^2 = 64 \implies CF = 8$

Now let $y = \angle BFC$.  Law of Cosines on triangle BFC gives

$BC^2 = BF^2 + CF^2 - 2BF\cdot CF \cos y$

$3^2 = 7^2 + 8^2 - 2\cdot 7\cdot 8 \cos y$

$\cos y = \frac{13}{14}$

Since $\angle AFB = \angle BFC$, Law of Cosines on AFC yields

$AC^2 = AF^2 + CF^2 - 2AF \cdot CF \cos 2y$

$AC^2 = 5^2 + 8^2 - 2\cdot 5\cdot 8 (\cos^2 y - \sin^2 y)$

$AC^2 = 89 - 80(2\cos^2 y - 1)$

$AC^2 = 89 - \frac{2840}{49}$

$AC^2 = \frac{1521}{49}$

$AC = \frac{39}{7}$

Similarly, let $z = \angle EBF$.  Law of Cosines on triangle EBF gives

$EF^2 = BF^2 + BE^2 - 2BF\cdot BE \cos z$

$5^2 = 7^2 + 8^2 - 2\cdot 7\cdot 8 \cos z$ (note that CF = BE)

$\cos z = \frac{11}{14}$

Since $\angle EBF = \angle FBA$, Law of Cosines on AFC yields

$AE^2 = AB^2 + BE^2 - 2AB \cdot BE \cos 2z$

$AE^2 = 3^2 + 8^2 - 2\cdot 3\cdot 8 (\cos^2 y - \sin^2 y)$

$AE^2 = 73 - 48(2\cos^2 y - 1)$

$AE^2 = 73 - \frac{552}{49}$

$AE^2 = \frac{3025}{49}$

$AE = \frac{55}{7}$

Note that CE = BF = 7.  Ptolemy's Theorem on ACDE yields

$AC\cdot DE + AE\cdot CD = AD\cdot CE$

$\frac{39}{7}\cdot 5 + \frac{55}{7}\cdot 3 = AD\cdot 7$

$\frac{360}{7} = AD\cdot 7$

$\frac{360}{49} = AD$

Of course there are other solutions, but I imagine most of them lie along these lines.

[quote]Let the hexagon be ABCDEF with AB = BC = CD = 3 and DE = EF = FA = 5. Angle BAC is 120,[/quote]

I think this should be angle BAF = 120, no?

Yakko


abstract algebra
superior algebra
superior algebra unsolved

Trying LaTeX....

Let $ K$ be the splitting field of $ X^p$ - $ X$ - $ 1$ over $ k$ = $ Z_p$.  Show that $ k$ $ \in$ $ K$ is Galois and compute the Galois group.

Also, is every algebraic extension of a finite field Galois and Abelian?


I'm pretty sure this problem is well known by now, but here it is:

Find x such that $ cos(cos(cos(cos x))) \equal{} sin(sin(sin(sin x)))$
I have made [hide="A start of a start"]
let $ y_1\equal{}cos(cos(cos x))$ and $ z_1 \equal{} sin(sin(sin x))$,
$ cos(y_1) \equal{} sin(z_1)$
$ cos(y_1) \equal{} cos(90\minus{}z_1)$
$ y_1\equal{}90\minus{}z_1$
$ cos(cos(cos x))\equal{}90\minus{}sin(sin(sin x))$
Now, if we repeat the same steps, the added 90 will make things too complicated.
How should I proceed?
[/hide]

Using the identity $ \cos x \equal{} \sqrt {1 \minus{} \sin^2 x}$ and alternatively $ \sin x \equal{} \sqrt {1 \minus{} \cos^2 x}$ in $ RHS$ and $ LHS$ respectively.

Equation is equivalent with:

$ \sqrt {1 \minus{} \cos^2 \sqrt {1 \minus{} \cos^2{\sqrt {1 \minus{} \cos^2 \sqrt {1 \minus{} \cos^2 x}}}}}$ = $ \sqrt {1 \minus{} \sin^2 \sqrt {1 \minus{} \sin^2{\sqrt {1 \minus{} \sin^2 \sqrt {1 \minus{} \sin^2 x}}}}}$

I was wondering if there's a clever way to solve this equation ?

[hide="@ Poincare"] I tried the way you showed but I did stuck on it.
It was almost impossible to continue because of too many calculations!
Hope that this helps you a bit  :(
[/hide]

If you graph them, there aren't any intersections.

How you'd prove this, I have no idea.

First note that $ - 1\le\cos{x}\le1$, so $ \cos(\cos(\cos(\cos{x})))$ is nonnegative. Furthermore, if $ \sin{x}$ is negative, then since $ - 1\le\sin{x} < 0$, $ \sin(\sin(\sin(\sin{x})))$ will also be negative. So we assume that $ 0\le x\le\pi$ because of periodicity.

The key idea is that $ - 1\le\sin{x},\cos{x}\le1$. When $ \sin{x}$ takes in positive values of $ x$, it gets smaller. So each time we take the $ \sin$ it gets smaller. But when $ \cos{x}$ takes in "small" values of $ x$, it becomes larger. And even if $ \cos{x}$ gets too large so that it will get smaller by taking its cosine, we apply $ \cos$ four times which intuitively will balance it out again. This makes us think that $ \cos(\cos(\cos(\cos{x}))) > \sin(\sin(\sin(\sin{x})))$.

The logic in that hand-waviness also applies to just $ \cos(\cos{x})$ and $ \sin(\sin{x})$ by parity, so we conjecture that $ \cos(\cos{x}) > \sin(\sin{x})$ as well. Indeed, this is equivalent to
\begin{align*} \cos(\cos{x}) - \cos\left(\frac\pi2 - \sin{x}\right) & = 2\sin\left(\frac12\left(\frac\pi2 - \sin{x} - \cos{x}\right)\right)\sin\left(\frac12\left(\frac\pi2 + \sin{x} - \cos{x}\right)\right) \\
& > 0..\end{align*}
Note that $ - \sqrt2\le\sin{x} + \cos{x},\sin{x} - \cos{x}\le\sqrt {2}$, and
\[ 0 < \frac12\left(\frac\pi2 - \sqrt2\right) < \frac12\left(\frac\pi2 + \sqrt2\right) < \pi,\]
from which our desired inequality $ \cos(\cos{x}) > \sin(\sin{x})$ follows, because sine is positive on $ (0,\pi)$.

Then
\begin{align*} \cos(\cos(\cos(\cos{x}))) & > \sin(\sin(\cos(\cos{x}))) \\
& > \sin(\sin(\sin(\sin{x}))),\end{align*}
where the second inequality is because $ \sin(\sin{y})$ is monotonically increasing on the interval $ [ - \pi/2,\pi/2]$, which contains $ [ - 1,1]$, the interval of $ \cos(\cos{x})$ and $ \sin(\sin{x})$.


1)Sa se rezolve sistemul:

 x^2+4xy+6y^2=11
 y^2+3yz+2z^2=0
 z^2+4zx+12x^2=9

2) Fie polinomul f=X^4-5X+3 din Z[X].
    Sa se arate ca:
    (i) a apartine lui Z <=> f(a) apartine lui Z.
    (ii) multimea {b din R\Q | f(b) din N} este infinita.

[b]1)[/b] Sa se rezolve sistemul:

 $x^{2}+4xy+6y^{2}=11$
 $y^{2}+3yz+2z^{2}=0$
 $z^{2}+4zx+12x^{2}=9$

[color=blue]Indicatie.[/color] A doua ecuatie a sistemului se scrie $(y+z)(y+2z)=0$ si apar doua cazuri: a) $y=-z$; b)$y=-2z$.
Inlocuim $y$ in prima ecuatie si avem de rezolvat sistemele omogene:
a)
$x^{2}-4xz+6z^{2}=11$
$z^{2}+4zx+12x^{2}=9$
b)
$x^{2}-8xz+24z^{2}=11$
$z^{2}+4zx+12x^{2}=9$
[b]2)[/b] Fie polinomul $f=X^{4}-5X+3$ din $Z[X]$.
        [b](i)[/b] Daca $a\in Q$, atunci $a \in Z$ $\Longleftrightarrow$ $f(a) \in Z$.
       [b](ii)[/b] Sa se arate ca multimea $\{b\in{R-Q}| f(b) \in N\}$ este infinita.
[color=blue]Indicatie.[/color] [b](i)[/b] "=>". Evidenta.
"<=". Folosim teorema: "Daca un polinom cu coeficienti intregi are o radacina rationala $a=\frac{m}{n}, m,n\in Z, n\neq 0, (m,n)=1$, atunci $m$ divide termenul liber si $n$ divide coeficientul dominant."
[b](ii)[/b] In mod evident, daca $a\in Z$, atunci $f(a)$ este numar intreg impar.
In plus, din $f(1)=-1$, $f(\infty)=\infty$ si faptul ca functia polinomiala $f$ atasata este continua, rezulta ca $f$ parcurge toate valorile reale de la -1 la $\infty$.
Prin urmare, valorile naturale pare (adica o infinitate de valori), nu pot fi atinse decat pentru argumente din $R-Q$.
Daca mai tinem cont ca $f$ este strict crescatoare pe $[2;\infty)$, concluzia este imediata.

Apreciezi promptitudinea voastra,merci de ajutor.


HMMT
probability
inequalities
calculus
number theory
graph theory

What exactly is in this category?
Sample problems?

Actually, because we have a separate Number Theory test, our Advanced Topics isn't exactly the same as other existing ones, though similar. If you want to practice, Rice AT tests and HMMT combinatorics tests should be good, and here's a general overview of the topics we consider to be "advanced topics":

Graph Theory, Combinatorics and Probability, Sequences and Series, Algebraic Inequalities

None of the questions will require knowledge of calculus, but it may be possible to use calculus to solve some of the problems (this is true on most contests, actually).


Find all finite subset $A \subset N^{*}$ such that there exists a finite subset $B \subset N^{*}$ containing $A$ so that the sum of the numbers in $B$ is equal to the sum of the squares of the numbers in $A$

Let $S=\sum_{k\in A}(k^{2}-k).$ Yuor condition is $\sum_{k\in B/A}k=S$. Therefore if $A\not =\{1\}$, then  $S>max_{k\in A}k$. It mean work $B=A+\{S\}$.

I don't understand.
Can you post full solution?

Rust, there is a slight problem with your solution. Sometimes $k^{2}-k$ might also be in A. Of course, we could just take $B = A \cup \{ \sum_{k \in A}k^{2}-k \}$. But the problem is trivial right now. toan hoc, did you make a typo somewhere in the problem statement?

Wait. I'm confused too now.

If you take B as the union of those two, wouldn't the same problem occur if k^2-k is already in the A?

A = {2,3}

B = {2,3} union {2,6} = {2,3,6}

sum of B elements = 11.

sum of A elements squared = 13.

Well, I think Rust was right. $S$ is a set of only $1$ element, not a set of all $k^{2}-k$ where $k \in A$.
If $A = \{ 2,3 \}$, then $S = (2^{2}-2)+(3^{2}-3) = 8$ and $B = (2,3,8)$.
He's wrong in $1$ condition : If $A = \{ 2 \}$.

But still, maybe it's not all possible solution..  :(

[quote="Rust"]I don't undestand, why my solution is not full?
Answer $A\not =\{1\}$.[/quote]

Sorry  :blush:  I misread your solution. But toan hoc, what is the actual problem?

Oh whoops. I misinterpreted the question/answer...

Wow, both of them.


geometry
incenter
complex numbers
geometry theorems

ABC is a triangle positioned in the complex plane.Let $Z_1$,$Z_2$,$Z_3$ be the complex numbers representing A,B,C.
Could someone tell me the representations of the circumcentre incentre fermat point etc. Of the triangle?
for example the orthocentre  is $Z_1 + Z_2 + Z_3$.

Hello,
    These are standard results available in any book on complex numbers.The orthocentre is clearly not Z1 + Z2 + Z3. Its is rather (Z1 tan A + Z2 tan B + Z3 tan C)/(tan A + tan B + tan C). The cicrcumcentre can also shown to be similar as the expression, excepting the change that tan A is replaced by sin 2A and so on.
Sathej