function
logarithms
algebra unsolved
algebra

Do there exist function f:R-->R and g:R-->R such that
$ f(g(x))=x^{2}$ and $ g(f(x))=x^{4}$ for all $ x\in R$

[quote="boxedexe"]Notice that $ f(x^{4})=f[g(f(x))]=[f(x)]^{2}$ and, similarly, $ g(x^{2})=[g(x)]^{4}$.  Set $ f(x)=a^{P(\log_{b}x)}$ and $ g(x)=u^{Q(\log_{v}x)}$.  It follows that $ f(x)=a^{\sqrt{\log_{b}x}}$ and $ g(x)=u^{(\log_{v}x)^{2}}$ are possible solutions.  It is now easy to find suitable values for $ a,b,u,v$ for a solution.[/quote]

Be careful with the different situations ($ x$ versus $ -b,-v,0,v,b$)


superior algebra
superior algebra unsolved

Hello,

Can any one tell me a good book or website to study how to determine the character table of $ S_6$?

I use Dummit-Foote as my algebra book but it doesn't show how to determine the character table of $ S_6.$

I heard that the table will includes representations such as exterior powers $ \Lambda^k V$ and symmetric powers $ \text{Sym}^k V$ of known representations $ V$, as well as tensor products $ V \otimes W$ of a pair of known representations.

If anyone knows a good resource for this one, I would appreciate it!

Some very good sources:

1. I. M. Isaacs, Character Theory of Finite Groups

2. Fulton & Harris: Representation Theory

3. J. P. Serre: Linear Representations of Finite Groups

Ref 1 focuses on characters, and it provides numerous methods and tricks to identify irreducible characters. It works out the case of S_4 and although S_6 is certainly more complicated you may be able to get it if you're patient enough.

Ref 2 should be helpful as well but I don't remember much about it; I do remember it spends some time on symmetric and exterior powers of representations and how they decompose into irreducible ones.

Hope this helps.

Ref 3, if I remember correctly, has a part written by Serre's wife which focuses on applications for chemistry, and as such I think it has a lot of practical information on character table calculation for small groups. I don't know if it covers S_6 specifically.


Anyone know how much they cost?

EDIT: If you want to more about this rare and fantastic instrument, go to http://www.finkenbeiner.com/GLASSHARMONICA.htm and look around a bit. They have some nice pictures and sound samples, as well as articles on history and information about Glass Harmonicas.

That would suck if you dropped yours :P .

the cost is the number of pitches u need and multiply that number by the cost of a wine glass. (Assuming that you are not buying the water aswell)

Mozart used this instrument in one of his orchestral pieces or was it a concerto I believe :D That Mozart!  good luck with it.

Forgive me if I'm being dense, but didn't Benjamin Franklin make the glass harmonica?


AMC AIME geometry circumcircle trigonometry

ABC is a scalene triangle. The circle with diameter AB intersects BC at D, and E is the
foot of the altitude from C. P is the intersection of AD and CE. Given that AP = 136,
BP = 80, and CP = 26, determine the circumradius of ABC.

What is the solution to this problem?

let $r$ denote the circumradius. label the circumcenter of the circle $O$, and label the midpoints of $BC$, $CA$, $AB$ as $T$, $U$, $V$, respectively. note that $P$ is the orthocenter of the triangle, and so as one of the results of the nine-point-circle theorem, $OT=AP/2=136/2=68$, and similarly $OU=40$, $OV=13$. denote by $\alpha$, $\beta$, $\gamma$ the measures of the angles of the triangle at $A$, $B$, $C$ respectively. then

$\cos \alpha = \frac{68}r$

$\cos \beta = \frac{40}r$

$\cos \gamma = \frac{13}r$

from the fact that $\alpha+\beta+\gamma=\pi$, we have $\cos (\alpha+\beta) =-\cos \gamma$, so

$\frac{68}r\frac{40}r-\sqrt{\frac{r^{2}-68^{2}}{r^{2}}}\sqrt{\frac{r^{2}-40^{2}}{r^{2}}}=-\frac{13}r$.

now solve this equation.


superior algebra
superior algebra solved

List all elements of order 8 in Z_8000000. ( Z subscript 8000000)

How do u know your list is complete?

$k.10^{6}$, where $k \in \{1,3,5,7\}$ (solve $8k=8*10^{6}t$, and then check the orders of the solutions).


geometry
probability
MATHCOUNTS
AMC
AIME
number theory

Just a small idea...

just like there is an AMC problems/solutions section, i believe there should be a:

Introductory/Intermediate/Advanced Geometry Problems/Solutions

Introductory/Intermediate/Advanced Counting and Probability Problems/Solutions

Introductory/Intermediate/Advanced Number Theory Problems/Solutions

Introductory/Intermediate/Advanced <whatever else> Problems/Solutions.

Introductory should come from AMC and Mathcounts ... etc
Intermediate should come from AIME and things of similar diffuculty
Advanced should come from USAMO.

These can just be copied and transferred from the relative contest pages, and organized into types of problems.

I might have misunderstood, but perhaps you're looking for [[Category:Introductory Algebra Problems]], [[Category:Intermediate Geometry Problems]], [[Category:Olympiad Number Theory Problems]], and in general [Category:<level> <topic> Problems]?

Further, you can find problems relating to specific topics via [[Special:WhatLinksHere/<topic name>]], like [[Special:WhatLinksHere/Power of a Point Theorem]]  :)

:blush:

thanks for embarrasing me, and for pointing this out

=D

It's very clever to respond to a one week old revival with "don't revive"...  :wink:
But theoretically, yes, don't revive for no reason.


There are 1991 planets in some irregular arragement of planets such that no two planets are equidistant. Each alien planet is observing the closest planet to it. Prove that there is a planet which is not being observed.

Each planet may only observe one other planet, so each planet must be observed only once for all planets to be covered.

Since no two planets are equidistant, there must be a minimum distance in such $ {1991 \choose 2}$ distances, corresponding to pairs of planets.

The two planets involved in that pair will observe each other.

If any planets observe such a planet in this pair, then we are done because that means at least one planet in the set is not observed.

Now, since those two planets do not interact with the remaining, we remove them from consideration.

Repeat this process of removing the closest pair, until we reach the case where n = 3. But the third planet MUST observe one of the two other planets. So we are done, and at least one planet is unobserved.


Let $ p,q,r,b,c$ be five different non zero real numbers. Determine the sum $ x\plus{}y\plus{}z$ provided that numbers $ x,y,z$ satisfy the system

\[ \frac{x}{p} \plus{}\frac{y}{p\minus{}b} \plus{}\frac{z}{p\minus{}c}\equal{}1\]
\[ \frac{x}{q} \plus{}\frac{y}{q\minus{}b} \plus{}\frac{z}{q\minus{}c}\equal{}1\]
\[ \frac{x}{r} \plus{}\frac{y}{r\minus{}b} \plus{}\frac{z}{r\minus{}c}\equal{}1\]

we take $ f: \mathbb{R}\minus{}\{0,b,c\}\to \mathbb{R}\ s.t: \ f(X)\equal{}\frac{x}{X}\plus{}\frac{y}{X\minus{}b}\plus{}\frac{z}{X\minus{}c}$
we have $ f'(X)\equal{}\minus{}(x\plus{}y\plus{}z)(\frac{1}{X^2}\plus{}\frac{1}{(X\minus{}b)^2}\plus{}\frac{1}{(X\minus{}c)^2})$

if $ x\plus{}y\plus{}z\neq 0$ then $ f$ is injective
but we have $ f(p)\equal{}f(q)\equal{}f(r)\equal{}1$ impossible

then $ x\plus{}y\plus{}z\equal{}0$


analytic geometry
graphing lines
slope
Pythagorean Theorem
geometry

[quote="linzk3"]A circle has center (12,3).  Line q is tangent to the circle at (6,5).
A tangent to a circle is perpendicular to the radius drawn to the point of tangency.

a.  Find the equation of the tangent shown.

b.  Find the radius of the circle.[/quote]

[hide="a"]
The slope of the radius is $\frac{3-5}{12-6}=\frac{-1}{3}$.  Therefore, the tangent line has slope 3.  So the equation of the tangent is $y=3x+m$, where m is some number.  Substituting x=6 and y=5, we find that $m=-13$.  The equation is [b]y=3x-13[/b]. (I bet that's wrong)[/hide]

[hide="b"]
Draw the coordinate plane and graph the points 12,3 and 6,5.  Draw a line segment from point (6,5) to the line y=3.  This segment has length 2.  Draw another line segment from point (12,3) straight to the line x=6.  This has length 6.  We have a right triangle with legs of length 2 and 6 and we want to find the hypotenuse.  From the pythagorean theorem, we have
$\sqrt{2^{2}+6^{2}}=\sqrt{40}=2\sqrt{10}$.

The answer is $2\sqrt{10}$.
We could have used the distance formula, but it's the same thing as the pythagorean theorem.[/hide]

[quote="CircleSquared"][hide="a"]
The tangent is perpendicular to the line through (12,3) and (6,5), so its slope is 3.  Also, it goes through (6,5).  So the equation is $y=3x-13$

[/hide]
[hide="b"]
This is just the distance from (12,3) to (6,5), which is equal to $\sqrt{(12-6)^{2}+(5-3)^{2}}=2\sqrt{10}$
[/hide][/quote]
why is it (-13)?

Basically, the slope of the line through (12,3) and (6,5) is $\frac{-1}{3}$.  Since the tangent is perpendicular, its slope is the negative reciprocal, which turns out to be 3.  So far, the equation in slope-intercept mode is $y=3x+b$.  Plug in (6,5), and you get $b=-13$


Prove that :

$ \mathop {\lim }\limits_{n \to \infty } \left[ {\left( {\sum\limits_{k \equal{} 0}^n {\frac{{n!}}{{(n \minus{} k)!}}\frac{1}{{n^k }}} } \right) \minus{} \sqrt {\frac{{\pi n}}{2}} } \right] \equal{} \frac{2}{3}$

Let $ Q(n)\equal{}\sum_{k\equal{}1}^{n}\frac{n!}{(n\minus{}k)!n^k}$. Then looks this [url]http://www.mathlinks.ro/viewtopic.php?t=197712[/url].
Easy to see that:
$ Q(n)\plus{}R(n)\equal{}\frac{n!}{n^n}\left(\left(1\plus{}n\plus{}...\plus{}\frac{n^{n\minus{}1}}{(n\minus{}1)!}\right)\plus{}\left(\frac{n^n}{n!}\plus{}\frac{n^{n\plus{}1}}{(n\plus{}1)!}\plus{}...\right)\right)\equal{}\frac{n!e^n}{n^n}$
And now $ Q(n)\equal{}\frac{n!e^n}{n^n}\minus{}R(n)$

More general:
$ Q(n)\equal{}\sum_{k\equal{}1}^{n}\frac{n!}{(n\minus{}k)!n^k}\equal{}\sqrt{\frac{\pi n}{2}}\minus{}\frac{1}{3}\plus{}\frac{1}{12}\sqrt{\frac{\pi}{2n}}\minus{}\frac{4}{135n}\plus{}\frac{1}{288}\sqrt{\frac{\pi}{2n^3}}\plus{}O(n^{\minus{}2})$

Your sum will be $ Q(n)\plus{}1$.


geometry
3D geometry
sphere
topology
induction

Show that any combinatorial surface can be changed to a combinatorial sphere by a finite number of surgeries.

[quote="Nukular"]I'd like to look at your question, but what is a combinatorial (surface/sphere)? A surgery?[/quote]
Yeah, sorry these aren't exactly common terms.
this is just a lemma when proving the classification theorem of surfaces. Basically a combinatorial surface is just a closed surface $ S$ with a triangulation $ h: |K|\to S$. It has the nice properties: Has dimension 2, any two vertices can be joined by an edge path, each edge lies in exactly two triangles and each vertex lies in at least three triangles which fit around it to form a cone with apex at the given vertex and base a polygonal simple curve.
A combinatorial surface is a combinatorial sphere if $ |K|$ is homeomorphic to a sphere.
There's a wikipedia article on what [hide="surgery"]http://en.wikipedia.org/wiki/Surgery_theory[/hide] is

isn't orientability required for your thesis to be true?

if i'm not mistaken, surgery is just cutting along a closed path and filling the two holes with two disks (or the inverse of this move), and this does not change orientability.. so for example i don't think a combinatorial $ \mathbb{RP}^2$ can be made into a sphere by surgeries.. right?

if you have orientability, on the other hand, it seems that you can proceed by induction on $ \rk H_1(S)$. maybe i'll write something down later on, if none steals my job (and i hope someone does)...

I think it works for any combinatorial surface. the inverse of the move you described could be one of two things  :)
[quote]maybe i'll write something down later on, if none steals my job (and i hope someone does)...[/quote]
 :lol:


analytic geometry
trigonometry
geometry
geometric transformation
rotation

Hello
I have a bar of lenght $ \ell$ and mass m. My reference frame is such that the bar forms an angle $ \theta$ with the x axis and its center of mass has coordinates $ \vec r_{C}\equal{}\frac{\ell}{2}\sin\theta\hat\j$. Let A and B the extrema of the bar. At the beginning it is $ \vec r_{A}\equal{}\minus{}\frac{\ell}{2}\cos\theta_{0}\hat\i$. The bar (no friction with ground) falls from initial angle $ \theta_{0}$. Suppose the rotating angle speed is  $ \vec\omega \equal{}\dot\theta\mathbf{\hat k}$

I computed the speed of A and B when the bar falls ($ \theta \equal{} 0$) and are $ \vec v_{A}\equal{} 0$ and $ \vec v_{B}\equal{}\ell\dot\theta\hat\j$. since $ \theta(t)$ is decreasing it must be $ \dot\theta < 0$.

Now I want to find $ \dot\theta(t)$.

The only force here is the weight which is conservative so at the beginning the mechanical energy is

$ E_{i}\equal{}\frac{\ell}{2} g\sin\theta_{0}$ and no kinetic energy, this must be equal to the final mech energy which is

$ E_{f}\equal{}\frac{1}{2}\left ( m\|\vec v_{CM}\|^{2}\plus{}I\dot\theta^{2}\right )$. I know that $ \vec v_{CM}\equal{}\frac{\ell}{2}\dot\theta\cos\theta\hat\j$ so i put it in $ E_{f}$ and solve $ E_{i}\equal{} E_{f}$ for $ \dot\theta$ which is such that

$ \dot\theta^{2}\equal{}\frac{3}{2\ell}g\sin\theta_{0}\Rightarrow\dot\theta \equal{}\minus{}\sqrt{...}$

is this right or I missed something?

[quote="dphexen"]
I computed the speed of A and B when the bar falls ($ \theta \equal{} 0$) and are $ \vec v_{A}\equal{} 0$ and $ \vec v_{B}\equal{}\ell\dot\theta\hat\j$. since $ \theta(t)$ is decreasing it must be $ \dot\theta < 0$.
[/quote]

when the bar falls $ \theta \equal{}\theta_{0}$ and $ t \equal{} 0$

I assume that the pivot center is the center of mass, but, since your calculations $ \vec v_{A}\equal{} 0$ at the initial moment is true only if the pivot point is the point $ A$.

Assuming all this, there is a much more easy way to make those equations.
Assuming initial velocity zero.

$ S \equal{}\left\{0,\minus{}\frac{g t^{2}}{2}\right\}\plus{}\{L\cos (\theta (t)),L\sin (\theta (t))\}\equal{}\left\{L\cos (\theta (t)),L\sin (\theta (t))\minus{}\frac{g t^{2}}{2}\right\}$

In this case, $ L$ is not necessary constant, but it dont varies with time. the interval is $ [0,L]$

[quote="dphexen"]
E_{i}=\frac{\ell}{2} g\sin\theta_{0} and no kinetic energy, this must be equal to the final mech energy which is
[/quote]

not true, the Center of mass have an velocity.

and your formula for $ I$ is false, since it is rotation in another axis of rotation

and $ \vec v_{CM}\equal{}\frac{\ell}{2}\dot\theta\cos\theta\hat\j$ is false, since it have a component in direction of the $ x$ axis. this is only true if the pivot point ins the CM,

so your calculations are controversial.

making A point the frame of reference.

$ \frac{1}{2}m (g t)\plus{}\frac{1}{2}g m\left(g t^{2}\right)\plus{}\frac{1}{2}\text{MI}\theta'(t)\equal{}\frac{\text{MI}\omega_{0}}{2}$

To compute the speeds I assumed $ \vec r_{CM}\equal{}\frac{\ell} 2\sin\theta\hat{\mathbf j}$ so I can find its speed (just vertical since teh weight is applied to the CM) $ \vec v_{CM}\equal{}\frac{\ell} 2\dot\theta\cos\theta\hat{\mathbf j}$. At $ \theta\equal{}0$ I have $ A \equal{}\minus{}\frac{\ell} 2\hat{\mathbf j}$ and $ B \equal{}\frac{\ell} 2\hat{\mathbf j}$ so I can find their speeds using the equation for the speed field of a rigid body.

err... dude. use just one point.

for the velocity you use CM, for rotation use A, for the initial condictions use B

and for miracle you want a right result? impossible

it could really easy if u could put ur problem in the form of a statement and then explain ur procedure than doing both simultaneously

table with no friction, pen of mass m and length $ \ell$, initial angle $ \theta_{0}$
[code]
         B
       //
      //
     //
    //
   // angle of theta_0
  //)________
A

[/code]

I let the pen fall. When it does ($ \theta \equal{} 0$) find the speed of A and B[/code]

so is the pen on the table? do u give i some initial speed

yes it is on the table, the extremum A is in contact with it and makes an angle $ \theta_{0}$. I keep ii it that position then I let it fall.

so isn't equivalent to let allow a rod fall freely that makes an angle $ \theta_{0}$ with the horizontal?

Of course it isn't. You are not considering the normal force of the table.

how come a bar kept on the table which i suppose i shorizontal starta falling automatically

[quote="dphexen"]I don't understand your procedure... the ploblem is: put a pen on a frictionless table inclinated with angle $ \theta_{0}$ and let it fall. It doesn't rotate around A, the point on table, but around the CM :)[/quote]

er?!?!

you say it fall. but it dont fall! it rotate!

is differente. Your problem is not someting falling, is rotations.

you never made the problem correctly. if the problem is just that. the problem is just an Physical Pendulum, how's ODE is very hard.

Conservation of mechanical energy:

  \[ mg{\ell\over2}(sin\theta_{0}-sin\theta) ={1\over2}(mV^{2}+I_{cm}\dot\theta^{2})\]

where V is the CM velocity (negative) and $ I_{cm}={{m\ell^{2}\over{12}}}$ the moment of inertia.

As point A has no vertical component of velocity, we have $ V ={\ell\over 2}cos\theta\;\dot\theta$  and $ v_{A}={\ell\over2}sin\theta$, sliding along the table.

For $ \theta = 0$ we get straight $ v_{a}= 0$  and ${ \dot\theta =\sqrt{{3g sin\theta_{0}}\over\ell}}$

  So,  $ v_{B}= V+{\ell\over2}\dot\theta =\ell\dot\theta =\sqrt{3\ell g sin\theta_{0}}$

[quote="Immanuel Bonfils"]

For $ \theta = 0$ we get straight $ v_{a}= 0$  and ${ \dot\theta =\sqrt{{3g sin\theta_{0}}\over\ell}}$  [/quote]

$ \theta$ is decreasing so you must pick $ \dot\theta =-\sqrt{...}$ and why there isn't a 2 in denominator?

and what's different between your energy equation and mine?

the mech energy conservation equation should be:
(it was $ CM \equal{} (0,\frac{\ell}{2}\sin\theta)$ so it is $ V \equal{} (0,\frac{\ell}{2}\dot\theta\cos\theta)$ and also $ \|V\|^{2}\equal{}\frac{\ell^{2}}{4}\dot\theta^{2}\cos^{2}\theta |_{\theta \equal{} 0}\equal{}\frac{\ell^{2}}{4}\dot\theta^{2}$

$ mg\frac{\ell}{2}\sin\theta_{0}\equal{}\frac{1}{2}m\frac{\ell^{2}}{4}\dot\theta^{2}\plus{}\frac{1}{2}I\dot\theta^{2}$ iff $ mg\ell\sin\theta_{0}\equal{}\left (\frac{\ell^{2}}{4}m\plus{}I\right )\dot\theta^{2}$

which gives

$ \dot\theta \equal{}\minus{}\sqrt{\frac{3 g\sin\theta_{0}}{\ell}}$
without the 2 at denominator... I must have missed something in the computation in first post.

This is the same Immanuel found but actually it is $ \dot\theta(t_{f})$ if $ t_{f}$ is the time when the pen already fell. How can I find, if possible, $ \dot\theta(t)$ for all t?? solving some DE like

$ mg\frac{\ell}{2}\sin\theta_{0}\equal{}\frac{1}{2}m\frac{\ell^{2}}{4}\dot\theta^{2}\plus{}\frac{1}{2}I\dot\theta^{2}\plus{}mg\frac{\ell} 2\sin\theta$ maybe?

No need to solve the differential equation.

Just keep the general angle in the energy conservation equation, like I've post before,

 \[ mg{\ell\over2}(sin\theta_{0}\minus{}sin\theta) \equal{}{1\over2}(mV^{2}\plus{}I_{cm}\dot\theta^{2}),\]
 and in the center of mass velocity
\[ V \equal{}{\ell\over2}cos\theta\;\dot\theta,\]to get

\[ \dot\theta \equal{}\sqrt{{{12g(sin\theta_{0}\minus{}sin\theta)}\over{\ell (3 cos\theta\plus{}1)}}}\]

[quote="dphexen"]No, it's not a physical pendulum.No points are fixed here![/quote]

The velocity of the point A is zero, you said that. so. the point A is stationary.

so it is a fixed point, mister paradox

[quote="lordykelvin"]
The velocity of the point A is zero, you said that. so. the point A is stationary.

so it is a fixed point, mister paradox[/quote]

no, it's not fixed. At time 0 we have it is $ A(0)\equal{}(\minus{}\frac{\ell} 2\cos\theta_{0},0)$ at time $ t_{f}$ when the pen lying on table it is $ A(t_{f})\equal{}(\minus{}\frac{\ell} 2,0)$ and of course $ \theta_{0}\neq 0$

Before I found that $ v_{A}(t_{f})\equal{}0$ and [i]not[/i] that $ v_{A}(t)\equiv 0$ forall t

pen!? there is a pen now?!

in the begining of the problem there wasnt table or pen, now we have pen and table.

woow.

are you buying all those things by yourself?!

hauhuauha :rotfl:

I made an example to describe the problem because someone didn't get the hypoteses

[quote="lordykelvin"]you is the one how doesnt understand...

If there is no pivot point, with the need for a table?

dump![/quote]
right point.
dphexen can u clearly describe what's theree in the problem.forget about ur euations.just give the initial  conditions just the way it was presented in the book from where u got this (if any)

Sorry again:

\[{ \dot\theta =-\sqrt{{12g(sin\theta_{0}-sin\theta)}\over{\ell(3cos\theta+1)}}}\]

Proble (don't know the book, it was made in class):

There is a bar (or a pen, or whatever you like) of mass m and length $ \ell$. It is positioned on a frictionless table. The low extremum which we call A is i contact with table, the other is in air. If we consider a plane which contains the bar and orthogonal to the table we call $ \theta_{0}$ tha angle that the pen forms with the table.

Imagine a wire fixed in the other extremum B keeps the bar in this position. The wire get burnt and the pen falls.

Find:

 :arrow: The speeds of the extrema [i]when the pen is all on table[/i]. If the pen starts falling at time 0 and arrives on table at time $ t_{f}$ we want $ v_{A}(t_{f})$ and $ v_{B}(t_{f})$.

This was made in class, what I asked for, and tried myself, is
 :arrow: Find $ \dot\theta$

Ah, that's better! A clear text!

I got:

$ v_{A}(t_{f})=0$

$ v_{B}(t_{f})=\sqrt{3g\ell\sin\theta_{0}}$

${ \dot\theta=-\sqrt{{12g(\sin\theta_{0}-\sin\theta)}\over{\ell(1+3\cos^{2}\theta)}}}$

[quote="pedroomagico"]
$ v_{A}(t_{f}) = 0$

$ v_{B}(t_{f}) =\sqrt{3g\ell\sin\theta_{0}}$
[/quote]
the same we got but:
[quote]
${ \dot\theta =-\sqrt{{12g(\sin\theta_{0}-\sin\theta)}\over{\ell(1+3\cos^{2}\theta)}}}$[/quote]

why $ \cos^{2}\theta$ at denominator?


geometry
trigonometry
Olimpiada de matematicas

Dado un triangulo $\triangle{ABC}$, la tangente por $A$ al circuncirculo corta a $BC$ en $D$. La perpendicular a $BC$ por $B$ corta la mediatrez de $AB$ en E, y la perpendicular por $BC$ en $C$ corta a la mediatriz de $AC$ en $F$. Demuestre que $D$, $E$,$F$ son colineales.

Try searching for it in the Geometry section. I'm sure it's been posted before.

its ok grobber, I already solved this problem, i just posted it to enrich the spanish comunity content  :D

Esta Facil,

Tenemos el triangulo ABC, con angulos A, B, C => A+B+C=180
Notemos que angulo EBC= 90-B => BEC=B
por lo tanto
BE= AB/(2senB), idem FC=AC/(2senC)

Veamos que basta ver que EB/FC=DB/DC, para ver que D,F,E son colineales [un argumento seria EBD semejante con FCD, por razon-angulo-razon]

Por un lado EB/FC= [i]ABsenC / ACsenB[/i]

Luego con Teo. Bisectriz generalizado, en ABC con AD tenemos:  (Nota, como AD es tangente, DAB=C)
BD/DB= ABsenC / ACsen (A+C) =[i]ABsenC / ACsen (B)[/i]

[u]Y acabamos[/u]

[/hide][/list][/code]

Supongamos que $ D' \equiv{} EF \cap BC$ siendo $ D$ distinto de $ D'.$ Por semejanza tenemos que:

$ \frac {DC}{DB} \equal{} \frac {FC}{EB}. $ Facilmente aplicando teorema del seno, reescribimos:

$ \frac {DC}{DB} \equal{} \frac {FC}{EB} \equal{} \frac {b \sin B}{c \sin C}$

Aplicando Ley del seno en los triángulos $ \triangle AD'C$ y $ \triangle AD'B,$ teniendo presente que $ \angle BAD' \equal{} \angle ACD',$ tenemos

$ \frac {D'C}{D'B} \equal{} \frac {b \sin B}{c \sin C}$  $ \Longrightarrow$ $ D \equal{} D'$ y con esto demostramos la colinealidad de $ E,F,D$


The denominator of a fraction is 7 less than 3 times the numerator.  If the fraction is equivalent to $ 2/5$, what is the numerator of the fraction?

Let $ x$ be the numerator.

$ \frac{2}{5} \equal{} \frac{x}{3x \minus{} 7}$

Cross multiply.

$ 6x \minus{} 14 \equal{} 5x$

$ \minus{}14 \equal{} \minus{}x$

$ \boxed{14} \equal{} x$


inequalities
function
inequalities proposed

[i] A friend gave me this problem( he created it) and wants to see your way  for approximating the solution.[/i]

If $ a , b , c$ are numbers in the interval  $ ( \minus{} 1,2)$ and

$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \equal{} \frac {1}{2 \minus{} a} \plus{} \frac {1}{2 \minus{} b} \plus{} \frac {1}{2 \minus{} c}$ , prove that

 $ \frac {1}{5 \minus{} a} \plus{} \frac {1}{5 \minus{} b} \plus{} \frac {1}{5 \minus{} c} \geq \frac {2}{3}$

My ugly proof  :(
If we take $ x \equal{} \frac {a \minus{} 1}{2 \minus{} a}$ and so on, then we have to prove $ \sum{x} \equal{} \sum{\frac {1}{x}}$ then $ \sum{\frac {1}{2 \plus{} x}} \le 1$.
If we take $ S_{1} \equal{} \frac {x \plus{} y \plus{} z}{3},S_{2} \equal{} \frac {xy \plus{} yz \plus{} zx}{3},S_{3} \equal{} xyz$ then we have to prove
$ S_{1}S_{3} \equal{} S_{2}$ then $ S_{3} \plus{} 3S_{2} \ge 4$, which is equivalent to $ \sqrt {\frac {S_{2}S_{3}}{S_{1}}} \plus{} 3S_{2} \ge 4\frac {S_{3}S_{1}}{S_{2}}$.
Because it's homogeneous we make a condition $ S_{2} \equal{} 1$, then inequality transformed into $ \sqrt {\frac {S_{3}}{S_{1}}} \plus{} 3\ge4S_{3}S_{1}$.
Since $ S_{1} \ge S_{2}^{1/2} \ge S_{3}^{1/3}$, $ S_{1} \ge 1 \ge S_{3}$. Also, since $ S_{2}^{2} \ge S_{1}S_{3}$, $ 1 \ge S_{1}S_{3}$.
If we consider $ f(S_{3}) \equal{} \sqrt {\frac {S_{3}}{S_{1}}} \plus{} 3 \minus{} 4S_{3}S_{1}$, $ f'' < 0$, and since $ f(0) \equal{} 3, f(1) \ge 0$ for all $ S_{1}\ge 1$, $ f \ge 0$ in $ S_{3} \in [0,1]$. Therefore, original inequality is proved.
Equality occurs when $ x \equal{} y \equal{} z$, which is $ a \equal{} b \equal{} c$.

But this is ugly version, and I believe there is a better solution.

[quote="msceok"]
If we take $ x \equal{} \frac {a \minus{} 1}{2 \minus{} a}$ and so on,
[/quote]
You have meant that $ x \equal{} \frac {a \plus{} 1}{2 \minus{} a}.$ :wink:
Beautiful idea!
[quote="msceok"] which is equivalent to $ \sqrt {\frac {S_{2}S_{3}}{S_{1}}} \plus{} 3S_{2} \ge 4\frac {S_{3}S_{1}}{S_{2}}$[/quote]
Yes! It's true!
[quote="msceok"]
If we consider $ f(S_{3}) \equal{} \sqrt {\frac {S_{3}}{S_{1}}} \plus{} 3 \minus{} 4S_{3}S_{1}$, $ f'' < 0$, and since $ f(0) \equal{} 3, f(1) \ge 0$ for all $ S_{1}\ge 1$, $ f \ge 0$ in $ S_{3} \in [0,1]$. Therefore, original inequality is proved.
[/quote]
I think, here your mistakes. Why $ f'' < 0$ $ ?$ ( $ S_3$ is function of $ S_1.$  :wink:  )
Why $ f(1) \ge 0$ for all $ S_{1}\ge 1$ $ ?$ Explain pleas. Thank you.

[quote="arqady][quote="msceok"]
If we consider $ f(S_{3}) \equal{} \sqrt {\frac {S_{3}}{S_{1}}} \plus{} 3 \minus{} 4S_{3}S_{1}$, $ f'' < 0$, and since $ f(0) \equal{} 3, f(1) \ge 0$ for all $ S_{1}\ge 1$, $ f \ge 0$ in $ S_{3} \in [0,1]$. Therefore, original inequality is proved.
[/quote]
I think, here your mistakes. Why $ f'' < 0$ $ ?$ ( $ S_3$ is function of $ S_1.$  :wink:  )
Why $ f(1) \ge 0$ for all $ S_{1}\ge 1$ $ ?$ Explain pleas. Thank you.[/quote]
I had a little mistake  :blush:
I meant, there exist some positive reals $ x,y,z$ satisfying $ S_{1}\equal{}a,S_{2}\equal{}1,S_{3}\equal{}b$, where $ a$ and $ b$ are arbitrary positive reals with $ a\ge1$ and $ b \le 1$ and $ ab \le 1$ since $ S_{1}S_{3}^{1/3} \le S_{2}$, so I thought that if we fix $ S_{1}$ then $ S_{3} \in [0,\frac{1}{{S_{1}}^{3}}]$ can be variable independent from $ S_{1}$. So, $ f''\equal{}\minus{}\frac{1}{4} \sqrt{\frac{1}{{S_{3}}^{3}S_{1}}}<0$ in $ S_{3} \in [0,1]$, hence we can conclude that f is concave downward. Therefore, if we evaluate extreme values, $ f(0)\equal{}3\ge0, f(\frac{1}{{S_{1}}^{3}})\equal{}3\minus{}3\frac{1}{{S_{1}}^{2}}\ge 0$, so we can conclude that $ f \ge 0$ in $ S_{3} \in [0,\frac{1}{{S_{1}}^{3}}]$. Done!

[quote="msceok"] $ f'' \equal{} \minus{} \frac {1}{4} \sqrt {\frac {1}{{S_{3}}^{3}S_{1}}}$
[/quote]
I think, it's wrong because $ S_3$ is function of $ S_1.$  :wink:

I fixed $ S_{1}$ and I considered $ S_{3}$ as just a variable in interval $ [0,\frac{1}{{S_{1}}^{3}}]$
I only considered the function $ f(x)\equal{}\sqrt{\frac{x}{a}}\plus{}3\minus{}4ax$ with $ a \ge 1$ and $ x \in [0,\frac{1}{a^{3}}]$. Because this case can cover all of the possible cases, $ f \ge 0$.

[quote="msceok"]I fixed $ S_{1}$ [/quote]
No! You can't make it. $ S_3$ is function of $ S_1$ and $ S_1$ is function of $ S_3.$
Hence, you need to derive $ S_1$ too. :wink:
For example:
let $ f(x) \equal{} \sqrt x$  and $ y \equal{} x.$
Hence, $ f(x) \equal{} \frac {y}{\sqrt x}.$ "I fixed" $ y > 0.$ We obtain $ f''(x) \equal{} \frac {3y}{4\sqrt {x^5}} > 0.$
Thus, $ f$ is convex function. :wink:

But your example is related to only one variable, which is the case $ x$ is fixed when $ y$ is fixed.
But, it the original problem, there are three variables; thus there is an interval for $ S_{3}$ when $ S_{1}$.
$ S_{3}$ and $ S_{1}$ are related to each other only for determining the interval of $ S_{3}$. So, I can fix $ S_{1}$.

Consider that if we reversely get the values of $ x,y,z$ from the values of $ S_{1},S_{2},S_{3}$, they will exist under some conditions of $ S_{1},S_{2},S_{3}$. So, I considered $ S_{1},S_{2},S_{3}$ with looser conditions of existence of $ x,y,z$ and I considered $ x,y,z$ are determined by $ S_{1},S_{2},S_{3}$. After the moment of considering the function $ f(S_{3})$, $ x,y,z$ is not considered, and all of the possible values of $ (S_{1},S_{3})$ with the condition $ S_{2}\equal{}1$ is covered by looser condition, $ S_{3} \in [0,\frac{1}{{S_{1}}^{3}}]$, when $ S_{1}$ is fixed.

See also here for a way of solution  :)
http://www.mathlinks.ro/viewtopic.php?t=172516


[quote="t0rajir0u"]I've always found it more profitable just to read a lot.  I've never had to actually study vocabulary lists for the SAT.[/quote]

reading for fun any types of books? or only challenging ones.

I can understand many books, but when it comes to SAT time, I fail miserably.

just read AOPS alot i'm sure reading forums is mORE than enough

Like those Hardcover Large Print books like The Da Vinci code ?

[quote="t0rajir0u"]Read more books, and I'm not talking about "Young Adult" novels or anything in the Teen section of Barnes & Noble, I mean [b]books[/b].  Go to the library and find something written for adults that looks interesting.[/quote]

Can you provide a few examples?

[url=http://www.amazon.com/Time-Travelers-Wife-Audrey-Niffenegger/dp/015602943X]The Time Traveler's Wife[/url]:  Actually, this might be a little R-rated for an 8th grader.  Time travel and true love, if you're into that.

[url=http://www.amazon.com/Wicked-Life-Times-Witch-West/dp/0060987103]Wicked[/url]:  Might also be a tad risque.  Good book.  Lots of words.  Very deep.

[url=http://www.amazon.com/Atlas-Shrugged-Ayn-Rand/dp/0451191145]Atlas Shrugged[/url]:  Long.  Certified to turn you into a complete jerk for at least a few weeks.  It took me about eight months to shake it off   :rotfl:

I hear reading [url=http://www.amazon.com/Grapes-Wrath-Centennial-John-Steinbeck/dp/0142000663]The Grapes of Wrath[/url] immediately after helps to counteract the effects  :P

I didn't memorize vocab words, I memorized various prefixes and suffixes.  Also do practice tests, they ask the same styles of questions over and over again, and often you just need to get used to recognizing what kind of word they're looking for (a happy word, an angry word etc.) and that's where the prefixes and suffixes are good.  They won't tell you what a word mean, but they will tell you what it implies.

Atlas Shruggeds? I am so going to read that!!!

Randroids in the world? that sounds totally Awesome!  :lol:

Dostoevsky's Notes from the Underground is good just for the chance to read one of the most interesting characters ever

[quote="Elemennop"]Oh god please don't subject yourselves to reading atlas shrugged we dont need anymore randroids in the world[/quote]
Though more capitalists is always a good thing, especially when fighting bubka and his Communists of Evil :rotfl:

The writing in that book is horrible, anyway.

Make sure that when you read you look up words whose meaning you don't know.

[quote="Elemennop"]lol

I got 410 when i took the sat verbal in 7th grade[/quote]

SUREEE you didd.............

You're too smart for that. The SAT is terrified of you. So is the AMC.

Getting a score 600 and lower is totally uncool haha. it makes me feel stupid.

[quote="K81o7"][quote="Piggypi314"][quote="K81o7"]Dostoevsky books are bound to get you thinking...[/quote]

Dostoyevsky, right?  That would be true if thinking = sleeping.  Blah blah blah.[/quote]

Have you read his good books?  I suppose that's just you, but I personally like Crime and Punishment once it started getting good...
Sure there's lighter reading around...[/quote]

Yahh, don't laugh at the retarded kid who's borderline illiterate.  [u]The Brothers Karamazov[/u] was okay.  Kinda boring, but The Grand Inquisitor was hilarious.

[quote="Go Around the Tree"][quote="moogra"]hey I was worse than you at first in 7th grade (500). Now after a year and a half of this, My score went up 200. It really helps, but you have to work hard on this[/quote]

after a year and a half of what?

did your score go up 200.

I hope it's not too late for me to improve my reading.[/quote]

A year and a half of vocab study. Turns out, I didn't know half the things I was reading

Read Fitzgerald. For Pete's sake. I've seriously never read better writing. Hmm... who else.

Kurt Vonnegut is pretty much amazing at satire.
I also recommend Ayn Rand as t0rajir0u did. I read "The Fountainhead" and yes, I became quite a bit of a jerk - aka. completely socially irresponsible - for a bit. Her books are very intelligent.

Basically ever since I've started writing short stories and poetry, and reading good material, my SAT reading has been good. I used to score in the high 600s in 9th grade, but I got a 78 on the PSAT on five hours of sleep. So it's all ok in the end. And I definitely went to a party the night before I took my SATs this month. Rofl...

heh old post lol.

740 M
550 CR
600 W

 :(  :(  :(

[quote="now a ranger"]Will memorizing Vocabulary increase my Verbal/critical reading score on the SAT?

I'm in the 550-590  range, which is not a good thing.

I'm in 8th grade, and I need to improve it really badly.

Anyone have any advice?[/quote]
don't worry you're in 8th grade.

Benito Cereno  by Herman Melville will help you expand your vocabulary. However, you are in eighth grade so you really don't need to worry about SATs for another two years.

[quote="1=2"]My last SAT(which was my first), I got a 430 on my verbal and 460 on critical reading. I'm in 7th grade, but I skipped a grade.

Read Raymond Smullyan, Issac Asimov, and study latin. That's the key.  :rotfl:

edit: My math score was 790. :([/quote]
HA! i got 480 on verbal and 490 on CR. (also 7th) But still i suck.
Would reading anything other than fiction help. My parents are sugesting some vocabulary boosters like Wordly Wise and others. Will it accually be possible just to memorize words and words and words forever untill you take the test and forget all of them because you memorized too much?

You people worry too much.  You guys are in 7th/8th grade!  Enjoy it!

I didn't prepare explicitly for the SAT at all, I just read a lot throughout my life and took it as soon as I could to get it over with.  Result? 760 on reading, 770 on writing.  My advice, then: don't study for the SAT in 7th grade, for God's sake.  If you don't read much, start.  If you do, keep at it.

Oh - and if you detest the "classics" people are telling you to read, don't read them!  Read what you enjoy - it'll help [sure, pulp fantasy won't help your vocabulary as much as Dostoyevsky] but you'll have more fun and will probably not burn yourself out.

Just my two cents.


[color=darkblue]Solve this equation in $ R^ \plus{}$:

$ \sqrt {\frac {4x \plus{} 9}{28}} \equal{} 7x^2 \plus{} 7x$

I have a solution for this problem:

Equation $ \Longleftrightarrow \frac {4x \plus{} 9}{28} \equal{} (7x^2 \plus{} 7x)^2 \Longleftrightarrow 1372x^4 \plus{} 2744x^3 \plus{} 1372x^2 \minus{} 4x \minus{} 9 \equal{} 0$

$ \Leftrightarrow (98x^2 \plus{} 112x \plus{} 9)(14x^2 \plus{} 12x \minus{} 1) \equal{} 0$

Then it's easy.

But I don't like this way. Can you give me another solution for this problem?

Thank you very much.[/color]

Let :  $ y\plus{}1\frac{1}{2}\equal{}\sqrt{ \frac{4x\plus{}9}{28} }$
$ \Rightarrow 7y(y\plus{}1)\equal{}x\plus{}\frac{1}{2}$
And :$ 7x(x\plus{}1)\equal{}y\plus{}\frac{1}{2}$


find all of the fuction $fR...R$ such that
$f(xf(x)+f(y)) = f(x)^2+y$

The solution of tµtµ is not sufficient ! :D  Actually there are two solutions : $f_1(t)=t ,\ f_2(t)=-t$
Manifestly $f$ is surjective, so there is $x_0$ such that $f(x_0)=0$, so $\forall y,\ f(f(y))=y$
$x=f(t) \Longrightarrow f(f(t)t + f(y))= t^2 + y$.
But  $f(tf(t) + f(y))= f^2(t) + y$, so $f^2(t) +y=t^2 + y \Longrightarrow f^2(t)=t^2 \Longrightarrow f(t)\in\{-t,t\}$
$(f(f(t)t + f(y)))^2= (t^2 + y)^2 \Longrightarrow$$t^2f^2(t)+2tf(t)f(y)+ f^2(y) = t^4 +2t^2y +y^2 \Longrightarrow f(t)f(y)= ty$
Finally there are two solutions : $f(1)=1 \Longrightarrow f_1(t)= t$ and $f (1)=-1\Longrightarrow f_2(t) =-t$


inequalities
calculus
geometry solved
geometry

Here is an other one . Hope at least this one is corect.

Let ABC be a triangle.Prove that

R* \sum (sinAcosA/(sinA+cosA)) \leq  \sqrt (p(R+r)/4)

I am sorry I made a mistake in my ineq
    The corect ineq is
    R* \sum sinAcosA/sinA+cosA \leq p+R+r/4

Now it seems strange! You mean the inequality is so:
R*  sum(sinAcosA/(sinA+cosA)) <=  p+R+r/4?

okay .. done this in a hurry.. hope it's good!

R sum(sinAcosA\(sinA+cosA)<=p+R+r/4

we know that 2sinAcosA/(sinA+cosA)<=sinA+cosA

so we have that R/2 sum(2sinAcosA\(sinA+cosA)<=R/2 (sum (sinA+cosA))
all we have to prove now is that:
R/2 (sum (sinA+cosA))<=p+R+r/4 we divide it now by R
and we know that: sum(sinA)=p/R;
prod(sinA/2)=r/(4R) but sum(cosA)=1+4 prod(sinA/2) which gives: r/(4R)=(sum(cosA)-1)/4

so rewriting the inequality we get that:
1/2(sum (sinA+cosA))<=sum(sin A)+(sum(cosA)-1)/4+1
after some calculus we have: 0<=2(sum(sinA))-sum(cosA)+3 which is true cause
1<sum(cosA)=<3/2 and sin in a triangle is always >0
so I proved that:
R sum(sinAcosA\(sinA+cosA)<=R/2 (sum (sinA+cosA))<=p+R+r/4.

cheers! hope it's true!  :D

Reply 4 Lagrangia
    I think that your solution is good but you could use the fact that
    sinAcosA/sinA+cosA<(sinA+cosA)/4
    So \sum sinAcosA/sinA+cosA< (\sum sinA+ \sum cosA)/4
    We know that \sum sinA=p/R and  \sum cosA=1+r/R
    After some calculs we will have  \sum sinAcosA/sinA+cosA<(p+R+r)/4R
    And if you multiply with R the solution is over.
    Hope my solution is good!


Can someone please help me with this problem?

You transferred 20. mL of acetic acid solution to an Erlenmeyer flask and added 2 drops of phenolphthalein solution. You filled your buret with 0.10 M NaOH so that the initial buret reading was 1.47 mL. After titrating your sample to the faint-pink phenolphthalein endpoint, the final buret reading was 17.41 mL.
a) How many moles of NaOH were delivered to the flask?

b) What is the concentration of the acetic acid solution?

First, welcome to this forum. Now, regarding your question:

[quote="Planke"]...so that the initial buret reading was 1.47 mL. (...) the final buret reading was 17.41 mL.[/quote]

How can that be possible? Fill a buret with only 1.47 mL of NaHO? And ending with more NaHO that you have started?

[quote="Carcul"]First, welcome to this forum. Now, regarding your question:

[quote="Planke"]...so that the initial buret reading was 1.47 mL. (...) the final buret reading was 17.41 mL.[/quote]

How can that be possible? (...) ending with more NaHO that you have started?[/quote]

Heat it up, a lot!

Which would not actually give you more $ \mbox{NaOH}$, just a larger volume of it...

I think the type of buret Planke is talking about starts with a zero at the very top and increases reading down the side of the buret. If this is the case, then, to solve part A:

[hide="Solution"]$ 17.41 \mbox{ mL} \minus{} 1.47 \mbox{ mL} \equal{} 15.94 \mbox{ mL}$ of $ \mbox{NaOH}$ were added to the solution.

Then, since the NaOH solution is .10 M (or moles per liter), we can use a conversion factor:

$ 15.94 \mbox{ ml NaOH} * \frac {1 \mbox{ L}}{1000 \mbox{ mL}} * \frac {.10 \mbox{ mol NaOH}}{1 \mbox{ L}} \equal{} .001594 \mbox{ mol NaOH}$

With the correct number of significant figures, the answer is $ .0016 \mbox{ mol NaOH}$.[/hide]

That makes sense KrazyNerd. Now, to find the concentration of acetic acid, start by writing the titration reaction's equation:

$ CH_3CO_2H \plus{} NaHO \longrightarrow NaCH_3CO_2 \plus{} H_2O$

and we can see that, in the equivalence point, the number of moles of acid equals the number of moles of base, [hide=" and so"]

$ n_a \equal{} n_b \equal{}> C_aV_a \equal{} C_bV_b \equal{}> C_a \equal{} \frac{C_bV_b}{V_a} \equal{} \frac{0.10 \cdot 15.94}{20} \equal{} 0.080\,M$.[/hide]


For those of you who were in the world of mathematics before Fermat's Last Theorem was proven, was it believed to be true?
What did the large part of the mathematical community believe?

Well, it had been verified true for all exponents $n$ up to $4 \times 10^{6}$ by $1993$.

Almost everyone believed that it was true. There were a few outside skeptics who thought that it wasn't provable within the standard axioms (rules) of mathematics. But there were very few people who thought that it was false. There was a great deal of circumstantial evidence to support it.

The majority believed it to be true, note how asking for integer solutions for x^n + y^n = z^n is the same as asking for rational cosines and sines of t for
cos^n t + sin^n t = 1
where cos t = y/z or x/z and similarly for sin t

just a fresh take on it, neat implications in trigonometry but hardly used in elementary work.

but it has been proofed by andrew wiles in 1994

[quote="davidyko"]We know. Read the first post.[/quote]

thanx i dont read all the post


trigonometry symmetry Gauss quadratics search

I think you wanted to write $ \boxed{\cos\frac{2\pi}{13}\plus{}\cos\frac{6\pi}{13}\plus{}\cos\frac{18\pi}{13}\equal{}\frac 14(\sqrt{13}\minus{}1)}$

D u already know the answer or r u lookin for help

This should probably be moved to Pre-Olympiad.

A similar problem was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215608]here[/url]. But not with a full solution, only with some hints.

BTW, does anyone know a book with more problems like this?

I don't know Answer.
Plz solve it step by step.  :!:  :!:  :!:

I know a way but i dont  think it will lead us anywhere..
let $ z \equal{} \cos \frac {\pi}{13} \plus{} i\sin \frac {\pi}{13}$
so now using Demoivre's theorem we get $ \cos \frac {2\pi}{13} \equal{} \frac {1}{2}(z^{2} \plus{} \frac {1}{z^{2}})$,
 $ \cos \frac {6\pi}{13} \equal{} \frac {1}{2}(z^{6} \plus{} \frac {1}{z^{6}})$ and $ \cos \frac {18\pi}{13} \equal{} \frac {1}{2}(z^{18} \plus{} \frac {1}{z^{18}})$ also note that $ z^{13} \equal{} \minus{} 1$
[hide]Not able to proceed further[/hide]

[hide="Following the hints from the other thread"]Let $ z \equal{} e^{i\frac {2\pi}{13}}$.
We seek the value of $ P \equal{} \cos\left(\frac {2\pi}{13}\right) \plus{} \cos\left(\frac {6\pi}{13}\right) \plus{} \cos\left(\frac {18\pi}{13}\right) \equal{} \frac {z \plus{} \bar{z}}{2} \plus{} \frac {z^3 \plus{} \bar{z}^3}{2} \plus{} \frac {z^9 \plus{} \bar{z}^9}{2}$
$ \equal{} \frac {1}{2}(z \plus{} z^{12} \plus{} z^3 \plus{} z^{10} \plus{} z^9 \plus{} z^4)$

So $ P \equal{} \frac {1}{2}(z \plus{} z^3 \plus{} z^4 \plus{} z^9 \plus{} z^{10} \plus{} z^{12})$

$ 4P^2 \equal{} z^2 \plus{} z^6 \plus{} z^8 \plus{} z^{18} \plus{} z^{20} \plus{} z^{24} \plus{} 2z^4 \plus{} 2z^5 \plus{} 2z^7 \plus{} 2z^{10} \plus{} 2z^{11} \plus{} 2z^{12} \plus{} 6z^{13} \plus{} 2z^{14} \plus{} 2z^{15} \plus{} 2z^{16} \plus{} 2z^{19} \plus{} 2z^{21} \plus{} 2z^{22}$

$ \equal{} z^2 \plus{} z^6 \plus{} z^8 \plus{} z^5 \plus{} z^7 \plus{} z^{11} \plus{} 2z^4 \plus{} 2z^5 \plus{} 2z^7 \plus{} 2z^{10} \plus{} 2z^{11} \plus{} 2z^{12} \plus{} 6 \plus{} 2z \plus{} 2z^2 \plus{} 2z^3 \plus{} 2z^6 \plus{} 2z^8 \plus{} 2z^9$

$ \equal{} z^2 \plus{} z^5 \plus{} z^6 \plus{} z^7 \plus{} z^8 \plus{} z^{11} \plus{} 6 \plus{} 2(z \plus{} z^2 \plus{} z^3 \plus{} z^4 \plus{} z^5 \plus{} z^6 \plus{} z^7 \plus{} z^8 \plus{} z^9 \plus{} z^{10} \plus{} z^{11} \plus{} z^{12})$

$ \equal{} z^2 \plus{} z^{ \minus{} 2} \plus{} z^5 \plus{} z^{ \minus{} 5} \plus{} z^6 \plus{} z^{ \minus{} 6} \plus{} 6 \plus{} 2( \minus{} 1)$

$ \equal{} 2\left(\cos\left(\frac {4\pi}{13}\right) \plus{} \cos\left(\frac {10\pi}{13}\right) \plus{} \cos\left(\frac {12\pi}{13}\right)\right) \plus{} 4$

$ \equal{} 2\left( \minus{} \frac {1}{2} \minus{} P\right) \plus{} 4$

Therefore $ 4P^2 \equal{} 3 \minus{} 2P$
Solving gives $ P \equal{} \frac { \minus{} 1 \pm \sqrt {13}}{4}$
Taking the positive sign because of size considerations yields the answer.[/hide]

How the step (-0.5-P) came  ??

 :help:  :help:  :help:

Because $ 1 \plus{} e^{i\frac {2\pi}{13}} \plus{} e^{i\frac {4\pi}{13}} \plus{} e^{i\frac {6\pi}{13}} \plus{} \ldots \plus{} e^{i\frac {22\pi}{13}} \plus{} e^{i\frac {24\pi}{13}} \equal{} 0$, we know by looking at the real parts

that $ \cos\left(\frac {2\pi}{13}\right) \plus{} \cos\left(\frac {4\pi}{13}\right) \plus{} \cos\left(\frac {6\pi}{13}\right) \plus{} \ldots \plus{} \cos\left(\frac {22\pi}{13}\right) \plus{} \cos\left(\frac {24\pi}{13}\right) \equal{} \minus{} 1$

Then because of the symmetry of cosine; $ \cos\left(\frac {2\pi}{13}\right) \equal{} \cos\left(\frac {24\pi}{13}\right), \cos\left(\frac {4\pi}{13}\right) \equal{} \cos\left(\frac {22\pi}{13}\right), \ldots$ and so on,

$ \cos\left(\frac {2\pi}{13}\right) \plus{} \cos\left(\frac {4\pi}{13}\right) \plus{} \cos\left(\frac {6\pi}{13}\right) \plus{} \cos\left(\frac {8\pi}{13}\right) \plus{} \cos\left(\frac {10\pi}{13}\right) \plus{} \cos\left(\frac {12\pi}{13}\right) \equal{} \minus{} \frac {1}{2}$

So $ \cos\left(\frac {4\pi}{13}\right) \plus{} \cos\left(\frac {10\pi}{13}\right) \plus{} \cos\left(\frac {12\pi}{13}\right) \equal{} \minus{} \frac {1}{2} \minus{} \cos\left(\frac {2\pi}{13}\right) \minus{} \cos\left(\frac {6\pi}{13}\right) \minus{} \cos\left(\frac {8\pi}{13}\right)$
$ \equal{} \minus{} \frac {1}{2} \minus{} P$

Where did you get this problem from?

I got this problem from the book "PLANE TRIGONOMETRY"- by S.L. Loney.
Thank you for giving the full solution.

[quote="Poincare"]Hmmm... 13 is yet another number that cannot be split to do the double angle and half angle bash...[/quote]
This is a fundamentally different problem; the evaluation of the [url=http://en.wikipedia.org/wiki/Quadratic_Gauss_sum]quadratic Gauss sum[/url] is classic but nontrivial.  Kent Merryfield has an interesting partial approach [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=286361238&t=273797]here[/url].


geometry
geometric transformation
homothety
trigonometry
function
circumcircle
similar triangles

Let $I$it is insentre of triangle $ABC$.

${A_1}\in{BC}$and${IA_1}\perp{BC}$,${B_1}\in{AC}$and${IB_1}\perp{AC}$,${C_1}\in{BA}$and${IC_1}\perp{BA}$.${N}\in{C_1B_1}$,${C_1N}={B_1N}$${A_2}\in{C_1B_1}$,and ${A_1A_2}\perp{C_1B_1}$

${M}\in{AA_2}$and$AM=MA_2$.

Let $l_A$ it is line pass trough $M$and $N$,similiar $l_B$ and $l_C$.

Prove that $l_A$, $l_B$ and $l_C$ intersect in the same point.
 ;)

This problem also has an easy trigonometric solution, which requires nothing else but the sine, cosine and Ceva's theorems. While the lines $A_1A_2, B_1B_2, C_1C_2$ are the altitudes of the contact triangle $\triangle A_1B_1C_1$, the lines $AN, BP, CR$ are the altitudes of its medial triangle $\triangle NPR$. Let $D \in PR, E \in RN, F \in NP$ be the feet of these altitude lines and let the cevians $l_A \equiv MN, l_B \equiv OP, l_C \equiv QR$ of the triangle $\triangle NPR$ meet its opposite sides at points $X \in PR, Y \in RN, Z \in NP$. The segments XP, XR are easily calculated. The right angle triangles $\triangle ANA_2 \sim \triangle NDX$, hence

${XD = ND \cdot \frac{AA_2}{AN} = \frac{A_1A_2}{2} \cdot \frac{\frac{B_1C_1}{2} - C_1A_1 \cos \widehat C_1}{\frac{B_1C_1}{2} \tan \widehat A_1}} = \frac{C_1A_1 \sin \widehat C_1}{2} \cdot \frac{B_1C_1 - 2\ C_1A_1 \cos \widehat C_1}{B_1C_1 \tan \widehat A_1}$

$DP = \frac{A_1B_1}{2} \cos \widehat B_1,\ \ DR = \frac{C_1A_1}{2} \cos \widehat C_1$. Denote $a_1 = C_1B_1, b_1 = C_1A_1, c_1 = A_1B_1$. Assuming the points P, X, D, R follow on the line PR in this order (the other cases are similar and we can always use directed line segments),

$XP = XD + DP = \frac{b_1 \sin \widehat C_1}{2} \cdot \frac{a_1 - 2b_1 \cos \widehat C_1}{a_1 \tan \widehat A_1} + \frac{c_1}{2} \cos \widehat B_1$

$XR = DR - DX = \frac{b_1}{2} \cos \widehat C_1 - \frac{b_1 \sin \widehat C_1}{2} \cdot \frac{a_1 - 2b_1 \cos \widehat C_1}{a_1 \tan \widehat A_1}$

Skipping the useless indices 1, we have

$\frac{XR}{XP} = \frac{ab \tan \widehat A \cos \widehat C - b \sin \widehat C (a - 2b \cos \widehat C)}{ca \tan \widehat A \cos \widehat B + b \sin \widehat C (a - 2b \cos \widehat C)} = \frac{ab \sin \widehat A \cos \widehat C - b \cos \widehat A \sin \widehat C (a - 2b \cos \widehat C)}{ca \sin \widehat A \cos \widehat B + b \cos \widehat A \sin \widehat C (a - 2b \cos \widehat C)}$

Substituting the following sine formulas in both the numerator and denominator, respectively,

$ab \sin \widehat A \cos \widehat C = a^2 \sin \widehat B \cos \widehat C,\ \ ca \sin \widehat A \cos \widehat B = a^2 \sin \widehat C \cos \widehat B$

and the following cosine theorem formula in both the numerator and denominator,

$2b^2 \cos \widehat A \sin \widehat C \cos \widehat C = 2bc \cos \widehat A \sin \widehat B \cos \widehat C = (b^2 + c^2 - a^2) \sin \widehat B \cos \widehat C$

$\frac{XR}{XP} = \frac{a^2 \sin \widehat B \cos \widehat C - ab \cos \widehat A \sin \widehat C + (b^2 + c^2 - a^2) \sin \widehat B \cos \widehat C}{a^2 \sin \widehat C \cos \widehat B + ab \cos \widehat A \sin \widehat C - (b^2 + c^2 - a^2) \sin \widehat B \cos \widehat C} =$

$= \frac{(b^2 + c^2) \sin \widehat B \cos \widehat C - ab \cos \widehat A \sin \widehat C}{a^2 \sin \widehat A - (b^2 + c^2) \sin \widehat B \cos \widehat C + ab \cos \widehat A \sin \widehat C} =$

Using the cosine theorem formula for $a^2$, the sine theorem formula $ab \cos \widehat A \sin \widehat C = bc \cos \widehat A \sin \widehat A$ and the formula for $\sin \widehat A = \sin (\widehat B + \widehat C)$ again:

$= \frac{(b^2 + c^2) \sin \widehat B \cos \widehat C - bc \cos \widehat A \sin \widehat A}{(b^2 + c^2 - 2bc \cos \widehat A) \sin \widehat A - (b^2 + c^2) \sin \widehat B \cos \widehat C + bc \cos \widehat A \sin \widehat A} =$

$= \frac{(b^2 + c^2) \sin \widehat B \cos \widehat C - bc \cos \widehat A \sin \widehat A}{(b^2 + c^2) \cos \widehat B \sin \widehat C - bc \cos \widehat A \sin \widehat A} =$

Now it is the time to get rid of all trigonometric functions by substituting $\sin \widehat A = \frac{a}{2r}, \sin \widehat B = \frac{b}{2r}, \sin \widehat C = \frac{c}{2r}$, where r is radius of the circumcircle (I) of the contact triangle $\triangle A_1B_1C_1$, and the cosine theorem formulas for $\cos \widehat A, \cos \widehat B, \cos \widehat C$:

$= \frac{(b^2 + c^2) b \cos \widehat C - abc \cos \widehat A}{(b^2 + c^2) c \cos \widehat B - abc \cos \widehat A} = \frac{(b^2 + c^2) \frac{a^2 + b^2 - c^2}{a} - a(b^2 + c^2 - a^2)}{(b^2 + c^2) \frac{c^2 + a^2 - b^2}{a} - a(b^2 + c^2 - a^2)} =$

$= \frac{(b^2 + c^2) (a^2 + b^2 - c^2) - a^2(b^2 + c^2 - a^2)}{(b^2 + c^2) (c^2 + a^2 - b^2) - a^2(b^2 + c^2 - a^2)} = \frac{(b^2 + c^2) (b^2 - c^2) - a^4}{(b^2 + c^2) (c^2 - b^2) - a^4} =$

$= \frac{b^4 - c^4 - a^4}{c^4 - b^4 - a^4}$

By the cyclic exchange of a, b, c, we similarly obtain

$\frac{YN}{YR} = \frac{c^4 - a^4 - b^4}{a^4 - c^4 - b^4},\ \ \frac{ZP}{ZN} = \frac{a^4 - b^4 - c^4}{b^4 - a^4 - c^4}$

$\frac{XR}{XP} \cdot \frac{YN}{YR} \cdot \frac{ZP}{ZN} =$

$= \frac{b^4 - c^4 - a^4}{c^4 - b^4 - a^4} \cdot \frac{c^4 - a^4 - b^4}{a^4 - c^4 - b^4} \cdot \frac{a^4 - b^4 - c^4}{b^4 - a^4 - c^4} = 1$

so that by Ceva's theorem, the cevians $\l_A, l_B, l_C$ of the triangle $\triangle NPR$ are concurrent.

[quote="yetti"]Let N, P, R be the midpoints of the sides $B_1C_1, C_1A_1, A_1B_1$ of the contact triangle $\triangle A_1B_1C_1$ of the triangle $\triangle ABC$ and M, O, Q the midpoints of the segments $AA_2, BB_2, CC_2$. The lines $A_1A_2, B_1B_2, C_1C_2$ are the altitudes of the contact triangle and $\triangle A_2B_2C_2$ is its orthic triangle. This orthic triangle is centrally similar to the triangle $\triangle ABC$, because their corresponding sides are parallel by the inverse transversal theorem, for example, $\angle A_2B_1A \equiv \angle C_1B_1A = 90^\circ - \frac{\angle A}{2}$ and $\angle C_2A_2B_1 = \angle C_1A_1B_1 = \frac{\angle B + \angle C}{2} = 90^\circ - \frac{\angle A}{2}$. Therefore, the lines $AA_2, BB_2, CC_2$ concur in their homothety center. The lines through the vertices A, B, C parallel to the lines $l_A \equiv MN, l_A \equiv OP, l_A \equiv QR$ are the isogonal conjugates of the lines $AA_2, BB_2, CC_2$, hence, they are also concurrent. The triangles $\triangle MOQ$ is centrally similar to the centrally similar triangles $\triangle A_2B_2C_2 \sim \triangle ABC$ with the same homothety center. The lines MN, OP, QR are parallel to the concurrent isogonal conjugates of the lines $AA_2, BB_2, CC_2$ through the corresponding vertices of 2 centrally similar triangles $\triangle MOQ \sim \triangle ABC$, therefore, they are also concurrent.[/quote]
Wery nice solution YETTI  :clap2: :D

[quote="silouan"]The problem which posted by Tiks reminds me [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5588]the BMO 2004 problem 3[/url][/quote]
SILOUAN I think you don't undrstend MY problem :o  :rotfl:

[quote="yetti"]Show that the given lines $l_A, l_B, l_C$ cut the sides BC, CA, AB of the original triangle $\triangle ABC$ at their midpoints.

Yetti[/quote]
Yetii that is  Lemme in my solution.
You can easy see that$\triangle{BC_1A_2}\sim\triangle{CB_1A_2}$ and we know this problem
 ABCDE is a  pentagon such that angleBAC=angleDAE and angleBCA=angleDAE. The lines BC and DE intersect in F.

Prove that the midpoints of BE, CD and AF are collinear.See this
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=237403&highlight=#237403]www.mathlinks.ro/Forum/viewtopic.php?p=237403&highlight=#237403[/url]=> $l_A, l_B, l_C$ cut the sides BC, CA, AB of the original triangle $\triangle ABC$ at their midpoints.

[quote="silouan"]I said that it REMINDS me this problem. :mad:  :mad:  :!: This is only an opinion .And your opinion is different . No problem man, but don't laugh so easy with the others opinion. Thank you[/quote]
Don't worry silouan,I have joked   ;)

It follows from Yetti's computations that the common point of la, lb, lc is the homothetic of the Gergonne point in (G,1/4) with G = centroid


Here's a short one:
Find all primes $p$ such that $p + 2$ and $p + 4$ are also primes.

I think this is rather something for Getting Started.

Quite easy:
We check whether $p$ can be $2$ or $3$. Now assume, that$p>3$, so we have two possibilities:
$p\equiv{1}\pmod{3}$ or $p\equiv{2}\pmod{3}$
in the first case $p+2$ is divisible by $3$, and in the second $p+4$ is divisible by $3$
The answer: $p=3$

This is one reason why there are 'twin' primes but no 'triplets' primes higher than 3,5,7 :)


Given a convex polyhedron $F$. Its vertex $A$ has degree $5$, other vertices have degree $3$. A colouring of edges of $F$ is called nice, if for any vertex except $A$ all three edges from it have different colours. It appears that the number of nice colourings is not divisible by $5$. Prove that there is a nice colouring, in which some three consecutive edges from $A$ are coloured the same way.
[i]D. Karpov[/i]

D. Karpov is my teacher in Graph theory  :roll:
I don't remember I thinc we have solvee this problem in our lessons.

The difficulty of this problem relies on using the conditions in the right way at the right time. The most important thing is that if we have a nice-coloring, then by beginning a "road" from the $ A$ vertex, suppose on an edge colored $ 1$, we can go on a road $ 1,2,1,2,1,2\dots$ for example. It's easy to see that this road must end, and the only place it can end is $ A$. From every edge from the $ 5$ connected to $ A$ there are $ 2$ roads beginning from it (for example from an edge colored $ 1$ there is a road $ 1,2,1,2..$, call it a $ 1-2$ road, and $ 1,3,1,3...$, a $ 1-3$ road, we consider $ 1-3$ road the same as $ 3-1$ road). If we take all roads from the $ 5$ edges, because they all end in $ A$ each road has to appear an even number of times(there are $ 5$ roads that appear $ 10$ times in total). Therefore it's easy to see we must have $ 3$ edges colored the same(if not we would have smth like:$ 1,1,2,2,3$, but the road $ 2-3$ appears $ 3$ times($ 2$ of them from the $ 2$-s  and one from the $ 3$) which is odd, contradiction). After with same reasonings we can show that we can't have only $ 2$ colors. Now assuming we don't have $ 3$ consecutive edges colored the same, we will have $ 5$ cases(we consider two cases equal if  by switching numbers from one case we obtain the other one):
$ 1,1,2,1,3$
$ 1,2,1,3,1$
$ 1,2,1,1,3$
$ 2,1,3,1,1$
$ 2,1,1,3,1$

We have the other observation that in a road ($ 1-3$ for example) we may switch the $ 2$ numbers (so $ 1,3,1,3...$ will become $ 3,1,3,1...$ or vice versa) and we'll still have a nice coloring. Using this we can show that in every case every road uniquely connects $ 2$ edges (else we would have $ 3$ consecutive edges colored the same by switching). If we take one case, there are $ 5$ roads, in each one by switching we will get another case (except switching the $ 2-3$ road, it's the same as switching the $ 2$ colors), so from $ 1$ case we get all other $ 4$ cases. Assuming that the first case appears the most, in each different first-case-coloring we do the switchings(and these will be all different) and we will get the number of colorings with the other cases$ \ge$ than the first one, which we assumed that appears the most, so all the $ 5$ cases appear the same number of times (we don't have to show it this way actually, it's pretty clear). So the number of nice colorings is divisible by $ 5$, contradiction. :)

[quote="N.T.TUAN"]
Prove that there is a nice colouring, in which some three consecutive
edges from $ A$ are coloured the same way.[/quote]

Well, I wonder, what "three consecutive edges from $ A$ means"...


$ \left( x\plus{}y\plus{}z \right)  \left(  \left( y\plus{}z \right) ^{\minus{}1}\plus{} \left( z\plus{}x
 \right) ^{\minus{}1}\plus{} \left( x\plus{}y \right) ^{\minus{}1} \right) \minus{}9/2\equal{}{\frac {y}{z\plus{}x}}
\plus{}{\frac {z}{x\plus{}y}}\plus{}{\frac {x}{y\plus{}z}}\minus{}3/2$


What're your opinions on Harvey Mudd for a potential compsci major?

It is good for  custruction/moving engineering. :)

see [url=http://people.bu.edu/fmri/somers/cannon.html]This link[/url]
:)

I'm a math major (not a CS major) at Mudd, but from what I do know of the CS program here it is a good one.  There are some excellent teachers in the CS department, and the CS majors seem to get a lot out of the program.    Additionally, there's a Math/CS joint major (which is becoming more and more popular), containing large portions of both the math and CS majors.

For more info, you can check out the Harvey Mudd CS website: http://www.cs.hmc.edu/

Gyan -- and it looks like MIT copied us: http://howeandser.com/

Setofallsets, thanks for the reply. What's your opinion of the college as a whole though? Are you enjoying your time there? Do you like the honor system? Etc.

I am enjoying life at Mudd.  The small size is a big plus -- one doesn't feel lost in the crowd when there are fewer than 200 students in each entering class -- and much of the faculty are both excellent teachers and very accessible.  The other Claremont Colleges (excellent schools in their own rights) offer good opportunities to take classes in areas outside of science.

For the most part, I like the community here at Mudd.  The dorms have a lot of "personality," which tends to encourage dorm loyalty and social interaction within each dorm.  The small size of the school (and of each dorm -- there are eight total) means that you know a large proportion of the people at the school, which is something I really like.

On the honor code:  the system seems to work very well.  People generally tend to take it seriously.  It's because of the honor code that a large proportion of the upper-division classes here can offer take-home midterms and expect people not to cheat on them.  It's a lot nicer to have a 3-hour time limit take-home midterm to do sometime over the weekend than to have a 1.25 hour in-class midterm.

[quote]Gyan -- and it looks like MIT copied us: http://howeandser.com/[/quote] I know, my son IM'd me :). The story is  now  in Boston Globe (as well as in LA Times). Interesting execution, but the idea was copied from Mudd :)
And a place like Mudd, has to be fun!

My friend was actually visiting caltech while the cannon was being moved. The tourguide apparently had no idea what was going on because he had told the visitors that the cannon was being "taken in for maintenence or something."

Nigma- Here is one of the many  [url=http://www-tech.mit.edu/V126/N16/16cannon.html]stories[/url] but according to this the moving was done early in the morning....

One more question setofallsets: what's the part scene like at Mudd?

Oh, and Gyan, what's your story? If you're a college student, where do you go? If not, where would you like to go? Major? Etc.

[quote="nigma"]
Oh, and Gyan, what's your story? If you're a college student, where do you go? If not, where would you like to go? Major? Etc.[/quote]
FWIW - I am just a parent,  and   recent interest is due to my sons  (and many nephews, nieces, friend's kids etc)  :) . Two son's (One in MIT, other graduating  from Duke and is unsure where he will go for graduate studies - Cornell, Duke or MIT).

Did my college shopping LONG ago. Physics. A little bit familiar with (gone for undergraduate/graduate/taught/visited_friends/met_ wife/drove_kids_and_their_stuff_to_their_dorms etc ) some Indian schools (IIT), some  American schools. :)

And yes,  would recommend Harvey Mudd. Heard nice things from people who ought to know  while investigating colleges. (There is another college,  Rose Hullman  in Indiana  which is similar in many respects for Engineering students)

I have just visited Harvey Mudd College this weekend, and as a potential math or math/cs major, I feel confident in reccomending HMC as a great school. The overall environment is really great, very laid back and definitely not anal about anything, even though there is a TON of work, from what I hear from the current students it's very engaging and "fun" work as opposed to feeling like "a really big chunk of work." Plus the faculty are second to none. Third, no grad students so you get all the labs, faculty attention, and ersearch opportunities. Also, the honor code is regarded and upheld very seriously. I am considering HMC as my top choice right now, even over Stanford and CalTech. I would definitely argue that HMC is worth at least a visit and an application.

1 last thing. They also have an automatic merit scholarship for applicants who meet certain minimum sat scores (750 math and crit reading, 700 writing I think) it's like 10k a year for 4 years, very very nice and you don't have to do anything to get it (aside from the test scores)

Thanks for the reply Iasagnaman! The merit scholarship is, I believe, for 5k ([b]Edit: this is incorrect[/b]) a year -- which is still quite substantial.

The merit scholarship is in fact now 10k / year. It used to be 5k / year, but they increased it fairly recently.

And nigma:  about the party scene -- I don't go to parties (and in fact generally tend to stay as far from them as I can), so I wouldn't be the person to ask about their quality.  North, West, and Linde (three of the Mudd dorms -- there is no Greek system here) hold parties fairly regularly, though, and there are also parties on the other 5C campuses.

[i]Edited to fix my silly addition of dollar signs.[/i]

I just recieved a booklet from them and figured it would be recent, oh well...


Given that $f(x)=2^{x}$, sequence $\{ a_{n}\}$ satisfies $a_{n}= n \cdot f(n)$, $S_{n}$ is the sum of the first $n$ terms of $\{ a_{n}\}$. Find the sum of all the terms of the sequence: $\frac{1}{S_{2}-2}, \frac{2}{S_{3}-2}, \cdots \frac{n-1}{S_{n}-2}$.


calculus
integration
function
calculus computations

If you mean $ e^{x^2}$... then this is one of the most known integrals which has no closed form representation by common functions.
We say that $ \int{e^{x^2}\,dx}=-\frac{i\,\sqrt {\pi }}{2}{\it{erf} }(i\,x)$, where $ {\it{erf} }$ is the error function which is defined ${ {\it{erf} }(x)=\frac{2}{\pi}\,\int_0^{x}{e^{-t^2}}\,dt}$, where $ x\in\mathbb{C}$.
So its no so interesting from the first look...

P.S. Write, if its possible, more clearly.

Since it has no closed form, we can use Taylor series to approximate it.

$ e^x\equal{}\sum_{n\equal{}0}^\infty\frac{x^n}{n!}$

So $ e^{x^2}\equal{}\sum_{n\equal{}0}^\infty\frac{x^{2n}}{n!}$

And $ \int{e^{x^2}\;dx}\equal{}\int\sum_{n\equal{}0}^\infty\frac{x^{2n}}{n!}\;dx\equal{}\boxed{C\plus{}\sum_{n\equal{}0}^\infty\frac{x^{2n\plus{}1}}{(2n\plus{}1)n!}}$


function
combinatorics unsolved
combinatorics

use generating functions to solve the following problem:
How many 4-tuples of integers (x1, x2, x3, x4) have the properties:

a) x1+x2+x3+x4=27
b) each Xi is at least 3 and at most 8

Hint: (1-t^n)/(1-t) =  1 +  t^2 + t^3 +...+ t^(n-1)

here's a suggestion; though probably not the best one.  so you want the coefficient of $x^{27}$ in the expansion of $(x^3+...+x^8)^4 = (\frac{x^9-x^3}{x-1})^4$.  now change this expression to $(\frac{(x+1)^9-(x+1)^3}{x})^4$.  the coefficient of $x^{27}$ is easy to compute in this new expression... then find how it differs from the coefficient of $x^{27}$ in the old expression...


factor out 1/3. $ 1\plus{}1/2\plus{}1/3\plus{}.....1/2005$ now i am stguck

[quote="abacadaea"]factor out 1/3. $ 1 \plus{} 1/2 \plus{} 1/3 \plus{} .....1/2005$ now i am stguck[/quote]

That's the hard part!  :rotfl:

There is no "nice" expression for a sum of this type.

[quote="abacadaea"]factor out 1/3. $ 1 \plus{} 1/2 \plus{} 1/3 \plus{} .....1/2005$ now i am stguck[/quote]

So what do you do after you get factor out 1/3?

Just solve or do something else?

[quote="abacadaea"]you can do it like the sum formula. i dont think there is a way besides punching the numbers into a calculator.[/quote]

Oh..duh... :lol:

how do you search things like this on AoPS wiki?

On any Wiki page, on the left hand side, you should see a box in which you type in your topic.

[quote="abacadaea"]i know but if you want a certain thing and you dont know what its called like this, should i just type in "sums consecutive fractions"?[/quote]
:read:
That describes a search.  :lol:
Just type it in and it will search every AoPS wiki article.  (You might want to be more specific or else it will give you [i]lots[/i] of results.)
[i][b]-minicon[/b][/i]

Sums of this form are called [url=http://en.wikipedia.org/wiki/Harmonic_number]harmonic numbers[/url], and they are the partial sums of the [url=http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)]harmonic series[/url].


name one person that you like/admire the most
also who here likes mafia?

STOP SPAMMING. Delete ALL his USELESS pots please!

these are just harmless questions and nice little polls????? :?:

maybe all these could be kept in one single thread.

Lets not be mean to  him, he could be new.
@Shadow...Remember, just because this is the Fun Factory, doesnt mean you get to spam. You dont need two threads. One is enough. The other one is just superfluous. If you want to start a game, add more info to your post. Your first post has like two words in it. People who look at it wont gain interest, and will be treated as spam.
So whats this thread about? Name a person who's nice? I'll say me! :P
Also, [b]yes[/b] I like mafia. This game doesnt seem very interesting though...I think we should just let this die.

Thats pretty easy. I could pm you cuz i dont want people to know.

i bet he is new because I haven't seen him before.

um. ShadowBlast, these polls because they are related to Mafia, and a little bit like a game, could you ask a mod to transfer it to the fun factory or game forum please?Thanks!
EDIT: oops, sorry it's alreay there!
Thanks anyway

I think Shadow is banned now. and shadow is really annoying (he's a 6th grader @ my school) :mad:

Alright, Cricket let's not be mean. Just because he is at your school, and you know him, just don't treat him unfairly. Also, please don't gloat over if he has been banned or not. It doesn't really matter. Also, don't say he's really annoying, because that is quite mean. And, let's just lock this thread.


Is there easy method to solve this problem.?

$\frac{a}{b}=\frac{(100^3-1)(101^3-1)}{(101^3+1)(102^3+1)}$
where $gcd(a, b) = 1$ and $a, b$ be any positive integer.

find $a + b$  :blush:

[hide]
Well, can't you just simplify the fraction? Giving you $\frac{1650}{1751}$ those two have a gcd of 1, so the answer would be 3401.
[/hide][/hide]

Now show how you simplified it without a calculator :P

I guess you could factor 101^3 as (100+1)^3 or 102^3 as (100+2)^3 then simplify...thats what I would do at least.  Its not extremely fast, but the arithmetic shouldn't take too long.

Yeah. Thanks everyone . Now I know the best way to solve it.  :D

[hide]

The trick is to realize that
(a^3-1)/((a+1)^3+1) =  [b](a-1)/(a+2)[/b]

(This is from:
 a^3-1  =  (a-1)(a^2+a+1) and  (a+1)^3+1 = (a+2)(a^2+a+1)

Thus (100^3-1)/(101^3+1) = 99/102  = 33/34   etc..

So the expression is (33*100)(34*103)  = (33*50)/(17*103)

Hope this helps.

Some more comments:

BTW Another helpful hint,  (may be helpful in some other cases),  here (if the  numbers were larger   ) to  realize that only common factors of  (a-1) and (a+2)  can be 3... (and    say of  a and a+2 can be 2 .. etc.. thus   in the original
 (99*100)/(102*103) only numbers  that you have to check out  to make sure the gcd is 1 are the  factors  2 and 3 [/hide]

[i]Mod Edit - Hidden text appreciated.[/i]


geometry
inequalities
circumcircle
similar triangles
inequalities solved

If a,b,c are the sides of a triangle of area A and x,y,z are arbitrary real numbers prove the inequality

((ax+by+cz)/4A)^2 \geq yz/bc+zx/ca+xy/ab

Let P be a point inside of ABC, PX, PY and PZ are perpendiculars to BC, AC and AB.
We can find P such that PX:PY:PZ=x:y:z. Then a*PX+y*PY+z*PZ=2SABC.
So we must prove 1/4 \geq PX*PY/ab+PY*PZ/bc+PZ*PX/ca  <==> abc/4R \geq \sum PX*PY*c/R (where R is a radius og circumcircle) <==>
SABC \geq \sum 2*PX*PY*sin(C) <==> SABC \geq \sum 4SPXY = 4SXYZ where XYZ is a pedal triangle.
There is a formula for area of pedal triangle: SXYZ = 1/4*|1-PO2/R2|*SABC, where O is circumcenter.

The second line contains small misprint. The correct identity is a*PX+b*PY+c*PZ=2SABC.

On proof of pedal triangle's area.

Let X1, Y1, Z1 are points of intersection of AP, BP, CP and circumcircle.
Let S1=SXYZ and S2=SX1Y1Z1.
It is easy to prove:
1) XY=AB*CP/2R, ...
2) X1Y1=AB*X1P/BP
3) XYZ and X1Y1Z1 are similar triangles
4) X1P*AP=|OP2-R2|
5) S2/SABC=X1Y1*Y1Z1*Z1X1/AB*BC*CA
6) X1Y1/AB=X1P/BP=|OP2-R2|/(AP*BP)
Using 3) we have S1/S2=k2, where k=XY/X1Y1=BP*CP/(2R*X1P).
Using 4) we obtain S1/S2=(AP*BP*CP)2/4R2(OP2-R2)2.
Using 5) and 6) we obtain S2/SABC=|OP2-R2|3/(AP*BP*CP)2.
Finally S1/SABC=(S1/S2)*(S2/SABC)=|OP2-R2|/4R2.


inequalities
logarithms
LaTeX
Irrational numbers

Prove that for any $ p, q\in\mathbb{N}$ with $ q > 1$ the following inequality holds:
\[ \left\vert\pi\minus{}\frac{p}{q}\right\vert\ge q^{\minus{}42}.\]

Wrong part of 'Reference' deleted.
(K. Mahler, 1953) AMM, Problem 10630, Richard Strong should changed to (K. Mahler, 1953)

Remarks from PEN book: This is a deep theorem in transcendental number theory. Note that it follows from this result that $ \pi$ is irrational! In fact, it's known that for suciently large q, the exponent 42 can be replaced by 30. Here is a similar result due to A. Baker: for any rationals $ \frac{p}{q}$, one has
\[ \left\vert\ln 2\minus{}\frac{p}{q}\right\vert\ge 10^{\minus{}100000}q^{\minus{}12.5}.\]

[quote="Peter"]Why do we need such "deep theorems" into PEN? Probably none of our readers can prove it, and it is not elementary either?[/quote]

When I saw it first, it was very impressive for me. So that's why I include it into PEN.
As Peter mentioned, the proof is highly non-elementary, so let's remove those difficult problems.
(Peter, but let's keep the latex source of those deleted problems for the cases.)


So I installed all the needed files for LaTeX, and when I open Texnic Center, a Configuration Wizard pops up.
It asks to enter the full path of the executables (latex, tex, etc.) of my TeX-distribution.
So I entered Program Files->Miktex 2.7->miktex->bin.
However it says that "The application 'LaTeX' has not been found in the directory of your LaTeX-distribution". What do I do?
By the way, my OS is Windows XP.

To check that you have entered things correctly:
1. Check that latex.exe is in the C:\Program Files\MiKTeX 2.7\miktex\bin directory
2. Rerun the Wizard (Build->Define Output Profiles->Wizard) then press the ... button to browse to C:\Program Files\MiKTeX 2.7\miktex\bin which will enter the directory automatically and ensures that there are no typos.


where are your ideas. it is very nice and maybe hard.

Indeed, without Peron's criterion the problem is hard. But using this criterion (which, I think, has already been extensively discussed in theorems section) the problem is obvious.

Yes, that's right harazi :)
Valentin said it here : http://www.mathlinks.ro/Forum/viewtopic.php?t=24007
Here's the Perron's criterion thread : http://www.mathlinks.ro/Forum/viewtopic.php?t=150


[hide="answer"]Simply look at a pattern $mod 3$ and we see that the answer is $7$[/hide]

[quote="easyas3.14159..."][hide]You only need to worry about the last digit, 3. Units digits of powers of 3 go in cycles of 3,9,7,1,3,9...
2003/3 has a remainder of 2, so the units digit is 9.[/hide][/quote]

Your suppose to divide 2003 by four to get a remainder of three.

[quote="easyas3.14159..."][hide]You only need to worry about the last digit, 3. Units digits of powers of 3 go in cycles of 3,9,7,1,3,9...
2003/3 has a remainder of 2, so the units digit is 9.[/hide][/quote] ur wrong the answer is 7.

When working with problems like this, its helpful to find a power of 13 that leaves a remainder of 1 or -1 (this follows from the fact that if we know a is congrument to b modulo m, then a^n is congruent to b^n modulo m)
Euler's Thm. gives you a power very quickly. It gives 13^4 is congruent to 1 modulo 10. Then 13^2003 is congruent ot (13^4)^500 * 13^3, which is in turn congruent to 1*13^3, congruent to 9*3, congruent to 7.
Let me know if you need more explanation with the number theory reasoning.


Determine all triples $\left( x,y,z\right)$ of positive integers for which the number \[ \sqrt{\frac{2005}{x+y}}+\sqrt{\frac{2005}{y+z}}+\sqrt{\frac{2005}{z+x}} \] is an integer .

None I suppose?
Each of 2005/(x+y), 2005/(y+z), 2005/(x+z) must be an integer.
Since 2005 is odd, x+y, y+z, x+z must be odd, so their sum 2(x+y+z) should be odd, contradiction.

Well, maybe I am wrong but you used the following statment :
[color=blue]If $\sqrt{a} +\sqrt{b} +\sqrt{c}$ is an integer with $a,b,c \in \mathbb Q^{+}$ then each of $a,b,c$ must be an integer.[/color]
This is false, because for example : Take $a=\frac{9}{4}$ , $b=\frac{25}{4}$ and $c=\frac{36}{4}$
We have $\sqrt{a} +\sqrt{b} +\sqrt{c}=7$ is an integer , but $a$ and $b$ aren't integers.

Simple.
one can easily show that for the sum of the three square roots to be an integer,
each of the terms should be a perfect square.
So 2005/(x+y) = (p/q)^2 ,...
Now 2005= 5(401) none of them being perfect squares.
Hence x+y = 2005 n1^2
y+z=2005n2^2
z+x=2005 n3^2
and finally 1/n1 + 1/n2 +1/n3 is an integer.
Note that the triple (n1,n2,n3) satisfying the above is a solution for (x,y,z) iff n1+n2+n3 be an even number.
checking the cases we see easily that (n1,n2,n3) = (2,2,4),(2,4,2),(4,2,2)

-Ali

what do u think of : sqrt((2-sqrt(3))^2)+sqrt((2+sqrt(3)^2))+sqrt(4)=8 ??
and u also forget to check in the expression the values u found !!!

here is what i think of your example:
[2-sqrt(3)]^2 = 2005 / (x+y) is impossible!!
Since each term is a rational it should be the perfect square of the rational.
And i don't see exactly what i need to check? :?
Anyways, i have to go now!

I think there is one triple (14*2005, 2*2005, 2*2005). Is it true?

Yes, it is :)
BTW: I've seen some theorems about sums of square roots being an integer, but don't know where to find it now... Do you know?

[quote="ali"]
one can easily show that for the sum of the three square roots to be an integer

-Ali[/quote]
 i cannot show

I didn't understand exactly what Ali meant, but here is the key idea [I just hope it's not the same thing what Ali said]:

[i]If $a,b,c\in\mathbb{Q}$ and $\sqrt{a}+\sqrt{b}+\sqrt{c}\in\mathbb{Q}$ then $\sqrt{a},\sqrt{b},\sqrt{c}\in\mathbb{Q}$.[/i]

[b]Proof:[/b]

Because $s=\sqrt{a}+\sqrt{b}+\sqrt{c}\in\mathbb{Q}$, by squaring ,subtracting $a+b+c\in\mathbb{Q}$ and dividing by 2, we get $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\in\mathbb{Q}$. Let $\sqrt{abc}=p$. Repeating again the operation we get $\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})=s\sqrt{p}\in\mathbb{Q}$, that is $\sqrt{abc}\in\mathbb{Q}$.

Now, $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\sqrt{a}(\sqrt{b}+\sqrt{c})+\frac{\sqrt{abc}}{\sqrt{a}}=\sqrt{a}(s-\sqrt{a})+\frac{\sqrt{p}}{\sqrt{a}}\in\mathbb{Q}$. It results $\sqrt{a}\left(s+\frac{\sqrt{p}}{a}\right)-a\in\mathbb{Q}$. It results $\sqrt{a}\in\mathbb{Q}$. The same way we get that $\sqrt{b}$ and $\sqrt{c}$ are rational.
___
From now, the solution is easy, with the solution indeed being $(2005\cdot2,2005\cdot2,2005\cdot14)$ and its permutations.


this one is probably easy but:

Sum (i=1..n) (1+1/2^i)<3

It is well-known that for x  \geq  0, we have e x   \geq  1 +x.
Let P_n be the product on the lhs.
Then :
P_n  \leq  e 1/2 + 1/4 + ... 1/2  n    < e 1/2 + 1/4 + 1/8 + ...   = e < 3.

Pierre.


who plays connect 4? I just play on a piece of paper against a friend if I don't have anything else to do. (sometimes during class :D )

Oh what are your tictactoe results? I played 26 games in one day once, tied 26

Bah gets sort of boring. But I like to play it when nothing else to do, or  :lol: during class, HEALTH CLASs especially

Ha,
I used the actual set when I was in Elem.... it was very, very popular. For some reason...
Tic-tac-toe...
I play that at restaurants :)

I used to pwn at this game. Then I stopped playing for some reason.

Actually now Im wondering why I'm still Playing it !

In middle school they don't have any more board games and stuff... sigh, I want to release my inner irresponsible kid again :P

No one ever wins when I play tic-tac-toe, which is probably why I stopped playing it. I get a tie every single time.


linear algebra
matrix
linear algebra unsolved

Given A and B are two commutive matrix in M(R) (i.e AB=BA).Prove:
           $det(A^2+B^2)\geq 0$

Work complex
If you have $AB=BA$ you have $(A+i B)(A - i B)=A^2+B^2$

so the determinant on the right side is the product of the determinants of the two factors, but those two determinats are complex conjugates, so the determinant of $A^2+B^2$ is positive


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Let $ G$ be a finite , non-cyclic group with the property every subgroup of $ G$ is abelian prove $ G$ has a non-trivial normal subgroup

I have not been in contact with Algebra for a while, so you should be critical of what I am going to say here;-

1. Let H be a non-trivial proper subgroup of G. Note that every subgroup of H is normal since H is abelian.

2. Let $ \phi : G \rightarrow H$ be a homomorphism and $ \psi: H \rightarrow H/N$ another homomorphism, where N is a normal subgroup of H.

3. Show that $ \phi^{\minus{}1} (N) \leq G$.

4. Show that $ \phi^{\minus{}1} (N)$ is the kernel of the map $ \psi \circ \phi: G \rightarrow H/N$ [a group is normal if and only if it is the kernel of some homomorphism onto some group].

5. What if $ \phi^{\minus{}1} (N)$ is trivial? We look back at the hypotheses to see if we can rule this out.

first of all thanks for the reply. in number 2:

if the homomorphism $ \varphi$ is not trivial  since  $ G$ is a finite group

and $ |G|>|H|$

the $ \ker\varphi$ would not be trivial hence it would be a normal subgroup of $ G$

I don't see how this works.  Why are you guaranteed a homomorphism of G into some proper subgroup of G? And why are you guaranteed the kernel is non-trivial, i.e., ker != G?

The result of the OP is an old theorem from about 100 years ago.  A modern treatment, easily understood, is given by Ken Brown at:
http://www.math.cornell.edu/~kbrown/4340/simple.pdf

Steve

no i was not  trying to prove it was  working the contrary... i meant if we "had" this homophrism we would have finished since
 we dont explain its existence something is wrong....

Ken Brown's exposition is quite nice for this specific question.  If one were interested in more information about groups all of whose proper subgroups were abelian, then one should look for "minimal non-abelian groups" and the synonym "Miller-Moreno groups" (the people who classified them in [1]).  Not only do they have a normal subgroup, but if they are not a p-group, then they have exactly one non-identity proper normal subgroup; it is a vector space over a finite field, and the quotient is a cyclic group acting irreducibly on the vector space.

[1] G.A. Miller and H.C. Moreno, Non-abelian groups in which every subgroup is abelian, Trans. AMS 4 (4) (1903), pp. 398–404, http://dx.doi.org/10.2307/1986409

I wonder if this is correct.
Denote $ A_x$ by all the elements that comunicate with x. This is a subgroup. If Z(G) is non trivil, we are done. Suppose Z(G)={1}. Then $ A_x\neq G$ for all $ x\neq 1$.

1. for any x,y, $ xy\neq yx$ then $ A_x\bigcap A_y\equal{}{1}$.  If else, set $ u\in A_x\bigcap A_y$ consider $ A_u$, since all elements of $ A_x,A_y,A_u$ comunicate with u (all of these subgroups are abelian), then $ A_x,A_y\subseteq A_u$(by definition). But this contradict with the fact that $ A_u$ is abelian and $ x,y\in A_u$ and $ xy\neq yx$.

2.If $ x\in C(y)$, $ A_x\neq A_y$.  Consider $ N_G(A_x)$, if $ G\subsetneq N_G(A_x)\subsetneq A_x$, since it's abelian, all of its elements comunicate with x, so $ N_G(A_x)\equal{}A_x$, a contradiction. If $ N_G(A_x)\equal{}G$ we are done. Suppose $ N_G(A_x)\equal{}A_x$, that is to say, any $ z\notin A_x$, $ zA_xz^{\minus{}1}\bigcap A_x\equal{}{1}$, so x,y are in different "A-class", which prove the assertion.

3.Count number and get contradiction. Choose $ x\neq 1$, consider $ A_{x_i}$, $ x_i\in C(x)$. $ |C(x)|\equal{}\frac{|G|}{|N_G(A_x)|}\equal{}\frac{|G|}{A_x}$. Set $ H\equal{}G\minus{}\bigcup_i A_{x_i}$, |H|=|C(x)|-1>0, so H contains at least one non-trivil element, say h. we have $ C(h)\subseteq H$,so $ |C(h)|<|C(x)|$. However, x is choosen arbitrarily (and by 2, x can't be contained in any $ A_{h'}, h'\in C(h)$), so change h,x we get the oposite ineq, a contradiction.


Divide both sides by 27 and then apply Power Mean-Arithmetic Mean inequality.

I have two solutions. I like this one since it is very short. Is it correct?

\[ \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\geq \frac{a+b+c}{3} \]\[ \Leftrightarrow\frac{a^3+b^3+c^3}{3}\geq \frac{(a+b+c)^3}{27} \]\[ \Leftrightarrow9(a^3+b^3+c^3)\geq (a+b+c)^3 \]

[quote="yassin88"]Let a,b,c be positive real numbers, prove:

\[ 9(a^3+b^3+c^3)\geq (a+b+c)^3  \][/quote]
Another method is to apply the Generalized Holder Inequality,
$(1^3+1^3+1^3)(1^3+1^3+1^3)(a^3+b^3+c^3)\geq (a+b+c)^3$. ;)

Anyway,I don't know why you need such a very strong instrument to prove this one since from Chebyshef we have a stronger one:

                                            \[ 3(a^{3}+b^{3}+c^{3})\geq (a+b+c)(a^{2}+b^{2}+c^{2}). \]

[quote="Virgil Nicula"]I am satiated with these very easy inequalities here.
Here is the [b]Olympiad Section[/b] !!!
The manager Valentin Vornicu & Co must send these problems to other sections.[/quote]
agree. These problems should exist in pre-olympiad section.

[quote="yassin88"]Let a,b,c be positive real numbers, prove:

\[ 9(a^3+b^3+c^3)\geq (a+b+c)^3  \][/quote]

I know generalization of this problem her it $(a^k+b^k+c^k)/3\geq ((a+b+c)/3)^k$ for any real $k\geq 0$
And this problem has very easy solution with jensen ineq. :D

Virgil do you want that I solve it ? :roll:  :D

No, Tiks ! This inequality is classically. It is a generalization of "your generalization" !
I think now you understand me.

I understand you,but this isn't my generalization,this problem is well-know a long time :)

I just want to use as simple idea as possible. This is a homogenous inequality. Without loss of generality, we may assume that $a+b+c=3$, so that it suffices to show that $a^3+b^3+c^3\geq3$. By AM-GM inequality, we have $a^3+1+1\geq 3a$, adding two other inequalities $b^3+1+1\geq3b$, and $c^3+1+1\geq 3c$, we have $a^3+b^3+c^3+6\geq3(a+b+c)$, which is $a^3+b^3+c^3\geq 3$, since $a+b+c=3$. This completes the proof.


Simplify $f(x)=\sec(2\sin^{-1}\frac1x)$ and find its domain.

[hide]
The domain  is $-1\leq 1/x\leq 1$ $implies x\leq -1$ , $x\geq 1$

Letting $\sin^{-1}\frac{1}{x}=y$ $\implies \sin y=\frac{1}{x}$

So $\sec 2y=\frac{1}{1-2\sin^2y}=\frac{x^2}{x^2-2}$

[/hide]

[hide]$x^2-2\neq 0\Longleftrightarrow x\neq\pm\sqrt2$
So, the domain should be $(-\infty,-\sqrt2)\cup(-\sqrt2,-1]\cup[1,\sqrt2)\cup(\sqrt2,\infty)$.[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra theorems

One active member on mathlinks ask more explanation on Mac Coys theorem

http://mathworld.wolfram.com/McCoysTheorem.html

[quote="posabogdan"]Please explain me this theorem http://mathworld.wolfram.com/McCoysTheorem.html
.I know this theorem but for a polynomial with 1 variable.For example :if A have the eigenvalue 0 and 1 then $ A^2\plus{}I_{n}$ has eigenvalue $ f(0)$ and $ f(1)$ where $ f(x)\equal{}x^2\plus{}1$....Can you give me a exemple with this teorem http://mathworld.wolfram.com/McCoysTheorem.html ?Maybe i can't understand it because i don't know very much about polynomial with 2 variables but an example will make me understand
Thx![/quote]

Can someone give more explanation on this theorem ?

Eventually some situation where we an use this ?

Here is an excerpt from Prof Olga Taussky's [i]How I Became a Torchbearer for Matrix Theory[/i] which is published in AMM, Nov $ 1988$, pp. $ 801\minus{}812$. (I strongly recommend you to read the article in full, in the case you haven't done so already.)

"[b]The theorem of McCoy[/b]. John Todd and I spent the first year of World War II, $ 1939\minus{}40$, in Ireland on leave from our positions in the University of London. I had no assigned duties during the first term, and found the library at Queen's University very appealing. Among items that I had not seen before was the work of McCoy, in particular his well known characterization of pairs of matrices which can be transformed simultaneously to upper triangular form.

Precisely, here is McCoy's result. Let $ A, B$ be two $ n\times n$ matrices, and suppose they have the following property: for every polynomial $ f$ in two variables, the matrix $ f(A, B)$ has for its eigenvalues the numbers $ f (\lambda, \mu)$, where $ \lambda, \mu$ are suitably ordered eigenvalues of $ A$ and $ B$, respectively. McCoy showed that for $ A, B$ to have this property it is necessary and sufficient that a matrix $ T$ exists such that both$ T^{\minus{}1}AT$ and $ T ^{\minus{}1}BT$ are upper triangular.
Since this work was kind enough to jump off of the library shelves into my hands, I started a correspondence with McCoy, and my interest in matrices was further strengthened. Drazin, Dungey, and Gruenberg $ [1951]^{\dagger}$ gave a more elementary proof of McCoy's theorem, but my $ 1957$ proof, via the radical in abstract algebras, takes only a few lines. See also Flanders' treatment."

PS.$ \dagger$ The same article is cited at the end of Prasolov's[i] Problems and Theorems in Linear Algebra[/i].


integration
algebra
polynomial
function
real analysis
calculus
real analysis solved

[quote="liyi"]$f\in L([0,1])$ and it holds $\int_0^1 x^n f(x)dx = \frac{1}{n+2}$ for all $n=0,1,2,\dots$.
Show that $f(x)=x, \mbox{ a.e. } x\in[0,1]$.[/quote]

 It suffices to prove that if $\int_0^1 x^n g(x)dx = 0, (1)$ then $ g= 0$ a.e and then apply to $f(x) - x$, right?
 $(1)$  holds for every polynomial instead of $x^n$.
 You then can utilize the fact that the polynomials are dense in $C\left([0,1]\right)$, which in turn is dense in $L\left([0,1]\right)$.
 I think so!

[quote="liyi"]Yes, it is dense in $L^1$ sense: $\forall \epsilon>0$, there exists a polynomial $p(x)$ such that $\int_a^b |f(x)-p(x)| < \varepsilon$...

but i need $f$ to be bounded to solve the problem ....  :?[/quote]

 Why you need $f$ to be bounded?

if $f$ is bounded,
$\left|\int_a^b f^2(x)dx - \int_a^b f(x)p(x) dx\right| \leq \int_a^b |f^2(x) - f(x)p(x)| dx \leq \int_a^b |f(x)||f(x)-p(x)|dx \leq M\int_a^b |f(x)-p(x)|dx$
because $\int_a^b f(x)p(x)dx = 0$ we have
$|\int_a^b f^2(x)dx|\leq M\epsilon \to 0$
hence $f(x)=0$ a.e.

This is much harder than I thought! :D
 You have $\int_{0}^{1}f(x)p(x)dx = 0$ for all polynomials $p$
 Then you have $\int_{0}^{1}f(x)g(x)dx = 0$ for all continuous functions $g$
 Then you have $\int_{0}^{1}f(x)h(x)dx = 0$ for all simple functions $h$ ( using Lusin's theorem, and this is a step where the absolute continuousness of the Lebesgue integral is needed )
 From that it can be easily deduced that $f = 0$ $ a.e$

Thank you, dickchimney.
This problem can be moved to "Solved problems" section.

i have another approach.

Let $F(x) = \int_a^x f(t)dt$ and then $F(b)=0$. Integrate by part (this is valid)
$\int_a^b x^n f(x)dx = x^nF(x)|_a^b - n\int_a^b x^{n-1}F(x) = 0 \quad (n\geq 1)$
hence
$\int_x^n F(x)dx = 0$ for n=0,1,...

F(x) is continuous  :lol: $\forall \epsilon>0$, there exists a polynomial $P(x)$ s.t. $|P(x)-F(x)|<\epsilon \forall x\in[a,b]$. $\int_a^b P(x)F(x)dx = 0$ hence $\int_a^b F^2(x)dx = \int_a^b F(x)(F(x)-P(x))dx$ then we have $\int_a^b F^2(x)dx \leq \epsilon\int_a^b |F(x)|dx$. $\epsilon$ can be arbitrary small, therefore $F(x)=0$ and $f(x)=0$ a.e.


during the class, latex stuff loads very slowly.
Is it just my computer?

Probably - are you on a very slow internet connection?

My computer is slow...but yesterday, during intermediate algebra, it didnt load at all.   :(

This sounds like a Java problem on your computer.  I suggest trying a few things:

1) Use Opera (http://www.opera.com)

2) Re-install Java.  There's a link on the classroom page for where to go to do this.


limit
logarithms
inequalities
calculus
calculus computations

Let $\displaystyle \left( x_n \right)_{n \geq 1}$ be a sequence of numbers defined by \[ x_{n+1} = \frac{1}{\displaystyle \sum_{k=1}^n \frac{x_k}{k}}, \ \forall n \in \mathbb N^{\ast}, \ x_1 \in \left( 0,\infty \right) .  \]

(a) Prove that $\displaystyle \left( x_n \right)_{n \geq 1}$ is convergent and compute its limit.

(b) Let $\displaystyle a \in \mathbb R$. Compute $\displaystyle \lim_{n \to \infty} n^a \cdot x_n$.

[i]Romanian MO 2006, First Round, 11th Grade[/i]

It seems $x_n \sim \frac{1}{\sqrt{2\ln n}}$
Could someone check if this equivalent is reasonable ?
Thank you

Playing with it numerically:

Let $y_n=x_n\sqrt{2\ln n}$

It looks like $y_n<1$ for $n\ge 3.$ It also looks like $y_n$ is eventually increasing, probably increasing for $n\ge 4.$

If $x_1$ is not near $1$ - either large or small - then $y_n$ quickly becomes much smaller than $1$ and then grows very slowly.

Examples: if $x_1=.1,$ then $y_{1000}=.603,$ approximately. And if $x_1=.001,$ then $y_{1000}=.0074,$ approximately.

None of this is inconsistent with Moubinool's conjecture, although it does point out the need to account for the effects of an early large term.

Of course, if Moubinool's conjecture turns out to be correct, or even if my somewhat weaker empirical observation (that $y_n$ is less than $1$ but increasing) were to be true, then the answers to perfect_radio's questions would be (a) zero, and (b) infinity for $\alpha>0.$

[quote="Kent Merryfield"][quote="kunny"]$x_{n+1}\leq \frac{\sqrt{n}}{2}.$[/quote]
I assume that's for $n\ge 2,$ as one can produce sequences with arbitrarily large $x_2.$

This inequality is far weaker than Moubinool's conjecture, and in fact, too weak to be used in answering the original questions.[/quote]

You are right.

kunny

Moubinool's conjecture seems very nice. I'd like to see a proof of that. (supposing that it's true)

For the problem I used just the fact that the sequence is (eventually) strictly decreasing.

I proved Mubimool conjecture:

Let $y_n=\frac 1 {x_n}$. After writing the equality in the problem for $n$ and $n+1$ we obtain that:
\[ y_{n+1}=y_n+\frac 1 {ny_n}\text{ (1) } \]
We easily get that $y_n$ is increasing and goes to infinity as $n\to infinity$.
Indeed, assume that it is bounded, and denote by $\ell$ it's limit. we have that $y_{n+1}\ge y_n+\frac 1 {n \ell}\Rightarrow y_{n+1}\ge\frac 1 \ell \sum \frac 1 n \sim \ln n \to \infty$. Contradiciton.
From the fact that $y_n$ is incresing we get that, $0\le \frac{y_{n+1} }{y_n}-1=\frac 1{ny_n^2}\le \frac 1 {y_1^2}\frac 1 n$    (2)

From (1) we get that
 \[ \frac{y_{n+1}^2-y_n^2}{y_{n+1}+y_n}=\frac 1 {ny_n}\Rightarrow \frac{y_{n+1}^2-y_n^2}{\frac{y_{n+1}}{y_n}+1}=\frac 1 {n}\Rightarrow {y_{n+1}^2-y_n^2}=(\frac{y_{n+1}}{y_n}+1)\cdot \frac 1 {n} \]
Using (2) we get that:
\[ 2 \frac 1 n\le y_{n+1}^2-y_n^2 \le (2+\frac 1 {y_1^2}\frac 1 n)\frac 1 n \]
Summing up with respect to $n$ from $0$ to $p-1$ we get:

$2\sum \frac 1 n+y_1^2\le y_p^2\le 2 \sum \frac 1 n +\frac 1 {y_1^2}\sum \frac 1{n^2} +y_1^2$

And now both right and left term are $\sim 2 \ln p$. This implies that $y_n^2\sim 2 \ln n\Rightarrow x_n\sim \frac 1 {\sqrt{2\ln n}}$ .

What about the next term could it be  like

$x_n-\frac{1}{\sqrt{2\ln n}} \sim \frac{b}{n(\ln n)^{3/2}}$ ?


inequalities
function
absolute value
algebra unsolved
algebra

Let f:Q* to R satisfied:
f(x+y) \leq f(x)+f(y)
f(xy)=f(x)*f(y)
f(1997)=1
Prove:f(2000) \leq 1

[quote="galois"]doesnt the condition f(x)f(y)=f(xy) ==> f(x)=x^c for some
constant c \varepsilon R[/quote]

No. Smth else must be known about f.

It's similar to the famous Cauchy eqn f(x+y)=f(x)+f(y): this isn't enough to prove that f(x)=ax for some a \in R, we need to know something extra: f continuous, f monotonic, f is bounded on on a bounded interval etc.

Does anyone have a solution to this beautiful problem?

I also solved it after 1 year from ThanhBinh12T's post.
Here the link to my post. I hope it is more readable then this ;) http://www.mathlinks.ro/Forum/viewtopic.php?t=23869

For pedagogical reasons, I post the general result, due to Ostrowski: any absolute value on the set of rational numbers is equivalent either to the usual absolute value on $R$ or to the $p$-adic absolute value for some prime number $p$. The proof is of course identical.


assume that $p>3$ be a prime.and let $a_{1}+\dots+a_{p}=2p$ $n<2p$,prove that the polynomial $x_{1}^{a_{1}}.\dots.x_{p}^{a_{p}}+x_{1}^{n}+\dots+x_{p}^{n}-1$ is irreducible.

Sam-n, have you found any proof till now? Because I looked in Prasolov's book and I think he used a previous problem, which is clearly wrong: he says that if $f(x,x,...,x)$ is irreducible then $f(x_{1},...,x_{n})$ is also irreducible. What about $f(x,y,z)=(x^{2}-yz+1)(y^{2}-zx+1)$? Therefore his approach is completely wrong. Unfortunately, in my many attemps to solve this problem, I failed. I do not deserve your gift.   :blush:

It is irreducible, because irreducible $x^{2p}+px^{n}-1$.

Indeed, $f(x,x,...,x)$ is irreducible. But, as I said before, this obviously does not imply the conclusion of the problem.

We can consider $x_{i}=b_{i}x$ or $b_{i}x+c_{i}$, were $b_{i}, c_{i}$ constants.

As usual, I have no idea what you want to say...


For each positive integer $ n$, the mean of the first $ n$ terms of a sequence is $ n$. What is the $ 2008$th term of the sequence?

$ \textbf{(A)}\ 2008 \qquad
\textbf{(B)}\ 4015 \qquad
\textbf{(C)}\ 4016 \qquad
\textbf{(D)}\ 4,030,056 \qquad
\textbf{(E)}\ 4,032,064$

[hide]The average of the first $ n \minus{} 1$ numbers is $ n \minus{} 1$ and therefore the sum of the first $ n \minus{} 1$ numbers is $ (n\minus{}1)^2\equal{}n^2 \minus{} 2n \plus{} 1$. Likewise, the sum of the first $ n$ numbers is $ n^2$. Therefore, the $ n$th number is $ n^2\minus{}(n^2\minus{}2n\plus{}1) \equal{} 2n \minus{} 1$, so the $ 2008$th term is $ \boxed{4015}$, (B).[/hide]

It is easy to see that $ a_n$ is an odd number by induction. Hence the answer must be $ \boxed{B}$.

[hide]

the sum up to the 2008th term is equal to the sum up to the 2007th term plus the 2008th term.

since the average is equal to n, we have

sum/n=n and therefore sum = n^2

so the 2008th term would be:

(sum 2008) - (sum 2007) = (2008^2)-(2007)^2 = 4015

Therefore the answer is (B)

[/hide]

[hide="yaargh"]All numbers in the sequence are odd because of perfect squares, so the answer is B.[/hide]


Find the [b]sum of all real values [/b]of k such that the equation
    $x^{2}-(k-2)x+1$ [b]= 3X   [/b]
[Here X denotes the modulus of x]

sorry, the question actually reads
Find the sum of all real values of k such that the equaton
$x^{2}-(k-2)x+1 = 3X$
has three real roots.
Here $X$ denotes the modulus of $x$.

By the RHS, you could have just written 3modx.

The would make it easier to see.

The ultimate question is x^2-(k-2)*x+1=3 mod x. Assuming x is positive integer and less than or equal to 3 i.e 0<= x <=3 we have three values of x viz. 1,2,3 as 0 is not permitted as a mod operation's right hand operator.Now for each values of x from the above equation we get
for x=1,k=4;x=2,k=4;x=3,k=16/3.So the summations of this values of k i.e 4+4+16/3=40/3.This should be the answer.


Consider the complex equation $ (x\plus{}2003i)(x\plus{}2004i)(x\plus{}2005i)\equal{}2002i$, it has a root $ x\equal{}a\plus{}bi$, where $ a,b\in\mathbb{R^\plus{}}$. Find the value of $ a$.

Let $ x \equal{} \minus{} i(y \plus{} 2004)$, then $ a \equal{} Im(y)$ and $ y^3 \minus{} y \equal{} 2002$. If $ y_0 \equal{} 2b$ is real root, then $ y_{1,2} \equal{} \minus{} b\pm ai$ fnd can be calculate by Cardano formul or $ a\equal{}\sqrt{3b^2\minus{}1}$.