$ \text{3 \minus{} bromo \minus{} 5,5 \minus{} dimethylcyclohexene}$ or $ \text{2 \minus{} bromo \minus{} 4,4 \minus{} dimethylcyclohexene}$

$ \text{3,3 \minus{} dimethylcyclohexane}$ or $ \text{2,2 \minus{} dimethylcyclohexane}$ etc.

How do you name cycloalkanes/cycloalkenes?

In the case of cycloalkanes we number each carbon of the ring in order to give the lowest possible numbers to the substituents. For cycloalkenes, the numbering of the ring is made in such a way to give the lowest possible numbers, first to the double bond, and then to the substituents (regardless of their priority).

In your first compound, the first option is the correct one, since for the other alternative (6-bromo-4,4-dimethylcyclohexene) the sum of the numbers (14) is greater than for the first case (13).

In your second case, the name 2,2-dimethylcyclohexene does not exist, since carbon 2 (in any numbering of the ring) is double bond carbon.

Oops, that was a typo. It's cyclohexane sorry.

My book seems to have a lot of problems (Organic Chemistry- John McMurry).


Hi,

I'm working on a project right now that requires a background in mathematics.
Something I don't have.

I searched on the net for help and found this forum.  :)

Please consider the following equation.

y = 0,00003x2 + 0,0086x + 0,466

Where x is, in this case, -12
So it would become:

Y = 0,00003*-12^2 + 0,0086*-12 + 0,466

I cannot come to the right answer.  That is, I get two answers and do not know which one is the right one.

If I calculate it in excel the answer is 0,36712.
If I put in google calculator the answer comes out as 0,35848.

Again for another equation:

y = -0,00007*-12^2 - 0,0016*-12 + 0,2795

Excel returns 0,28862
and Google: 0,30878

And yet another:

y = 0,00004*-12^2 - 0,007*-12 + 0,2545

Excel returns: 0,34426
And Google: 0,33274

---

Who is Right???

I have more faith in Excel but I'm far from sure.
Something that I need to be for this project.

If you know the answer, please let me know.
Your help is greatly appreciated.

Thanks
Gijs

Excel's right.  Google is evaluating $ ( \minus{} 12)^2$ as $ \minus{} (12)^2$, which is not what you want.

Different calculators might use different methods of calculating. For example, one might use intermediate rounding, which would return a different result than a calculator that doesn't round until the end. To remove this possible error, I suggest making all coefficients of an equation integral, and then proceeding to computation.

The way you entered the equations, Google might take $ \minus{}12^2$ to be $ \minus{}(12)^2$ instead of $ (\minus{}12)^2$.

[quote="gijsbijs"]y = 0,00003x2 + 0,0086x + 0,466

Excel: 0,36712
Google: 0,35848[/quote]
[hide]$ y \equal{} 0.00003x^2 \plus{} 0.0086x \plus{} 0.466\quad\Rightarrow\quad100000y \equal{} 3x^2 \plus{} 860x \plus{} 46600$

Now, plug in $ x \equal{} \minus{} 12$:

$ 100000y \equal{} 432 \minus{} 10320 \plus{} 46600 \equal{} 36712\quad\Rightarrow\quad y \equal{} \frac {36712}{100000} \equal{} \boxed{0.36712}$

[url=http://www.google.com/search?hl=en&q=0.00003%28-12%29%5E2%2B0.0086%28-12%29%2B0.466&btnG=Search]Google[/url][/hide]
[quote="gijsbijs"]y = -0,00007*-12^2 - 0,0016*-12 + 0,2795

Excel: 0,28862
Google: 0,30878[/quote]
[hide]$ y \equal{} \minus{} 0.00007x^2 \minus{} 0.0016x \plus{} 0.2795\quad\Rightarrow\quad100000y \equal{} 7x^2 \plus{} 160x \plus{} 27950$

Now, plug in $ x \equal{} \minus{} 12$:

$ 100000y \equal{} \minus{} 1008 \plus{} 1920 \plus{} 27950 \equal{} 28862\quad\Rightarrow\quad y \equal{} \frac {28862}{100000} \equal{} \boxed{0.28862}$

[url=http://www.google.com/search?hl=en&q=-0.00007%28-12%29%5E2-0.0016%28-12%29%2B0.2795&btnG=Search]Google[/url][/hide]
[quote="gijsbijs"]y = 0,00004*-12^2 - 0,007*-12 + 0,2545

Excel: 0,34426
Google: 0,33274[/quote]
[hide]$ y \equal{} 0.00004x^2 \minus{} 0.007x \plus{} 0.2545\quad\Rightarrow\quad100000y \equal{} 4x^2 \plus{} 700x \plus{} 25450$

Now, plug in $ x \equal{} \minus{} 12$:

$ 100000y \equal{} 576 \plus{} 8400 \plus{} 25450 \equal{} 34426\quad\Rightarrow\quad y \equal{} \frac {34426}{100000} \equal{} \boxed{0.34426}$

[url=http://www.google.com/search?hl=en&q=0.00004%28-12%29%5E2-0.007%28-12%29%2B0.2545&btnG=Search]Google[/url][/hide]

Apart from simple four-function calculators, most calculators (especially all graphing and scientific calculators and major mathematical programs) will follow the [[order of operations]]. So, if you enter "-12^2", it will do the exponentiation first (that is, square the 12), and then evaluate the sign, which denotes multiplication by -1. To fix that, you need to group the - and the 12 first, via parentheses, and then square: (-12)^2. This also applies to when you are writing things by hand.

Thanks a lot for your help guys!

You don't have to help me, yet yo do....
It is appreciated.

Indeed, that must be it.  Because when I use a possitive number instead of -12, 3 for example, Google and Excel return the same answer.

I'm in the process of learning elementary mathematics. But since I quit school early I had to start in the beginning and thus are not into this stuff yet.

However, thanks to you guys, my work can continue now.

Thanks again
Gijs


function
calculus
derivative
vector
complex analysis

how do you solve
$ \minus{}z''z^3\equal{}|z|$ where z is a complex valued function of a real variable,
|z| is the magnitude of z and z'' is the second dirivative?

I see nothing preventing the adaptation of Euler's method to differential equations of a complex variable although I've never done so until now. The derivative, interpreted as an "amplitwist", still applies as I see it.  For:

$ z' \equal{} f(w,z);\quad z(w_0) \equal{} z_0$

$ z_{n \plus{} 1}\approx z_n \plus{} \Delta w_nf(w_n,z_n)$

where $ \Delta w$ is a small vector emanating from $ w_n$ in the direction of the path of a contour over which the solution is desired.

For example, consider:

$ z' \equal{} 2w;\quad z(1 \plus{} i) \equal{} 2i$

and consider estimating the solution along a straight-line contour from $ w \equal{} 1 \plus{} i$ to $ w \equal{} 2 \plus{} 2i$. I used the expression above and the Mathematica code below. I then plotted the imaginary component of the solution of the numerical results superimposed against the imaginary component of the actual results, $ z(w) \equal{} w^2$ which I posted below.  Pretty good for my money. Granted it's not as complicated as:

$ z'' \equal{} \minus{} \frac {|z|}{z^3}$

but that's where I'd start if it were my problem.

[code]Subscript[z, 0] = 2*I;
Subscript[w, 0] = 1 + I;

etable = Table[Subscript[w, n] = 1 + I + n*(0.01 + 0.01*I);
        Subscript[z, n + 1] = Subscript[z, n] + (0.01 + 0.01*I)*2*
              Abs[Subscript[w, n]]*Exp[Pi*(I/4)];
        {1.01 + n*0.01, Im[Subscript[z, n + 1]]}, {n, 0, 100}];
p1 = ListPlot[etable, PlotRange -> {{0, 10}, {0, 10}},
     AspectRatio -> 1, PlotStyle -> Green]
p2 = Plot[Im[(x + x*I)^2], {x, 1, 2},
     PlotRange -> {{0, 10}, {0, 10}}, AspectRatio -> 1,
     PlotStyle -> Blue]
Show[{p1, p2}][/code]

[img]http://img144.imageshack.us/img144/1442/test1fv1.jpg[/img]


Let $z_{1},\ z_{2},\ z_{3}$ pairwise distinct numbers satisfying
$\left|z_{1}\right|= \left|z_{2}\right|= \left|z_{3}\right|=1$ and

$\frac{1}{2+\left|z_{1}+z_{2}\right|}+\frac{1}{2+\left|z_{2}+z_{3}\right|}+\frac{1}{2+\left|z_{3}+z_{1}\right|}=1$

If the points $A(z_{1}),\ B(z_{2}),\ C(z_{3})$ are vertices of an acute angled triangle prove that this triangle is equilateral .

THe condition is equivalent to :
$\frac{1}{2+2OM}+\frac{1}{2+2ON}+\frac{1}{2+2OP}= 1$ (1)
where M,N and P are the midpoints of the sides of the triangle.
By Erdos-Mordel inequality we get:
$2(OM+OM+OP) \leq OA+OB+OC=3$ (here we use the fact that triangle $ABC$ is acute and so its circumcentre is interior point).
But from here and (1)(after clearing denominators) we get $OM=ON=OP$ so the triangle is equilateral.

Very , very nice , great solution  :lol:
Thank you very much for sharing with us . :)

The condition is equivalent to $\frac{1}{2+2cos(A)}+\frac{1}{2+2cos(B)}+\frac{1}{2+2cos(c)}= 1$.Cauchy obtein that $\frac{3}{2}\leq cos(A)+cos(B)+cos(C)$...because cos is concave $cos(A)+cos(B)+cos(C)\leq\frac{3}{2}$...so A=B=C


Can someone please tell me how I go about simplifying the following:

$ \dfrac {1}{(x^4 \plus{} 4)^{1/2}} \minus{} \dfrac {2x^4}{(x^4 \plus{} 4)^{3/2}}$

to a single fraction please?

Thanks for any help! :)

You need a common denominator, which in this case is $ (x^{4}\plus{}4)^{3/2}$.  So multiply the first fraction by $ \frac{x^{4}\plus{}4}{x^{4}\plus{}4}$.  Then we get
$ \frac{x^{4}\plus{}4\minus{}2x^{4}}{(x^{4}\plus{}4)^{3/2}}\equal{}\frac{4\minus{}x^{4}}{(x^{4}\plus{}4)^{3/2}}$

hello
[hide="solution"]
just write $ x^4\plus{}4\equal{}y$
Thus the question now is
$ \frac{1}{\sqrt{y}}\minus{}\frac{2(y\minus{}4)}{y^(3/2)}$
=$ \frac{4\minus{}x^4}{(x^4\plus{}4)^(3/2)}$
[/hide]
Thank u.

hello, the result is
$ \frac{4\minus{}x^4}{(4\plus{}x^4)^{\frac{3}{2}}}$.
Sonnhard.

Thanks guys! Suddenly it looks embarrassingly simple with
the solution JRav points out with the common denominator :)


Given a positive integer $ n\geq 2$, consider a set of $ n$ islands arranged in a circle. Between every two neigboring islands two bridges are built as shown in the figure.

Starting at the island $ X_1$, in how many ways one can one can cross the $ 2n$ bridges so that no bridge is used more than once?

WLOG the crossing starts clockwise and then multiply by two.

Define a returning island as an island $ i$ such we go from island $ i$ to island $ i \plus{} 1$ and returning back to island $ i$. (Same with $ n$ to $ 1$ and back to $ n$).

Case 1: Path with no returning island, $ 123 \cdots n123\dots n1$ there $ 2n$ steps to take with two options on each case then $ 2^{n}$ crossings

Case 2: Path with one returning island $ i$, $ 12 \cdots i \cdots 21 n(n \minus{} 1) \cdots (i \plus{} 1)i(i \plus{} 1) \cdots (n \minus{} 1)n1$ then $ 2^{n} \times n$ crossings.

Case 3: Path with $ 2$ or more returning islands, no possibilities.

To sum up total crossings $ \equal{} 2 \times (2^{n} \plus{} 2^{n} \times n) \equal{} (n \plus{} 1) 2^{n \plus{} 1}$

Edited with correction...from next post.

I think there is a mistake in your solution...
Where it is $ 2n$ it should be $ n$...
For example, in case 1, on the route $ X_1 \minus{} X_2 \minus{} \cdots \minus{} X_n \minus{} X_1$, there are $ 2^n$ possibilities, because from one island to another there are always two options.
But when we do the same route again, from one island to another, we can't pass through one of the bridges, because we have already passed through it, so there is only one possibility...
In case 2 you have the same mistake.
Regardless of that, my solution is similar to yours...

But... I have a question, can you return?, I say if it is possible that you can go from X_1 to X_2, and then return from X_2 to X_1?

I am waiting for an answer.

First of all, where it is written $ n\ge2$ it should be written $ n>2$. In Portuguese it's that. Anyway, it doesn't matter.
Now answering to the question... Yes, it is. But you can only return once because otherwise you won't be able to do the entire route as you want. I think it's easy to see that...

Own correction: when you return once you return twice, even though the returning island is the same, as shown by mszew.


Hoy la "Spanish Speaking Organization" aqui en la universidad donde estudio presento una pelicula latina ([b]Password: Una mirada en la Obscuridad[/b]) y pues he quedado anonadado de ver lo intolerantes a la verdad que son en ocasiones los gringos. La pelicula es medio cruda, pero trata cosas reales (trafico humano para explotacion sexual), y muchos salieron espantados de la pelicula, una muchacha casi se descompone! (vale que no presentamos La Noche de Los Lapices, hubiesemos tenido muertos seguro!).

En fin, el post era simplemente por que me interesaba saber que peliculas enteramente latinoamericanas de, digamos, los ultimos cinco anios recomendarian ustedes. He visto recientemente Diarios de una Motocilceta (bueno, han comprado los derechos, pero la pelicula es buenisima) y me ha parecido buena, aunque no genial, aunque he visto a Gael Garcia en mucho mejores peliculas.

Algun nombre escucho por alli. Yo tengo varios en mente, pero quiero ver cuales recomiendan ustedes.

Bueno yo piensoque las pelicula latinoamericanas y europeas son mil veces mejor que hollywood,  y si hay muy buenas latinomaericanas y españolas, pero los gringos aveces no entienden el crudo realismo que recrean estos filmes:

Y tu mama tambien
La mala educación
Diarios de una motocycleta
Mar adentro
El encanto discreto de un burgues (del maestro surrealista  Luis Buñuel)

Hace poco vi la pelicula 'mar adentro'. muy buena¡¡

Antier vi una de (a mi gusto personal) una de las mejores peliculas mexicanas recientes que haya visto en mi vida (y Gael Garcia no salia alli!!). Si quieren verla, se las recomiendo, se llama [b]Amar te Duele[/b].

Por otro lado, ayer vi una pelicula Ecuatoriana, jamas habia visto una pelicula de alli, se llama [b]Fuera de Juego[/b], una trama algo desordenada, pero no esta mal. Por cierto, me ha parecido bastante peculiar el acento ecuatoriano de Quito (o de los barrios bajos de Quito), la practica totalidad de la gente que conozco de Ecuador es de Guayaquil o indigenas de la parte andina, y todos tienen un acento bastante distinto.

Amores Perros

otra pelicula latinoamericana muy buena

Anoche ve [b]B-Happy[/b], una pelicula chilena, pues yo la catalogaria de buena, cruda, pero si estan deprimidos, no les recomendaria que la vieran.

Por cierto, alguien ha visto [b]Nadie Conoce a Nadie[/b] o [b]Abre los Ojos[/b], o en su defecto, [b]Tesis[/b]. Todas son peliculas españolas, y a mi todas me han parecido muy buenas.

Una pelicula que me parecio EXCELENTE, es "La vida de una mentira", es una pelicula española que la pasaron aca en un festival de cine español, pero no la he logrado conseguir ni en video, ni dvd ni nada. Otras peliculas que me parecieron buenas son "Maria llena eres de Gracia" ,aunque no es enteramente latina...el director es gringo, pero la trama es excelente. Tambien vi "Allende", una especie de documental sobre Salvador Allende, tambien muy buena.

Hoy pasaron [b]Ojos que no ven[/b], pelicula peruana del 2003, muy buena si la quieren ver. Cuenta los deschavez de Montesinos.

Por otro lado, estoy viendo en estos momentos [b]La Celestina[/b], pelicula española (desgraciadamente doblada, dado que estoy en gringolandia!) pero esta bastante entretenida!

Hay una película Argentina: [b]Moebius[/b] no la pude ver cuando la pasaron hace poco en la ciudad en la que estudio pero me parece que no es mala pues creo recibió varios premios. Mete algo de Topología. Veré si la puedo ver próximamente

Recomendada solo para audiencias maduras: [b]Sexo con Amor[/b], es chilena, una sexicomedia, a mi gusto bastante buena para una comedia (la trama no es muy buena, pero uno se rie durante toda la pelicula), pero el humor es algo adulto. No es que tenga escenas demaciado rajadas (aunque si tiene algunas) si no que yo creo que un muchacho (o muchacha) de 13 o 15 años no la entenderia muy bien, y podria mas bien encontrarla aburrida.

[quote="RaMlaF"]Una de las mejores peliculas que vi es una argentina que se llama "Nueve Reinas". Si no la vieron, no pierdan mas el tiempo que es lo mejor que hay en cine argentino![/quote]

Ramiro Lafuente: Tenés toooda la razón!!!!! Anoche la vi, y me pareció fantástica... una de las mejores que he visto!!!!!! (Nueve Reinas: elegida mejor película de la década)...
Se las recomiendo; vayan corriendo a un Video Club y alquílenla! no se van a arrepentir!!!!!!...

Yo acabo de ir a ver la de "Un mundo maravilloso", que acaba de salir aqui en méxico, y la verdad es que todo me gustó excepto el final, que me dejó con un sabor de boca asi como que medio extraño, y con eso de que soy bien sensible me quedé medio traumada el resto de la semana, pero pues por ahi se las recomiendo. :)

Otras dos muy buenas: El juego del miedo I y II

[quote="jasamper88"][quote="José"]Perdón por volver a postear yo, pero otras tres muy buenas son Titanic, Dragón Rojo, y el hombre de la máscara de hierro.....[/quote]

TItanic y El hombre de la mascara de hierro ( y no se si Dragon Rojo) no son cine latinoamericano...

Ciudad de Dios es una de las mejores peliculas que me e visto.. :D[/quote]

Tenés razón, ni lo pensé a eso que distraído....

[b]Todo sobre mi madre[/b] Es bastante buena, la ví hace poco y está muy interesante

[b]El laberito del Fauno[/b]  (película española-mexicana) muy buena, entremezcla dos mundos de una forma sutil.

[b]Machuca[/b] (Película chilena) muestra la realidad chilena del golpe militar mirada por los ojos de un niño.

Una mexicana muy chévere es [b]Matando Cabos[/b]. Ampliamente recomendada.

La " Las malas intenciones" película peruana que ha recibido buena critica.


To find the relation between the tension at the ends of a rope which are $T_{1}$ and $T_{2}$ which is connected to masses  by passing it  over pulley with frictionional coeffecient $k$ in the limiting case .
i am differing by a factor $\frac{1}{2}$ from the answer

[hide="the answer"]$\frac{T_{2}}{T_{1}}=e^{\pi k}$[/hide]

I've already solved this problem, try searching for it...

well let me post my approach ...after all i am just missing the factor $\frac{1}{2}$
isn't the normal forc on any small part of the rope $F_{n}=2Tsin(\Delta \theta)$ :maybe:
but it seems for this answer to be right it has to be
$F_{n}=Tsin(\Delta \theta)$

[url=http://www.mathlinks.ro/viewtopic.php?search_id=2032861136&t=80502]Here[/url] is ur solution.

but i couldn't understamnd why u took the component due to tension on one side only
but i got mu mistake corrected actually
[quote="pardesi"]isn't the normal forc on any small part of the rope $\F_{n}=2T\sin(\Delta \theta)$
[/quote]
actually that should be $\frac{\Delta \Theta}{2}$


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac{\pi}{3}} \frac{1}{\cos ^ 7 x}\ dx$.

I use an iteration method and got $ \frac{16 \sqrt{3}}{3} \plus{} \frac{5}{16}[2\sqrt{3} \plus{} \ln(2 \plus{} \sqrt{3})$

After using $ 1 \plus{} \tan^2{x} \equal{} \sec^2{x}$ I broke the integral down, etc. and got:

$ \frac{61 \sqrt{3}}{8} \plus{} \frac{ 5 \ln(2\plus{}\sqrt{3})}{16}$

That's right! :)

Let $ I_n\equal{}\int_0^{\frac{\pi}{3}} \frac{1}{\cos ^ n x}\ dx\ (n\geq 1)$, we have $ I_{n\plus{}2}\equal{}\frac{nI_n\plus{}\sqrt{3}\cdot 2^n}{n\plus{}1}.$


geometry
3D geometry
probability
MATHCOUNTS
USAMTS

Hmm...this is a pretty beast of a problem, are you sure that there's an elementary Mathcounts solution? It reminds me of USAMTS 18.3.1, which took painful casework.

[quote="happyme"]Bob decides to color the faces of a cube indigo, maroon, and black, one at a time, with equal probability. What is the probability that the resulting cube has at least one vertex such that all three faces containing that vertex and colored indigo?[/quote]
That problem looks suspiciously familiar...  Are you sure you thought of it all by yourself? Somehow, it reminds me of Harker Math Invitational #30 :D

Woah, I didn't even bother to see the 7th and 8th grade contests of the Harker Math Invitational  :ninja: . My colors are much cooler though! I would never think that Harket would get that long, and confusing (should start preparing)!

The answer I'm getting is, hugely small?

[quote="happyme"]
Woah, I didn't even see the 7th and 8th grade contests of the Harker Math Invitational :ninja: .

The answer I'm getting is, hugely small?
[/quote]
I finally got an answer after a while, it's not insanely small, but you have to add some insanely small values

yeah i've seen one like that in a textbook from europe

dude the prob. is very puny and i do not like small, so i am not going to answer.   :bye:  :gathering:  :laugh:

[hide]is it 125/729?  Just tell me what Im doing wrong[/hide]


For $ a,b,c$ positive reals prove

$ \sum \frac{a^2\plus{}b^2}{c(a\plus{}b)} \geq (\sum a^2) \sqrt {\frac{3}{abc(a\plus{}b\plus{}c)}}$


function
advanced fields
advanced fields unsolved

Consider all subsets of the Toplogic space $\tau$ like $A$. Suppose there we have 2 functions and we can change our subsets with the functions $A\rightarrow \overline{A}$ and $A\rightarrow \tau\backslash A$

a) Prove with this functions we can reach to at most 14 different sets starting from $A$.
b) Make an example in the standard toplogy of $\mathbb R$ that that we can reach to exactly 14 distinct sets.


Find the ratio of the radius of the inner of two concentric circles to the radius of the outer circle given that the figure is composed of 12 sections, with each the same area. Express your answer as a common fraction in simplest radical form.
[asy]import olympiad; size(150); import geometry_dev; import graph; defaultpen(linewidth(0.8)); draw(Circle(origin,10));draw(Circle(origin,5*sqrt(2))); pair[] rods; for(int i = 0; i < 6; ++i){ rods[i] = 10*dir(30 + 60*i); draw(origin--rods[i]);}[/asy]

Let r be the radius of the inner circle and R the radius of the outer one.
Then, $ R^2\pi - r^2\pi = r^2\pi$, so $ R^2\pi = 2r^2\pi$ and $ R^2 = 2r^2$
Square rooting gives $ R = \sqrt2 r$, and $ \frac rR = \frac{\sqrt2}{2}$


solve in positive integers

$x^p+y^p=p^z$, where $p$ is a prime

All integer solutions is:
1.$x=0,y=p^n,z=pn, \ or \ y=0,x=p^n,z=pn.$
2.$p=2,x=y=2^k,z=2k+1.$

[quote="Rust"]All integer solutions is:
1.$x=0,y=z=p, \ or \ y=0,x=z=p.$
2.$p=2,x=y=2^k,z=2k+1.$[/quote]
OK can you post your solution???
Actually this problem is from Bulgarian National Olymiad 1994 ;)
See http://www.imo.org.yu/othercomp/Bul/BulMO944.pdf

It is obviosly $x+y=p^n$. If p>2 x=0 or y=0.


I provide this by way of comparison, since the 1st course is the same pitch as a guitar's first string, and the 8th course is the same pitch as a guitar's sixth string.  The lute needs much lower tension, so both strings are visibly thinner than their counterparts on the guitar.  Also, the 6th through 13th course are strung in octaves, so that the upper string is an octave above the lower string.  As you can see, the nylon and wire-wrapped strings are close to the same diameter.


find all positive real numbers x,y,z
such that
$x=\frac{1+z}{1+y}$
$y=\frac{1+x}{1+z}$
$z=\frac{1+y}{1+x}$

Partial solution..

[hide]
one obvious solution is..
$x=y=z=1$ (assuming cosinator is right)

if you rearrange equation 1 and 2...you get

$1+z=x\left(1+y\right)$
$1+z=\frac{1+x}{y}$
$\frac{1+x}{y}=x\left(1+y\right)\Rightarrow{y}\left(x+xy\right)=1+x\Rightarrow{xy^{2}+xy-(1+x)}=0$

Solve for y using the quadratic formula...
$y=\frac{-x \pm\sqrt{x^{2}+4(x)(1+x)}}{2x}=\frac{-x \pm\sqrt{5x^{2}+4x}}{2x}=\frac{-x+\sqrt{5x^{2}+4x}}{2x}$*Since it asks for positive reals*

Not sure what to do next..i bet theres a nicer way to do this...
[/hide]

hint: rewrite the system as $x=\frac{c+z}{b+y}$, $y=\frac{a+x}{c+z}$, and $z=\frac{b+y}{a+x}$, then solve for $a$, $b$, and $c$ in terms of $x$, $y$, and $z$.

We're looking for $(x,y,z)\in\mathbb{R}^{+}^{3}$ such that \[\left\{\begin{array}{l}x = \frac{1+z}{1+y}\\ y = \frac{1+x}{1+z}\\ z = \frac{1+y}{1+x}\end{array}\right.\]Multiply the three equations to get $xyz = 1$. So there exist $a,b\in\mathbb{R}^{+}$ with $x = a$, $y = \frac{b}{a}$ and $z = \frac{1}{b}$. The equations become $b(a+b) = 1+b = a(a+1)$

Suppose $a>b$. Then $1+b = a(a+1) > a(b+1)$, so $1 > a$. But then $1+b = b(a+b) < b(1+b)$, so $1<b$. This means $a < 1 < b$, contradicting $a>b$.

Suppose $a<b$. Then $1+b = a(a+1) < a(b+1)$, so $1<a$. But then $1+b = b(a+b) > b(1+b)$, so $1>b$. This means $b < 1 < a$, contradicting $a<b$.

So $a = b$. But then $1+b = a(a+1) = b(b+1)$, so $a=b=1$, so $x=y=z=1$.


In a cyclic quad ABCD inscribed in a unit circle
if AB.BC.CD.DA >= 4

Prove that ABCD is a square
.

[hide]
Ptolemy's Theorem
AC*BD=AB*DC + AD*BC
Max of BD*AC=4 (unit circle)
AB*DC + AD*BC>=2(AB*BC*CD*DA)^(1/2)>=4

Therefore,
(AB*DC + AD*BC)=4
Therefore
BD and AC are diamewtres.
Then angles in semicircle
=> ABCD is square.[/hide]

[hide="Almost"]We have AB = 2sinA
BC = 2sinB
CD = 2sinC
DA = 2sinD

where cyclic 2A is the angle subtended by cyclic AB at the center.

A+B+C+D = $ \pi$

Given relation gives $ sinAsinBsinCsinD \geq \frac{1}{4}$

Now Jenson's gives $ sinA \plus{} sinB \plus{} sinC \plus{} sinD \geq \frac{4}{\sqrt{2}}$

Now all that is left is to prove $ sinAsinBsinCsinD \leq \frac{1}{4}$ and equality holding when A = B = C = D in which case our cyclic quad becomes a square. [/hide]

Can anyone help me further.


calculus
derivative
function
calculus computations

How would I go about finding the derivative of this function with respect to x?

$ \sum^n_{k\equal{}0} {2n\choose2k}  (2k\plus{}1)^{\minus{}x}$

X is real, greater than 0, and less than 1.

$ f(x)\equal{}\sum^n_{k\equal{}0} {2n\choose2k} (2k\plus{}1)^{\minus{}x}\equal{}\sum^n_{k\equal{}0} {2n\choose2k} e^{\minus{}x \ln(2k\plus{}1)}\equal{}\sum^n_{k\equal{}0} f_k(x)$

$ f_k (x)\equal{} {2n\choose2k} e^{\minus{}x \ln(2k\plus{}1)}$
$ f_k'(x) \equal{} \minus{}{2n\choose2k} e^{\minus{}x \ln(2k\plus{}1)}\ln(2k\plus{}1)$

$ \to f'(x)\equal{} \sum^n_{k\equal{}0} f_k'(x) \equal{}\minus{} \sum^n_{k\equal{}0} {2n\choose2k} (2k\plus{}1)^{\minus{}x}\ln(2k\plus{}1)$

unless youre looking for something more pretty?


function
real analysis
real analysis unsolved

let $f$ be a function such that :
1)$f: R \rightarrow R$
2)$f(f(x))=x 3)f(f(x)+x)=f(x)+x$
is this function continuous?

I think you have a typo. What's the first condition in 2)?


My name is Benjy Hirsch, and I was on team Pig from NYC. We just received the packet containing our power test, which it seems you misgraded. You also seem to misgraded the power test of team Colbert (also from NYC). On theirs, the point breakdown was:
Part 1: 8 pts
2: 12 pts
3: 12 pts
4: 12 pts
5: 2 pts
6: 5 pts
Total: 49 Pts.
That does not add up (in case you didn't notice). I think the problem was that the 8 for part 1 looked like a 6, but the points given for the parts of part 1 added up to 8--(1,1,1,2,3).

Now, on our (team Pig's) power test, it seems as if you simply did not grade part 4e). What must have confused you is that we typed up most of it, and then hurriedly finished writing up parts 4b and 4e on the train (Adam figured them out as he was falling asleep the night before). We stapled in the handwritten pages, so there was a blank spot for part e, and then on the page immediately after it, there was our solution.

I assume that part 4e) was not worth 8 pts, so we wouldn't have tied or beaten Quagga [b](edit: beaten Quagga on the power test only, obviously not overall)[/b], (but I assume it is worth at least 2 points, so we still have the bragging rights that we beat the top city team (Colbert) on the power round).

I don't really see what can be done about this at this point, but in the future, it might be a good idea to have the power round done completely ahead of time, so we email you our solutions, and you grade it before we get there, so if we have an appeal or other scoring issue, it can be addressed before everything is finalized.

Besides that, the competition went very well. The power round was a great idea, though I'm biased since we did well, the problems were great, and the Math Bowl was entertaining, even if we got knocked out in the third round.

[b] Edit: Also, it would be nice to have a breakdown of how much each part of each section was worth. [/b]

[quote="yjneb"]in the future, it might be a good idea to have the power round done completely ahead of time, so we email you our solutions, and you grade it before we get there, so if we have an appeal or other scoring issue, it can be addressed before everything is finalized.[/quote]
Good idea!

[quote="yjneb"]Edit: Also, it would be nice to have a breakdown of how much each part of each section was worth. [/quote]
This exists, and will be posted eventually.

Will we get solutions to the Power Test soon!!! Please???


1.Show that fo rany a and b naturel numbers ab(a+b)(a-b) is a multiple of 3
2.Find all  the naturel numbers n s.t:n+2 and n-6 are coprime.

1.
[hide] this is actually pretty trivial.
if $a$ or $b\equiv 0 \ \mod3$, the case is obvious
if $a\equiv b\equiv 1,2 \mod3 \ \ \ \Rightarrow a-b\equiv 0 \mod3$
if $a\equiv 1 \mod3$ and $b\equiv 2 \mod3$ then $a+b\equiv 0 \mod3$ (same when you replace $a$ with $b$)[/hide]

2.
[hide]$\gcd(n+2,n-6)=1$
$\Leftrightarrow \gcd(n-6,n+2-(n-6))=1$
$\Leftrightarrow \gcd(n-6,8)=1$
$\Leftrightarrow n-6\not\equiv 0 \mod2$
$\Leftrightarrow n$ is odd.
edit-and of course: $n>6$  ;) [/hide]


The sum of all the numbers in a 15-term arithmetic sequence with first term 7 is -210. Find the common difference.

How can you quickly find the answer?

well, to find the middle number, you divide -210 by 15 to get -14. -14 is the 8th term.
-14=7-(8-1)d
-2=1-d
d=3
so the common difference is -3.

Let $d$ be the common difference. Then the last term is $7 + (15-1)d = 7+14d$. The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms, so the sum of the series is \[\frac{7 + (7 + 14d)}{2} \cdot 15 = 15(7d + 7) = 105d + 105.\]We are told that this sum equals $-210$, so we have $105+105d = -210$, from which we find $d=\boxed{-3}$.

Note: $\boxed{3}$ is also accepted as an answer.


function
complex numbers
superior algebra
superior algebra unsolved

Let $ t$ be an indeterminate element, consider the field of rational functions $ K\equal{}\mathbb{C}(t)$. Take complex numbers $ p,\ q\in\mathbb{C}$, set $ L\equal{}\mathbb{C}(t^{3}\plus{}pt\plus{}q)$, prove that $ K/L$ is an algebraic extension, then find the Galois closure of $ K/L$ and the extended dimension over $ L$.

It seems there are two cases to consider: $ p\equal{}0$ and
$ p\neq 0$, isn't it?
 because we have to add a degree $ 2$ extension above $ \mathbb{C}(t)$
 if $ t^{2}\plus{}p$ is not a square, i.e. if $ p\neq 0$.


articles
videos
search
probability
expected value
Support

I think the Monty Hall Paradox should be included in one of the videos. Also, where should I post other suggestions?

That's nice to know.  :P

If I were you, I would directly PM the moderators/administrators.

[quote="PowerOfPi"]Moderators can't do anything, and administrators say they don't pay attention to PM's.[/quote]We pay attention to PMs sent to us.

[quote="Valentin Vornicu"]That's a really dumb advice. DO NOT PM us with these requests, we will simply ignore them.[/quote]
I don't know. :?:

[quote="PowerOfPi"][quote="Valentin Vornicu"]That's a really dumb advice. DO NOT PM us with [b]these [/b]requests, we will simply ignore them.[/quote]
I don't know. :?:[/quote]

do you know what the word "these" means in this situation?

edit: just looked it up
Mr. Vornicu was referring to changing usernames

[url=http://en.wikipedia.org/wiki/Monty_Hall_problem]Google is your friend.[/url]

[quote="maybach"]What is the Monty Hall Paradox?
Anyways, isn't Monty Hall the host of that really old game show "Lets Make a Deal"?[/quote]

You should ask that in a problem-specific-related forum, not in a forum for the general workings of a part of a site. Anyway, just read that Wikipedia article nikeballa96 provided; it has the information.

I know what it is (learned it in seventh grade), I am suggesting that a video could be made about it (expected value?).

[quote="r15s11z55y89w21"]You should ask that in a problem-specific-related forum, not in a forum for the general workings of a part of a site. [/quote]

[quote="PowerOfPi, in his first post,"]Also, where should I post other suggestions?[/quote]

EDIT: Sorry  :blush:

[quote="PowerOfPi"]I know what it is (learned it in seventh grade), I am suggesting that a video could be made about it (expected value?).

[quote="r15s11z55y89w21"]You should ask that in a problem-specific-related forum, not in a forum for the general workings of a part of a site. [/quote]

[quote="PowerOfPi, in his first post,"]Also, where should I post other suggestions?[/quote][/quote]

Notice I was talking to maybach, not you.

They have a section for 'Expected Value' already

[quote="thinker8"]You could search the Monty Hall Problem on YouTube for videos.[/quote]

yes you can, but creating videos take a long time to prepare and take
P.S. try not to revive old posts


quadratics
induction
group theory
combinatorics proposed
combinatorics

We call a permutation $ \left(a_1, a_2, ..., a_n\right)$ of $ \left(1, 2, ..., n\right)$ [i]quadratic[/i] if there exists at least a perfect square among the numbers $ a_1$, $ a_1 \plus{} a_2$, $ ...$, $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n$. Find all natural numbers $ n$ such that all permutations in $ S_n$ are quadratic.

[i]Remark.[/i] $ S_{n}$ denotes the $ n$-th symmetric group, the group of permutations on $ n$ elements.

[quote="seshadri"]again as i was looking at older pages this one caught my attention and in my opinion this should have been solved long ago.
i claim the answer is the set of those $n$ for which $\dfrac{n(n+1)}{2}$ is a perfect square(of course there are infintely many such $n$ that can be found by the so-called Pell's method or whatever)
to start with let $K_n=\{k\le n|\dfrac{k(k+1)}{2} = \square\}$.(i.e is a square)
suppose $k,k+1\in K_n$. Then $k^2+k=2r^2,k^2+3k+2=2s^2$ where $r,s$ are positive integers. Then $k+1=s^2-r^2$. Plugging this back into the two equations and dividing, we have $\dfrac{s^2-r^2-1}{s^2-r^2+1}=\dfrac{r^2}{s^2}$ so that we have $(s^2-r^2)^2=(s^2+r^2)$. Now we know that $(s^2-r^2)^2+(2rs)^2=(s^2+r^2)^2$ so that we have $(r^2+s^2)(r^2+s^2-1)=(2rs)^2$ and that is a contradiction since product of two consecutive integers cannot be a perfect square.
hence $K_n$ has 'gaps' and we an use these gaps to get a sequence without any partial squares appearing.
start with the word $21$ and add all consecutive integers till you hit an element $k$ of $K_n$; then we throw in $k+1$ before $k$ and then $k$. for example: $2134567\underline{98}\ldots$. we just exchange at the occurence of the squares and by what we have just seen, this exchange will not create any partial squares till the integer $k+1$ and we can carry on the same way.[/quote]
It is not clear how we can create sequance $2134567\underline{98} \ldots$.Words "we can carry in the same way" doesn't make sense.

The idea is that he will create a sequence $a_i$ such that if $\frac{n(n+1)}{2}$ is a square then the only square in the sums $a_1,...,a_1+...+a_n$ is $a_1+a_2+...+a_n=\frac{n(n+1)}{2}$, and $a_n=n$, and if that number is not a square then none of the sums will have a square.
This can be done inductively. Base case is clear. Suppose for $k$ such a sequence exist, for $k+1$, if $\frac{k(k+1)}{2}$ is not a square then just add $a_{k+1}=k+1$. Then none of sums $a_1,a_1+a_2,...,a_1+...+a_k$ is a perfect square, and $(a_1,a_2....a_{k+1})$ is a permutation of $1$ to $k+1$. (and if $\frac{(k+1)(k+2)}{2}$ is a square it has $a_{k+1}=k+1$). If $\frac{k(k+1)}{2}$ is a square then change $a_k=k$ to $a_k=k+1$ and then let $a_{k+1}=k$. Clearly now there cannot be any square in the sums ($a_1,...,a_1+...a_k$(which is some square minus $1$) is clear, and $a_1+...+a_{k+1}$ is not a square from elements of $K_n$ are not consecutive.). By induction we can construct the necessary sequences.


Why doesn't USAMTS accept word documents?
It says security reasons, but what do they mean in specific?
Just wondering...

[quote="Treething"][quote="rrusczyk"][quote="Anonymous"]It says you won't accept MS word documents for security reasons.  Is it possible to transmit viruses or something through Word?[/quote]

Yes.  You should never open a Word document sent by anyone you don't know or trust completely.[/quote]

That's only because you can embed viruses in an .rtf ...[/quote]

RTF stands for rich text format.  When you save a word file you can save it as a .rtf file instead of a .doc file.  I don't really know what the point of using .rtf files is, they might take up less space or something.

If you want to use Word to write up your solutions, you can use a PDF printer to convert your Word document into a .pdf document.  Instructions for using a PDF printer can be found on the LaTeX page on this site (click the link on the sidebar).

However, it is much easier to use LaTeX for the USAMTS.  You can get a template right off the USAMTS site that basically does all the formatting for you.  All you have to do is type up your solutions where it tells you to.  Plus, you can just look up the symbols you need on the LaTeX page on this site as you need them.

That's right.  I find LaTeX much easier to use than word.

Word is very cumbersome to type math in... :noo:

Yeah. MathType is really confusing compared to LaTeX... :(

It's not like LaTeX is un-confusing though

It does take time to learn it but when you understand it the math will look a lot more better.


Could anyone tell why

$ \sum{n^3} \equal{} (\sum{n})^2$  ?

I know the algebraic proof (induction). I want a physical significance.Why are they equal??
Like in a problem, you count something in two different ways and get these two expressions. Could anyone give such an example?

Thanks.

Easy Arjun. You have $ 1 * 1^2 \equal{} 1^3
                              2*2^2 \equal{} 2^3$ and so on.
What u do is for each number form $ 1$ to $ n$, u make $ n$  squares of area $ n^2$ so that you separate each square into unit squares and now you arrnage those unity squares in order to form a bigger square with side length $ 1\plus{}2\plus{}3\plus{}4\plus{}...\plus{}n$ and therefore equating the areas we get that $ (1\plus{}2\plus{}3\plus{}...\plus{}n)^2 \equal{} (1^3\plus{}2^3\plus{}3^3\plus{}....n^3)$.

@t0rajir0u and Aravind:
Thank you very much.
It was awesome. I never knew that this had been studied SO much.


function
calculus
derivative
Putnam
algebra unsolved
algebra

It seems a lot difficult to me:
   Give a function $ f:R->R$ which is three times differentiable.
    Prove that there exists a real $ a$ such that:
         $ f(a).f'(a).f''(a).f'''(a) \geq 0$
Please help!

Hey [b]darkmaster[/b],a long time to see you.What the matter with you? :D

[quote="grobber"]And I think it was solved before somewhere around here.[/quote]
Yep,it's was posted by [b]romano[/b],solved by[b] liyi[/b] in "Calculus proposed and own section".Here is the link:
http://www.mathlinks.ro/Forum/viewtopic.php?t=24592

Hi,pigply. Falling in HCM city team selection contest makes me so disappointed.It takes time to encourage. Thanks for your attention.
And also thanks for the link :)


inequalities
induction
rearrangement inequality
inequalities proposed

For given $n \geq 4$ and positive real numbers $x_{1},x_{2}, \cdots, x_{n}$ which satisfy $x_{1}+x_{2}+\cdots+x_{n}=n$, prove that \[x_{1}x_{2}\cdots x_{n}( \frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{2}}+\cdots+\frac{x_{1}}{x_{n}}) \leq n.\] For $n=4$, Doosung Park proposed the problem at this post: http://www.mathlinks.ro/Forum/viewtopic.php?t=100960 . And if $n=3$, then the maximum is equal to 4, which is a problem from Canada. So it can be paraphrased like:

For given $n \geq 3$ and positive real numbers $x_{1},x_{2}, \cdots, x_{n}$ which satisfy $x_{1}+x_{2}+\cdots+x_{n}=n$, prove that \[x_{1}x_{2}\cdots x_{n}( \frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{2}}+\cdots+\frac{x_{1}}{x_{n}}) \leq \max(n,4).\]

I also proved that the inequality is true for $n$ is odder  :) ...
But it seems not very useful.
You said in another topic that you have proved if "2n is true,then 2n+1 is true."
But yours and mine can't combine :(

Well, I proved the two statements: "$2n$ is true" and "if $2n$ is true, then so is $2n+1$". If you proved that $2n+1$ is true, then the second statement is needless, as the proof can be obtained by combining our proofs.

[quote="Sung-yoon Kim"]For given $n \geq 4$ and positive real numbers $x_{1},x_{2}, \cdots, x_{n}$ which satisfy $x_{1}+x_{2}+\cdots+x_{n}=n$, prove that \[x_{1}x_{2}\cdots x_{n}( \frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{2}}+\cdots+\frac{x_{1}}{x_{n}}) \leq n.\] [/quote]

Nice proof. (I think that the lemma should be modified like: There exists $k$ such that $x_{k}\geq x_{k+4}, x_{k+1}\geq x_{k+5}$ or $x_{k}\leq x_{k+4}, x_{k+1}\leq x_{k+5}$.)

Here is my proof. The LHS is equal to $x_{2}\cdot x_{2}x_{3}\cdots x_{n}+\cdots+x_{1}\cdot x_{1}x_{2}\cdots x_{n-1}$, which is the sum of products with one in $x_{i}$ and one in $\frac{x_{1}x_{2}\cdots x_{n}}{x_{i}}$. So if we rearrange $x_{1},x_{2},\cdots, x_{n}$ into $y_{1}\geq y_{2}\geq \cdots \geq y_{n}$, as $y_{2}y_{3}\cdots y_{n}\leq \cdots \leq y_{1}y_{2}\cdots y_{n-1}$, by rearrangement inequality, the LHS is smaller than or equal to \[(y_{1}y_{2}\cdots y_{n-1})y_{1}+(y_{1}y_{3}\cdots y_{n})y_{2}+\cdots+(y_{2}y_{3}\cdots y_{n})y_{n}.\] Now I will prove that \[(y_{1}y_{2}\cdots y_{n-1})y_{1}+(y_{1}\cdots y_{n-2}y_{n})y_{2}+\cdots+(y_{2}y_{3}\cdots y_{n})y_{n}\leq n(\frac{y_{1}+y_{2}+\cdots+y_{n}}{n})^{n}.\] [b]Step 1.[/b] If $n$ is even, then the inequality holds.
I will prove this by induction. For the initial case, namely $n=4$, it was proved in another post: http://www.mathlinks.ro/Forum/viewtopic.php?t=100960 . Now assume that $n=2k$ holds. Then, in case of $n=2k+2$, \[y_{1}^{2}y_{2}\cdots y_{2k+1}+y_{1}y_{2}^{2}\cdots y_{2k}y_{2k+2}+\cdots+y_{1}y_{2}\cdots y_{k}^{2}y_{k+2}\cdots y_{2k+2}\]  \[+y_{1}y_{2}\cdots y_{k-1}y_{k+1}^{2}\cdots y_{2k+1}+\cdots+y_{2}y_{3}\cdots y_{2k+1}y_{2k+2}^{2}\]  \[=(y_{1}^{2}y_{2}\cdots y_{2k+1}+y_{2}y_{3}\cdots y_{2k+1}y_{2k+2}^{2})+y_{1}y_{2k+2}(y_{2}^{2}\cdots y_{2k}+\cdots+y_{3}\cdots y_{2k+1}^{2})\]  \[\leq y_{2}y_{3}\cdots y_{2k+1}(y_{1}^{2}+y_{2k+2}^{2})+y_{1}y_{2k+2}(2k)(\frac{y_{2}+y_{3}+\cdots+y_{2k+1}}{2k})^{2k}\]  \[\leq (y_{1}^{2}+y_{2k+2}^{2}+2ky_{1}y_{2k+1})(\frac{y_{2}+y_{3}+\cdots+y_{2k+1}}{2k})^{2k}\]  \[\leq (2k+2)(\frac{y_{1}+y_{2k+2}}{2})^{2}(\frac{y_{2}+y_{3}+\cdots+y_{2k+1}}{2k})^{2k}\]  \[\leq (2k+2)(\frac{2\cdot \frac{y_{1}+y_{2k+2}}{2}+2k \cdot \frac{y_{2}+y_{3}+\cdots+y_{2k+1}}{2k}}{2k+2})^{2k+2}\]  \[=(2k+2)(\frac{y_{1}+y_{2}+\cdots+y_{2k+2}}{2k+2})^{2k+2}\] which completes the proof.

[b]Step 2.[/b] If $n=2k$ holds, then so does $n=2k+1$.
For $n=2k+1$, \[y_{1}^{2}y_{2}\cdots y_{2k}+y_{1}y_{2}^{2}\cdots y_{2k-1}y_{2k+1}+\cdots+y_{1}y_{2}\cdots y_{k}^{2}y_{k+1}y_{k+3}\cdots y_{2k+1}\]  \[+y_{1}y_{2}\cdots y_{k}y_{k+1}y_{k+2}\cdots y_{2k+1}+y_{1}y_{2}\cdots y_{k-1}y_{k+1}y_{k+2}^{2}\cdots y_{2k+1}+\cdots+y_{2}y_{3}\cdots y_{2k}y_{2k+1}^{2}\]  \[=y_{k+1}(y_{1}^{2}y_{2}\cdots y_{k}y_{k+2}\cdots y_{2k}+\cdots+y_{1}y_{2}\cdots y_{k}^{2}y_{k+3}\cdots y_{2k+1}+y_{1}y_{2}\cdots y_{k}y_{k+2}\cdots y_{2k+1}\]  \[+y_{1}y_{2}\cdots y_{k-1}y_{k+2}^{2}\cdots y_{2k+1}+\cdots+y_{2}y_{3}\cdots y_{2k}y_{2k+1}^{2}\]  \[\leq y_{1}y_{2}\cdots y_{k}y_{k+2}\cdots y_{2k+1}+y_{k+1}(2k)(\frac{y_{1}+\cdots+y_{k}+y_{k+2}+\cdots{}_{y}{}_{2k+1}}{2k})^{2k}\]  \[\leq y_{k+1}((\frac{y_{1}+y_{2}+\cdots+y_{k}+y_{k+2}+\cdots+y_{2k+1}}{2k})^{2k}+2k(\frac{y_{1}+y_{2}+\cdots+y_{k}+y_{k+2}+\cdots+y_{2k+1}}{2k})^{2k})\]  \[=(2k+1)y_{2k+1}(\frac{y_{1}+y_{2}+\cdots+y_{k}+y_{k+2}+\cdots+y_{2k+1}}{2k})^{2k}\]  \[\leq (2k+1)(\frac{y_{k+1}+2k \frac{y_{1}+y_{2}+\cdots+y_{k}+y_{k+2}+\cdots+y_{2k+1}}{2k}}{2k+1})^{2k+1}\]  \[=(2k+1)(\frac{y_{1}+\cdots+y_{2k+1}}{2k+1})^{2k+1}\] which completes the proof.

By [b]step 1[/b] and [b]step 2[/b], the inequality holds for any integer $n \geq 4$. :)

PS. Actually the induction method used in step 1 can be applied to the odd cases. So if we prove $n=5$ easily, then the solution can become shorter.


Let $ABCD$ be a quadrilateral such that all sides have equal length and $\angle{ABC} =60^o$.  Let $l$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D$).  Let $E$ and $F$ be the points of intersection of $l$ with $AB$ and $BC$ respectively.  Let $M$ be the point of intersection of $CE$ and $AF$.

Prove that $CA^2 = CM \times CE$.

The triangle $ABC$ is equilateral, because $AB=BC$ and $\angle{ABC}=60$, and the triangle $ACD$ is equilateral too, because $AD=DC=AC$. Now, $CD\|AB$ and $BC\|AD$, so $ADE$ and $DCE$ are similar, so $\frac{AE}{AD}=\frac{DC}{CF}$, but $ADC$ is equilateral, so $\frac{AE}{AC}=\frac{AC}{CF}$. Also, we note that $\angle{EAC}=\angle{ACF}=120$, so $EAC$ and $ACF$ are similar, so $\angle{AEC}=\angle{CAF}$, so $CA$ is tangent to the circuncircle of $AME$. Hence $CA^{2}=CM\times{CE}$.

[hide="Here is a equivalent enunciation."]
[color=darkred]Let $BEF$ be a triangle with $m(\widehat {EBF})=60^{\circ}$. Denote the points: $D\in EF$, $\widehat {DBE}\equiv \widehat {DBF}$; $C\in BF$, $DC\parallel BE$; $A\in BE$, $DA\parallel BF$; $M\in AF\cap CE$. Prove that $CA^2=CM\cdot CE$.[/color]

......Later I shall offer here the my proof. [/hide]

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1993Problem1

Vo Duc Dien

This is actually very hard for an APMO 1. Here is a very long solution which uses only the most ultra basic techniques:

We draw the line segment connecting points $A$ and $C$. Since quadrilateral $ABCD$ is a rhombus with $\angle ABC=60^\circ$, $\triangle ABC$ and $\triangle ADC$ are both equilateral. Hence, $AB=BC=AC=AD=DC$. Let the intersection of $\overline{AD}$ and $\overline{EC}$ be $H$ and let the intersection of $\overline{AF}$ and $\overline{CD}$ be $I$.

[b]Claim:[/b] $HD=IC$.

[i]Proof.[/i] Since $AD \parallel CF$ and $AE \parallel CD$ and $E,D,$ and $F$ are colinear, $\triangle EAD \sim \triangle DCF$ by AA Similarity. Therefore, $\frac{AD}{CF} = \frac{ED}{DF} \Rightarrow CF \cdot ED = AD \cdot DF$.

Similarly, $\triangle EHD \sim \triangle ECF$ by AA Similarity. Therefore, $\frac {HD}{CF} = \frac{ED}{EF}\Rightarrow HD = \frac{CF \cdot ED}{EF}$.

We substitute $CF \cdot ED = AD \cdot DF$ into $HD = \frac{CF \cdot ED}{EF}$ to get $HD = \frac{AD \cdot DF}{EF}~~~(1)$.

Since $CD \parallel BE$, $\triangle FCD \sim \triangle FBE$ by AA Similarity. Therefore, $\frac{FC}{FB} =\frac{FD}{FE}$.

Similarly, $\triangle FIC \sim \triangle FAB$ by AA Similarity. Therefore, $\frac{IC}{AB} = \frac{FC}{FB} \Rightarrow IC=\frac{FC\cdot AB}{FB}~~~(2)$.

Equations $(1)$ and $(2)$ are equal since $AD=AB$ from the equilateral triangles and since $\frac{FC}{FB}=\frac{FD}{FE}$ from $\triangle FIC \sim \triangle FAB$. Hence, $HD = \frac{AD \cdot DF}{EF}=\frac{FC\cdot AB}{FB}=IC$. $\square$

Because $\triangle ACD$ is equilateral, we have $AC=CD$ and $\angle ACD = \angle CDA$. But we also found that $HD=IC$. Thus, $\triangle CIA \cong \triangle DHC$ by SAS Congruence.

Let $\angle ECD = x$. It follows that $\angle ECA=180^\circ-\angle BCA - \angle ECD - \angle DCF=180^\circ-60^\circ-x-60^\circ=60^\circ-x$. But it also follows that $\angle CAI=\angle CAM=x$ because $\triangle CIA \cong \triangle DHC$. Therefore, $\angle AMC = 180^\circ-\angle CAM - \angle ACM =180^\circ-x-(60^\circ-x)=120^\circ$. So $\triangle AMC$ has angle measures $\angle A=x, \angle M=120^\circ,$ and $\angle C = 60^\circ-x$.

Now we look at $\triangle EAC$. We already know that $\angle ACE=60^\circ-x$. Because $\triangle EHA \sim \triangle CHD$, we have $\angle HEA=x$ and $\angle HAE=60^\circ$. It follows that $\angle CAE=120^\circ$. Hence, $\triangle EAC$ has angle measures $\angle E=x, \angle A=120^\circ,$ and $\angle C=60^\circ-x$.

By AAA Similarity, we have $\triangle EAC \sim \triangle AMC$ so $\frac{CA}{CM}=\frac{CE}{CA}\Rightarrow CA^2=CM\cdot CE$ as desired. $\blacksquare$

[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMi9mL2U0NTY5OWM5ODcyNzIxNWZkMDI2YzY1YzA3MTNkMzg5MzIzMjM3LnBuZw==&rn=T2x5bXBpYWQgR2VvbWV0cnkgMS4xIERpYWdyYW0ucG5n[/img]

[quote=Sailor]Let $AB=1$ and consider the inversion $I$ of pole $C$ and power 1.
Then $I(E)=E'$ is the intersection of line $CE$ and the circumcircle of $\triangle{ABC}$.
Since $\triangle{EAD}\sim{\triangle{DCF}}$ it follows that $AE=\frac{1}{CF}=CF'$, where $F'$ is the inverse of point $F$. Thus $\triangle{BEF'}$ is equilateral and consequently points $A,E',F$ are collinear,i.e. $E'=M$.
Considering this we get $AC^2=1=CM\cdot{CE}$.[/quote]

15 years later...................:rotfl:

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pair F = extension(E,D,B,C);
pair M = extension(E,C,A,F);
filldraw(A--B--C--D--cycle,opacity(1)+lightcyan, blue);
draw(A--E--C--cycle,paleblue+linewidth(1.0pt));
draw(C--A--F--cycle,lightgreen+linewidth(1.0pt));
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filldraw(anglemark(A,D,E),purple,heavymagenta+linewidth(0.5));
filldraw(anglemark(C,F,D),purple,heavymagenta+linewidth(0.5));
filldraw(anglemark(D,E,A),orange,red+linewidth(0.5));
filldraw(anglemark(F,D,C),orange,red+linewidth(0.5));
filldraw(anglemark(E,A,C),pink,magenta+linewidth(0.5));
filldraw(anglemark(A,C,F),pink,magenta+linewidth(0.5));
dot(A^^B^^C^^M);
[/asy]

It is clear that because $ABCD$ is a rhombus, $AB=BC$ combined with $\angle ABC=60^\circ$ yields that $\triangle ABC$ is equilateral, and analogously that $\triangle ADC$ is equilateral. So, $\angle EAC=180^\circ-\angle CAB=120^\circ,$ and by symmetry $\angle ACF = \angle EAC = 120^\circ.$

By the converse of Power of a Point, proving $AC^2 = CE \cdot CM$ is equivalent to proving that the circumcircle of $\triangle AME$ is tangent to $AC,$ which is equivalent to proving that $\angle AEC=\angle CAF.$

Note that $\angle EDA=\angle DFC$ since $\overline{AD} \parallel \overline{BF}$, and $\angle AED= \angle CDF$ since $\overline{EB} \parallel \overline{DC}.$ So $\triangle DAE \sim \triangle FCD$ by AA similarity, which implies that$$\frac{DA}{FC}=\frac{AE}{CD}.$$Let the side length of the rhombus be $s.$ Since $DA=CD=AC=s,$ the previous equations becomes$$\frac{s}{FC}=\frac{AE}{s} \implies \frac{AC}{FC}=\frac{AE}{AC}.$$However, $\angle EAC=\angle ACF,$ so $\triangle AEC \sim \triangle CAF$ by SAS similarity, which yields the desired $\angle AEC=\angle CAF.$

[b]Claim:[/b] Triangles $EAC$ and $AMC$ are similar.

[color=#f00][b]Proof:[/b][/color] Note that by proving this claim, we have completed the problem, since \[\frac{CM}{AC} = \frac{AC}{CE} \implies CM \cdot CE = AC^2.\] By construction, we know rhombus $ABCD$ is formed from joining equilateral triangles $\triangle ABC$ and $\triangle ACD.$ Furthermore $AD \parallel CF$ and $AE \parallel CD$ since $CF$ is an extension of side $BC$ and $AE$ is an extension of side $BA.$ This implies $\angle AED = \angle CDF$ and $\angle ADE = \angle CFD,$ so by AA similarity, $\triangle EAD \sim \triangle DCF.$ However, recall that $AC=AD=CD$ due to properties of an equialteral triangle, so $\tfrac{EA}{AC}=\tfrac{AC}{CF}.$ Furthermore, as $\angle EAC = \angle ACF = 120^\circ,$ it follows $\triangle CAE ~ \triangle FCA.$ From this similarity, we deduce $\angle AEC = \angle CAF.$

Once again we see another pair of similar triangles. Noting $\angle AEC = \angle CAF$ and triangles $EAC$ and $AMC$ both share angle $\angle ACM,$ we obtain $\triangle EAC ~ \triangle AMC,$ as desired.


Let $a,b$ be arbitrary.
Then $(a^2+b^2)^3=$
$= \| (a+bi)^3 \| = \| (a^3-3ab^2) + i (3a^2b-b^3) \| =$
$= (a^3-3ab^2)^2 + (3a^2b-b^3)^2$
(this can also be checked by manual computing).
Thus we have infinetely many solutions.

[quote="pinky"]Show that the equation X^2+Y^2=X^3 has infinitely many solutions for x,y,z positive integers.[/quote]

If you say integers x,y,z, so maybe equation should be $x^2 + y^2 = z^3$?

very interesting solution ZetaX,could that solution be possible without using complex numbers ?

Of course it could. Just omit these lines where they are used and check the corresponding identities by clearing the parenthesis  :lol:  Actually the solution is based on an identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ which arises when one figures out that norm $||\cdot ||$ is multiplicative in $\mathbb{Z}[i]$.


you already posted in another forum, please don't post one problem into multiple forums  :spam:
link is here
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=275519&highlight=#275519]http://www.mathlinks.ro/Forum/viewtopic.php?p=275519&highlight=#275519[/url]

[quote="Pork_Chop8"]Well, pal, since no one has yet solved it, I don't think there is anything wrong with trying to get someone to do it. I am trying to right now and if I do I will post it right there---->.[/quote]
please stop calling me "pal", whether you meant the first 3 letters of my login or the real word 'pal' (i dunno who you are so im not your pal)
and im advising ALGEBRA not to do it in the future because it is SPAM, you can do whatever you want, but SPAM is the issue here. of course he can get someone to do it, but not this way.

can you answer this problem "palytoxin" :D

i am stuck frm here onwards::

a^4 + b^4 + c^4 + d^4 + 2abcd > 0

thus, ( A + B + C + D) ^2 + 2abcd > 2(ÂB + CD + AC+ BC+ AD + BD)

caps rep ^2. which means A = a^2 B= b^2 etcetera (i lazy sry)

then i am stuck..


[quote="hxy09"]$ a,b,c,d\ge 0$
Prove that $ \sum a^4 \plus{} \sum abc(a \plus{} b \plus{} c)\ge 2\sum_{sym} a^2b^2 \plus{} 4abcd$[/quote]
[b]Problem. [/b]If $ a,b,c,d$ are nonnegative real numbers, then

$ \sum a^{4}\plus{}\sum abc(a\plus{}b\plus{}c)\geq 2\sum_{sym}a^{2}b^{2}\plus{}4abcd.$

[i][b]Proof. [/b][/i]Without loss of generality, assume that $ d\equal{}\min \{a,b,c,d\}.$ Using the fourth degree Schur's Inequality, we have

$ \begin{aligned} \sum_{a,b,c}a^{4}\plus{}abc(a\plus{}b\plus{}c) &\geq \sum_{a,b,c}ab(a^{2}\plus{}b^{2}) \\ &\equal{}2\sum_{a,b,c}a^{2}b^{2}\plus{}\sum_{a,b,c}ab(a\minus{}b)^{2} \\ &\geq 2\sum_{a,b,c}a^{2}b^{2}\plus{}d^{2}\sum_{a,b,c}(a\minus{}b)^{2}.\end{aligned}$

Therefore, it suffices to prove that

$ d^{2}\sum_{a,b,c}(a\minus{}b)^{2}\plus{}d^{4}\plus{}d\sum_{a,b,c}ab(a\plus{}b)\plus{}d^{2}\sum_{a,b,c}ab\geq 2d^{2}\sum_{a,b,c}a^{2}\plus{}4abcd,$

or

$ d^{3}\plus{}\sum_{a,b,c}ab(a\plus{}b)\geq d(ab\plus{}bc\plus{}ca)\plus{}4abc.$

By the AM-GM Inequality, we have

$ \sum_{a,b,c}ab(a\plus{}b)\geq 6abc,$

so it is enough to check that

$ d^{3}\plus{}2abc\minus{}d(ab\plus{}bc\plus{}ca)\geq 0,$

which is equivalent to the obvious inequality

$ d(b\minus{}d)(c\minus{}d)\plus{}c(a\minus{}d)(b\minus{}d)\plus{}b(a\minus{}d)(c\minus{}d)\geq 0.$

The proof is completed. On the assumption $ d\equal{}\min \{a,b,c,d\},$ equality holds for $ a\equal{}b\equal{}c\equal{}d,$ and again for $ a\equal{}b\equal{}c$ and $ d\equal{}0.$

See also here: http://canhang2007.wordpress.com/2009/11/26/inequality-83-unknown-author/

[quote="can_hang2007"][quote="hxy09"]$ a,b,c,d\ge 0$
Prove that $ \sum a^4 \plus{} \sum abc(a \plus{} b \plus{} c)\ge 2\sum_{sym} a^2b^2 \plus{} 4abcd$[/quote]
[b]Problem. [/b]If $ a,b,c,d$ are nonnegative real numbers, then

$ \sum a^{4} \plus{} \sum abc(a \plus{} b \plus{} c)\geq 2\sum_{sym}a^{2}b^{2} \plus{} 4abcd.$

[i][b]Proof. [/b][/i]Without loss of generality, assume that $ d \equal{} \min \{a,b,c,d\}.$ Using the fourth degree Schur's Inequality, we have

$ \begin{aligned} \sum_{a,b,c}a^{4} \plus{} abc(a \plus{} b \plus{} c) & \geq \sum_{a,b,c}ab(a^{2} \plus{} b^{2}) \\
& \equal{} 2\sum_{a,b,c}a^{2}b^{2} \plus{} \sum_{a,b,c}ab(a \minus{} b)^{2} \\
& \geq 2\sum_{a,b,c}a^{2}b^{2} \plus{} d^{2}\sum_{a,b,c}(a \minus{} b)^{2}.\end{aligned}$

Therefore, it suffices to prove that

$ d^{2}\sum_{a,b,c}(a \minus{} b)^{2} \plus{} d^{4} \plus{} d\sum_{a,b,c}ab(a \plus{} b) \plus{} d^{2}\sum_{a,b,c}ab\geq 2d^{2}\sum_{a,b,c}a^{2} \plus{} 4abcd,$

or

$ d^{3} \plus{} \sum_{a,b,c}ab(a \plus{} b)\geq d(ab \plus{} bc \plus{} ca) \plus{} 4abc.$

By the AM-GM Inequality, we have

$ \sum_{a,b,c}ab(a \plus{} b)\geq 6abc,$

so it is enough to check that

$ d^{3} \plus{} 2abc \minus{} d(ab \plus{} bc \plus{} ca)\geq 0,$

which is equivalent to the obvious inequality

$ d(b \minus{} d)(c \minus{} d) \plus{} c(a \minus{} d)(b \minus{} d) \plus{} b(a \minus{} d)(c \minus{} d)\geq 0.$

The proof is completed. On the assumption $ d \equal{} \min \{a,b,c,d\},$ equality holds for $ a \equal{} b \equal{} c \equal{} d,$ and again for $ a \equal{} b \equal{} c$ and $ d \equal{} 0.$

See also here: http://canhang2007.wordpress.com/2009/11/26/inequality-83-unknown-author/[/quote]
very nice  :D

How amazing your proof  :first:
And I will post an alternative proof next week :lol:

Hi canhang2007

The general case for this ineq is:$ x_1,...,x_n\ge0$ denote $ S_k \equal{} \sum x_{i_1}x_{i_2}...x_{i_k}$
Prove that:$ \sum( \minus{} 1)^kk^2S_kS_1^{n \minus{} k}\ge0$

It is called YangXuezhi inequality :D

Sorry the ineq should be $ \sum(\minus{}1)^{k\minus{}1}k^{2}S_{k}S_{1}^{n\minus{}k}\ge0$
Can anyone post a direct and thinkable proof? :) (The following one is very strange...)


I will look and I'll have a couple of badges and then later they are all gone.   Why is that?  :huh:

You will only be able to see badges from subjects in which you have passed or mastered. If you begin to answer problems incorrectly in a subject or related subject from which you have previously passed or mastered, or others start answering problems correctly that you previously answered incorrectly, it may be determined that you have no longer passed or mastered a subject. At that time you will no longer be able to see the badges. Once you have sufficiently answered problems in the subject and related subjects, you will again be flagged as "passed" or "mastered" in that subject and the badge will appear.


algebra
polynomial
complex analysis
linear algebra
linear algebra unsolved

Let $ A,B$ be $ n\times n$ Hermitian matrices and $ m\in \mathbb{N}$.
Show that the polynomial $ p(t) \equal{} \mathrm{Tr} \bigl( (A \plus{} tB)^m \bigr)$ has all real coefficients.

Correction; the questions are different !
Here it is not difficult.
Let $ t$ be a real.
$ p(t)\equal{}tr((A\plus{}tB)^m)\equal{}tr((A^T\plus{}tB^T)^m)$.
$ \bar{p(t)}\equal{}tr((A^*\plus{}tB^*)^m)\equal{}p(t)$ and $ p(t)$ is real.
$ p$ is a polynomial of degree $ \leq{m}$, thus a holomorphic function. Then the coefficients of the series giving $ p$ are real.

[quote="loup blanc"]Hi JC,
it is an open problem.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=189941[/quote]
Hi loup blanc,
Temptation was too high to get you back on the comment you made here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=270737
"before sending a post, please read and reread it."  :wink:


Find all positive integer solutions of the equation:

$\frac{1}{x}+\frac{2}{y}-\frac{3}{z}=1$

${1\over x}=1-{2z-3y\over yz}\iff x={yz\over yz-2z+3y}=1+{2z-3y\over yz-2z+3y}$

For $x$ to be natural, we must have either ${2z-3y\over yz-2z+3y}=0$ or ${2z-3y\over yz-2z+3y}\geqslant 1$. The first possibility yields $y=2k, z=3k, k\in\mathbb{N}$. The second one, assuming $y\neq 1$, gives

$2z-3y\geqslant yz-2z+3y$ [size=59][note: the denominator factors to (y-2)(z+3)+6 and is always positive if y is greater than one][/size] $\iff yz-4z+6y\leqslant 0\iff (y-4)(z+6)\leqslant-24$

Therefore $y\leqslant 3$. Casework.

1. $y=2\iff x=1+{2z-6\over 6}={z\over 3}$. Hence $x=k, z=3k,k\in\mathbb{N}$

2. $y=3\iff x=1+{2z-9\over z+9}=3-{27\over z+9}$. The only natural possibility for $z$ is $z=18\iff x=2$.

Now return to $y=1\iff{3\over z}-{1\over x}=1\iff (z-3)(x+1)=-3$, which yields one solution only and that is $x=2,y=2$

Therefore all the solutions are $(1,2k,3k),k\in\mathbb{N},(k,2,3k),k\in\mathbb{N},(2,1,2),(2,3,18)$

Similar to my solution. Casework for $x=1,2,3$. Then $x \geq 4$ yields $y=1$ or $2$


quadratics algebra function domain ratio AMC AIME

Find $A^2$, where $A$ is the sum of the absolute values of all roots of the following equation: \begin{eqnarray*}x &=& \sqrt{19} + \frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{x}}}}}}}}}\end{eqnarray*}

[hide="Solution"]
First note that if $x$ is a solution to the above equation, it must also satisfy the infinite equation  \begin{eqnarray*}x &=& \sqrt{19} + \frac{91}{\sqrt{19}+\frac{91}{\sqrt{19} + \cdots}}.\end{eqnarray*} In other words, \[ x = \sqrt{19} + \frac{91}{x}. \] Now we can solve this using the quadratic formula.

\begin{eqnarray*} x^2 - \sqrt{19}x-91 &=& 0\\ x &=& \frac{\sqrt{19}\pm \sqrt{19+4(91)}}{2}\\ x &=& \frac{\sqrt{19}\pm \sqrt{383}}{2}\\ |x_1|,|x_2| &=& \frac{\sqrt{19}+\sqrt{383}}{2}, \frac{\sqrt{383}-\sqrt{19}}{2}\\ A = |x_1|+|x_2| &=& \sqrt{383}\\ A^2 &=& \boxed {383}, \end{eqnarray*}

where $x_1$ and $x_2$ are the two roots of the equation.
[/hide]

Same solution here.  Seems like the easiest way to do it.

Slight derail:
I am not complaining about this post as such, but in these forums (Intermediate and Pre-Olympiad) I see "infinite" fractions and radicals thrown around and treated in a very superficial way. I don't know much about the two, and Mathworld doesn't say that much about the issue. "Infinite" fractions are defined as limits, right? Is it obvious that all such fractions then converge? I mean, the solution to the problem in this post simply assumes that the limit will exist. I see this all the time, and also on competitions there will be "infinite" expressions that aren't really talked about whether they converge or not, but just assumed to be "infinite" as if this is a straight-forward concept. What is the deal with this?

edit: To clarify my confusion: chess64 says that x is equal to some infinite fraction, and then finds that x can have two possible values. That doesn't really agree with the whole limit concept.

Well, from what I can see, one value of x is the limit you're talking about, while the other isn't.  For the other value of x, it seems that if you put it in, and then plug and chug the expression backwards, each thing under the 91 is the same, but negative.

It's like the idea of domain stretching.  One solution to this is $x=\sqrt{19}+\frac{91}{\sqrt{19}++\frac{91}{\sqrt{19}+...}}$, while the other is the sort of "negative complement".  True it's very weird...

I mean similarly, find the value of x in each of the following
\[ x^{x^{x^{x^{...}}}}=2  \]
\[ x^{x^{x^{x^{...}}}}=4  \]

In the first one, you get $x^2=2 \implies x=\pm\sqrt{2}$, while, by the same method, you get the same thing in the second one.  If you actually did it out, you'd get that it's 2, and I don't know if one can really do it for $-\sqrt{2}$.  Nevertheless, quite weird...domain stretching in action.  We similarly have \[ 1+x+x^2+x^3+...=n  \] \[ x+x^2+x^3+...=nx  \]
so \[ 1=n(1-x) \implies n=\frac{1}{1-x}  \] So what is it when $x \ge 1$?  Negative?  We know that $1+x+x^2+...+x^a=\frac{1-x^{a+1}}{1-x}$.  Another strange thing.
Domain stretching for things zeta function also makes things very strange.  People cheat infinity...

The fact is, in the original problem here, both x's satisfy the equation...We technically can't say that it's an infinite fraction, but that's just to make it easy...

[hide="Answer"]Factoring out the problem, we have $x^2-x\sqrt{19}-91=0$. By Pythagorean, the roots are $\frac{\sqrt{19}\pm \sqrt{383}}{2}$. Summing the absolute values of the roots, we have $\sqrt{383}=A$, so $A^2=\boxed{383}$.[/hide]

[quote="K81o7"]Well, from what I can see, one value of x is the limit you're talking about, while the other isn't.  For the other value of x, it seems that if you put it in, and then plug and chug the expression backwards, each thing under the 91 is the same, but negative.

It's like the idea of domain stretching.  One solution to this is $x=\sqrt{19}+\frac{91}{\sqrt{19}++\frac{91}{\sqrt{19}+...}}$, while the other is the sort of "negative complement".  True it's very weird...
[/quote]
What chess64 is actually doing is finding all fixed points of the recursively defined sequence which makes up the continuous fraction. However, I don't feel that the argument that those two fixed points are precisely the solutions to the original equation is a very convincing argument.

[quote="Kalle"]What chess64 is actually doing is finding all fixed points of the recursively defined sequence which makes up the continuous fraction. However, I don't feel that the argument that those two fixed points are precisely the solutions to the original equation is a very convincing argument.[/quote]

Well, the correct argument would be like this: If we simplify the continued fraction, the equation will have the form ${{ax+b}\over{cx+d}}=x$. Hence, it's quadratic and has exactly two solutions. If we now put $x=\sqrt{19}+{91\over x}$, using recursion we can easily prove that any root of this equation is also a root of the initial equation. Since the simple one has 2 roots, the complicated one must also have those same 2 roots, since it can't have more than 2 roots.

See the generalization of the fifth problem at
http://www.mathlinks.ro/Forum/viewtopic.php?t=70055&postorder=asc&start=40

[quote="JesusFreak197"]Factoring out the problem...[/quote]

:P

I'll redo the solution, paying attention to terminology this time. :roll:

[hide="Solution"]
Let \begin{eqnarray*}f(x) = x &=& \sqrt{19} + \frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{x}}}}}}}}}\end{eqnarray*} If $f(x) = x$, then $f(f(x))=x, f(f(f(x)))=x, f(f(f(f(x))))=x$, and it is easy to see that if $x$ is a solution to the equation $f(x)=0$, then it must also be a solution to the equation  \begin{eqnarray*}x &=& \sqrt{19} + \frac{91}{\sqrt{19}+\frac{91}{\sqrt{19} + \cdots}}.\end{eqnarray*} In other words, \[ x = \sqrt{19} + \frac{91}{x}.  \] Now we can solve this using the quadratic formula.

\begin{eqnarray*} x^2 - \sqrt{19}x-91 &=& 0\\ x &=& \frac{\sqrt{19}\pm \sqrt{19+4(91)}}{2}\\ x &=& \frac{\sqrt{19}\pm \sqrt{383}}{2}\\ |x_1|,|x_2| &=& \frac{\sqrt{19}+\sqrt{383}}{2}, \frac{\sqrt{383}-\sqrt{19}}{2}\\ A = |x_1|+|x_2| &=& \sqrt{383}\\ A^2 &=& \boxed {383}, \end{eqnarray*}

where $x_1$ and $x_2$ are the two roots of the equation.
[/hide]

Better?

That does not finish the problem.

$ f(x)\equal{}x\implies f(f(...f(x)...))\equal{}x$, but the converse is not always true.

However, this is true for $ f(x)\equal{}\frac{ax\plus{}b}{cx\plus{}d}$ [which can be proven easily...].

So then only one root can be accepted in general?

$ x \equal{} \sqrt {19} \plus{} \frac {91}{\sqrt {19} \plus{} \frac {91}{\sqrt {19} \plus{} ...}}$ is clearly non-negative, so wouldn't the negative root be considered extraneous (and thus cannot be considered as a solution)?

The fraction in the question isn't infinite.

OK but how would you prove that $ f^n(x)\equal{}x\implies f(x)\equal{}x$ iff $ f(x)\equal{}\frac{ax\plus{}b}{cx\plus{}d}$ where $ f^n$ is the nth iteration?

Neither implication holds in general; I'm not sure what proof Altheman had in mind.  In particular, there exist $ f$ of the given form such that $ f^n(x) \equal{} x$ holds identically; an obvious example is $ f(x) \equal{} \frac {1}{x}, n \equal{} 2$, but there are less trivial ones for larger $ n$.  You need to show that this particular $ f$ isn't one of them.  In the other direction, there exist functions with that property that aren't of the given form.

I had Farenhajt's post in mind. But you are right that other cases need to be considered.

In the case that $ f(x)\equal{}\frac{ax\plus{}b}{cx\plus{}d}$, then $ f^n(x)$ has $ 0,1,2$ or infinite solutions.

In some sense the case of $ 2$ solutions is the 'most common' in that when $ f^n(x)$ reduces, it is usually a ratio of linear polynomials.

I suppose that we should consider the $ n$th power of matricies in the form:

[a,b]
[c,d]

As another comment, the case of infinite solutions is trick because the function is undefined for $ x$ such that $ f^k(x)\equal{}\minus{}\frac{d}{c}$ (if $ c\ne 0$, $ 0\le k<n$)

In this light, I am now not sure that the following statement is true:
if $ S_n$ is the solution set to $ f^n(x)\equal{}x$, then $ S_a\subset S_{ka}$ ($ k$ an integer).

I was doing this AIME for practice and for whatever reason (this is my weakest subject) found this question to be one of the most difficult on this year's particular test.  I've read through the comments above but i can't quite grasp what they're trying to say, can anyone explain this to me as simply as possible?

Thanks

"Please explain the entire solution to me" is not a useful question.  Be precise about what parts you understand and what parts you don't understand, and then maybe someone can help you.  [Knowing how to ask useful questions is a really important skill that many people lack, despite the fact that it's often a prerequisite for getting useful answers.]

[quote="zhuzhu88"]

Why does vieta's not work here? SUm of roots is c/a in this case its -91. Square that get 8281, obviously incorrect.

thanks if you can answer my questions.[/quote]

The question asks for the sum of the absolute values, not merely the sum of the values.

[quote="zhuzhu88"][quote="chess64"]
First note that if $x$ is a solution to the above equation, it must also satisfy the infinite equation  \begin{eqnarray*}x &=& \sqrt{19} + \frac{91}{\sqrt{19}+\frac{91}{\sqrt{19} + \cdots}}.\end{eqnarray*} In other words, \[ x = \sqrt{19} + \frac{91}{x}. \] Now we can solve this using the quadratic formula.
[/quote]

Why is it if $x$ is a solution to the original equation, then it must satisfy the infinite equation as well. ? I understand all the math and everything else, just not why the roots are part of the infinite version.

[quote="chess64"]
Now we can solve this using the quadratic formula.

\begin{eqnarray*} x^2 - \sqrt{19}x-91 &=& 0\\ x &=& \frac{\sqrt{19}\pm \sqrt{19+4(91)}}{2}\\ x &=& \frac{\sqrt{19}\pm \sqrt{383}}{2}\\ |x_1|,|x_2| &=& \frac{\sqrt{19}+\sqrt{383}}{2}, \frac{\sqrt{383}-\sqrt{19}}{2}\\ A = |x_1|+|x_2| &=& \sqrt{383}\\ A^2 &=& \boxed {383}, \end{eqnarray*}

where $x_1$ and $x_2$ are the two roots of the equation.
[/quote]

Why does vieta's not work here? SUm of roots is c/a in this case its -91. Square that get 8281, obviously incorrect.

thanks if you can answer my questions.[/quote]

EDIT: Deleted an incorrect answer to the first question.

For your second question, first of all, Vieta's Formulas imply that the sum of the roots is $-\frac{b}{a}$, not $\frac{c}{a}.$ The latter expression is the product of the roots.

But the reason we cannot use Vieta here is because the problem asks for the [i]absolute values[/i] of the roots. Vieta's says nothing about the absolute values of the roots, just the roots themselves. Therefore, we have to compute the individual roots and take the absolute value of each one. Only then can we add them.

[quote="Dukejukem"]Therefore, we conclude that $x$ is equal to the infinite fraction.
[/quote]
$x$ does not [i]equal[/i] the infinite continued fraction. The continued fraction has only one value.

[quote="Dukejukem"]But the reason we cannot use Vieta here is because the problem asks for the [i]absolute values[/i] of the roots. Vieta's says nothing about the absolute values of the roots, just the roots themselves. Therefore, we have to compute the individual roots and take the absolute value of each one. Only then can we add them.[/quote]

We can still use Vieta's ;) [hide]If the roots are $a, b,$ then we want to find $|a| + |b|.$ But notice that we can eliminate the absolute values by squaring them: \[\begin{aligned} (|a|+|b|)^2 &= |a|^2 + |b|^2 + 2\cdot|a|\cdot|b| \\ &= a^2 + b^2 + 2|ab| \\ &= (a^2+2ab+b^2) - 2ab + 2|ab| \\ &= (a+b)^2 - 2ab + 2|ab| \\ &= (\sqrt{19})^2 - 2(-91) + 2|-91|  \\ &= \boxed{383}. \end{aligned}\][/hide]

Yeah, that's a nice way to employ Vieta's :). I just meant we can't [i]directly [/i] plug in the formula for sum of the roots here.

Plug in the value of $x$ given in the equation to the RHS repeatedly, which ends up giving an infinite equation. We therefore get $x=\sqrt{19}+\frac{91}{x}$, and the absolute values of the solutions are $\frac{1}{2}\sqrt{383} \pm \frac{1}{2}\sqrt{19}$, for a sum of $\sqrt{383} \implies \boxed{383}$.

Because this is a continuous fraction, just substitute $x$ for the value $\sqrt{19} + \frac{91}{\sqrt{19} + ...}$. This yields $x = \sqrt{19} + \frac{91}{x}$. Multiply both sides by $x$ gives $x^2 - \sqrt{19}x - 91 = 0$ and use quadratic formula to get the two roots $x_1$ and $x_2$, which are $\frac{\sqrt{19} + \sqrt{383}}{2}$ and $\frac{-\sqrt{19} + \sqrt{383}}{2}$, respectively. Finding the sum $A$ of these two roots is $\sqrt{383}$. Therefore, $A^2 = 383$.


limit
integration
function
calculus
trigonometry
calculus computations

Here are three nice limits.

Find the limits:

1) $L=\lim_{n \rightarrow \infty}\int_{0}^{\infty}e^{-x^{n}}dx$.

2) $\lim_{n\rightarrow \infty}n\left(L-\int_{0}^{\infty}e^{-x^{n}}dx\right)$, the value of this limit is a special number.

3) $\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{1+x^{n}}{1+x^{2n}}dx$.

Hmmm...number 1 looks vaguely like the gamma function, probably an easier way though.

yes, in general we have
$\int_{0}^{\infty}x^{m}e^{-ax^{n}}dx=\frac{1}{n}a^{-(m+1)/n}\Gamma(\frac{m+1}{n})$.

1) By previous post, we have for $m=0$ and $a=1$ that
$\int_{0}^{\infty}e^{-x^{n}}dx=\frac{1}{n}\Gamma(\frac{1}{n})=\Gamma(1+\frac{1}{n})\to 1$.

2) $\lim n(1-\Gamma(1+\frac{1}{n}))=\lim\frac{(1-\Gamma(1+\frac{1}{n}))}{1/n}=\gamma$
where $\gamma$ is Euler's costant.
This follows by the limit
$\lim_{x\to 0}\frac{(1-\Gamma(1+x))}{x}=\lim_{x\to 0}\frac{-\Gamma^{'}(1+x))}{1}=\gamma$.

To get
$\int_{0}^{\infty}e^{-x^{n}}dx=\frac{1}{n}\Gamma(\frac{1}{n})$
we can change variable $x = t^{n}$ and then put $a=\frac{1}{n}$ in
$\Gamma(a) = \int_{0}^{\infty}x^{a-1}e^{-x}dx$

In the third exercise I have used residue theorem, and you didilica what have you used?

There's such way to evaluate 3rd integral ($t=x^{2n}$)
\[\int_{0}^{\infty}\frac{1+x^{n}}{1+x^{2n}}\,dx=\frac1{2n}\int_{0}^{\infty}\frac{1+\sqrt t}{1+t}\,t^{\tfrac1{2n}-1}dt=\frac1{2n}(\frac{\pi}{\sin\frac{\pi}{2n}}+\frac{\pi}{\cos\frac{\pi}{2n}})\]
because
\[\int_{0}^{\infty}\frac{x^{s-1}}{1+x}\,dx=B(s,1-s)=\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}\]
for $\Re s\in(0;1)$

Looking at (3) in a completely different way:

$\lim_{n\to\infty}\frac{1+x^{n}}{1+x^{2n}}= \begin{cases}1,&0\le x\le 1\\0,&x>1\end{cases}.$

Furthermore, for $n\ge 2,$ we have that this function is no bigger than $2$ on $[0,1]$ and $\frac2{x^{2}}$ on $[1,\infty).$ Since that bound is an integrable function, we may invoke the dominated convergence theorem to say that

$\lim_{n\to\infty}\int_{0}^{\infty}\frac{1+x^{n}}{1+x^{2n}},dx = \int_{0}^{\infty}\lim_{n\to\infty}\frac{1+x^{n}}{1+x^{2n}}\,dx$

$=\int_{0}^{1}1\,dx=1.$


calculus
integration
function
algebra
rational function
calculus computations

[quote="Abhishek"]Find the condition on a & b such that the integral is a rational function

$\int\frac{x^{2}-1}{(x-a)^{3}(x-b)^{2}}\,dx$[/quote]

If $a=b,$ the integral is always a rational function, if $a\neq b,$ then the condition $a^{2}+4ab+b^{2}=6$ is neccessary and sufficient for the integral to be a rational function.

  --Ben


Show that $X^{1/2} + Y^{1/2} = A^{1/2}$

  For some constant $A$.

I don't understand your question. Are they all integers? Is A a certain constant or any constant? Please elaborate.


parameterization
algebra proposed
algebra

find all f:$ \mathbb{R}$ --> $ \mathbb{R}$ ,,all continious fructions $ f(0)\equal{}0$, $ f(xcosB)\minus{}f(x(cosB)^2)\equal{}x\minus{}x^2cosB$ ,,for avery $ x\in \mathbb{R}$ and $ B \in \mathbb{R} \minus{} \{ B: B \equal{} k \pi ,,or,, B \equal{} k \pi \plus{} \frac {\pi}{2} \ , k \in \mathbb{Z}\}$

[quote="prester"]double post...however in this section you may post your own solution...[/quote] yeah , i put them here beacause i solved it so
$ f(xcosB)\minus{}f(x(cosB)^2)\equal{}x\minus{}x^2cosB ,$ so $ f(x(cosB)^2)\minus{}f(x(cosB)^3)\equal{}xcosB\minus{}x^2(cosB)^3,$ so $ f(x(cosB)^3)\minus{}f(x(cosB)^4)\equal{}x(cosB)^2\minus{}x^2(cosB)^5,
......................................................................
f(x(cosB)^n\minus{}1)\minus{}f(x(cosB)^n)\equal{}x(cosB)^n\minus{}2\minus{}x^2(cosB)^{2n\minus{}3}$
and if we Sum it we take ,$ f(xcosB)\minus{}f(x(cosB)^n)\equal{}x \frac {(cosB)^{n\minus{}1}\minus{}1}{cosB\minus{}1}\minus{}x^2cosB \frac{(cosB)^{2n\minus{}2}\minus{}1}{(cosB)^2\minus{}1}$ beacause f is continious if we take the lim for n->+oo we take :
$ f(xcosB)\equal{} \frac {x}{1\minus{}cosB}\minus{}\frac{x^2cosB}{1\minus{}(cosB)^2}$ and beacause $ 0<|cosB|<1 \equal{}> \liminf(cosB)^n\equal{}0$ so if we put $ x \equal{} \frac {x}{cosB}$ we take $ f(x)\equal{} \frac{x}{(1\minus{}cosB)cosB}\minus{}\frac{x^2}{(sinB)^2cosB}$  :maybe:

[quote="pco"][quote="prester"]double post...however in this section you may post your own solution...[/quote]

No, it's not a double post.

The other post was : "find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u]"

So two changes :

1) Now, $ f(x)$ is continuous
2) the equation is no longer true $ \forall B$ but only for a given parameter $ B$

And now the problem has a meaning.

So KapioPulsar made an error in the other post and lies when he says he copied exactly the original problem. A polite attitude would have been to apologize, but ....[/quote]

Cconcerning (1) you're right, but this add a constraint and so it excludes some other solutions...
Concerning (2), also the previous post was for a given parameter but it was not clear for me and mathias_dk, I think...maybe the statement description could be more explicative...I wish to know the original statement
Anyway, this problem has solution as shown the author before...thanks

[quote="pco"][quote="prester"]double post...however in this section you may post your own solution...[/quote]

No, it's not a double post.

The other post was : "find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u]"

So two changes :

1) Now, $ f(x)$ is continuous
2) the equation is no longer true $ \forall B$ but only for a given parameter $ B$

And now the problem has a meaning.

So KapioPulsar made an error in the other post and lies when he says he copied exactly the original problem. A polite attitude would have been to apologize, but ....[/quote] sorry i apologise  :(  :blush:  ( the original statment of this problem is exactly is i post it in this version :S )


geometry
trigonometry
ratio
angle bisector
area of a triangle
Heron\u0027s formula
geometry proposed

this is problem createdby me yesterday.have a look at it.

In a triangle $ABC$, the angle bisector of the angle $CAB$ meets the side $BC$ at $D$. Let $O$ be a point on the segment $AD$. The lines $CO$ and $BO$ meet the sides $AB$ and $AC$ at two points $P$ and $Q$, respectively. The line $PQ$ intersects the line $AD$ at a point $E$. If $R_{1}$ and $R$ are the circumradii of triangles $APQ$ and $ABC$, and $AB\neq AC$, then prove that

$\frac{R_{1}}{R}=\frac{AD}{AE}\cdot\frac{PE-QE}{BD-DC}$.

Denote a = BC, b = AC, c = AB, d = AD the sides and angle bisector of the triangle $\triangle ABC$ and o = PQ, p = AQ, q = AP, e = AE the sides and the angle bisector of the triangle $\triangle APQ$. Let $s = \frac 1 2 (a + b + c),\ z = \frac 1 2 (o + p + q)$ be the semiperimeters and $\triangle_{ABC},\ \triangle_{APQ}$ the areas of these 2 triangles.

$\frac{AD}{AE} \cdot \frac{PE - QE}{CD - BD} = \frac d e \cdot \frac{PE}{CD} \cdot \frac{1 - \frac{QE}{PE}}{1 - \frac{BD}{CD}} = \frac{d}{e} \cdot \frac{oq}{p + q} \cdot \frac{b + c}{ab} \cdot \frac{1 - \frac p q}{1 - \frac c b} =$

(using the angle bisector formula $d = \sqrt{bc \left(1 - \frac{a^2}{(b + c)^2}\right)}$ and a similar one for the angle bisector e)

$= \frac{d}{e} \cdot \frac o a \cdot \frac{q - p}{q + p} \cdot \frac{b + c}{b - c} = \sqrt{\frac{bc}{pq} \cdot \frac{1 - \frac{a^2}{(b+c)^2}}{1 - \frac{o^2}{(p + q)^2}}} \cdot \frac o a \cdot \frac{q - p}{q + p} \cdot \frac{b + c}{b - c} =$

$= \sqrt{\frac{bc}{pq} \cdot \frac{(b + c)^2 - a^2}{(p + q)^2 - o^2}} \cdot \frac o a \cdot \frac{q - p}{b - c} = \sqrt{\frac{bc}{pq} \cdot \frac{s(s - a)}{z(z - o)}} \cdot \frac o a \cdot \frac{q - p}{b - c} =$

(using Heron's formula for the triangle areas $\triangle_{ABC},\ \triangle_{APQ}$)

$= \frac{\triangle_{ABC}}{\triangle_{APQ}} \cdot \sqrt{\frac{bc}{pq} \cdot \frac{(z - p)(z - q)}{(s - b)(s - c)}} \cdot \frac o a \cdot \frac{q - p}{b - c} =$

(using the triangle area formula $\triangle_{ABC} = \frac{abc}{4R_{ABC}}$ and a similar one for the area $\triangle_{APQ}$)

$= \frac{4R_{APQ}}{4R_{ABC}} \cdot \frac{abc}{opq} \cdot \sqrt{\frac{bc}{pq} \cdot \frac{(z - p)(z - q)}{(s - b)(s - c)}} \cdot \frac o a \cdot \frac{q - p}{b - c} = $

$= \frac{R_{APQ}}{R_{ABC}} \cdot \frac{bc}{pq} \cdot \sqrt{\frac{bc}{pq} \cdot \frac{(z - p)(z - q)}{(s - b)(s - c)}} \cdot \frac{q - p}{b - c}$

Using Ceva's theorem for the triangle $\triangle APQ$ with the concurent cevians AE, PB, QC,

$\frac{EP}{EQ} \cdot \frac{BQ}{BA} \cdot \frac{CA}{CP} = \frac q p \cdot \frac{p - c}{c} \cdot \frac{b}{q - b} = 1$

$pqb - bcq = pqc - bcp$

$pqb - pqc = bcq - bcp$

$pq(b - c) = bc(q - p)$

$\frac{bc}{pq} \cdot \frac{q - p}{b - c} = 1$

We could use Ceva's theorem for the triangle $\triangle ABC$ with the concurent cevians AD, BP, CQ to obtain the same result. Hence, it remains to be shown that

$(?)\ \ \ \frac{bc}{pq} \cdot \frac{(z - p)(z - q)}{(s - b)(s - c)} = 1$

$(?)\ \ \ \frac{(s - b)(s - c)}{bc} = \frac{(z - p)(z - q)}{pq}$

Denote $\alpha = \angle CAB = \angle PAQ$. The triangle areas are equal to $\triangle_{ABC} = \frac{bc}{2} \sin \alpha,\ \triangle_{APQ} = \frac{pq}{2} \sin \alpha$ and their ratio is $\frac{\triangle_{ABC}}{\triangle_{APQ}} = \frac{bc}{pq}$. Substituting this and using Heron's formula again,

$(?)\ \ \ \frac{\triangle_{ABC}^2}{s(s - a) \triangle_{ABC}} = \frac{\triangle_{APQ}^2}{z(z - o) \triangle_{APQ}}$

$(?)\ \ \ \frac{\triangle_{ABC}}{s(s - a)} = \frac{\triangle_{APQ}}{z(z - o)}$

Using the formulas $r_{ABC} = \frac{\triangle_{ABC}}{s},\ r_{APQ} = \frac{\triangle_{APQ}}{z}$ for the inradii $r_{ABC},\ r_{APQ}$ of the triangles $\triangle ABC, \triangle APQ$,

$\frac{r_{ABC}}{s - a} = \tan \frac \alpha 2 = \frac{r_{APQ}}{z - o}$

Since the last equation is an obvious identity, we are done. Not bad.

Yetti


function
algebra
domain
topology
geometry
complex analysis
complex analysis theorems

Let f be an analytic function in domain D = C\{z}, for some point z in C.

Prove the following assertion:

"If f(D) is a subset of C\[1,inf) then f is constant."

Without loss of generality $f$ is analytic in $\mathbb C\setminus\{0\}$. Let $g(z)=f(e^{z})$. The function $g$ is entire and $g(\mathbb C)\subset \Omega$, where $\Omega= \mathbb C\setminus [1,+\infty)$. Now you can either invoke Picard's theorem or map $\Omega$ onto a bounded domain and use Liouville's theorem.

Prove the same assertion if z is now a discrete sequence {z_1, z_2, ...} in C.

Let's simplify the problem by replacing $f$ with $\varphi\circ f$, where $\varphi$ maps $\mathbb C\setminus [1,\infty)$ onto the unit disk. You want to know for which domains $D\subset\mathbb C$ there is a nonconstant bounded analytic function defined on $D$. The [non-]answer is: this happens if and only if $\mathbb C\setminus D$ has positive analytic capacity. (Pretty much by the [url=http://en.wikipedia.org/wiki/Analytic_capacity]definition[/url]).

Sets of Hausdorff dimension less than $1$ have zero analytic capacity; sets of Hausdorff dimension greater than $1$ have positive analytic capacity... and a lot of stuff happens in between. Since major developments in this direction occurred recently (a lot of which had to do with [url=http://mat.uab.es/~xtolsa/xtolsa.html]Xavier Tolsa[/url]), there seems to be no up-to-date reference on this topic. Mattila's book on geometry of measures and [url=http://www.math.u-psud.fr/~biblio/pub/1998/abs/ppo1998_90.html]this survey[/url] describe what was known in the 1990s.

If the complement of $D$ is discrete, then any bounded analytic function on $D$ extends to $\mathbb C$ (points are removable singularities for bounded analytic functions), and therefore is constant.


limit
trigonometry
calculus
calculus computations

A very easy problem - here's a solution.


inequalities
limit
number theory solved
number theory

Let $a$ be a positive real number and $(x_n)_{n\geq 1}$ be a sequence of pair-wise distinct natural numbers such that $x_n\leq a\cdot n$. Prove that:

 a) if $a<5$ then there exist infinitely many numbers $k$ such that the sum of digits of number $x_k$ is divisible by 5;

 b) if $a=5$ the statement from a) is no longer true.

-- this problem was used to choose ROM6 between 3 contestants, ranked 6,7 and 8.

Hey, hey, this is my problem! Well, a variant of it, since the problem I proposed was to show that for infinitely many n the sum of digits of $ x_n$ is not a multiple of 5. This problem was also discused in the jury of the Balkan Mathematical olympiad. It's very nice. But I will have to think at this problem, which, sincerely, looks more difficult than my initial problem.

Well, are you sure that a) was given like this in the contest? I made a program to show the sequence of numbers with sum of digits not a multiple of 5. And the program shows that this sequence verifies $ x_n<4n $ and it doesn't contain infinitely many numbers with sum of digits multiple of 5. So, maybe it was my initial problem: sum of digits not a multiple of 5. And I have to say that i couldn't solve b) in this case and I let it as a conjecture. It seems that our teachers have solved it and gave it. And if this was the easy problem, what was the hard one?

my mistake; you are right harazi, i didn't even think about the problem, just put it up here (i heard about it yesterday and i missheard it probably). So the problem must be as you say:

Let $a$ be a positive real number and $(x_n)_{n\geq 1}$ be a sequence of pair-wise distinct natural numbers such that $x_n\leq a\cdot n$. Prove that:

a) if $a<5$ then there exist infinitely many numbers $k$ such that the sum of digits of number $x_k$ is [b]NOT[/b] divisible by 5;

b) if $a=5$ the statement from a) is no longer true (so there exist a sequence with only finitely many $x_k$ whose sum of digits are [b]not[/b] divisible by 5).

sorry again;

Maybe it's an idiot problem, with no idea, or maybe it's a trivial one, I dont know. But could anyone please post a solution for b)? I really don't find any.

consider in the sequence of the numbers which have their digit sum divisible by 5. Prove that this sequence fulfills the property in the statement. :)

Well, thanks a lot, I knew that. I made a Pascal program which shows that this sequence doesn't verify $ x_n<5n $ for all n. It seems that $\frac{x_n}{n} $ tends to 5 as n tends to infinity (in fact that was my conjecture), but very very slow. So, I'm still waiting.

Gosh, I find point b) much easier than point a), how can it be?!?

Consider the sequence starting from $0$, and containing all the numbers whose digit sum is divisible by $5$. We have $x_1=0<5$, $x_2=5<10$, $x_3=14<15$. Then we notice that, among the $10$ consecutive integers of the form $d_1 d_2 \ldots d_k$, with $d_k$ ranging from $0$ to $9$, exactly two fall into our sequence. Hence, we have $x_{2n}< 10n=5(2n)$, and $x_{2n-1}=x_{2n}-5<10n-5=5(2n-1)$.

Well, I know this example for b) and it's not that hard. But initially I asked for sequences that contain only natural numbers, not 0. This is the problem I had in mind. But I guess they gave the problem with non-negative integers. In this case, yes, a) is much harder than b). Or am I missing something? I don't know, I behave strangely when I have idiot exams.

Oh, I've always thought that 0 was a natural number. However, if you don't want 0, you may take 1 in its place, then go on as before. Indeed, 1<5.  :cool:

Ok, it's clear, I should rest a little bit before saying stupidities. Sorry, I'm an idiot today and always.

[quote="harazi"]
Sorry, I'm an idiot today and always.
[/quote]
 :?       :?       :?

Come on, my friend!! Everyone has some bad days!   :lol:

Now point a) isn't that hard, either.

Let $(y_n)$ be the sequence of the non-negative integers whose digit sum is divisible by $5$, with $y_1=0$. From the solution of point b) above, we derive $y_n \geq 5(n-1)$. Now, let $x_n \leq a \cdot n$, with $a<5$, and suppose $(x_n)$ has just a finite number of elements not in $(y_n)$. Then, there exists $j$ such that $x_{j+i} \geq y_i$ for every non-negative integer $i$. In particular, taking $i > \frac{a \cdot j + 5}{5-a}$, we have $x_{j+i} \leq a(j+i) < 5(i-1) \leq y_i \leq x_{j+i}$, a contradiction.

Wow! This is a hell of a solution! It's simply magnificient. It seems that I have nothing to do with proposing hard problems, but rather easy ones, tough I have long and difficult solutions to them. Anyway, I hope it wasn't very trivial. I must be quite idiot since my solution is long and uses complex numbers. But yours is so nice. I must have been blind when sending this problem for the TST!

Many thanks!!   :D
But I've implicitly made the assumption that $(x_n)$ was monotone... just a few more words are needed to fix my proof, however.

I saw that you used the monotony, but I realizd the proof can be adapted as follows:
  Put $ x_n=y_{k_n} $ for $n>M $. Then we have $ k_n\leq 1+\frac{an}{5} $ for $ n>M $. But $ k_n $ is injective and the inequality cannot be true for all $ n>M $.

Right. That's what I meant for "a few more words"...  :lol:

wow, MindFlyer, you solution is really short and to-the-point! It's great.

still i want to put up another solution for a), to explain why i named the topic "density":

we compute the "density" of the $y_k$'s over $\mathbb N$, that is: $\rho=\lim_{n\to\infty}\frac{\#\{y_k\ |\ y_k\leq n\}}n\leq 1$.
we count the number of $y_k$'s with $t$ digits: choose the first digit in $9$ ways, the rest except the last in $10$ ways and the last in $2$ ways so $18\cdot 10^{t-1}$ ways.  Summing after $t$ and doing the math (is it enough to evaluate the limit in the points $10^k$?...) we get $\rho=\frac 15$ !!!

 suppose that from a rank on every $x_k$ is a $y_{i_k}$ but then we will (eventualy) get $\rho\geq\frac 1a>\frac 15$, contradiction...

I must say that this doesn't look like a proof, does it? but i hope you get the idea...

Oh harazi, i don't know if $\frac{y_n}n$ tends to $\frac 15$, but this density stuff is similar, isn't it?


search function articles AMC USA(J)MO USAMO AIME

I'm currently a freshmen taking Algebra 2 and Geometry. I discovered the AMC contests online, and I talked to my teacher about taking them next year. I do very well in my math classes. I want to take the AMC 10, then go as high as I can if I do well on it.

But I'm not quite sure how to prepare. I've seen two sets of recommendations [url]http://www.unl.edu/amc/a-activities/a4-for-students/reading-list.shtml[/url] and [url]http://www.artofproblemsolving.com/Wiki/index.php/How_should_I_prepare%3F[/url]. I don't understand the first one's Background reading on particular topics. Do I need to know about each of these topics, and are those the books I should use?

Could someone piece together these recommendations for me?

First off, the search function is your greatest friend with these kinds of questions...
OK, the 1st article (from unl.edu..) is talking about preparation for Olympiads for the USAMO and the IMO. I wouldn't worry about that too much.
The second one is the one you should be looking at..
a very rough sequence of books is aops1, aops2, acops, pss

Thanks. That answer my question quite well. And I didn't search because I was asking about two specific webpages, not general recommendations.

ACoPS = Art and Craft of Problem Solving by Paul Zeitz.

PSS = Problem Solving Strategies by Arthur Engel.

what level is Acops for (AIME,AMC, or USAMO).
I no PSS has IMO problems

ACoPS is mainly for people learning proofs, and PSS is for Olympiad prep.

Can the AaCoPS book help me ace the AMC 10&12 tests?

If you're thorough with the AoPS Volume 1 and 2, you should be absolutely fine on the AMC's.

eh... I read AoPS 1 and got much better... I'm saving AoPS 2 for later.


For each positive integer n, let S_n denote the total number of squares in an n×n square grid. Thus S_1 = 1 and S_2 = 5, because a 2×2 square grid has four 1×1 squares and one 2×2 square. Find a recurrence relation for S_n, and use it to calculate the total number of squares on a chess board

[hide="hint"]Use the recurrence by finding the number of n x n squares on an (n+1) x (n+1) board and then subtract out the number of shared (n-1) x (n-1) squares.  Or just explicitly calculate the number of k x k squares in an n x n board.[/hide]

A more interesting version of the problem was posted a while back:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155463]Here[/url]


Hello. Fu Yunhao has already had two perfect scores 2002 and 2003 at IMO. Will he be returning next year in 2004 IMO, Greece. If he manages
 another perfect score , Ciprian would not be only one to hold the record of three perfect scores at IMOs.
                                                  THANKS

the difference is that Ciprian managed to be the only one to score maximum of points once, and he also went 1st, 2nd with the Romanian Team. I belive that he is still much more better than any other IMO contestant (up untill now).

But then of course, I'm Romanian, therefore I'm subjective. :)

hey! no doubt ciprian is the best.no1 comes close ( i dont know why i like ciprian !! :D mayb bcoz he won the perfect score when imo was in india in 1996 and i was in so much in awe of him then :D

Fu Yunhao is not bad. He solved all four of the very difficult IMO problems: 2002/3, 2002/6, 2003/3, 2003/6 and get perfect scores in IMO 2002 and 2003. More than 400 contestansts got 0 point in both problem 2002/6 and 2003/3 and more than 400 contestants go 0 or 1 point in both problem 2002/3 and 2003/6. It's more difficult to get perfect scores in IMO 2002 and 2003 than IMO 1995 and 1997.

I think Fu Yunhao is as great as Ciprian because he got perfect scores in two very difficult IMO, 2002 and 2003.

nobody said Fu Yanhao isn't a great problem solver, but rather that the difference between the difficulty level of the problems 7-8 years ago is lower than this of our days. Also in 1997 Ciprian was the only one to score maximum points, which is in a sort of a contradiction with your saying that 1997 was an easy IMO.

Surely, the 10th IMO 1969 was a hard IMO (I know that because Prof. Radu Gologan, the Romanian IMO Team Leader was a student in the Romanian Team Back then) for the contestants, but to us now it seems extremely easy.

In plus, these days we have a huge amount of materials accesible than back then (at least in Romania there was a deficit of problem solving math books - in fact there still is one deficit, although I have wrote a book on that subject, which has proven to be a best - seller in the math series :D - enough braggin' )

I surely belive that each IMO medalist is a very good mathematician, and most probably, if he, unlike Freddy :), studies further mathematics in college, will have important articles/reasearch in maths.

Long talk, lot's of time wasted :)

so i hear yunhao fu went back and won another perfect score

http://users.pandora.be/frederic_bel6/china.jpg

this is pic on my site, and I know the guys right next to me are the perfect scorers ( no I didn't arrange them like that, it just turned out that way)

i've always wondered which one was yunhao fu en which one was botong wang? :?  :D

Hey Fredbel6 - Frdric,
Yunhao Fu (Fu Yunhao ?) is the guy in the middle of the picture.


algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Find all possible classes of isomorphism of Galois groups of polynomials $X^{4}-a$ over a field $F$, where $a\in F$ and $F$ vary.

You can write down the Galois group of $X^{n}-a$ (if separable) as a subgroup of the upper triangular $2\times 2$ matrices with entries in $\mathbb{Z}/n\mathbb{Z}$---at least if $n$ is prime. I haven't thought about it in general.


[quote="sstutter"]What did 50 Cent say when Eminem gave him a sweater?[/quote]

er, gee u knit?

question:
What is long and thin and goes "Hith Hith"?

[hide="punchline:"]A snake with a lisp!  :rotfl:  :D [/hide]


calculus
integration
trigonometry
complex numbers
complex analysis
complex analysis unsolved

Use residues to evaluate the improper integral.
$ \int^{\infty}_{\minus{}\infty} \frac{\cos x dx}{(x\plus{}a)^2 \plus{} b^2}$ ($ b>0$).

I do not see how to evaluate this integral now. In this section, we use $ M_R$ and $ C_R$. I do not see how to use this here. I need hints on how to do this. Thanks in advance.

$ \cos x$ goes to $ \infty$ at both $ i\infty$ and $ \minus{} i\infty$, so we don't want to build it into a big loop. How do we fix it? $ \cos x$ is the real part of $ e^{ix}$. Look at the integral of $ \frac {e^{ix}}{(x \plus{} a)^2 \plus{} b^2}$ around an appropriate contour. If necessary, consider the real part.

I hope by answering this  I can check out whether  I am right. Sorry for canberra1454 , I am working out on my own. If you are trying on your own . Avoid my post.. :)

Choose $ C$ to be the regular contour as we choose for calculation of real definite integrals .

$ C$ is the  semicircle of radius $ R$ in the upper half plane big enough to enclose the zeroes of the denominator.

Assume $ f(z) \equal{} \int_{C}\frac {e^{iz} dz}{(z \plus{} a)^{2} \plus{} b^{2}}$

Thus $ \int_{C}f(z).dz \equal{} \int_{ \minus{} R}^{R}f(x).dx \plus{} \int_{\gamma}f(z).dz$

$ \minus{} a \plus{} ib$ and $ \minus{} a \minus{} ib$ form the simple poles. $ \minus{} a \plus{} ib$ lies inside the assumed contour, hence

evaluation of residues at this pole gives $ \frac {e^{i( \minus{} a \plus{} ib)}}{2ib}$

Thus by Cauchy's Residue Theorem we get,

 $ \int_{C}f(z).dz \equal{} 2 \pi i \frac {e^{i( \minus{} a \plus{} ib)}}{2ib} \equal{} \frac {\pi e^{i( \minus{} a \plus{} ib)}}{b}$

As $ R \to \infty$ by Jordan's Lemma $ \int_{\gamma}f(z).dz \to 0$

Thus this leads us to $ \int_{C}f(z).dz \equal{} \int_{ \minus{} \infty}^{\infty}f(x).dx$ , thus

Equating the real part we get $ \int^{\infty}_{ \minus{} \infty}\frac {\cos x dx}{(x \plus{} a)^{2} \plus{} b^{2}} \equal{} \frac {\pi e^{ \minus{} b}cos(a)}{b}$ :maybe: