[color=blue] [size=150]$5^{th}$  Iberoamerican Olympiad[/size]
Valladolid, ESPAÑA. [1990]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 2.[/b] En un triángulo $ \triangle ABC$ sean $ D,E,F$ los puntos de tangencia de su incirculo $ (I)$ con $ BC,CA,AB.$ La recta $ AD$ corta por segunda vez a $ (I)$ en $ P.$ Si $ M$ es el punto medio de $ EF,$ probar que los puntos $ P,D,I,M$ son concíclicos o están alineados.[/quote]
Como es claro que $ AM$ es mediatriz de $ EF,$ por teorema del cateto en el triángulo rectángulo $ \triangle AEI$ en $ E$ se tiene $ AE^2 \equal{} AM \cdot AI.$ Por otro lado de la potencia de $ A$ con respecto al incirculo $ (I)$ resulta $ AE^2 \equal{} AP \cdot AD.$ Entonces $ AE^2 \equal{} AM \cdot AI \equal{} AP \cdot AD$ $ \Longrightarrow$ $ P,D,I,M$ son concíclicos o están alineados si evidentemente $ AB \equal{} AC.$

[quote][b]Problema 4.[/b] Se da un circunferencia $ \mathcal{C}_1$ y $ AB$ un diametro fijo de ella. $ t$ es la tangente a $ \mathcal{C}_1$ en $ B$ y $ M$ es un punto en $ \mathcal{C}_1$  diferente de $ A,B.$ $ \mathcal{C}_2$ es la circunferencia tangente a $ \mathcal{C}_1$ en $ M$ y a $ t$ en un punto $ P.$
a) Construir el punto de tangencia $ P$ de $ \mathcal{C}_2$ con $ t$ y hallar el lugar del centro de $ \mathcal{C}_2$ al variar $ M.$
b) Probar que existe una circunferencia ortogonal al haz $ \mathcal{C}_2.$ [/quote]
a) La inversión positiva con respecto a la circunferencia centrada en $ B$ con radio $ BA$ transforma a $ t$ en si misma y a $ \mathcal{C}_1$ en la tangente $ t'$ a $ \mathcal{C}_1$  por $ A.$ Entonces, el haz $ \mathcal{C}_2$ se transforma en una cadena de círculos iguales tangentes al par de paralelas $ t$ y $ t'.$ El rayo $ BM$ corta pues a $ t'$ en el inverso $ M'$ de $ M.$ La circunferencia inversa $ \mathcal{C}'_2$ de $ \mathcal{C}_2$ que pasa por $ M$ es pues construible ya que es tangente a $ t'$ en $ M'$ con su centro en la mediatriz $ \ell$ de $ AB.$ Hallado el punto de tangencia $ P'$ de  $ \mathcal{C}'_2$ con $ t,$ que es el inverso de $ P,$ entonces bastará trazar la circunferencia que pasa por $ M,M'$ y $ P',$ esta cortará a $ t$ nuevamente en el punto buscado $ P.$ Por otro lado, para observar el lugar del centro $ O_2$ de $ \mathcal{C}_2$ basta considerar la simétrica $ \ell'$ de la mediatriz $ \ell$ de $ AB$ con respecto a $ t.$ Entonces si $ O_2P$ corta a $ \ell'$ en $ Q,$ se tiene $ O_2Q \equal{} O_2O_1 \equal{} r_1 \plus{} r_2$ $ \Longrightarrow$ $ O_2$ equidista de $ \ell'$ y $ O_1.$ Su lugar es pues la parábola con foco $ O_1$ y directriz $ \ell'.$

b) Es sabido que la inversión es una transformación conforme, es decir que preserva ángulos entre curvas. Note que el lugar geométrico de los centros $ O_2'$ del haz $ \mathcal{C}_2'$ inverso del haz $ \mathcal{C}_2$ es al mediatriz $ \ell$ de $ AB.$ El ángulo entre $ \ell$ y $ \mathcal{C}_2'$ es recto, por tanto el ángulo entre $ \mathcal{C}_2$ y la circunferencia inversa $ \mathcal{L}$ de $ \ell$ será también recto $ \Longrightarrow$ $ \mathcal{L} \perp \mathcal{C}_2.$

[quote][b]Problema 5.[/b] Sean $A$ y $B$ vértices opuestos de un tablero cuadriculado de $n$ por $n$ casillas $(n \ge 1),$ a cada una de las cuales se añade su diagonal de dirección $AB,$ formándose así $2n^2$ triángulos iguales. Se mueven una ficha recorriendo un camino que va desde $A$ hasta $B$ formado por segmentos del tablero, y se coloca, cada vez que se recorre, una semilla en cada uno de los triángulos que admite ese segmento como lado. El camino se recorre de tal forma que no se pasa por ningún segmento más de una vez, y se observa, después de recorrido, que hay exactamente dos semillas en cada uno de los $2n^2$ triángulos del tablero. Hallar tal valor de $n.$[/quote]
Primero observamos que como cada triángulo debe tener dos semillas necesariamente se debe pasar por sólo dos lados de cada triángulo. Además, cualquier vértice distinto de $A$ y de $B$ debe pertenecer a una cantidad par de aristas del camino, puesto que, por no ser puntos terminales, cada vez que se llega se tiene que salir. Por el contrario, los vértices $A$ y $B$ pertenecen a un número impar de aristas del camino.

Para el caso en que $n = 1,$ sean $P$ y $O$ los otros vértices del cuadrado. Para que haya 2 semillas en $\triangle AQB$ y 2 semillas en $\triangle APB,$ por lo menos un cateto de $\triangle AQB$ y un cateto de $\triangle APB$ deben pertenecer al camino. Entonces deben estar los dos catetos de los dos triángulos porque $P,Q$ pertenecen a un número par de aristas del camino; y no puede estar la diagonal $AB$ porque ya hay 2 semillas en cada triángulo. En consecuencia, el vértice $A$ pertenece a dos aristas del camino. Contradicción.

Si $n = 2,$ una solución es la secuencia de aristas de 1 a 10 como se muestra en la figura 2. Para el caso $n = 3,$ llamamos $P$ y $Q$ a los otros dos vértices extremos del tablero, numeramos las aristas como se indica en la figura 3.

Vemos lo siguiente: Como $P$ no es un punto terminal debemos pasar por las aristas 1 y 2. Como en cada triángulo no se pueden utilizar más de dos aristas, no se usa la arista 3 y por lo tanto se deben usar la 4 y la 5. Por paridad, no se pueden utilizar ni la 10, ni la 6, lo cual obliga a usar la 11 y la 7. Por paridad, las aristas 12 y 9, o están ambas en el camino o no está ninguna. Para que haya 2 semillas en el triángulo del que son lados, deben estar ambas en el camino, lo cual obliga a que la 14 y la 8 no se utilicen. Ahora, Como 8 no se usa, estamos obligados a usar 18 y 17.

Razonando analogamente, se obliga a usar las aristas 20 y 15 y a no pasar por la 19, con lo cual el vértice $B$ no puede ser un punto terminal, pues hay exactamente 2 aristas del camino que inciden en él. Con esto, el caso $n = 3$ no es posible.

Para $n \ge 4,$ utilizando la misma notación que en el caso 3, pero llamando $H$ a lo que era el vértice $B.$ Como el 19 no se usó se debe utilizar el 24 (ver figura 4). Así, el vértice $H$ no es terminal y sin embargo hay un número impar de aristas del camino que inciden en él. Con esto se ve que el caso $n \ge 4$ tampoco es posible. Así se concluye que el único caso posible es $n = 2.$

[quote][b]Problema 6.[/b] Sea $f(x)$ un polinomio de grado 3 con coeficientes racionales. Probar que si el gráfico de $f$ es tangente al eje $x,$ entonces $f(x)$ tiene sus 3 raíces racionales.[/quote]
$f$ es tangente al eje $x$ $\Longleftrightarrow$ $f$ tiene una raíz doble real $\alpha.$ Si $\beta$ denota la otra raíz, entonces $f(x)= a(x -\beta)(x-\alpha)^2.$ siendo $a$ el coeficiente de $x^3,$ entonces $a \in \mathbb Q,$ exceptuando $0.$

Así, $(x-\beta)(x -\alpha)^2 = x^3 - (2\alpha+ \beta) x^2 + (\alpha^2 + 2\alpha\beta)x - \alpha^2\beta$ tiene coeficientes racionales, i.e. $2\alpha+ \beta,$ $\alpha^2 + 2\alpha\beta$ y $\alpha^2\beta$ son racionales. Por otro lado, sabemos que en el polinomio $P(x)=x^3+bx^2+cx+d$ con $b,c,d$ racionales, la sustitución $x \longrightarrow x-\frac{_b}{^3}$ es una traslación que elimina el término de 2do grado. Como $\frac{_b}{^3}$ es racional, las raíces racionales corresponden a raíces racionales, y las raíces irracionales corresponden a raíces irracionales. Por ende, no se pierde generalidad si consideramos $2\alpha+ \beta=0,$ i.e. $\beta=-2\alpha.$ Así, $\alpha^2+2\alpha\beta \in \mathbb Q$ $\Longrightarrow$ $\alpha^2 \in \mathbb Q$ y $\alpha^2\beta \in \mathbb Q$ $\Longrightarrow$  $\alpha^3 \in \mathbb Q.$ De ésto deducimos pues que $\alpha,\beta \in \mathbb Q.$


Define the sequence $a_1, a_2, a_3, ...$ by $a_1 = 1$, $a_n = a_{n-1} + a_{[n/2]}$. Does the sequence contain infinitely many multiples of $7$?

can someone please do this problem? i've spent much time on this but i found no idea.

Thank you.

[hide]
The answer is yes.

$ a_1 \equal{} 1$
$ a_2 \equal{} 2$
$ a_3 \equal{} 3$
$ a_4 \equal{} 5$
$ a_5 \equal{} 7$

So we know at least one multiple of 7 appears in the sequence. Suppose there were a finite number of multiples of 7; then let $ a_k$ be the largest multiple multiple of 7.

Now $ a_{2k\plus{}1}\equal{}a_{2k}\plus{}a_{k}\equal{}a_{2k\minus{}1}\plus{}2a_{k}$, so $ a_{2k\minus{}1}\equiv a_{2k}\equiv a_{2k\plus{}1}\mod 7$. These three values are nonzero mod 7, because otherwise we would have a multiple of 7 greater than $ a_k$. Let these three values be $ b$ mod 7. Next,

$ a_{4k\minus{}3} \equiv a_{4k\minus{}3}\mod 7$.
$ a_{4k\minus{}2} \equiv a_{4k\minus{}3} \plus{} b\mod 7$
$ a_{4k\minus{}1} \equiv a_{4k\minus{}3} \plus{} 2b\mod 7$
$ a_{4k}    \equiv a_{4k\minus{}3} \plus{} 3b\mod 7$
$ a_{4k\plus{}1} \equiv a_{4k\minus{}3} \plus{} 4b\mod 7$
$ a_{4k\plus{}2} \equiv a_{4k\minus{}3} \plus{} 5b\mod 7$
$ a_{4k\plus{}3} \equiv a_{4k\minus{}3} \plus{} 6b\mod 7$

Since $ b$ is nonzero mod $ 7$, one of these seven values must be zero mod 7. Hence there are are infinitely many multiples of 7 in the sequence.
[/hide]


algebra
polynomial
greatest common divisor
superior algebra
superior algebra unsolved

Supposed that F_q(q=r^f) is a finite field and k<f. What's the solutions of t^{r^k}=t for t in F_q ?

To prove what -oo- said, you need to look at the splitting polynomial for finite fields.  For simplicity, let me rewrite $ r$ as $ p$, so we know it is prime.

All the roots of the polynomial $ t^{p^{k}}-t$ are precisely the elements of the field $ \mathbb{F}_{p^{k}}$, and so we are looking for the set $ \mathbb{F}_{p^{f}}\cap\mathbb{F}_{p^{k}}$.  Because the extensions $ \mathbb{F}_{p^{f}}$/$ \mathbb{F}_{p}$ and $ \mathbb{F}_{p^{k}}$/$ \mathbb{F}_{p}$ are Galois, we know that their intersection is also.  Hence the intersection of these two fields is the splitting field of the $ gcd(x^{p^{f}}-x,x^{p^{k}}-x)$.  Because $ x^{p^{m}}-x|(x^{p^{n}}-x)$ iff $ p^{m}-1|p^{n}-1$ iff $ m|n$, we can with a little work deduce that the greatest common divisor mentioned above must be $ x^{p^{gcd(f,k)}}-x$.  Thus since $ \mathbb{F}_{p^{gcd(f,k)}}$ is the splitting field of $ x^{p^{gcd(f,k)}}-x$ over $ \mathbb{F}_{p}$, we are finished.


calculus
integration
function
calculus computations

Hello, I have some calculus problems I need help with:

1. A sprinkler distributes water in a circular pattern, supplying water to a depth of e^(-r) feet per hour at a distance of r feet from the sprinkler.

A. What is the total amount of water supplied per hour inside of a circle of radius 18 in ft^3 per hour?

B. What is the total amount of water that goes throught the sprinkler per hour?

2. Electric charge is distributed over the disk x^2+y^2<=2 so that the charge density at (x,y) is  o(x,y)=9+x^2+y^2 coulombs per square meter.
Find the total charge on the disk.

3. A lamina occupies the part of the disk x^2+y^2<=4 in the first quadrant and the density at each point is given by the function p(x,y)=3(x^2+y^2).

A. What is the total mass?

B. What is the moment about the x-axis?

C. What is the moment about the y-axis?

D. Where is the center of mass?

E. What is the moment of inertia about the origin?

[quote]Do you want to see a solution without the application of a double integral?[/quote]
Yes, that would help as well

1)   The amount of water in the circle per hour is $\int_0^{2\pi}\int_0^{18} e^{-r} \ dr d\theta = 2\left(1-e^{-18}\right)\pi$ and the total amount of water through the hose is $\int_0^{2\pi}\int_0^\infty e^{-r} \ dr d\theta =2\pi$

2) The total charge is given by $\iint_R 9+x^2+y^2 \ dxdy = \int_0^2\int_0^{2\pi} r(9+r^2) \ d\theta dr = 44\pi$.

Here is a site that should help you with the lamina problems: http://ltcconline.net/greenl/courses/202/multipleIntegration/MassMoments.htm

Edit: I corrected #1

[quote="223rem"][quote]

1) The amount of water in the circle per hour is  1-exp(-18)
and the total amount of water through the hose is  1

2) The total charge is given by 44*pi.
[/quote]
I tried these answers but they are not correct, is there a mistake somewhere?[/quote]

I fixed number 1, but I'm not sure what's wrong with number 2.

[quote]The amount of water in the circle per hour is 2*(1-e^(-18))*pi and the total amount of water through the hose is 2*pi[/quote]
Thanks, that is correct. The total charge from problem 2 is 20*pi, which can be obtained by integrating r from 0 to sqrt(2) and theta from 0 to 2*pi. Now as far as the third problem:

3. A lamina occupies the part of the disk x^2+y^2<=4 in the first quadrant and the density at each point is given by the function p(x,y)=3(x^2+y^2).

A. What is the total mass?

The total mass is the integral of 3*r^2*r dr with r from 0 to sqrt(2) and theta from 0 to 2*pi which is equal to 6*pi.

B. What is the moment about the x-axis?

In Maple: int(int(3*(x^2+y^2)*y,y=0..sqrt(4-x^2)),x=0..2);

96/5

C. What is the moment about the y-axis?

In Maple: int(int(3*(x^2+y^2)*x,y=0..sqrt(4-x^2)),x=0..2);

96/5

D. Where is the center of mass?

(moment about y-axis/total mass,moment about x-axis/total mass) = (96/5/(6*pi),96/5/(6*pi))

E. What is the moment of inertia about the origin?

int(int(3*(x^2+y^2)*(x^2+y^2),y=0..sqrt(4-x^2)),x=0..2);
16*pi

Q.E.D.


function
calculus
derivative
calculus computations

If f(x,y,z)=0 then how to show that $ \frac{\partial{x}}{\partial{y}}\frac{\partial{y}}{\partial{z}}\equal{}\minus{}\frac{\partial{x}}{\partial{z}}$

The title is misleading; "partial differential equations" are a specific advanced subject, and this isn't it.

Look carefully at the distinction between functions and variables. Here's how I would write that with this in mind:
We have a surface defined by $ f(x,y,z)\equal{}0$. Solving for each variable in terms of the others gives smooth functions $ X,Y,Z$ such that $ f(X(y,z),y,z)\equal{}0$, $ f(x,Y(x,z),z)$, and $ f(x,y,Z(x,y))\equal{}0$ identically. We want to show that \[ \left.\frac{\partial X}{\partial y}\right|_{(y,z)\equal{}(y_0,z_0)}\cdot \left.\frac{\partial Y}{\partial z}\right|_{(x,z)\equal{}(x_0,z_0)}\equal{}\minus{}\left.\frac{\partial X}{\partial z}\right|_{(y,z)\equal{}(y_0,z_0)}\]when $ (x_0,y_0,z_0)$ is a point on the surface (We actually want $ x_0\equal{}X(y_0,z_0)$, $ y_0\equal{}Y(x_0,z_0)$, $ z_0\equal{}Z(x_0,y_0)$; these functions are locally defined, and we want to allow for surfaces like spheres).

The actual calculation is a chain rule exercise. If we differentiate the defining relation $ f(X(y,z),y,z)\equal{}0$ with respect to $ y$, for example, we get $ 0\equal{}\frac{\partial f}{\partial x}\cdot \frac{\partial X}{\partial y}\plus{}\frac{\partial f}{\partial y}$, evaluated at the point $ (x,y,z)\equal{}(x_0,y_0,z_0)$, and we get that $ \frac{\partial X}{\partial y}\equal{}\minus{}\frac{\partial f/\partial y}{\partial f/\partial x}$ at our point. The rest works out similarly, and it all works out as a simple identity, independent of what the partial derivatives of $ f$ actually are.


Let be $ x_i\ ,\ i\in \overline{1,8}$ the roots of the equation : $ x^8\plus{}80x^2\minus{}36\equal{}0$ .

i)Prove that $ x_i\in (\minus{}\sqrt[4]{6},\sqrt[4]{6})$

i)Find the roots of this equation using only second degree equations .

hello, by the ansatz $ x^8\plus{}80x^2\minus{}36\equal{}(ax^4\plus{}bx^2\plus{}c)(dx^4\plus{}ex^2\plus{}f)$ we get the solution $ a\equal{}\frac{4}{e},b\equal{}\frac{\minus{}16}{e},c\equal{}\frac{72}{e},d\equal{}\frac{e}{4},e\equal{}e,f\equal{}\minus{}\frac{e}{2}$, setting $ e\equal{}4$ we have
$ x^8\plus{}80x^2\minus{}36\equal{}(x^4\minus{}4x^2\plus{}18)(x^4\minus{}4x^2\minus{}2)$ and solving the quadratic equation we find the real solutions in the form $ x_1\equal{}\sqrt{\minus{}2\plus{}\sqrt{6}}$ or $ x_2\equal{}\minus{}\sqrt{\minus{}2\plus{}\sqrt{6}}$.
Sonnhard.

It should factor out to be $ (x^{4}\minus{}4x^{2}\plus{}18)(x^{4}\plus{}4x\minus{}2)$ if you're setting $ e\equal{}4$...

hello your are wrong, when we expand your product we get
$ (x^4\minus{}4x^2\plus{}18)(x^4\plus{}4x\minus{}2)\equal{}x^8\minus{}4x^6\plus{}4x^5\plus{}16x^4\minus{}16x^3\plus{}8x^2\plus{}72x\minus{}36$.
Sonnhard.

[quote="J.Mayers"]It should factor out to be $ (x^{4} \minus{} 4x^{2} \plus{} 18)(x^{4} \plus{} 4x \minus{} 2)$ if you're setting $ e \equal{} 4$...[/quote]
$ (x^{4} \minus{} 4x^{2} \plus{} 18)(x^{4} \plus{} 4x^2 \minus{} 2)$

btw, what's "ansatz"???


I have a quick question somebody can hopefully clarify for me.

If there is a contest that runs every hour for 40 days. I enter 30 entries during one hour of a each day for 40 days what are my odds of winning during the course of 40 days?

Question already posted in multiple forums
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55548]http://www.mathlinks.ro/Forum/viewtopic.php?t=55548[/url]


[quote="karthik_k"]"The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom."

That's greek and latin to me...Anyone mind explaining what it means ?[/quote]
Yep... Alkali element $\text{Cs}$...perhaps it should be posted in the Chemistry forum?

i dunno exactly what it means, but thats how an atomic clock works.. something about measuring the decay of Cesium

A $\text{Cs-133}$ atom's electrons may transfer between two energy levels, emitting a certain frequency of radiation.  A second is defined as a certain multiple of the frequency of this radiation.  In otherwords, this radiation was defined to be at 9,192,631,770 Hz.  It is not actually the decay of cesium, if I remember correctly... just a transistion between two states, prompted by interaction between a nuclear dipole and an electron dipole.  I'm sure someone can explain it better than that, but its relation to the second is simply that the frequency of this particular type of radiation is used to define the second.

and that begs the question... how do we (now) define a meter?

The meter is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

For more units, see [url=http://physics.nist.gov/cuu/Units/meter.html]http://physics.nist.gov/cuu/Units/meter.html[/url]

[quote]...prompted by interaction between a nuclear dipole and an electron dipole...[/quote]

Huh ? :?


[b]1.[/b] Find the NUMBER of all the integral solutions $ (x,y)\in\mathbb Z^2$ of the equation

$ 9x^2 \plus{} 2xy \plus{} y^2 \minus{} 92x \minus{} 20y \plus{} 244 \equal{} 0.$

[b]2.[/b] Find the NUMBER of the integers $ n$ satisfying $ 3n \equal{} x(3x \minus{} 8)$ for some $ x\in[0,5]$.

[quote="Nora.91"]
$ x \in [0,5] \ \Rightarrow\ 3n \in [0, 35]$
[/quote]
I think it should be $ x \in [0,5] \ \Rightarrow\ 3n \in \left[\minus{}\frac{16}3, 35\right]$ since

$ 3\left(\left(\frac43\minus{} \frac {4}{3}\right)^2 \minus{} \frac {16}{9}\right)\equal{}\minus{}\frac{16}3$  and  $ \frac43\in[0,5].$

yeah, you are correct ifai,

so that gives 13 integral solutions.


This is a reminder that tonight's Math Jam is a level 5 Math Jam on the subject of sequences, series, and the use of sigma notation.  This Math Jam will focus on techniques that should help students interested in learning how to solve AMC and AIME level sequence and series problems.  The Math Jam will begin at 7:30 PM ET/4:30 PM PT.

Students interested in our online classes should note that tonight's material is partially taken directly from this summer's AMC/AIME Algebra class and should give student's an indication of the level of the class.  The full Algebra class does however build significantly deeper into the subjects we will be discussing tonight including a look into difference equations which tie together many of the individual methods students use to evaluate various kinds of series.

Could you be a bit more specific in telling what you will cover?

Well I think it s a bit too late for him to explain now. :wink: The math jam will start in a few hours so I think that we can wait until then to find out.


induction
number theory unsolved
number theory

Let $ \{a_k\}_{k \equal{} 1}^{\infty}$ be a sequence of real numbers such that $ a_1 \equal{} 3, a_2 \equal{} 7,$ and $ a_n^2 \plus{} 5 \equal{} a_{n \minus{} 1}a_{n \plus{} 1}$ for all $ n \ge 2$. Prove that if $ a_n \plus{} ( \minus{} 1)^n$ is a prime, then $ n \equal{} 3^m$ for some nonnegative integer $ m$

Let $ x\equal{}\frac{3\plus{}\sqrt{5}}{2},$ $ y\equal{}\frac{3\minus{}\sqrt{5}}{2}$. So $ x\plus{}y\equal{}3,$ $ xy\equal{}1$.
We have $ a_1\equal{}x\plus{}y$ and $ a_2\equal{}x^2\plus{}y^2$ so by inducton:
$ a_{n\plus{}1}\equal{}\frac{a_n^2\plus{}5}{a_{n\minus{}1}}\equal{}\frac{x^{2n}\plus{}y^{2n}\plus{}7}{x^{n\minus{}1}\plus{}y^{n\minus{}1}}\equal{}\frac{x^{2n}\plus{}x^2\plus{}y^2\plus{}y^{2n}}{x^{n\minus{}1}\plus{}y^{n\minus{}1}}\equal{}x^{n\plus{}1}\plus{}y^{n\plus{}1}.$
And $ a_{n\plus{}1}\equal{}(x\plus{}y)(x^n\plus{}y^n)\minus{}xy (x^{n\minus{}1}\plus{}y^{n\minus{}1})\equal{}3a_n\minus{}a_{n\minus{}1}$ is an integer.

Let $ n\equal{}3^ml$, where $ m$ is a nonnegative integer and $ l$ is a positive integer not divisible by $ 3$.

Let $ z\equal{}x^{3^m}$, $ t\equal{}y^{3^m}$. We will prove that $ a_{3^m}\minus{}1|a_n\plus{}(\minus{}1)^n$. Induction by $ l$:
For $ l\equal{}1$ it's obvious, for $ l\equal{}2$ we have $ \frac{a_n\plus{}1}{a_{3^m}\minus{}1}\equal{}\frac{z^2\plus{}1\plus{}t^2}{z\minus{}1\plus{}t}\equal{}z\plus{}1\plus{}t>1$.
If $ l>3$, then $ \frac{a_n\plus{}(\minus{}1)^n}{a_{3^m}\minus{}1}\equal{}\frac{z^l\plus{}(\minus{}1)^l\plus{}t^l}{z\minus{}1\plus{}t}$ $ \equal{}\frac{z^l\plus{}z^{l\minus{}3}}{z\minus{}1\plus{}t}\minus{}\frac{z^{l\minus{}3}\plus{}(\minus{}1)^{l\minus{}3}\plus{}t^{l\minus{}3}}{z\minus{}1\plus{}t}\plus{}\frac{t^{l\minus{}3}\plus{}t^l}{z\minus{}1\plus{}t}\equal{}$ $ z^{l\minus{}1}\plus{}z^{l\minus{}2}\plus{}t^{l\minus{}2}\plus{}t^{l\minus{}1}\minus{}\frac{z^{l\minus{}3}\plus{}(\minus{}1)^{l\minus{}3}\plus{}t^{l\minus{}3}}{z\minus{}1\plus{}t}\equal{}$ $ \equal{}a_{3^m(l\minus{}1)}\plus{}a_{3^m(l\minus{}2)}\plus{}\frac{a_{3^m(l\minus{}3)}\plus{}(\minus{}1)^{3^m(l\minus{}3)}}{a_{3^m}\minus{}1}$ is an integer greater than $ 1$.
So for every $ l$ greater than $ 1$ and not divisible by $ 3$, $ a_n$ has a divisor greater than $ 1$ and less than itself.


algebra
polynomial
MATHCOUNTS
floor function
factorial
Pascal\u0027s Triangle

Find a reasonable explanation for, and the next three terms of, the following sequence.

1, 2, 4, 8, 16, [b]31[/b]...

Maybe you should include another term, this looks very strange.

Obviously its not doubling. Its something complicated that produces something like doubling for the first couple terms.

it looks like doubling...but its not...can you include another term?

there's some geometrical counting reference, and I cant put my hand on it.

All you do is double it, but if its greater than or equal to 31, then you subtract 1, I don't know

yeah, that's it!  how do you prove though.. ?_?

How about this sequence:
$3, 5, 10, 19, ...$

WAIT!!!
What are the next three numbers to the first sequence?!

nat mc, include another term, because I see something but with only four terms, it can't make me certain.

how do you solve mathnerd's original question? I got the geometrical reference, but I don't know how to actually count and show that it's the sequence he wrote (and thus generate more terms)..

[quote="JesusFreak197"][quote="joml88"][hide]
Consider a circle.  Now for $n=1,2,3,\ldots$, the $n$-th term in the sequence is the number of regions that the circle will be divided into if you place $n$ points on the circumference and connect all possible segments.[/hide][/quote]

Looks right, but is there some formula for finding the next three numbers quickly?[/quote]

Counting is obviously inaccurate. If there is a polynomial formula for this one, it is at least the fifth degree (first term has variable $x^5$) but there isn't enough terms to determine a constant difference or such a pattern.

Hmm, I don't know the next three terms to the sequence that I defined in my earlier post  :blush:  It is actually, as I have heard, a very hard problem.  It is definitely beyond Mathcounts level.  I plan on sitting down and working on this problem so more sometime.  I'm sure someone on this site knows the formula as it is a well known problem (well known to advanced problem solvers that is).

Hahah that floor solution is awesome.  Yes the circle solution is the intended one as I remember this to be a mathcounts challenge question I think (at nats?)... I just dont know how to solve it...

[quote="yif man12"]Could be $2^{n-1}-\lfloor n/6 \rfloor$

Of course, using floors and ceilings are sort of cheating, and the intended solution is probably the one with the circle.[/quote]

Yes, we definetly need another term, because the sequence above isn't wrong if you round up to the next number. For now, I'm taking a trip to the high school forums.

We need another term to answer the questino joml88's is the intended solution.  can someone please figure out how to actually count that though ? as in how do you show that that is true... I figured it out last night in a dream but I forgot it now  :)

mathnerd's problem is within the level of State or National Masters Round material.

[hide="hint"]I think there are enough terms in the sequence.  With [b]patience[/b], you might be able to deduce a formula by considering what happens when you add the fifth point to a circle already containing 4 points.  [How many new regions are created -- and why?]  What happens when you add the 6th point.  What about the nth point?[/hide]

At first, it seems like each segment will add one new piece, and since there are $\displaystyle{n\choose2}$ lines (n=number of points), the formula now is $\displaystyle{n\choose2} + 1$, adding that one that was already there. When we get to four points, thisno longer works. One of the segments will generate two pieces. In five points, five extras are created. With 6 points, 15 extras are generated.

I can't seem to explain why, other than that the lines cut more than one secton up, so there are more pieces "made" at a time. I think that it follows a pattern $\displaystyle{n\choose4}$. Now the formula is $\displaystyle{n\choose4}+\displaystyle{n\choose2} + 1$. That should give us 57 pieces for 7 points, 99 pieces for 8 points, and 163 pieces for 9 points.

factorial, I believe your formula is correct.  I'll try to (partially) explain why it works in a minute.

This problem is *very hard* for MATHCOUNTS.  For a MATHCOUNTS student, the best way to solve this would probably be to look for a pattern, using differences.  Given the original sequence:
$1, 2, 4, 8, 16, 31$
the first differences are
$1, 2, 4, 8, 15$
The second differences are
$1, 2, 4, 7$
The third differences are
$1, 2, 3$
And the fourth differences are

$1, 1$

It would be a "hunch" that the differences are constant at this point, but it would be a good hunch.  You can then work backwards from the fourth difference to get the next term of the sequence.  $1+3=4,$ $4+7=11,$ $11+15=26,$ $26+31=57.$  And voila!  We have the seventh term of the sequence.

Let me try to explain why there are 26 more regions.

We had 6 points around the circle, and we are adding a 7th point.  From the new point, we can draw a line to each of the previous 6 points, which creates at least 6 more regions, even if we crossed no other lines.  For n=7, we add $n-1=6$ regions.

But we know we may cross some other lines -- maybe a lot of other lines.  There is a trick to counting these.  Let's consider all of the triangles that already exist, formed by connecting groups of 3 existing points.  A line from our new point to the two "outer" vertices of the triangle won't cross an existing line.  But a line from our new point to the "middle" vertex of the triangle will cross an existing line.  So, for every triangle that existed, we will cross one additional line.   So we add ${{n-1} \choose {3}}$ more regions.  For n=7, we add ${6\choose 3}=20$ regions.

For n=7, we have added a total of 26 new regions.

I don't wish to explain it in this forum, but a summation of the ${n-1}$ terms leads to the ${n\choose 2}$ term in factorial's formula.  A summation of the ${{n-1}\choose{3}}$ terms leads to the ${{n}\choose{4}}$ term in factorial's formula.

If you like looking for patterns, write out Pascal's triangle to about row 7 or 8, and see if you can see why this occurs.

[color=darkblue]A note on difficulty from moderator RCV:  The average level of problem in this Forum has gotten way too hard.  While most of the problems are within the level of MATHCOUNTS, many problems are only within the level of MATHCOUNTS for National competitors and sometimes just National Masters Round Winners.

We allowed the difficulty level to rise in preparation for the National Competition, but it's time to pull it back.  This forum should be appealing to many of the 500,000 MATHCOUNTS participants -- not just a few dozen elite National Countdown wannabe's.

I think it is legitimate to accept authentic MATHCOUNTS problems, even if they are hard.  But you must list the source of the problem -- E.g., "1995 National Masters Round".

The difficulty of other problems that are merely "similar" to MATHCOUNTS problems needs to be pulled back.  When school starts in August / September, the majority of problems in this forum should be at about the Chapter level.  (Not the *hardest* Chapter questions, but typical Chapter questions.)

For those of you who have been tackling these very difficult problems, you should consider moving up to the High School forums.  Congratulations!

I'll update the Forum Guidelines by the end of the month.  But please start pulling it back now.[/color]

Yes, I tried the differences pattern too, but I screwed up on one calculation and it gave some strange numbers, which was why I mentioned a polynomial representation to the fifth degree, when according to your differences, it should be to the fourth. Also, I believe that such a polynomial can be found right through multiplying out the combination formulas in the equation you and I agreed on.

But, this topic is on my book called "Advanced Algebra", for average tenth graders or above average ninth graders in the nation as a classroom text, and I'm not sure whether it would be appropriate to continue to discuss other ways in this forum.

And yes, I definetly think the forum level is getting slightly out of hand. Although it isn't too hard for Mathcounts as a whole, I'm definetly getting confused with some posts. I'm actually a very average Mathcounts student, getting seventh at my little chapter competition (I didn't know what Mathcounts was until summer of last year). Now, I can say that I've improved a lot, but still, it would be nice to see some problems on this forum that the majority of students can solve.

I have a feeling that the basic forum is actually a second forum for Mathcounts participants with lower capacities (chapter/early state level).

The next term is 57, and then 98.

It's a fourth order equation.


calculus
integration
calculus computations

hello, it gives $ \minus{}2\text{EllipticE}[\frac{1}{2}(\frac{\pi}{2}\minus{}x),2]$
Sonnhard.


Take a look.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=80515[/url]

Someone has posted PDF files related this theme.

As the latest of many, many manifestations on this site, see if you can get it from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107423]the problem I worked just this morning.[/url]

[quote="Kent Merryfield"]..see if you can get it from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107423]the problem I worked just this morning.[/url][/quote]
Nice :)
with  $f(x) = \{ \pi/4-x \text{ if }0\leq \pi/4,\qquad-\pi/4+x \text{ if }\pi/4\leq x<\pi/2$

$f(x)=\pi/8+\frac1{\pi}\sum_{{k=1}\above{k\text{ odd}}}^{\infty}\frac1{k^{2}}\cos(4kx) \Longrightarrow \pi/4 = \pi/8+\frac1{\pi}\sum_{{k=1}\above{k\text{ odd}}}^{\infty}\frac1{k^{2}}\\=\pi/8+\frac1{\pi}\sum_{k=1}^\infty1/k^{2}-\sum_{k=1}^\infty1/(2k)^{2}= \pi/8+\frac3{4\pi}\sum_{k=1}^\infty 1/k^{2}\\\boxed{\sum_{k=1}^\infty 1/k^{2}=\frac{\pi^{2}}{6}}$

edit: also check out this collection of proofs of this nice identity: http://www.secam.ex.ac.uk/~rjc/etc/zeta2.pdf


Good luck to everyone! 27 mins remaining :)

how did your ZCO go??
(if its over now :rotfl: )
I don't know anything about it :oops:


There are 15 student in a science class.Determine that atleast two of the share the same birthday.(every year has 365 days.)


When of my calculus assignments asks an interesting problem...
The problem is:

Find the limit as $ x$ approaches $ \minus{}2$ of:

$ \frac{3x^{2}\plus{}ax\plus{}a\plus{}3}{x^{2}\plus{}x\minus{}2}$

Is there a real value of a that makes this possible?
If so, what is the limit?

Solving it algebraically, I got the answer of $ a\equal{}15$ and the limit is $ \minus{}1$, which is correct. However, my teacher said there was a better and faster method, and I can't find it. I'm sure as soon as someone tells me, I'll think duh, but...I can't see it. Please help...

Thanks.

One way to find the limit is by the SUBSTITUTION METHOD.

more info : http://math.haiyanh.com/articles/how-to-find-the-limit.html

Oh... :blush:

I thought that $ \minus{}2$ would make the denominator $ 0$, but it doesn't really...

Thanks, I feel stupid.

EDIT: Um...now I feel even dumber. I copied the problem onto the board wrong, the denominator is $ x^{2}\plus{}x\minus{}2$, not $ x^{2}\minus{}x\minus{}2$, so the substitution method doesn't work...

Please use LaTeX by putting math in dollar signs.   :)

So, to clarify, $ \lim_{x\to\minus{}2}\frac{3x^{2}\plus{}(a\plus{}1)x\plus{}3}{x^{2}\minus{}x\minus{}2}$?

So you want a value of $ a$ such thatt $ \lim_{x\to\minus{}2}\frac{3x^{2}\plus{}(a\plus{}1)x\plus{}3}{x^{2}\plus{}x\minus{}2}$ is a real number. This can only happen if $ \minus{}2$ is a root of the numerator, since $ \minus{}2$ is a root of the denominator. So we plug in $ \minus{}2$ into $ 3x^{2}\plus{}(a\plus{}1)x\plus{}3 \equal{} 0$ and solve for $ a$.
$ 12\minus{}2a\minus{}2\plus{}3 \equal{} 0$
$ 13\minus{}2a \equal{} 0$
$ a \equal{}\frac{13}{2}$

We then have $ \frac{3x^{2}\plus{}\frac{15}{2}x\plus{}3}{x^{2}\plus{}x\minus{}2}\equal{}\frac{3(x\plus{}2)(x\plus{}\frac{1}{2})}{(x\plus{}2)(x\minus{}1)}\equal{}\frac{3(x\plus{}\frac{1}{2})}{x\minus{}1}$, so our limit is $ \frac{3(\minus{}2\plus{}\frac{1}{2})}{\minus{}3}\equal{} 2\minus{}\frac{1}{2}\equal{}\frac{3}{2}$.

Your answer doesn't seem to fit.

You're right. Unfortunately, there was another typo in my orginal problem, which I now fixed. However, I think your method is what I was looking for, and applying it to the actual equation gives $ a \equal{} 15$ and $ \minus{}1$ for the limit. Thanks Hamster.

And thanks Archimedes, the dollar sign was the step that I couldn't seem to figure out by myself...


In the figure below , all boxes are made by matchsticks. The 1st hast 4 , the second 10 , the third 18 , the next 28  matchsticks, ect.
 How many matchsticks has the n - th figure (with proof) ?

 Try - if possible-to write a simple solution , suitable for students in  classes 8-9.
 Thanks

Notice that in order to build the $ n$-figure from the $ (n\minus{}1)$-figure you have to add $ (n\plus{}1)$ horizontal matches and $ (1\plus{}n)$ vertical matches.

Then the general formula is:
$ 4\plus{}2 \cdot 3\plus{}\cdots\plus{}2 \cdot n\plus{}2 \cdot (n\plus{}1)$
$ 2 \cdot 2\plus{}2 \cdot 3\plus{}\cdots\plus{}2 \cdot n\plus{}2 \cdot (n\plus{}1)$
$ 2 \cdot 1\plus{}2 \cdot 2\plus{}2 \cdot 3\plus{}\cdots\plus{}2 \cdot n\plus{}2 \cdot (n\plus{}1)\minus{}2$
$ (1\plus{}(n\plus{}1))\plus{}(2\plus{}n)\plus{}\cdots\plus{}((n\plus{}1)\plus{}1)\minus{}2\equal{}(n\plus{}1)(n\plus{}2)\minus{}2$
$ n^2\plus{}n\plus{}2n\plus{}2\minus{}2$
$ n(n\plus{}3)$

[hide="Solution"]
Notice the number increases by an extra addition of two.
The following sequence:
$ 4, 10, 18, 28,...n,..$ can be rewritten as $ 4, 4 \plus{} 6, 4 \plus{} 6 \plus{} 8, 4 \plus{} 6 \plus{} 8 \plus{} 10,...$ which simplifies to
$ 2(2),2(2 \plus{} 3),2(2 \plus{} 3 \plus{} 4),...$ or $ 2\sum_{k \equal{} 1}^{n}(k \plus{} 1) \equal{} 2\left(\sum_{k \equal{} 1}^{n}k \plus{} \sum_{n \equal{} 1}^{n}1\right)$
$ \implies 2\left(\frac {n(n \plus{} 1)}{2} \plus{} n\right)$
$ a_n \equal{} \boxed{n(n \plus{} 3)}$
[/hide]
I'm not sure this proof I'm giving is appropriate for the scenario since it's based on the numeric sequence instead of the relation of the progressing structures.
[hide="My Very First Proof"]
I'll be very thankful to those that correct me, because this is my first proof I've ever given and I want to practice learning how to create a proof.

[u]Theorem(?)[/u]: The sequence $ 4, 10, 18, 28, ...a_n...$, which define the number of matchsticks needed to build each given structure and their later structures is defined by $ a_n \equal{} n(n \plus{} 3)$
[u][i]Proof[/i][/u]: This satisfies the first given of the sequence. $ a_1 \equal{} 4$
Looking back at the sequence, we look back at their difference
$ a_2 \minus{} a_1 \equal{} 6$
$ a_3 \minus{} a_2 \equal{} 8$
$ a_4 \minus{} a_3 \equal{} 10....$
$ a_{n \plus{} 1} \minus{} a_{n} \equal{} (n \plus{} 1)(n \plus{} 4) \minus{} n(n \plus{} 3)$
$ \equal{} (n^2 \plus{} 5n \plus{} 4) \minus{} (n^2 \plus{} 3n) \implies 2n \plus{} 4$ or $ 2(n \plus{} 2)$ if $ a_n \equal{} n(n \plus{} 3)$ holds true.
Looking at it independently as $ b_n$,
$ 6, 8, 10, 12, .... \implies b_n \equal{} 2(n \plus{} 2)$, which satisfies $ a_{n \plus{} 1} \minus{} a_{n}$.
[/hide]

That's a good way to start a proof, but it's not technically valid.  You assume that the pattern continues, which is not necessarily true.  What can seem like a pattern can break after a certain limit.

[url=http://www.stmarksschool.org/data/files//MathInstituteSolutions/Microsoft_Word__Newsletter_March06_Web_Version.pdf]Take a look at this.[/url]

Another approach
[hide]
Define $ f(n)$ as the number of squares in the $ n^{th}$ figure. Obviously $ f(n)\equal{}1\plus{}2\plus{}...\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$.
Define another function $ g(n)$ as the number of "overlapping" sides vertically. Each such side is given by two horizontally adjacent squares, from the bottom row to the top row, we have $ g(n)\equal{}0\plus{}1\plus{}2\plus{}..\plus{}(n\minus{}1)\equal{}\frac{n(n\minus{}1)}{2}$. This is also the number of overlapping sides horizontally.
Now, the total number of matches is: $ 4f(n)\minus{}2g(n)\equal{}n(n\plus{}3)$.
[/hide]

[quote="mountain.dew"]Or you could just use induction..[/quote]If I'm reading correctly, this is what stergiu said
[quote="stergiu"]Try - if possible-to write a simple solution , suitable for students in classes 8-9. [/quote]Induction would be the [b]LAST[/b] thing to use to find a pattern. It's used in proofs but it is in no way useful to find patterns.

[b]Associate [/b]vertical sticks with the square to the right (except the rightmost), and horizontal ones with the square above (except the topmost), creating an array of L's with no overlap.

Then, we have $ 2 \cdot \frac {n(n \plus{} 1)}{2} \equal{} n^2 \plus{} n$ matchsticks from L's.

Now, we need to add $ n$ horizontal bars to cover each column and $ n$ vertical bars to end each row, for a total of $ n^2 \plus{} 3n$ sticks.

Many thanks to all of you. Really nice and correct solutions!
 I first saw the problem to the junior level mathematics olympiad in South Africa. There it was asked only the number at 30 - th figure.
 The answer was $ 30 \cdot 33$ , since $ 4 \equal{} 1 \cdot 4 , 10 \equal{} 2 \cdot 5 , 18 \equal{} 3 \cdot 6 , ....$ .
 Of course this is not a rigorous proof . So I decide to put the problem for the general case. Nice work of you!

 [u]Babis[/u]


Let $ a,b,c>0$ and $ m \in N^*$ then prove:
$ \frac{ab}{m^2\plus{}ab}\plus{}\frac{bc}{m^2\plus{}bc}\plus{}\frac{ca}{m^2\plus{}ca} \le \frac{a\plus{}b\plus{}c}{2m}$.

$ \sum_{cyc}\frac{ab}{m^{2}\plus{}ab}\le\sum_{cyc}\frac{ab}{2m\sqrt{ab}}$     by AM-GM
$ \equal{}\frac{\sum_{cyc}\sqrt{ab}}{2m}\le \frac{\sum_{cyc}\frac{a\plus{}b}{2}}{2m}\equal{}\frac{a\plus{}b\plus{}c}{2}$ QED.

Let $ a,b,c \ge 0$ and $ m,n \in N^*, n>1$  then prove:
$ \frac {ab}{m^n\plus{}n\minus{}2\plus{}ab}\plus{}\frac {bc}{m^n\plus{}n\minus{}2\plus{}bc}\plus{}\frac {ca}{m^n\plus{}n\minus{}2\plus{}ca} \le \frac{\sqrt[n]{(ab)^{n\minus{}1}}\plus{}\sqrt[n]{(bc)^{n\minus{}1}}\plus{}\sqrt[n]{(ca)^{n\minus{}1}}}{nm}$.

Have $ \frac {ab}{m^n \plus{} n \minus{} 2 \plus{} ab} \plus{} \frac {bc}{m^n \plus{} n \minus{} 2 \plus{} bc} \plus{} \frac {ca}{m^n \plus{} n \minus{} 2 \plus{} ca}$
$ \equal{} \sum_{cyc}\frac {ab}{m^n \plus{} n \minus{} 2 \plus{} ab}\le \sum_{cyc}\frac {ab}{n\sqrt [n]{m^n(1)^{n \minus{} 2}ab}}$ by AM-GM
$ \equal{} \sum_{cyc}\frac {ab}{nm\sqrt [n]{ab}} \equal{} \frac {\sqrt [n]{(ab)^{n \minus{} 1}} \plus{} \sqrt [n]{(bc)^{n \minus{} 1}} \plus{} \sqrt [n]{(ca)^{n \minus{} 1}}}{nm}$ QED


Let $x_1, x_2, . . . , x_n$ be positive numbers. Prove that

   \[ \sum\limits_{i=1}^n \dfrac{x_i ^2}{x_i ^2+x_{i+1}x_{i+2}} \leq n-1 \]

Where $x_{n+1}=x_1$ and $x_{n+2}=x_2$.

I posted this a loooooooong time ago, it's a problem from the 1985 IMO Shortlist.

Induction.

1)For $n=3$ we get the inequality:
$\sum \dfrac{x^2}{x^2+yz} \leq 2$
which can be easily transformed into form:
$\sum \dfrac{yz}{x^2+yz} \geq 1$
and then from Cauchy in Engel's Form:
$\sum \dfrac{yz}{x^2+yz} = \sum \dfrac{y^2z^2}{x^2yz + y^2z^2} \geq \dfrac{(xy+yz+zx)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^2yz + xy^2z + xyz^2} = \dfrac{(xy+yz+zx)^2}{(xy+yz+zx)^2-x^2yz-xy^2z-xyz^2}$
which is clearly $\geq 1$

2)Suppose that inequality is true for $n$ variables, we will prove it for $n+1$ variables. So, we know that:
 $\sum\limits_{i=1}^n \dfrac{x_i ^2}{x_i ^2+x_{i+1}x_{i+2}} \leq n-1$
and we have to prove:
$\sum\limits_{i=1}^{n+1} \dfrac{x_i ^2}{x_i ^2+x_{i+1}x_{i+2}} \leq n$

Now, WLOG suppose that $x_{n+1}=max(x_1,x_2,...,x_{n+1})$.
If we prove:

$\dfrac{x_{n+1}^2}{x_{n+1}^2 + x_1x_2} + \dfrac{x_{n}^2}{x_{n}^2 + x_{n+1}x_1} + \dfrac{x_{n-1}^2}{x_{n-1}^2 + x_1x_2} - \dfrac{x_{n}^2}{x_{n}^2 + x_1x_2} - \dfrac{x_{n-1}^2}{x_{n-1}^2 + x_nx_1} \leq 1$
we will be done (sum up). But because $x_{n+1}$ is max among variables:

$\dfrac{1}{x_{n}^2+x_1x_2} \geq \dfrac{1}{x_{n}^2 + x_{n+1}x_1}$
$\dfrac{1}{x_{n-1}^2+x_nx_1} \geq \dfrac{1}{x_{n-1}^2 + x_nx_{n+1}}$

So, our inequality can be rewritten as:
$(1-\dfrac{x_{n+1}^2}{x_{n+1}^2 + x_1x_2}) + x_n^2(\dfrac{1}{x_{n}^2+x_1x_2} - \dfrac{1}{x_{n}^2 + x_{n+1}x_1}) + x_{n-1}^2(\dfrac{1}{x_{n-1}^2+x_nx_1} - \dfrac{1}{x_{n-1}^2 + x_nx_{n+1}}) \geq 0$
which is true.


This is a poll of, in general, peoples piano skills are on AoPS. I myself am playing op 10 no 12 of chopin's etudes. When you pick a catogory, note that you should judge each one by the average difficulty of its songs; so while there are bach inventions and whatevers that present a moderate challenge to me, I'd still pick the Chopin etudes because that song is in the midranges of difficulty.

Hmmm, I don't think I've ever heard or played Liszt's etudes....*shocked at myself*.  I should get on that, see if I can play any too!

I'd say a lot of these are based on what you are good at. Rachmaninoff feels more natural to me than Liszt. I think of Liszt as crazy technique, whereas Rachmaninoff is challenging due to the largeness of the chords. I've successfully learned things by Rachmaninoff, but nothing by Liszt. (I put myself at Chopin, though).

Hmm... I played La Campanella this June for a recital, and I'm actually not that much of a failure on that piece! I think it is average among Liszt's etudes, so I put myself at Liszt. :play_ball:

By the way, isn't it fascinating how so many math geniuses are also piano prodigies? :idea:  :!:

Of all the kids I know that are amazing at the piano, the three or four that are the best (and they really are amazing) aren't quite math geniuses. And it's not just a matter of them not being into math.

While I know very little about playing piano, I have a feeling that these people start from the artistic side of the music, and then develop a sense of rhythm and theory. I'd hypothesize that many people who are great at math can crack the rhythm and interpret written music very easily, and then try to develop the emotional feel.

Also, while I don't mean to fuel ethnic stereotypes, I have to say that many East-Asians have amazing dexterity, and, coincidentally (perhaps?), represent a large portion of those interested in mathematics in high school.

All  the Uber amazing pianists (below 21ish) I know personally are asian. Of course they're all girls too, so  :maybe: . Oh, and on the correlation between math and music, BIg math brain (generaly)= big brain in general, big brain in general (generally)=big creative side of brain=Good at music/art whatever. At least as far as I know

yeah, lots of math ppl play instruments, like i look through my nats albums, and so many ppl play instruments.

"chopsticks" is probably none dynamo... that means if someone sat you in front of a piano, you'd just try to play like a scale using index fingers or something...

i play easy stuff  :lol: . i used to play pretty well but then i stopped playing because my piano teacher moved away and i couldn't find a new one  :( but i can still play something at least. :)

is that too many emotes? I hope not.

Like 80% of my friends play piano, and they're good too(well that's what I think). One of my friend is like an international pianist, maybe you guys would know him(but I don't have permission to post his name). Hmm, I lot of people in my town play the piano....never really noticed that. Anyways I am horrible and can only play with one hand(it seems like I can't control both hands at once). I just know some basic easy songs.

[quote="mathelete9999"]All  the Uber amazing pianists (below 21ish) I know personally are asian. Of course they're all girls too, so  :maybe: .[/quote]

What about Lang Lang? Hes "uber" amazing. I got to hear him play, better than me  :D

shentang, I don't think the person you are talking about would care so much about you mentioning his name (though I don't know him personally, only...electronically), as he has like 75 Youtube videos...

This is probably a more fitting discussion for the Music & Art forum. But my technique pretty much sucks. I've never liked the idea of performing etudes, unless you're playing them as encores in which the point is usually to blatantly show off your technique. I strongly believe that one's "skills" in piano are not about his techinque, but his musicality.

Chopin Preludes are very pretty but not too difficult.  They rely more on rote understanding of fast finger positions to play effectively.  For example, Prelude 1 played with slightly faster tempo, has a very consistent rhythm (with a few fluctuated variations of course), and mastery only comes with muscle memory at such fast speeds.

the preludes do vary in lvl though, some are very easy, and some harder

I have noticed, unsuprisingly that most AoPSers are good at music.
I unfortunately am the only one in the Bach Invention level. :(

You should add a new choice called "Chopin scherzos." Seriously... 967 measures in scherzo no. 2. Darn.

I'm learning Chopin's Ballade no. 2...there's a lot to learn. I like the last two pages, the "agitato" part. :D

I quit piano and started violin instead, but I've taught myself a few songs. Like Hungarian Dance No. 5 and The Entertainer. Which is all easy stuff, though...:(

@Maxima,

ive played La Capanella, you just have to get over the wide jumps and learn how to play the chromatic and that wierd scale it has in it efficiently, once that's over, things come smoothly

I just finished Chopin Nocturne in E flat Opus 9 No 2 for a competition, so I think that would put me in Chopin etudes, is this correct??

[quote="anaverageaopser"]In my opinion, most etudes are harder than this Nocturne in particular.  You could ask someone else, but I don't think being able to play this piece means you can master the Etudes.[/quote]
I would agree with this statement.  There is a significant gap here in technical ability here.  For me, it was about 1-2 years in advancement.

I can't believe I haven't posted in this thread yet!! :o

I'm a little torn between the last two choices. I've played only one from each, and while I mastered the Rachmaninoff Etude (the e-flat minor one) quickly, Mazeppa (Transcendental Etude #4) took forever to learn to an acceptable level. I learned these two etudes about 4 months apart, so there shouldn't have been an appreciable change in my skill. However, I do believe that the author intends the Rachmaninoff etudes to be more difficult, so I don't know what to think. Anyway, I don't want to tear this issue any further, as I see plenty of discussion on the poll options earlier in the thread.

Has anyone played Islamey or Gaspard de la Nuit? I think those two pieces are harder than most of the Liszt/Rachmaninoff Etudes.

I currently am working on Chopin's first concerto and the third movement of Rachmaninoff's second. I just finished playing Beethoven's fifth in a competition. As you can see, I'm a huge fan of the concerto form and concept. The grandiosity of the piano/orchestra combination simply can't be beat. I will also be learning some Chopin for a recital this fall, and possibly for a "Chopin Musical" (student-written). Occassionally, I compose difficult etudes/pieces to improve technique, including a set of variations on Hot Cross Buns that helped me win my high school's talent show [/brag].

[quote="jasonzhang"]has anyone played the etude in thirds by chopin?  i just finished it and can play at half note 66, but are there any suggestions on how to start this etude without much warmup? It takes me like an hour before i get the hang of it.  :play_ball:[/quote]

This etude, in my experience, is ridiculous. I got the tempo to around 66 as well, but some of the passages were still uneven. I found relaxing the wrist and minimizing hand motion to be rather helpful (the second point might seem counterintuitive, but it worked - have you even seen those pianists that play at a crazy speed but their hands seem to be just floating on the keyboard?).

[quote="jasonzhang"]My piano teacher is making me do RACH 2!!!!! its scaring me.[/quote]

Yes, this is a monster. As I wrote above, I'm working on the third movement. I wrestled with the first movement for about 5 years (on and off, of course) and performed it more times than I care to remember, including twice with an orchestra, and now the piece will never leave my hand. While I don't have all the phrasing refined to perfection, I can confidently say that I "know" this piece inside-and-out. Despite what others say, I regard this concerto as one of the greatest pieces ever written.

Some of the passages are awkward and will just take time to sink in. For the first section, where the piano does over a minute of arpeggio-type stuff, tempo training might be helpful (start really slow, and build up to concert tempo).

i'm not good but not bad

@ people who think this is reviving: it'slike only 3 spots form the top thread

You can't call it reviving

[quote="leetnoob"]Currently, I play:
Fur Elise
Welcome to the Black Parade(Intro)
King of the Hill Theme
Lord of the Rings Theme
When the Saints Come Marching In

... and two and a half textbooks of songs.[/quote]
Well, you basically just stole my post.
Except I also play anime themes.


Let $ \minus{}1<a,b,c,d<1$. Prove that

$ \sum\sqrt[4]{1\minus{}a^2}\plus{}\sum\sqrt[4]{1\minus{}a}\plus{}\sum\sqrt[4]{1\plus{}a}\leq12$

-----------------------
 :blush:  Panagiote Ligouras

we have
$ LHS \le 4\plus{}\sum 2\sqrt[4]{\frac{1\minus{}a\plus{}1\plus{}a}{2}}\equal{}12$
q.e.d

[b]            tuandokim wrote:
[quote] we have
 $ LHS\le 4\plus{}\sum 2\sqrt[4]{\frac{1\minus{}a\plus{}1\plus{}a}{2}}\equal{}12$

q.e.d[/quote]

could you please write it a bit more detailed? thanks ahead.[/b]

Nothing is hidden here. We observe that the equality holds if and only if $ a\equal{}b\equal{}c\equal{}d\equal{}0$ and $ \sqrt{1\minus{}a^2} \le 1$
.
 Furthermore, it is trivial to show that $ \sqrt[4]{1\minus{}a} \plus{} \sqrt[4]{1\plus{}a} \le \sqrt{2\left(\sqrt{1\minus{}a} \plus{} \sqrt{1\plus{}a}\right)}$

Finally, our objective is to demonstrate the following

$ \sqrt{1\plus{}a} \plus{} \sqrt{1\minus{}a} \le \sqrt{2\left(1\minus{}a\plus{}1\plus{}a \right)}\equal{}2$

We finish our proof.


I need to solve this for my project can anyone help?

If x , y, and z are real numbers such that
x^2 + y^2 + z^2 -6x -8y -10z +50 = 0, then what is the value of xyz?

Thank you!

I know you're new, but try not to spam the board with threads :)

[hide="Answer"]
split the +50 constant up cleverly..
$x^{2}-6x+9+y^{2}-8y+16+z^{2}-10z+25 = 0$
$(x^{2}-6x+9)+(y^{2}-8y+16)+(z^{2}-10z+25) = 0$
$(x-3)^{2}+(y-4)^{2}+(z-5)^{2}= 0$
Sum of squares can only be zero if the squares are zero.
$x = 3$
$y = 4$
$z = 5$
$xyz = 60$
[/hide]


modular arithmetic
number theory proposed
number theory

Is the following statement true?

For each positive integer $n$, we can find eight nonnegative integers $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ such that $n=\frac{2^a-2^b}{2^c-2^d}\cdot\frac{2^e-2^f}{2^g-2^h}$.

Also the ideas of the following problem will help us
http://www.mathlinks.ro/Forum/viewtopic.php?t=35319

For odd $n$, the problem reduces to finding four nonnegative integers $a, b, c, d$ such that $(2^a-1)(2^b-1)=(2^c-1)(2^d-1)n$. Obviously, one of $a, b, c, d$ being $0$ doesn't yield anything, so assume they're all positive.

Take any $n \equiv 3 \pmod{8}$. Then, if all of $a, b, c, d$ are $\geq 2$, we have $(-1)(-1) \equiv n(-1)(-1) \pmod{4}$, so $n \equiv 1 \pmod{4}$, absurd. So one of $a, b, c, d$ must be $1$. Note that, since every odd number is its own inverse $\pmod{8}$, we know that, regardless of where the $1$ is, if the other $3$ numbers are $\geq 3$ then $n \equiv 7 \pmod{8}$, absurd; so another of $a, b, c, d$ is $\leq 2$. If the other two are $\geq 5$, then we can only get the values $3, 1, 11=1/3 \pmod{32}$. So, taking a further restriction such as $n \equiv 19 \pmod{32}$, we force $a, b, c, d$ to have just one number $\geq 5$. But the set of possible values will then have density $0$ and not cover all integers $\equiv 19 \pmod{32}$.

The following problem arises: does there exist a constant $c$ such that every integer can be written in the form $2^k\prod _{i=1}^c \frac{2^{a_i}-1}{2^{b_i} -1}$ for some nonnegative integers $k,a_1,...,a_c,b_1,...,b_c$?

[quote="silouan"]In the first equation take at the position of $n$ the number $121$.
then try to find numbers $a,b,c,d,e,f,g,h$  to write the number in the desired form . You will not find nothing. :rotfl: . So $121$ is a counterexample[/quote]

Then try to find a zero of $\zeta (s)$ which lies out of critical strip. You will find nothing  :rotfl: . Hence I proved Riemann Hypothesis.

Sorry if you received it as a mocking - it was not. I just wanted to show you that what you did is not a proof  :)

Ok. It is nothing :) I will try to find a proof to satisfay you :) Bye

[quote="silouan"]my friend told me that  the number  $121$  could not have this form....[/quote]
I think 11 is the smallest number which doesn't have this form.

If n is odd then must exist a,b,c,d, suth that
$n=\frac{2^{a}-1}{2^{b}-1}\frac{2^{c}-1}{2^{d}-1}$.
If odd number n can be represented in this form, then obviosly $2^{k}n$ can be represent in yuor form. All odd numbers less than 19 can be represent in this form. But $19$ can not.

Let \[K_{l}=\{n\in N|n=\prod_{i=1}^{l}\frac{2^{a_{i}}-1}{2^{b_{i}}-1}\}, K_{\infty }=U_{l=1}^{\infty }K_{l}.\] For example $19\not \in K_{2}$, but $19\in K_{3}$. Let $S_{p}=\{n |1<n<2^{p}-1, n|2^{p}-1\}$, were p is prime. It is easy to show, that $n\in S_{p}\Longrightarrow n\not \in K_{\infty }$. For example $2^{11}-1=23*89$ give $23\not \in K_{\infty }$.
My conjecture \[K_{\infty }U_{p}S_{p}\] is all odd numbers.

$11=\frac{2^{10}-1}{2^{5}-1}\frac{2^{1}-1}{2^{2}-1}.$
Minimal number, with has not presented in this form is 19. Minimal number, with has not presented in form: \[\prod_{k=1}^{n}\frac{2^{a_{k}}-2^{b_{k}}}{2^{c_{k}}-2^{d_{k}}}\] for any n is 23.

I cant find a simple way to prove your claim that $n\in S_{p}\Rightarrow n\not\in K_{\infty}$, but using zsigmondy's theorem ( http://mathworld.wolfram.com/ZsigmondyTheorem.html ), I could then give a proof for your result, but then also for the following more general statement:

Suppose $n$ has a prime factor $p$ and there is a prime $q\not | n$ with $ord_{n}(2)=ord_{p}(2)=ord_{q}(2)$. Then $n\not\in K_{\infty}$.

Sketch of the proof: We can assume that $a_{i}\neq b_{j}$ for all $i,j$. Then, we have to take $a_{1}$ with $2^{a_{1}}-1$ divisible by $p$. By Zsigmondys Theorem, there is then a prime factor $p_{2}$ of $2^{a_{1}}-1$ which does not divide $2^{b}-1$ for any $b<a_{1}$; if $a_{1}>ord_{n}(2)$, we have $p_{2}\not |n$. Otherwise, we can take $p_{2}=q$ to have $p_{2}\not |n$. In ordet to get rid of it, we have to choose $2^{b_{1}}-1$ with $a_{1}|b_{1}$ and $b_{1}>a_{1}$ in the denominator; this then gives us a new prime in the denominator. Continuing, we get the contradiction.

I suppose this also shows that your final conjecture is wrong.

[nonsense approach to show it more directly deleted]

The more interesting question is:
Are there infinitely many primes $p,q$ such that $\mathrm{ord}_{p}(2) = \mathrm{ord}_{q}(2)$¿

I think, that my statement is obviosly for needed to prove.
Let $S_{p}$ set of divisors greater than 1 number $2^{p}-1$ and less,than $2^{p}-1$, were p is prime. If p and q distinct, then $S_{p},S_{q}$ had not common terms. If conjuction of $S=U_{p}S_{p}$ is finite, then exist N, suth that for any prime p>N $S_{p}$ are empty. It mean, that Mersen's number $2^{p}-1$ are prime for all prime p>N. It is not true.

[quote="ZetaX"]Well, if $q|2^{r}-1$ ($p,q$ primes), then the order of $2 \mod q$ is $r$.
Thus $q| 2^{m}-1$ iff $r|m$ for all such primes $q$. If there are different ones for the same prime $r$, then they will always occure together. This proves Rust's statement more directly.
[/quote]

Well, they always occur together, but they do not always occur in the same power. For example $2^{23\cdot 11}-1$ is divisible exactly by $23^{2}$ and exactly by $89$. Therefore, an argument only considering these two primes cant work.

Let k>1 natural. Consider natural numbers can be represents in form: \[(1) \ \ x=\prod_{i=1}^{m}(k^{a_{i}}-1)^{\theta_{i}}\ , \theta_{i}=\pm 1.\] Let $M_{n}=\prod_{d|n}(k^{n/d}-1)^{\mu (d)},$ were $\mu(x)$ is mebius function. Number M_n>1 is natural, and for all divisor a>1
$a|M_{n}$ we have $a|k^{n}-1$, but $a\not |(k^{d}-1), \ d|n,d<n$.
Let $S_{n}$ denote all divisors $M_{n}$ less than $M_{n}$ and greater 1.
Statement. Numbers belongs $S_{n}$  can not be represent in form (1).
Prove. Let $a_{1}$ is maximal degree in form (1), then $\theta_{1}=1$, else x is not integer. It give $M_{a_{1}}|x$. If x is divisor $M_{n}$ it give $a_{1}=n$ and $x=M_{n}$.

[quote="Rust"]It give $M_{a_{1}}|x$. If x is divisor $M_{n}$ it give $a_{1}=n$ and $x=M_{n}$.[/quote]

I don't understand your reasoning(especially, why do we have $M_{a_{1}}|x$? some parts of $M_{a_{1}}$ can be cancelled through other terms...). However, using zsigmondy's theorem, one can prove that your statement is actually correct(as i have shown above).

[quote="Peter Scholze"] why do we have $M_{a_{1}}|x$? some parts of $M_{a_{1}}$ can be cancelled through other terms...). However, using zsigmondy's theorem, one can prove that your statement is actually correct(as i have shown above).[/quote]
Lemm. If $m\not =n$ then $(M_{m},M_{n})=1$.
Prove. For all divisor a of number $M_{n}$ we have $Per(a)=n$ ($a|k^{n}-1$ but $a\not |k^{d}-1$ for any divisor d|n , d<n). It proved Lemm.
$a_{1}$ is maximal degree in this form, thererefore for all $a_{i}$ with $\theta_{i}=-1$ we have $(M_{a_{1}},k^{a_{i}}-1)=1$. It give $M_{a_{1}}|x$.


[hide]Let $ x$ be the number of blue marbles in the bag. Then there are $ 2x$ red marbles. So $ 3x$=$ 21$ and $ x$=$ 7$. There are 7 blue marbles in the bag and 14 red marbles. If blue was chosen first, then the probability there's one blue and one red is (2/3)(7/20)=7/30. If red was chosen first, then it's (1/3)(7/10)=7/30. So the probability that one marble of each color is selected is 7/15.[/hide]
P.S. DonkeyKingKong, please don't post so many different topics with one problem per topic. It's like spam. Please post multiple problems in one thread, instead of 30 different separate threads. I also don't find it necessary to talk about EVERY single state problem from the sprint, either. I mean, seriously, I don't find it funny.


The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is \[ \textbf {(A) } 11\qquad \textbf {(B) } 12\qquad\textbf {(C) } 13\qquad\textbf {(D) } 14\qquad\textbf {(E) } 15 \]

[hide]Since the average of the eight integers is 8, then the sum of them is 64.  Since the median and the unique mode are 8, at least the 4th and 5th numbers are 8.  So, basically our set is a, b, c, 8, 8, d, e, a+8.  a+8 can't be 15 because then $a=7$ and then $b$ and $c$ would be either 7 or 8. But since $d$ and $e$ are greater than or equal to  8, this is impossible.  After trying $a=6$, it worked because the set 6, 6, 6, 8, 8, 8, 8, 14.  So, the answer is $\text{(D)}$.[/hide]

Not the best way to do it.  :D

[hide="Answer"]The middle term is an $8$, and the range is $8$. If we had 6 $8$'s, then the other two integers must sum $16$ to average to $8$, but this means that we have a $4$ and a $12$ to keep the range. However, we can obviously do better than this. Instead, we have 4 $8$'s (just enough to keep the mode), and the remaining numbers must sum to $32$. If we have 3 $6$'s and a $14$, this works just right, so our answer is $14\Rightarrow \boxed{D}$. Any smaller number of $8$'s would result in a minimum term smaller than $6$, and therefore a maximum term smaller than $14$.[/hide]


Let $\omega_{2}(n)$ denote the number of nonzero binary digits of a positive integer $n$. Let $B > 0$ be arbitrary. Prove that $\omega_{2}(n) \to \infty$ as $n$ runs through the odd integers with greatest prime factor $< B$.

In other words, [i]  for an odd integer[/i], few $1$'s in the binary expansion of $n$ imply that $n$ has a large prime factor.

And, in general, few nonzero digits of $n$ implies that $n$ has a large prime factor -- in any base that doesn't divide $n$.


Suppose $Y$ is a random variable with a Poisson distribution and mean $1.$ That means that $P(Y=n)=\frac{e^{-1}}{n!}.$ Then $E(Y^{k})=e^{-1}\sum_{n=0}^{\infty}\frac{n^{k}}{n!}.$

Now consider the moment-generating function:

$m(t)=E(e^{tY})=e^{-1}\sum_{n=0}^{\infty}\frac{e^{tn}}{n!}=e^{e^{t}-1}.$

The "moment-generating" part of the name has to do with the relationship between the $k$th moment and the $k$th derivative evaluated at zero:

$m^{[k]}(0)=E(Y^{k}).$

Now, calculating the derivatives of $e^{e^{t}-1}$ can get a little messy. Let's look at the power series:

$e^{t}-1=t+\frac{t^{2}}2+\frac{t^{3}}6+O(t^{4})$

Exponentiate that:

$e^{e^{t}-1}=1+\left(t+\frac{t^{2}}2+\frac{t^{3}}6+O(t^{4})\right)+\frac12\left(t+\frac{t^{2}}2+O(t^{3})\right)^{2}+\frac16(t+O(t^{2}))^{3}$

$=1+t+t^{2}+\frac56t^{3}+O(t^{4})$

Hence $m(0)=E(1)=1,m'(0)=E(Y)=1,m''(0)=E(Y^{2})=2,$ and $m'''(0)=E(Y^{3})=5.$

It follows that

$S=\sum_{n=0}^{\infty}\frac{C_{0}+C_{1}n+C_{2}n^{2}+C_{3}n^{3}}{n!}= e\left(C_{0}+C_{1}+2C_{2}+5C_{3}\right)$

We could compute higher moments by this method as well, but it will take more terms of the power series and hence more patience.

[color=green][Pointless quote removed by moderator.][/color]

Thank You! :lol:


vector
function
induction
combinatorics proposed
combinatorics

Zeroes and ones are arranged in all the squares of $n\times n$ table.
All the squares of the left column are filled by ones, and the sum of numbers in every figure of the form
[asy]size(50); draw((2,1)--(0,1)--(0,2)--(2,2)--(2,0)--(1,0)--(1,2));[/asy]
(consisting of a square and its neighbours from left and from below)
is even.
Prove that no two rows of the table are identical.

[i]Proposed by O. Vanyushina[/i]

[quote="mathmanman"]Zeroes and ones are arranged in all the squares of $n\times n$ table.
All the squares of the left column are filled by ones, and the sum of numbers in every figure of the form
[asy]size(50); draw((2,1)--(0,1)--(0,2)--(2,2)--(2,0)--(1,0)--(1,2));[/asy]
(consisting of a square and its neighbours from left and from below)
is even.
Prove that no two rows of the table are identical.[/quote]

Nice problem.
Let $v=\{v_{1},v_{2},\cdots,v_{n}\}$ a vector of $n$ $0$ or $1$ and $f_{v}(i,j)$, with $i,j$ integers $\in[1,n]$  the function defined by $f(1,j)=v_{j}$, $f(i,1)=f(i-1,1)$ for $i>1$ and $f(i,j)=f(i-1,j-1)\oplus f(i-1,j)$ $\forall$ integers $i,j\in[2,n]$ ($\oplus$ is XOR). The table of the problem is such a function $f()$ with $a_{1}=1$.

Let $u_{k}=\{0,0,\cdots,1,\cdots,0\}$, $k$ integer $\in[1,n]$ the vector with $1$ in $k^{th}$ position and zeroes elsewhere.

$\oplus$ is associative and so $f_{v}(i,j)=\bigoplus_{k=1}^{n}f(1,k)f_{u_{k}}(i,j)$
But $f_{u_{1}}(i,j)=p(\left(\begin{array}{c}i-1\\j-1\end{array}\right))$ where $p(n)$ is the "parity" function : $p(n)=0$ if $n$ is even and $p(n)=1$ if $n$ is odd.  and the convention that $\left(\begin{array}{c}n\\p\end{array}\right)=0$ if $n<p$
And $f_{u_{k}}(i,j)=f_{u_{1}}(i,j-k+1)$ if $j\geq k$ and $0$ if $j<k$

So $f_{v}(i,j)=\bigoplus_{k=1}^{j}f(1,k)f_{u_{1}}(i,j-k+1)$
And $f_{v}(i,j)=\bigoplus_{k=1}^{j}f(1,k)p(\left(\begin{array}{c}i-1\\j-k\end{array}\right))$
So $f_{v}(i,j)=p(\sum_{k=1}^{j}f(1,k)\left(\begin{array}{c}i-1\\j-k\end{array}\right))$

Now, assume lines $q$ and $r>q$ are equal. Then lines $p-1$ and $q-1$ are equal, ... and lines $1$ and $p=r-q+1$ are equal.

So $p(\sum_{k=1}^{j}f(1,k)\left(\begin{array}{c}p-1\\j-k\end{array}\right))=f(1,j)$ $\forall$ integer $j\in[1,n]$
So $p(\sum_{k=1}^{j-1}f(1,k)\left(\begin{array}{c}p-1\\j-k\end{array}\right))=0$ $\forall$ integer $j\in[2,n]$
So $\sum_{k=1}^{j-1}f(1,k)\left(\begin{array}{c}p-1\\j-k\end{array}\right)$ is even $\forall$ integer $j\in[2,n]$

For $j=2$, this means $f(1,1)\left(\begin{array}{c}p-1\\1\end{array}\right)$ is even, so $\left(\begin{array}{c}p-1\\1\end{array}\right)$ is even
For $j=3$, this means $f(1,1)\left(\begin{array}{c}p-1\\2\end{array}\right)+f(1,2)\left(\begin{array}{c}p-1\\1\end{array}\right)$ is even and, since $\left(\begin{array}{c}p-1\\1\end{array}\right)$ is even, $\left(\begin{array}{c}p-1\\2\end{array}\right)$ is even

And, by induction, it is easy to show that $\left(\begin{array}{c}p-1\\m\end{array}\right)$ is even $\forall$ integer $m\in[1,n-1]$

And this is impossible since $p-1\leq n-1$ and $\left(\begin{array}{c}p-1\\p-1\end{array}\right)$ is odd.

[u]Remark [/u]: it is possible to show that, for $n=2^{a}$, the $n+1^{th}$ line would be equal to the first one.

--
Patrick

[quote=VicKmath7]Something simpler?[/quote]

Solution by Amirhossein Bagheri Jebeli:
First define $L(n)$ be the three squares as below:
[asy]size(150); draw((12,12)--(13,12)--(13,13)--(12,13)--(12,12)); draw((6,12)--(7,12)--(7,13)--(6,13)--(6,12)); draw((6,12)--(7,12)--(7,13)--(6,13)--(6,12)); draw((12,6)--(12,7)--(13,7)--(13,6)--(12,6));[/asy]\\
Where the distance between the centre of squares in the same row or column is $2^n$ for example the following is $L(0)$:
[asy]size(50); draw((2,1)--(0,1)--(0,2)--(2,2)--(2,0)--(1,0)--(1,2));[/asy]
By putting $3$ of $L(n)$ from its central square in squares of $L(n)$ we can figure out inductively that sum of numbers in $L(n+1)$ is even.
imagine that two rows $e,d$ are identical, $e>d$ and $e-d=2^k.l$ where $l$ is odd, now put the left-most square of $L(k)$ on the left-most square of $e$ to conclude that the remaining two squares of this $L(k)$ are of different parity, now put the left-most square of $L(k)$ on the left-most square of $e-2^k$ to conclude that the remaining two squares of this $L(k)$ are of different parity, repeating this process easily implies that the parity of the $2^k$-th squares of $e,d$ are of different parity.


[quote="vlmathlv"]if abc>0, prove that
a^3/a^2+ab+b^2  +  b^3/b^2+bc+c^2  +  c^3/c^2+ca+a^2 >= a+b+c/3 [/hide][/list][/quote]
$ \sum_{cyc}\frac{a^3}{a^2\plus{}ab\plus{}b^2}\equal{}\sum_{cyc}\frac{a^4}{a^3\plus{}a^2b\plus{}ab^2}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^3\plus{}a^2b\plus{}a^2c)}\equal{}$
$ \equal{}\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}\geq\frac{a\plus{}b\plus{}c}{3}.$

we have $ \frac{a^3}{a^2\plus{}ab\plus{}b^2}\geq\frac{2a\minus{}b}{3}$
a generalization problem :$ \sum_{cyc} {\frac{a^3}{a^2\plus{}kab\plus{}b^2}}\geq\frac{a\plus{}b\plus{}c}{2\plus{}k} (\forall k\in N)$


modular arithmetic
vector
algebra
polynomial
combinatorics unsolved
combinatorics

For given integer m,n,k there`s a m*n grid and one is allowed to fill each grid with the number from 1 to k and numbers can get repeated as many times.
How many ways are there to fill the grid so that the sum of the numbers of each row and column is not multiple of k ( all m+n  sums are not multiple of k) ?

p.s i cannot find closed form solution. if the sum of the numbers of each row and column is multiple of k  then there are k^(m-1)(n-1) way to fill the grid.
but all the sum is not multiple of k.. then.. i cannot find nice solution.

I don't think that there are k^[(m-1)*(n-1)] ways to fill the grid so that each row/column sum is divisible by k. If we fill a (m-1)*(n-1) subgrid, there exist unique entries for the cells adjacent to the (m-1)*(n-1) subgrid that satisfy the property. But the sums of the incomplete row and column may not be congruent mod k, so there are strictly less than k^[(m-1)*(n-1)] ways to fill the grid.

(@mathophile953: that's not actually a problem. The sums of the incomplete last row and last column are both the sum of all the $ (m\minus{}1)(n\minus{}1)$ elements we filled initially, so in fact they are indeed the same in all cases. )

Given the modulus $ k$, let us denote by $ S(n,l)$ the following set:

$ S(n,l) \equal{} \{(a_1,a_2,\ldots,a_n) \,|\, a_i \in \{1,2,\ldots,k\minus{}1\}, \sum a_i \equiv l \pmod{k} \}$

In words, this is the set of row vectors mod $ k$ whose sum is $ l$ (mod $ k$) and in which no term is 0 (mod $ k$). Let $ s(n,l) \equal{} |S(n,l)|$ be the number of elements in $ S(n,l)$. It's obvious that $ s(n,l)$ is [i]about [/i]$ (k\minus{}1)^n/k$ since we have $ n$ entries, each of which has $ k\minus{}1$ options, and the requirement that the sum is $ l$ is "nearly uniform". In fact:

[b]Exercise[/b]: show that $ s(n,l) \equal{} \frac{(k\minus{}1)^n\minus{}(\minus{}1)^n}{k}$ when $ l \not \equiv 0 \pmod{k}$, and $ s(n,0) \equal{} \frac{(k\minus{}1)^n\plus{}(\minus{}1)^n (k\minus{}1)}{k}$.

Let's return to the original question: in how many ways we can fill an $ m \times n$ with elements in $ \{1,\ldots,k\}$ such that no row sum and no column sum is 0 mod $ k$. One way to generate such arrays is as follows:

1. Fill up the upper left $ (m\minus{}1) \times (n\minus{}1)$ subarray in an arbitrary fashion.

2. Complete the last row and last column in the unique way that satisfies the exact opposite of what we want: make all the row sums and column sums 0 mod $ k$. As noted above, this can always be done in a unique way. The last corner cell at the bottom right will always be uniquely determined by the requirement that the last row [i]and [/i]the last column are zero.

3. "Spoil" the first $ m\minus{}1$ rows by adding a vector in $ S(m\minus{}1,l)$ to the last column. The fact that all these vectors have entries which are never 0 ensures that non of the first $ m\minus{}1$ row sums will remain 0.

4. "Spoil" the first $ n\minus{}1$ columns by adding a vector in $ S(n\minus{}1,l')$ to the last row.

5. The last column sum is now $ l$ (mod $ k$), and the last row sum is now $ l'$. To make sure these are spoiled as well, we can set the bottom right cell to anything other than $ \minus{}l$ and $ \minus{}l'$ mod $ k$. If $ l\equal{}l'$ this excludes one option for that cell, while if $ l \neq l'$ this excludes two options.

Therefore, the desired number of such arrays can be calculated once the numbers $ s(n,l)$ are known. Denoting this number by $ N(m,n)$ we get

$ N(m,n) \equal{} k^{(m\minus{}1)(n\minus{}1)} \left( \sum_{l\equal{}0}^{k\minus{}1} \sum_{l'\equal{}0}^{k\minus{}1} s(m\minus{}1,l) s(n\minus{}1,l') (k\minus{}2\plus{}\delta_{l l'}) \right)$

Here, as usual, $ \delta_{l l'}$ is 0 if $ l \neq l'$ and 1 if $ l\equal{}l'$. We can now substitute the formulas for $ s(n,l)$ given in the Exercise and obtain an explicit, sum-free formula for $ N(m,n)$. I need more time than I have right now to do this calculation, but if I'm not mistaken the result is going to be something like this: letting $ X\equal{}(k\minus{}1)^{m\minus{}2}$ and $ Y\equal{}(k\minus{}1)^{n\minus{}2}$, we get

$ N(m,n) \equal{} k^{mn\minus{}m\minus{}n\minus{}1} \left( P(k) X Y \plus{} Q_1(k) X \plus{} Q_2(k) Y \plus{} R(k) \right)$
where $ P, Q_1, Q_2, R$ are certain polynomials in $ k$ of degree 3 or less whose coefficients depend on $ m$ and $ n$ only through $ (\minus{}1)^m$ and $ (\minus{}1)^n$.


hey guys, not realling calculus but just wondeting if any of you could help me with the following problems:

1) Find X inside the hexagon:
7 ____8
2/  54  \ 9
1\____/4

3 ____6
12/ X \ 2
5 \____/8

2) 2,7,11,20,38,69,127, X?
what should x be?

3) Someone asks 2 siblings (a brother and a sister), who is older
Sister: I am older
Brother: I am younger

If you know that atleast one of them was lying, who is older?

4) Two integers are chosen x,y such that 2 <= x <= y <= 20.
The sum x + y is given to a guy named A, and the product is given to a guy named B. They have the following conversation:
A: I see no way you can determine my sum.
B: Now I know your sum.
A: Now i know your product.
What are x and y?

This is really the wrong forum. Really.

Anyway, at a quick glance: 2) Sum of last three numbers 3) It's impossible for only one to be lying.

For number 3 I believe that they both must be lying so the brother is older.

[quote="cs_student"]4) Two integers are chosen x,y such that 2 <= x <= y <= 20.
The sum x + y is given to a guy named A, and the product is given to a guy named B. They have the following conversation:
A: I see no way you can determine my sum.
B: Now I know your sum.
A: Now i know your product.
What are x and y?[/quote]

This problem is similar to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37312&postorder=asc&start=41]problem[/url] I posted.

Suppose A has a number that can be written in form of (prime)+(prime), then B might figure out what the numbers are. Therefore, A has a number that cannot be written in form of (prime)+(prime). It is known that Goldbach's conjecture is true when the number is smaller than 20000000000 :D , so A cannot have an even number.

The smallest possible number that A can have is 11=2+9=3+8=4+7=5+6. Suppose A have this number. Possible numbers for B are 18=2*9, 24=3*8, 28=4*7, and 30=5*6. If B had 18, there are two possibility. (2,9) or (3,6).  Since 3+6=9=2+7, B can now deduce that the numbers are (2,9). If B had 24, he(she) can still find out the numbers. However, A cannot possibly figure out what B has, since B can have either 18 or 24. Therefore, A cannot have 11.

The second smallest possible number that A can have is 17. Suppose A have this number. Possible numbers for B are 30,42,52,60,66,70, and 72. If we do some casework, it is easy to see that B can only find out the numbers if and only if the number that B has is 52.

Therefore, the numbers are 4 and 13.

____________________________________________________
EDITED:
For #3, they might be twin, which means both of them are lying.


[quote="alexilic"]Suppose that we have complete binary tree on $ 2^n \minus{} 1$ vertices. If we add matching of size $ 2^{n \minus{} 1}$ to the leaves, what is minimal possible diameter we can achieve?[/quote]
I dont understand the problem. :!:
What is to be meant by "add matching of size $ 2^{n\minus{}1}$ to the leaves"
What are the definations of "leaves" and "diameter" :?:

Diameter is maximum distance between any two vertices in graph (for example in binary tree is $ 2n$).
Leaf is vertex with degree one and there are exactly $ 2^{n \minus{} 1}$ leaves in complete binary tree.
To add a matching means that you have to pair leaves with an edge such that every leaf is in exactly one pair.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

A,B \in M_3(R) such that

(AB-BA)^3 =
(a 0 0)
(0 b 0)
(0 0 c) where a,b,c \in R "nu toate egale" all different
Proove that AB-BA=
(a^1/3 0 0)
(0 b^1/3 0)
(0 0 c^1/3)
Cristinel Mortici

A solution for this one can be based on Hemilton Cayley's relation!

cheers!

For M matrix in M_3(R) the characteristic polynomial is
X^3-(trM)X^2+1/2((trM)^2-tr(M^2))X + detM
With M=AB-BA Hamilton-Cayley theorem gives
(AB-BA)^3 -1/2(tr(AB-BA)^2)(AB-BA) +det(AB-BA)I=0
det(AB-BA)=(abc)^(1/3)
1/2(tr(AB-BA)^2)(AB-BA) = (AB-BA)^3 +det(AB-BA)I =
(a 0 0 )    ((abc)^1/3 0 0)
(0 b 0 )    (0  (abc)^1/3 0)
(0 0 c)  + (0 0 (abc)^1/3)

Now i don't see how to get tr((AB-BA)^2

[quote="Moubinool"] the characteristic polynomial is
X^3-(trM)X^2+1/2((trM)^2-tr(M^2))X - detM

(AB-BA)^3 -1/2(tr(AB-BA)^2)(AB-BA) - det(AB-BA)I=0

1/2(tr(AB-BA)^2)(AB-BA) = (AB-BA)^3 - det(AB-BA)I =
(a 0 0 )    ((abc)^1/3 0 0)
(0 b 0 )    (0  (abc)^1/3 0)
(0 0 c)  - (0 0 (abc)^1/3)

Now i don't see how to get tr((AB-BA)^2[/quote]

I found a proof when a,b,c are all distincts
call C=
(a 0 0)
(0 b 0)
(0 0 c)
If M \in M_3(R) such that M^3=C
then M,C commute MC=M^4=CM
M=
(x y z)
(t u v)
(w s d)
a brute calculation MC=CM implies y=z=v=t=s=w=0
M is diagonal form
M=
(x 0 0)
(0 y 0)
(0 0 z)
Conversely M^3=C gives x^3=a,y^3=b,z^3=c
This gives
M=AB-BA=
(a^1/3 0 0)
(0 b^1/3 0)
(0 0    c^1/3)

But by solving the problem with this strategy i did not use the specific
for the unknown  AB-BA .

Any one have a proof ?


Let $ p\equal{}17$ and $ q\equal{}33488$.  We need to find $ \pm\frac{q}{p}$, but we need to test this for all possible factors, so that means $ \pm1/17$, $ \pm2/17$, $ \pm4/14$,...  You find one such that the equation is 0.  Then, you do some synthetic division to reduce it to a quadratic, which is easily solved.


ARML algebra polynomial trigonometry Vieta

I suppose that I'm a bit late on this (seeing as how the contest is over), but I thought that it would be nice to post solutions to my mock ARML. I wrote up the document, so why not share it. I'm pretty happy with how it turned out.

Yeah, nice problems you put.

I noticed that the roots to the polynomial $ f(x) \equal{} x^3 \minus{} 3x \plus{} 1$ are $ 2\cos{\frac {8\pi}{9}}, 2\cos{\frac {2\pi}{9}}, 2\cos{\frac {4\pi}{9}}$, so using ShoeLace, Vieta's and Manipulation, I got the answer to be $ \frac {9}{2}$


function
integration
calculus
real analysis
abstract algebra
logarithms
real analysis unsolved

$k$ natural number, How do I show that

$\int_{-\infty}^{\infty}|\frac{sin(2\pi kx)sin(2\pi x)}{(\pi x)^{2}}|dx \to \infty$  as $k\to \infty$?

$f_{k}(x)=\frac{\sin(2\pi k x)\sin(2\pi x)}{(\pi x)^{2}}$ is the Fourier transform (or inverse Fourier transform - it doesn't matter) of $g_{k}(\xi),$ where $g_{k}$ is the convolution of the characteristic function of $[-k,k]$ with the characteristic function of $[-1,1].$ That means that $g_{k}$ is a piecewise linear continuous function, compactly supported, with vertices at $(-k-1,0),(-k+1,2),(k-1,2),$ and $(k+1,0).$

If you're willing to look at weak (distributional) limits, then we see that $g_{k}$ tends weakly to the constant $2,$ which means that $f_{k}$ tends weakly to twice the Dirac delta.

By itself, that would not be a proof of what we want since sequences tending weakly to the Dirac delta often have bounded $L^{1}$ norms. But it gives some insight into what is going on. $g_{k}$ is the only slightly smoothed characteristic function of $[-k,k],$ which means that $f_{k}$ is essentially the Dirichlet kernel with some slight additional rate of decay as $|x|\to\infty.$ And that's essentially what jmerry used. Expect that $\|f_{k}\|_{1}=O(\ln k)$ as $k\to\infty.$

[quote="jmerry"]Going to $\pm\infty$ is irrelevant- the issue is behavior near zero. Near zero, we can approximate the integrand by $| sin(2\pi kx)|\cdot\frac2{\pi |x|}$, which is comparable to $\frac2{\pi}\cdot\frac2{\pi |x|}$ as $k\to\infty$, at least in integral. Since $\frac1x$ is not integrable near zero, our sequence of integrals should diverge to $\infty$.[/quote]

why $| sin(2\pi kx)|$ is compareble to $\frac2{\pi}$  as $k\to\infty$?
I understand as $k\to\infty$, the whole |sin(x)| function is squeezed towards 0, but why it became like a constant interms of integral near zero?

$\frac2{\pi}$ is the average value of $|\sin|$, and a rapidly oscillating function looks like its average value; that's the Riemann-Lebesgue lemma.

The comparison to $\frac{c}{x}$ should only be made for $|x|>\frac{1}{k}$ (or something like that). But that's enough to get the integral to be comparable to $c\ln k.$


geometry
perimeter
induction
modular arithmetic
geometry unsolved

Suppose we have n circles which all intersect in a region. Consider the number of circular arcs which form the boundary of the region; prove that this number is no more than 2n-2.

Select a point in the region and invert the figure in a circle with that point as centre. The image of the given region is the region outside all the circles. So we wish to find the maximum number of arcs that can make up the perimeter of the figure. Now note that a cycle of $k$ circles has a maximum perimeter of $k$ arcs, and if we add a circle which intersects more than one existing circle then we must form a cycle. Also, if we add one which intersects only one circle, then we create two points of intersection, and hence at most two more arcs. Noting also that a figure comprising two circles has a perimeter with at most two arcs, hence by induction the maximum is $2n-2$.

[quote="Fiachra"]Select a point in the region and invert the figure in a circle with that point as centre. The image of the given region is the region outside all the circles. So we wish to find the maximum number of arcs that can make up the perimeter of the figure.[/quote]I agree up to here, but I don't understand what the "cycles" are all about.

Anyway, we can strengthen the inverted version a little:
[quote]The boundary of the union of $n\ge2$ (not necessarily intersecting) disks consists of at most $2n-2$ arcs.[/quote]Induct on $n$. For $n=2$ it's obvious, so suppose $n>2$. If there are multiple "connected components" (two circles adjacent iff they intersect) then we can just use the inductive hypothesis with a bit of extra care for components of size $1$. Otherwise, let $u_1,u_2,\ldots,u_m$ (indices mod $m$) be the arcs on the boundary in clockwise order, and $f(u_i)$ the circle containing arc $u_i$.

Assume for the sake of contradiction that each circle has at least two arcs on the boundary. Then there exist distinct indices $j,k$ with $k-j\pmod{m}$ minimal such that $f(u_j)=f(u_k)$. But $u_j,u_k$ can't be adjacent arcs on the boundary, so by the choice of $j,k$, there exists $l\in(k,j)$ (intervals mod $m$) such that $f(u_l) = f(u_{j+1}) \ne f(u_j)$. On the other hand, $u_1u_2\ldots u_m$ is a simple path enclosing (and thus disjoint from) the interiors of the $n$ disks, so circles $f(u_j)$ and $f(u_{j+1})$ must intersect at $4$ or more points, which is absurd. (More precisely, if $P,Q,R,S$ are points from $u_j,u_{j+1},u_k,u_l$, then segments $PR\in f(u_j)$ and $QS\in f(u_{j+1})$ must intersect.)

Thus there exists a circle $C$ with at most one arc on the boundary. If $f^{-1}(C)=\emptyset$, then $m\le 2(n-1)-2$ by the inductive hypothesis. Otherwise, if $f^{-1}(C)=\{i\}$, then removing $C$ leaves arc $u_j$ intact for $j\ne i-1,i,i+1$ and replaces $u_{i-1}u_iu_{i+1}$ with a (nonempty) sequence of boundary arcs. Thus $(m-3)+1\le 2(n-1)-2$ by the inductive hypothesis, whence $m\le 2n-2$, as desired.


function
logarithms
calculus
calculus computations

Hi, im having a little trouble with something. i have the straight line function P=0.0647t+2.8253 and i wish to have it in the exponential form P=P0e^kt. the original function was obtained from a table of values.

Time(t)	1	2.5	5	7	12	15	18
P	18	20	23	27	37	46	56
lnP	2.89	2.996	3.135	3.296	3.611	3.829	4.025

any help is greatly appreciated. thanks

I understand that you have a list of values of $P$ and you want to approximate $P$ with an exponential function. Since you already have $\ln P$ tabulated, it seems reasonable to fit the best line to the logarithms ($\ln P(t)\approx kt+b$), which results in $P(t)\approx e^{b}e^{kt}$. Popular Spreadsheet Software can do such things.


For $p$ be a prime and $a_1,a_2...a_{p+1}$ be distinct positive numbers.Show that there exists a pair $(i,j)$,$1\leq i<j \leq p+1$ such that :$\frac{a_i}{gcd(a_i,a_j)}\ge p+1$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=graham&t=6334

Pierre.


Find all primes $(p,q)$ s.t. \[ p^3-q^5=(p+q)^2 . \]

Really nice and tough problem  :D

It's obvious that p>q,p^3>2^5,p>3,p^3≡p(mod3),q^5≡q(mod5),if q ≠3,
(1)3|p-q,3|p^3-q^5,p+q can't be divisible by 3,a contradiction.
(2)3 doesn't divide p-q,3|p+q,p^3-q^5 can't be divisible by 3,a contradiction.
So q=3,and we have p^3-p^2-6p-252=0,(p-7)(p^2+6p+36)=0,p=7.
So the only pair of prime satisfied the equation is (7,3). :)  :P

If you find the only solution,it's not a hard question. :P  :P


geometry
algorithm
geometric transformation
reflection
calculus
integration

There is an enormous amount of stuff that you need to learn before you can seriously do research on an important topic.

There are also many different levels of understanding of a subject.  At a very superficial level you may not know how to prove anything, or even be able to state the main results of a subject.  At a very deep level, you could write a book on the subject from scratch, proving all the theorems, and explaining the intuition behind the proofs, the strategies of the proofs, why the proofs aren't as hard as they look, what led someone to conjecture this theorem was true, and how someone may have thought of the proof.

A student who aspires to be a research mathematician (or a researcher who wants to do better research), is constantly feeling this tension between wanting to understand things deeply, and wanting to make progress through an enormous amount of material.

What is your advice about how to calibrate oneself as a math learner?  How well should you understand everything?  To what extent should you sacrifice a deep understanding for the sake of making progress?  Is it valuable to have a very superficial understanding of a very broad range of things, even if you know you'll never have time to come back and grok all of it?

There is a major question of calibration here that a student must answer.

I'm sure this question has been dealt with before.  If anyone knows of any good references or links, I'd appreciate it.

I think it's important to have that broad knowledge at least so when you are researching something, you have a general feel for what connections with other areas may arise.

[quote="Singularitarian"]There is an enormous amount of stuff that you need to learn before you can seriously do research on an important topic.[/quote]  I don't know what "important" means to you, but in fact there are many areas where this is not true.  Third-year undergraduate math majors are well-enough equipped to handle original research that lots of them run off and do [url=http://www.nsf.gov/crssprgm/reu/index.jsp]REUs[/url], for example.  Research is itself a learning process (just one in which the curriculum is not predetermined), and the set of "the things I have to learn before I can begin research" is not well-defined.

Thanks for your replies!

Any more comments from anyone would be welcome.

One approach to learning is: make sure you understand everything very deeply, and just work really, really hard so that you still have time to learn everything you need to know.  Another approach is: have a superficial understanding of a very broad range of stuff, spending significant amounts of time reading books at a superficial level, and be very selective about the things that you understand deeply.  If in the course of research some topic you have read about superficially seems important, then you can go back and learn it more carefully.

Another approach to learning is: don't worry about any of this stuff, just do what seems enjoyable at the moment.  If you're in the mood to understand something deeply, do that.  If you're in the mood to get a broad overview of a subject, do that.

Does anything about these approaches stand out as being obviously wrong, or obviously preferable?

I think it's great that someone wants to become a research mathematician.
I just completed year 12(high school in Australia) and throughout high school i too realised this conflict you speak of. Teachers I guess are a factor. One teacher I had was incredibly passionate about mathematics, and he always aimed for deep understanding and elegance, and he absolutely loved it when someone appreciated this sort of thing too. However, the majority of teachers don't exemplify this passion but rather emphasize a more dull mathematics where it's all about performing mere algorithms properly and quickly. The students are like this too here, ppl mainly study maths just to become doctors and lawyers etc. and earn big bucks, so 'wasting' their time with proof and anything mathematically intelligent that is irrelevant to their superficial dreams is obviously not on their agenda. Overall it makes math look like just a rat race, rather than the aesthetic art that it is.

How did I behave given such an environment?

Well first of all I'm one of those rare math-lovers that take things on a much deeper level. I guess having that passionate math teacher made me only concentrate on the fancy artistic math. THe next year I suffered a bit as I didn't have those simple skills liek performing algorithms quickly without making mistakes and exam technique etc. my math teacher then criticized me for this. I often shone in the class, solving tricky problems(most outside the syllabus) that other 'good' students had trouble with as well as even one that my teacher couldn't do!(he is a brillant teacher btw, intelligent man, he just criticized me a lot so i'll glorify myself with this little victory). However my test scores didn't reflect my ability. After all, what does performing algorithms quickly and flawlessly have anything to do with understanding math and being a clever problem solver?.
I guess to make a compromise, I had to focus on those more 'important' skills since I wanted to do well in high school. I sort of gave up artistic maths in year 12 and instead spent my time doin exam questions and shit like that. However that doesn't mean I had to sacrifice understanding math altogether! In year 12, we started finding definite integrals and so that fuelled my desire to begin 'understanding' calculus, rather than just merely memorizing the algorithms which the non-math-loving lawyer wannabes did. It was a disgrace to be honest, calculus wasn't really calculus, it was just algebraic manipulation of formulas that appear in calculus. Saying that u understand calculus from learning that is like saying u understand einstein coz u know that E=mc^2.
HOwever I stilll had time to develop a deeper understanding. I spent my free time trying to develop an intuitive understanding of the fundamental theorom of calculus and use tha as a basis for a formal proof of it using epsilon-delta definition of limit and notions of continuity etc. Effectively, I saw beautifull stuff that others did not, and it didn't really harm my exam skillz.

The moral of the story(and answer to ur question): There is no need to sacrifice! be proud of ur passion for mathematics. Conform to ur educational system as a means of achieving academically, but still pursue ur own individual path in understanding mathematics on the side =]. Appreciate the artistic nature of mathematics, do some interesting problems outside the syllabus when ur on the train back home or when u hav some free time. Treat it like a hobby... ppl can still play football and go to skool, just like u can 'understand' math and go to skool. Such an approach will benefit u as a mathematician in the future =]


Gauss
algebra
polynomial
Rational Root Theorem

ας πουμε δυο πραγματακια για αυτα ,κατι παρομοιο ειχε κανει και ο Σιλουανος και μου αρεσε

Factor theorem
1)Αν $a$ ειναι ριζα του $P(x)$ τοτε το $x-a$ διαιρει $P(x)$

Gauss Lemma
2)αν ενα πολυωνυμο με ακεραιους συντελεστες μπορει να παραγωντοποιηθει σαν γινομενο πολυωνυμων με ρητους συντελεστες, τοτε μπορει να παραγωντοποιηθει σαν γινομενο πολυωνυμων με ακεραιους συντελεστες,
αυτο οδηγει στο σημαντικο θεωρημα

Rational Root theorem
3)αν ενα πολυωνυμο με ακεραιους συντελεστες εχει μια ριζα ρητη $x=a/b$,me $(a,b)=1$ τοτε ο συντελεστης του μεγιστοβαθμιου ορου ειναι πολλαπλασιο του $b$, και ο σταθερος ορος ειναι πολλαπλασιο του $a$
εφαρμογη:

εστω πολυωνυμο $P$ οπου ο ορος  της μεγιστοβαθμιας δυναμης ειναι 1 (monic).Αν $ρ$ μια ριζα του $P$ τοτε αν το  $ρ$ αν δεν ειναι ακεραιος τοτε θα ειναι αρρητος( δηλ αν ειναι ρητη θα ειναι ακεραια (υποχρεωτικα) ριζα)...

Ορισμος:Αν $P(x)=a_{n}x^{n}+a_{n-1}+...+a_{0}=0$ και $k$ ριζες τις ταυτηζονται στην τιμη εστω $x=x_{k}$,τοτε λεμε οτι η ριζα αυτη εχει βαθμο πολλαπλαοτητας  $k$.

Προταση :
Εστω η εξισωση $f(x)=0$ με βαθμο $n$.Για να εχει την ριζα $x_{1}$ βαθμου πολλπλοτητας $k$,πρεπει και αρκει αυτη να ειναι ριζα των παραγωγων $1,2,...,K-1$ ταξεως της $f(x)$
Αποδειξη με τυπο Τaylor

Εφαρμογη : με βαση την προταση μπορουμε να βρουμε τον Μ.Κ.Δ ton $f(x)$ kai $f'(x)$

Ριζες πολυωνυμου Φραγμα Mac-Laurin
Δινεται η εξισωση:$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}=0$
με πρaγματικους $a_{j}$, j=1,2,...,n
Υποθετουμε οτι $a_{0}$ και οι  υπολοιποι συντελεστες ειναι ειτε αρνητικοι ειτε θετικοι. Αν P η μεγαλυτερη απολυτη τιμη των αρνητικων συντελεστων της ,τοτε η τιμη :$M=1+\frac{P}{a_{0}}$
ειναι ανωτερο φραγμα των θετικων ριζων της εξισωσης.
εξ ορισμου δεν υπαρχουν θετικες ριζες μεγαλυτερες αυτου του φραγματος

Φραγμα Lagrange
Εστω η εξισωση $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}=0$
για την οποια εχουμε $a_{n},a_{n-1},..,a_{n-r-1}>0,a_{n-r}<0$
ενω οι υπολοιποι συντελεστες ειναι ειτε θετικοι ειτε αρνητικοι.
Αν $P$ η μεγαλυτερη τιμη των αρνητικων συντελεστων τοτε το
$M=1+(\frac{P}{a_{0}})^{\frac{1}{r}}$αποτελει ανωτερο φραγμα των θετικων ριζων της εξισωσης

Descrates Law
αν εχουμε την πολυωνιμικη εξισωση $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$
με πραγματικους συντελεστες
θα λεμε οτι δυο διαδοχικοι συντελεστες παρουσιαζουν εναλλαγη αν τα προσημα αυτων αλλαζουν (ειναι διαφορετικα ) ή τηρηση(ν) αν ειναι ομοια .τοτε το θεωρημα του Ντεκαρτ αναφερει:
Το πληθος των θετικων ριζων  μιας αλγεβρικης εξισωσης$f(x)=0$ με πραγματικους συντελεστες δεν υπερβαινει τον αριθμο των εναλαγων αυτης , ενω η διαφορα των δυο αυτων αριθμων ειναι αρτια.

Προταση: let μια εξισωση με πραγματικους συντελεστες $f(x)=0$.Το πληθος των αρνητικων ριζων της δεν υπερβαινει τον αριθμο των εναλαγων της εξισωσης , ενω η διαφορα; των δυο αριθμων ειναι αρτια

Παρατηρηση-εφαρμογη

Εστω εξισωση $f(x)=0$ βαθμου $n$με πραγματικους συντελεστες .Αν u ο αριθμος των εναλαγων της και u' ο αριθμος των εναλλαγων της τοτε εχουμε :
$n-(u+u')=2k$
και ειναι ο ελαχιστος αριθμος των μιγαδικων ριζων της εξισωσης $f(x)=0$

Πολύ καλή δουλειά Κώστα!  :lol:  Αν έχεις και άλλα τέτοια, βάλε τα εδώ για να τα έχουμε συγκεντρωμένα..

ευχαριστω Στελάρα.εχω καποιες πολυ καλες ασκησεις , αν  ξερεις καλα αυτα τα θεωρηματα μπορεις να λυσεις εξισωσεις μεγαλου βαθμου χωρις δυσκολια , οσο αυτο μειναι δυνατο  :)

μια σημιωση : ομοιως οριζουμε το κατωτερο φραγμα θετικων ριζων,ανωτερο φργ αρνητικων,κατωτερο αρνητικων ριζων.
Ομως αν βρουμε το ανωτερο  φραγμα των θετικων ριζων της $f(x)=0$ μπορουμε να βρουμε και τα υπολοιπα φραγματα απλα κανουμε  την ιδια μεθοδο στις εξισωσεις :$f(\frac{1}{x})=0$,$f(-x)=0$ kai $f(-\frac{1}{x})$

μια ασκηση: αν $f(x)=x^{n}+a_{1}x^{n-1}++...+a_{n-1}x+a_{n}$ να δειχθει οτι στο εσωτερικο του κυκλου $|x|<\sqrt[n]{|a_{n}|}$ υπαρχει τουλαχιστον μια ριζα της εξισωσης $f(x)=0$

Μπορεί κανείς να δώσει ένα παράδειγμα για το κανόνα του Ντεκαρτ  :idea:

gianni  γεια , μαλλον θα απαντησω εγω σε αυτη την ωραια ερωτηση : πχ εστω η εξισωση : $3x^{3}-3x^{2}-5x-2=0$. κοιταμε οτι τις εναλλαγες των προσημων στη ακολουθια $3,-3,-5,-2$ ειναι μια ,αρα εχουμε το πολυ μια ριζα θετικη.Πραγματι αυτη  ειναι η $x=2$ και δεν υπαρχει αλλη θετιη ριζα.
η εξισωση $x^{3}+2x^{2}+7x+6=0$ δεν εχει θετικες ριζες αφου οι αλλαγες του προσημου ειναι 0  :D


There's this trick wherein there are 6 cards, with the numbers 1 to 64 written. Not all the numbers are written on one card, and a number can appear on more than one card. here's how it goes:

You think of a number from 1 to 64, and you are given the cards. you return the cards with your numberwritten on it, and the guy who tricked you instantly knows your number, to your astonishment.

How many ways can you make a set of 6 cards that will work the same way :?:

How can the guy know if there is more than one number on the card?

I'm not quite sure, but:
[hide]You can do this all by expressing everything in binary (base two). 64 in binary is 1000000. Using the 6 cards, you need to determine whether each digit is 0 or 1. So you start at the rightmost digit on the first card, displaying every number from 1 to 64 whose binary representation ends in a one. These are all the odd numbers. You do the same for the rest of the digits, and the trickster can determine from each individual digit what the binary representation of the number is.

However, for each of the 6 cards, you have the option of choosing whether you have all of the 0s or all of the 1s for the digit. So, for each card there are 2 possible options, for a final answer of $2^{6}= \fbox{64}$.
[/hide]

[quote="moogra"][hide]isn't it 1 way, because its binary system[/hide][/quote]

[hide]But you can choose whether, for each digit, you want the 0s or the 1s to be displayed on the card. For example, for the units digit card, you could choose to display all of the even numbers or all of the odd numbers, and you can still use it the same way.[/hide]


let $ a_0\equal{}A$, $ a_{n\plus{}1}\equal{}Pa_n\minus{}Qa_n^2$ whe $ \{a_n\}$, converge?

More generally, when does an iteration $ a_{n\plus{}1}\equal{}f(a_n)$ converge for a nice function $ f$? It can only converge to a fixed point $ x$ (satisfying $ f(x)\equal{}x$), and we get convergence for all nearby points when $ |f'(x)|<1$ and divergence away from that point when $ |f'(x)|>1$.
On the real line, we can use a few more tricks- if $ f$ is continuous and increasing, the sequence is increasing or decreasing depending on whether the graph of $ f$ lies above or below the line $ y\equal{}x$ at the starting point, and the sequence always either converges to a fixed point or diverges to $ \pm\infty$.