Let  $ x_{n} ,n\geq1$ be a sequence such that     $ [ x_{n+1}-\dfrac{1}{n}] =\lbrace x_{n}+\dfrac{1}{n}\rbrace$ Find $ \lim x_{n} n\rightarrow\infty$

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$
 :rotfl:

[quote="Pirkuliyev Rovsen"]Let  $ x_{n} ,n\geq1$ be a sequence such that     $ [ x_{n + 1} - \dfrac{1}{n}] = \lbrace x_{n} + \dfrac{1}{n}\rbrace$ Find $ \lim x_{n} n\rightarrow\infty$  [/quote]

LHS is an integer while RHS is in $ [0,1)$. So both are $ 0$.

So $ \{x_n+\frac 1n\}=0$ implies $ x_n+\frac 1n\in\mathbb Z$ and so $ x_n=a_n-\frac 1n$ for some $ a_n\in\mathbb Z$

Then $ [x_{n+1}-\frac 1n]=[a_{n+1}-\frac 1{n+1}-\frac 1n]=a_{n+1}-1$ $ \forall n>1$

So $ a_n=1$ $ \forall n>2$ and $ x_n=1-\frac 1n$ $ \forall n>2$

so $ \lim_{n\to+\infty}x_n=1$


I am very, very new to LaTeX, and I need some help.
I'm making my very first document on my own with LaTeX, and here is the code I have:

\documentclass{article}
\title{Unit 3 Matrix Project}
\author{My Name}
\date{October 20, 2007}
\usepackage{amsmath}
\begin{document}
\begin{flushleft}
\maketitle

\! 1)
Matrix M:
\hspace{1.05in} A
\ B
\ C
\ D
\ E
\[ \begin{bmatrix} 0 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \end{bmatrix}
\]
2a) $ 3$; The number of bridges you can cross from B to somewhere else.\\
\, b) $ 2 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 1 \equal{} 12$. It is twice the number of bridges on the map because each bridge is counted twice. For example, the bridge from D to E is shown in both the row for D and the row for E.
\end{flushleft}
\end{document}

When I build this document, I always have an extra blank page at the top, and page numbers at the bottom of each page. Can anyone tell me how to get rid of the page numbers and the blank page at the top?
Thanks so much!!!
BTW, if there's anything else wrong with my code, PLEASE tell me. I really need all the help I can get.

It's just the order of your commands that are causing the problem.
1. To avoid the blank page the flushleft environment must start after \maketitle
2. \pagestyle{empty} removes page numbers. The problem is that \maketitle puts them back again on the front page so you have to immediately follow it with \thispagestyle{empty}
3. Put packages at the top before title stuff. This is just good practice but won't affect your document.
Putting this together gives:
[code]\documentclass{article}
\usepackage{amsmath}
\title{Unit 3 Matrix Project}
\author{My Name}
\date{October 20, 2007}
\begin{document}
\pagestyle{empty}
\maketitle
\thispagestyle{empty}
\begin{flushleft}
\! 1)
Matrix M:
\hspace{1.05in} A
\ B
\ C
\ D
\ E
\[
\begin{bmatrix} 0 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \end{bmatrix}
\]
2a) $3$; The number of bridges you can cross from B to somewhere else.\\
\, b) $2 + 3 + 3 + 3 + 1 = 12$. It is twice the number of bridges on the map because each bridge is counted twice. For example, the bridge from D to E is shown in both the row for D and the row for E.
\end{flushleft}
\end{document}[/code](Note how using code tags on the forum means that it doesn't try and read the dollar sign and make a mess of the post).

wow, thanks so much stevem!
i am now done with my matrices project!!  :D

Dear Moderator-

Could you please delete this thread - "Extra Page at Top"?

Thank you so much.

Why? Wouldn't it better to leave it there for others who may have similar problems?

Well, I am just concerned that my full name is in the code for my document; I guess if you just edited that out, then it would be fine.

 :)

That's fair enough; I've removed your name.


algebra
polynomial
inequalities
algebra proposed

[u][i] Problem[/i][/u]
[i] Let $ P(x)$ be a polynomial with $ n$ real roots and the degree of $ P(x)$ is $ n$( $ n \geq 2$)

  Prove that
  $ n.P(x).P''(x) \leq (n\minus{}1).(P'(x))^2$ with all $ x\in R$[/i]
[i]
  Huỳnh Võ Phương An[/i]

[i]    Nayel , Scorpius119 . Let perform :lol: [/i]

If $ x_k$ are the roots of $ P$, then
\[{ \frac{P'(x)}{P(x)}=\sum_{k=1}^n \frac{1}{x-x_k}}\]
and
\[ \frac{P''(x)P(x)-(P'(x))^2}{P^2(x)}=-\sum_{k=1}^n \frac{1}{(x-x_k)^2}.\]
Now, the requested inequality follows from Cauchy-Schwartz:
\[{ n \sum_{k=1}^n \frac{1}{(x-x_k)^2}\ge ( \sum_{k=1}^n \frac{1}{x-x_k}})^2.\]


algebra
polynomial
function
linear algebra
linear algebra unsolved

Let $u \in V$ be fixed and $\zeta \in V^{*}$ such that $\zeta(u) \neq 0$. Define  an endomorphism $T$ of $V$ by $T(v)=u \zeta(v)$. Find a polynomial function of minimum degree $(\neq 0)$ such that $p(T) = 0$

someone? I can't figure it out

The rank of $T$ is one then $T^{2}=trace(T)T$

$P(X)=X^{2}-(trac(T))X$


abstract algebra
vector
group theory
superior algebra
superior algebra unsolved

Hi

1) Find a characterization for groups $G$ such that $Aut(G)$ is transitive on $G\backslash \{e\}$

(when the group is finite this is easy : elementary abelian groups, the point is that I can't do the infinite case)

2) If the latter cannot be done, such a nontrivial group either has trivial center or is abelian.  Can the first case happen?  If not, I will be very happy as it will have finally given me a proof for this :
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=quasifield&t=75184[/url]

All ideas in general are welcome.

Thanks,

If it may help I can provide a proof for the finite case? :?

a) I think the finite case is not that hard. It is easy to prove that every non-trivial element has order $p$ for a prime number $p$ and then the group is a $\mathbb Z/p\mathbb Z$ vector space and the conclusion follows.
[edit]: this was just the finite abelian case unfortunately
[edit2]: it can be corrected. If every element is of order $p$, then the group is a $p$-group and then the center is not trivial, so it must be the entire group by the second remark in the problem text.
===========================================================================================
b) Doesn't $(\mathbb Q,+)$  also work?

===========================================================================================
c) I also wonder if $G^{*_n}$ doesn't work if $G$ works. $G^{*_n}=\underbrace{G*\ldots*G}_{n\ times}$ is the free product of $G$ with itself taken $n$-times.
[edit:] No, it does not work. I don't think $1*1$ is a square in $\mathbb Q*\mathbb Q$. If $\mathbb Q$ works :)
===========================================================================================
These are just remarks:
Let $G$ be a group such that $Aut(G)$ acts transitively on $G\setminus\{e\}$.
Then either $G$ has a non-trivial element of finite order, in which case every non-trivial element of $G$ has order $p$ for some prime number $p$, or $G$ has no non-trivial elements of finite order.

If $G$ has no element of finite order, then $G$ is infinite and for every $n\in\mathbb Z^*$ and $x$ in $G$ there exists $y\in G$ such that $y^n=x$.

If $G$ has elements of finite order and $G$ is abelian then $G$ is a $\mathbb Z/p\mathbb Z$ vector space and it is easy to prove that $Aut(G)$ acts transitively on $G\setminus\{e\}$.

I asked my professor of group theory, while he couldn't immediately construct one, he was very certain that there a infinite p groups (all elements have as order a power of a same prime p) that are not abelian but on which the automorphisms group works transitive

So much for my idea... :blush:

[quote="fredbel6"]..such a nontrivial group either has trivial center or is abelian.  Can the first case happen?[/quote]

There are infinite torsion free groups with only two conjugacy classes (so yes, it can happen). There is a construction in Serre's "Arbres, Amalgames, $Sl_{2}$", for example, obtained through a transfinite inductive process as a consequence of the following theorem of Higman, Neumann and Neumann (HNN :)):

For every group $G$ and every automorphism $f$ of a subgroup $H$ of $G$, we can embed $G$ into a group $K$ such that $f$ is the restriction to $H$ of the conjugation by an element in $K$.

As for the abelian case, it's easy to settle: the abelian torsion groups with this property are the vector spaces over $\mathbb Z/p\mathbb Z$ for some prime $p$, and the abelian torsion free groups with the property are the vector spaces over $\mathbb Q$.

The torsion case was essentially solved by amfulger, since his argument works just fine for infinite abelian groups as well. In the torsion free case notice that the group has to be divisible. Indeed, if it's not, then we can find a non-zero integer $n$ and an element $x$ such that $ny=x$ has no solutions $y$ in our group. But then no automorphism of the group can ever send $x$ onto a multiple of $n$ (such as $nx$, for example). Since the divisible torsion free abelian groups are precisely the vector spaces over $\mathbb Q$, we're done.


calculus
integration
geometry
3D geometry
sphere
vector
real analysis

I need help solving this problem.  I've drawn out S, which is a sphere cut in half along the xz-plane.  I'm still not sure how to go about this problem :(

Problem:
Use Stokes' Theorem to evaluate the double integral over S of the curl of vector F dot dvectorS
Vector F(x,y,z)=<e^(xy)cos(z),x^2z, xy> and S is the hemisphere x=(1-y^2-z^2)^(1/2), oriented in the direction of the positive x axis.

I have that the curl of vector F=<x-x^2, -y-e^(xy)sinz, xe^ycos(z)-2xz>

Any help?  I am most distressed

You don't need to calculate the curl. Just apply the Stokes theorem first and get the line integral over the circumference on which $ x\equal{}0$ (contrary to what you said, the equation suggests that the sphere is cut by the $ yz$ plane).


133.5.
21 right. 1 wrong (#12). 3 blank(#16, 24, 25).
One thing I've definitely learned is that log problems are deceptively easy!  :lol:

[quote="tetrahedr0n"]12- B, 132.5  20 right, 5 blank.

Edit: For those who took the AMC A as individuals, where do you get results - do they send it to your home school, or along with the results at the visiting school?[/quote]My guess is home school, but I didn't end up taking the B as a Visitor as I meant to...  It's just that that registration fee must go to something.

132.  Three wrong.  I'm still kicking myself, too - I thought I had the right angle on them, but I made stupid, stupid, stupid mistakes.  It doesn't help that I had two tests and a major essay due that day and so got only two and a half hours of sleep.

But enough excuses.  On to the AIME!

Kicking myself at my stupidity...apparently i cant do basic arithmetic without messing something up...... anyways 127.5 on 12B, which should have been 139.5 were it not for my incompetency at arithmetic >.<

[quote="ShadowLord"]131....21 right,2wrong,2 blank,which also mean a min of 8-9 is needed on Aime,which is almost impossible.....[/quote]

I believe that an 8 will qualify you.


Auch wenn es nicht direkt mit den Übungsaufgaben zu tun hat - Welchen Zug nehmt ihr nach Oberwolfach?

Wenn ihr einen für mich halbwegs sinnvollen Zug nehmt würde ich auch mitfahren. (richte mich da ganz nach euch, muss es nur so eine Woche vorher wissen)

@Werwulff: warum hast du nicht auf mein
[quote]  :D  [/quote]
im Asso geantwortet¿  ;)  :lol:

[quote="ZetaX"]@Werwulff: warum hast du nicht auf mein
[quote]  :D  [/quote]
im Asso geantwortet¿  ;)  :lol:[/quote]
 :?:  :huh:  :?:  :?  :?:
[quote] :? [/quote]

[img]http://www.mathlinks.ro/Forum/templates/subSilver/images/topic_delete.gif[/img] [img]http://www.mathlinks.ro/Forum/templates/subSilver/images/topic_move.gif[/img] [img]http://www.mathlinks.ro/Forum/templates/subSilver/images/topic_lock.gif[/img] [img]http://www.mathlinks.ro/Forum/templates/subSilver/images/topic_split.gif[/img]  :?:  :?:  :!:  :!:  :D  :D [img]http://www.kiffopfer.de/bilder/red-banana.gif[/img] [img]http://www.kiffopfer.de/bilder/banana-bass.gif[/img] [img]http://www.kiffopfer.de/bilder/headbang-banana.gif[/img]

back to topic: fahrt ihr alle durch karlsruhe? wär einfach lustig... achja, und wollt noch sagen, nehmt ja nicht den zug der 15 minuten vor dem letzten bus in freudenstadt ankommt; der hatte bei mir letztes jahr 20 minuten verspätung, mit den ersichtlichen folgen ;) . aber da ihr wohl eh nicht durch freudenstadt fahrt ist es ja egal...

  dg

Hab mal auf bahn.de nach Verbindungen gesucht; die beiden sinnvollsten Resultate:

[b]I:[/b]

München Hbf 	28.05.05 	ab 	12:40 	17 		EC 64 	EuroCity
Fahrradmitnahme reservierungspflichtig, Fahrradmitnahme begrenzt möglich, BordRestaurant
München-Pasing 	  	ab 	12:48
Augsburg Hbf 	  	ab 	13:21
Ulm Hbf 	  	ab 	14:05
Plochingen 	  	ab 	14:49
Stuttgart Hbf 	  	ab 	15:08
Bruchsal 	  	ab 	15:37
Karlsruhe Hbf 	28.05.05 	an 	15:50 	5
Karlsruhe Hbf 	28.05.05 	ab 	16:01 	2 		ICE 877 	InterCityExpress
BordRestaurant
Offenburg 	28.05.05 	an 	16:28 	1
Offenburg 	28.05.05 	ab 	17:02 	7 		OSB82733 	Ortenau-S-Bahn GmbH
nur 2. Klasse, Ortenau-S-Bahn GmbH
Offenburg Kreisschulzentrum 	  	ab 	17:05
Gengenbach 	  	ab 	17:11
Biberach(Baden) 	  	ab 	17:17
Steinach(Baden) 	  	ab 	17:21
Haslach 	  	ab 	17:24
Hausach 	  	ab 	17:31
Wolfach 	28.05.05 	an 	17:35
Dauer: 4:55; fährt täglich, nicht 25., 26. Jun

[b]II
[/b]
München Hbf 	28.05.05 	ab 	13:26 	14 		ICE 516 	InterCityExpress
BordBistro
München-Pasing 	  	ab 	13:35
Augsburg Hbf 	  	ab 	14:04
Ulm Hbf 	  	ab 	14:51
Stuttgart Hbf 	  	ab 	15:51
Mannheim Hbf 	28.05.05 	an 	16:26 	3
Mannheim Hbf 	28.05.05 	ab 	16:44 	8 		ICE 77 	InterCityExpress
BordRestaurant
Karlsruhe Hbf 	  	ab 	17:09
Offenburg 	28.05.05 	an 	17:37 	1
Offenburg 	28.05.05 	ab 	18:03 	7 		OSB82835 	Ortenau-S-Bahn GmbH
Fahrradmitnahme begrenzt möglich, nur 2. Klasse, Ortenau-S-Bahn GmbH
Offenburg Kreisschulzentrum 	  	ab 	18:05
Gengenbach 	  	ab 	18:11
Biberach(Baden) 	  	ab 	18:17
Steinach(Baden) 	  	ab 	18:21
Haslach 	  	ab 	18:24
Hausach 	  	ab 	18:31
Wolfach 	28.05.05 	an 	18:35
Dauer: 5:09; fährt nicht täglich, Verkehrstage

Die erste Variante bietet den Vorteil, dass sie über Karlsruhe geht. Der Zug kommt zwar ziemlich früh an, aber ich glaube das ist ertragbar.

EDIT:

[b]III[/b]
München Hbf 	28.05.05 	ab 	10:26 	19 		ICE 108 	InterCityExpress
BordRestaurant
München-Pasing 	  	ab 	10:35
Augsburg Hbf 	  	ab 	11:04
Ulm Hbf 	  	ab 	11:51
Stuttgart Hbf 	28.05.05 	an 	12:47 	10
Stuttgart Hbf 	28.05.05 	ab 	13:11 	4 		EC 264 	EuroCity
Fahrradmitnahme reservierungspflichtig, Fahrradmitnahme begrenzt möglich, BordRestaurant
Bruchsal 	  	ab 	13:40
Karlsruhe Hbf 	28.05.05 	an 	13:53 	6
Karlsruhe Hbf 	28.05.05 	ab 	14:01 	2 		ICE 875 	InterCityExpress
BordRestaurant
Offenburg 	28.05.05 	an 	14:28 	1
Offenburg 	28.05.05 	ab 	15:02 	7 		OSB82825 	Ortenau-S-Bahn GmbH
Fahrradmitnahme begrenzt möglich, nur 2. Klasse, Ortenau-S-Bahn GmbH
Offenburg Kreisschulzentrum 	  	ab 	15:05
Gengenbach 	  	ab 	15:11
Biberach(Baden) 	  	ab 	15:17
Steinach(Baden) 	  	ab 	15:21
Haslach 	  	ab 	15:24
Hausach 	  	ab 	15:31
Wolfach 	28.05.05 	an 	15:35
Dauer: 5:09; fährt Sa, So, nicht 25., 26. Jun; auch 16., 26. Mai, 3. Okt, 1. Nov

ich tendiere gerade zu folgender verbindung (für die hinfahrt):

[code]Bad Aibling		28.05.05  	ab	10:42	2 		RB 31223 	RegionalBahn
Rosenheim		28.05.05  	an 	10:51	7
Rosenheim		28.05.05  	ab     	11:01	2 		IC 2298 	InterCity
(München Hbf ab)                       	    	11:45
Stuttgart Hbf		28.05.05  	an 	14:00	10
Stuttgart Hbf		28.05.05  	ab 	14:18 	3 		RE 19613 	RegionalExpress
Eutingen im Gäu		28.05.05	an 	15:05	2
Eutingen im Gäu		28.05.05	ab 	15:09 	  		Bus SEV 	Bus Richtung: Freudenstadt Hbf Schienenersatzverkehr
Freudenstadt Hbf	28.05.05	an 	16:03
Freudenstadt Hbf	28.05.05	ab 	16:43	4 		OSB82830 	Ortenau-S-Bahn GmbH
Wolfach			28.05.05	an	17:21[/code]

Dauer: 6:39; fährt nicht täglich, 26., 28., 29. Mai
(Dauer ab München Hbf: 5:36)

ist von münchen hbf aus nicht gerade die schnellste verbindung und man kommt auch früh an, aber in oberwolfach kann man bestimmt gut die zeit vertreiben...  [img]http://www.kiffopfer.de/bilder/red-banana.gif[/img][img]http://www.kiffopfer.de/bilder/banana-bass.gif[/img][img]http://www.kiffopfer.de/bilder/headbang-banana.gif[/img]

Von mir aus können wir die Verbindung gerne nehmen.

Nochmal zu den Aufgaben: Wie lautet denn Nummer 5 von Eric richtig? Muss man nur das erste C durch ein D ersetzen?

EDIT: Werde jetzt wahrscheinlich doch nicht obige Verbindung nehmen; eine Stunde Busfahrt ist mir zu viel...

[quote="Christian Hirsch"]EDIT: Werde jetzt wahrscheinlich doch nicht obige Verbindung nehmen; eine Stunde Busfahrt ist mir zu viel...[/quote]
und wie wärs damit:
[code]Bad Aibling	28.05.05 	ab 	10:17 	2 		RB 31221 	RegionalBahn
Rosenheim		28.05.05 	an 	10:26 	6
Rosenheim 		28.05.05 	ab 	10:31 	1 		RE 31018 	RegionalExpress
München Hbf Gl.5-10 	28.05.05 	an 	11:12 	10

München Hbf 	28.05.05 	ab 	11:26 	15 		ICE 518 	InterCityExpress
Mannheim Hbf 	28.05.05 	an 	14:26 	3
Mannheim Hbf 	28.05.05 	ab 	14:44 	8 		ICE 75 		InterCityExpress
Offenburg 	28.05.05 	an 	15:37 	1
Offenburg 	28.05.05 	ab 	16:02 	7 		OSB82829 	Ortenau-S-Bahn GmbH
Wolfach 	28.05.05 	an 	16:35[/code]

und mit so nem lidl-fahrschein ist das auch nicht so teuer :cool:

ich will eigendlich nicht erst um 18:35 ankommen... ein bisschen freizeit wär vorher nicht schlecht... dann könnten wir uns fröhlich über die wiesen von oberwolfach kringeln :rotfl:

[quote="Christian Hirsch"]Sorry, ich dachte du nimmst die Verbindung mit der langen Busfahrt schon sicher und habe mir deswegen ne Fahrkarte für den späten Zug gekauft  :blush:  :blush:[/quote]
und welchen zug nimmst du zurück *mal annehme, daß du für die rückfahrt auch schon ne fahrkarte hast*???
[img]http://sumo.goo.ne.jp/common/img/today.jpg[/img] <-- sumo ist lustig :rotfl:  :first:

Ich muss wegen dem dummen Kolloquium schon am Mittwoch fahren  :(  :(  :( :mad:  :mad:  (*Bitte noch einige gewalttätige Smileys einsetzen*)
[Mit späteren Zug meinte ich übrigens Variante I]

oberwolfach ist ein echt perverser ort, was die anfahrt mit der bahn betrifft... beziehungsweise nein, die bahn ist pervers. entweder man wartet ca. ne stunde lang auf einen bus, oder man hat in wolfach 4 minuten umsteigezeit bei 3 minuten fußweg vom bahn- zum busbahnhof; allerdings ist der fußweg nicht das problem (er ist rat ich mal in wirklichkeit locker in 1 minute zu schaffen), dafür aber die massigen verspätungen der regionalzüge (in freudenstadt hat letztes jahr der bus nicht gewartet). dann wartet man mit einer recht großen wahrscheinlichkeit 2 stunden auf den bus. :D wobei ich mittlerweile fest vorhab, ab wolfach den synthetischen/elementaren weg zu nehmen. mal sehen wer dann früher da ist...

  dg

[quote="darij grinberg"]oberwolfach ist ein echt perverser ort, was die anfahrt mit der bahn betrifft...[/quote]Rrichtich
[quote="darij grinberg"] beziehungsweise nein, die bahn ist pervers. entweder man wartet ca. ne stunde lang auf einen bus, oder man hat in wolfach 4 minuten umsteigezeit bei 3 minuten fußweg vom bahn- zum busbahnhof; allerdings ist der fußweg nicht das problem (er ist rat ich mal in wirklichkeit locker in 1 minute zu schaffen),[/quote]soweit ich mich erinnern kann gibt es eigendlich keinen fußweg... der bus fährt direkt am bahnsteig los...
[quote="darij grinberg"]dafür aber die massigen verspätungen der regionalzüge (in freudenstadt hat letztes jahr der bus nicht gewartet). dann wartet man mit einer recht großen wahrscheinlichkeit 2 stunden auf den bus. :D wobei ich mittlerweile fest vorhab, ab wolfach den synthetischen/elementaren weg zu nehmen. mal sehen wer dann früher da ist...

  dg[/quote]wenn du den bus verpasst solltest du auf mich warten...ich komm um 16:35 an :)

ps: deine verbindung ist fast der perverseste umweg den ich je gesehen hab... dauert mehr als ne halbe stunde länger, nur damit man in wolfach den bus erwischt...

[quote="Werwulff"]wenn du den bus verpasst solltest du auf mich warten...ich komm um 16:35 an :)[/quote]

oh cool, schon zwei leute... ab dreien dürfte sich ein taxi lohnen...

[quote="Werwulff"]ps: deine verbindung ist fast der perverseste umweg den ich je gesehen hab... dauert mehr als ne halbe stunde länger, nur damit man in wolfach den bus erwischt...[/quote]

argh, meinst du die verbindung, von der ich mal zeitweilig gedacht hatte, daß sie besser ist? die ist nicht nur ein perverser umweg, sondern ist auch sogar noch unrealistischer als die andere, da man dann in hausach 4 minuten umsteigezeit hat... also ich nehm schon noch die via freudenstadt...

  darij


geometry
3D geometry
sphere
ratio
search

OK   Here's the Question again
PLEASEEEEEEEEEEE... :(  solve it this time

 A sphere of radius R hanhgs from the ceiling by a thread of length l.It's Filled with water determine the ratio of oscillation before & after the water freezes.
IGNORE VISCOSITY &  CHANGE IN VOLUME

I clearly remember a russian site having lots of Irodov's solutions. Search for it. It's an easy problem.

[color=green]This is a question in the Weekly Physics Contest held recently.

[b]Before the water freezes[/b]
The law of conservation of energy
$\frac{1}{2}ml\dot{\theta}^2+\frac{1}{2}mgl\theta^2=const$
$\Longrightarrow T_0=2\pi\sqrt{\frac{l}{g}}$
[b]
After the water freezes[/b], it becomes
$\frac{1}{2}m(l+R)\dot{\theta}^2+\frac{1}{2}\frac{2}{5}mR^2\dot{\theta}^2+\frac{1}{2}mg(l+R)\theta^2$
$\approx\frac{1}{2}ml\dot{\theta}^2+\frac{1}{2}\frac{2}{5}mR^2\dot{\theta}^2+\frac{1}{2}mgl\theta^2=const$
(if $l>>R$)
$\Longrightarrow T=2\pi\sqrt{\cfrac{\cfrac{2}{5}\cfrac{R^2}{l^2}+1}{\cfrac{l}{g}}}$

Now, you can easily find the ratio.[/color]


induction
number theory unsolved
number theory

Consider the sequence:
$ x_1\equal{}1,x_2\equal{}1$
$ x_{n\plus{}1}\equal{}3x_n\plus{}x_{n\minus{}1}$
Easy to check that $ x_5\equal{}a_1,x_6\equal{}a_2$ so $ x_{n\plus{}4}\equal{}x_n$
So enough to prove for each m then exist $ n\in N$ such that $ n|x_n\minus{}1,x_{n\plus{}1}\minus{}1$
Consider the sequence $ r_n\equiv x_n(\mod m)$ where $ 0\leq r_n\leq m\minus{}1$
Because $ r_n$ take finite value so exist $ k>l$ such that $ r_k\equal{}r_l,r_{k\plus{}1}\equal{}r_{l\plus{}1}$
From relative we have $ r_{k\minus{}1}\equiv r_{l\minus{}1}(\mod m)$ so $ r_{k\minus{}1}\equal{}r_{l\minus{}1}$
And $ r_{k\plus{}1}\equal{}r_{l\plus{}1}$
Induction we have $ r_{n}\equal{}r_{n\plus{}k\minus{}l}$
Imply that $ r_n$ is period from the first term.
We have $ x_1\equiv x_{1\plus{}h(k\minus{}l)}\equal{}1(\mod m)$
              $ x_2\equiv x_{2\plus{}h(k\minus{}l)}\equal{}1(\mod m)$
So $ m|x_{h(k\minus{}l)\plus{}1}\minus{}1,x_{h(k\minus{}l)\plus{}2}\minus{}1$
This problem claim.

Excuse me, why there is [img]http://www.mathlinks.ro/Forum/latexrender/pictures/3/2/8/328387764c05e3fd6f120c62e48c4585bf63978c.gif[/img] such that    [img]http://www.mathlinks.ro/Forum/latexrender/pictures/c/5/b/c5bff434eebcfd6652ec601f46c12454ad245981.gif[/img]   ?

[quote="hessam"]Excuse me, why there is $ k > l$ such that $ r_k \equal{} r_l,r_{k \plus{} 1} \equal{} r_{l \plus{} 1}$?[/quote]

note that $ r_n\in\{0,1,...,m \minus{} 1\}$ so we have finitely many $ r_n$s...
now we can prove that in the sequence $ r_1,r_2,...$ there exists an ordered set of $ r_i$s which appears infinitely many times,for example $ r_2,r_5,r_6,r_{13}$ appears infintely many times...

the reason of this conclusion is as follow:
first of all note that the number of $ k \minus{}$tuples of $ r_i$s are finite (by [b]$ k \minus{}$tuple[/b] we mean a sequence like $ r_{i \plus{} 1},...,r_{i \plus{} k}$)...
we have exactly $ \binom{m}{k}k!$ such tuples,and also in the sequence $ r_1,r_2,...$ there exist infinitely many $ k \minus{}$tuples,thus there must exist at least one $ k \minus{}$tuple which appeared infinitely many times...
so for example the sequence $ r_{i \plus{} 1},...,r_{i \plus{} k}$ appears infinitely many times,so there exists a number $ j$ such that $ r_{j \plus{} 1} \equal{} r_{i \plus{} 1},r_{j \plus{} 2} \equal{} r_{i \plus{} 2},...,r_{j \plus{} k} \equal{} r_{i \plus{} 2}$ as wanted...


trigonometry ratio calculus algebra polynomial function

Ok I was reading this website http://www.grc.nasa.gov/WWW/K-12/airplane/sincos.html

And it explains sine, cosine, and tangent. I have a couple of questions regarding this.
For the first one, it gives the angle as 75.5 degrees. Above it, it has the sine and cosine of the angle as .25 and .967. How do you calculate the degrees of the angle from the sine and cosine ratios? The page doesn't say. It says .25 is 75.5 degrees, then below that it says .967 also translates to 75.5 degrees. I don't understand that.

Thanks.

Well, you can't find the angle value by yourself, its way to complicated. On a calculator though, there is a button that says $ sin^(\minus{}1)$. If you press it, and then type the ratiio value, you can find the angle. To go back, you just enter $ sin(75.5)$ to find the ratio value.

You have to be careful when using the sine inverse ($ \sin^{\minus{}1}$) on a calculator. For example, if you have the angle 150 degrees, the calculator will say the sine is 1/2. But if you put 1/2 in the calculator and press the sine inverse button, it will give you 30 degrees as the answer, because $ \sin 150^\circ \equal{} \sin 30^\circ \equal{} 1/2$.

For positive sine values and sine inverse, the calculator will always try to give you an angle between 0 and 90 degrees.

[quote="leroyjenkens"]Oh, I was afraid of that. Well thanks.
How did people do it before calculators existed?

Oh and one more question; does the sine ratio give you the angle of one corner and cosine give you the angle of the other corner of the triangle? And what is the tangent ratio used for? Seems like they're all used to find the same angle, which doesn't make sense to me.

Thanks again.[/quote]

You can do this by using a calculus topic known as [url=http://en.wikipedia.org/wiki/Taylor_series]Taylor Series[/url].  The idea behind Taylor Series is to create a polynomial that acts just like the function you want to evaluate and then evaluate that polynomial instead of evaluating the actual function.

[quote="Math Champion"]yes, they are all used for the same angle, but to find the other angle, just find the complement.[/quote]
Complement meaning the secant, the cosecant and the cotangent?

Man I feel so stupid, I'm confused about stuff apparently 12-16 year olds should know.

[quote]You can do this by using a calculus topic known as Taylor Series. The idea behind Taylor Series is to create a polynomial that acts just like the function you want to evaluate and then evaluate that polynomial instead of evaluating the actual function.
[/quote]
So it's a really long, complex polynomial that calculates the angle? That's why we use calculators to do it, since it does it instantly?

I think it's always good to know how to do it manually.
Like finding the square root of any number. Calculators can do it in an instant with any number, but if we were to do it by hand, it would take a long time and a lot of paper, depending on what the number is.

[quote="leroyjenkens"][quote]You can do this by using a calculus topic known as Taylor Series. The idea behind Taylor Series is to create a polynomial that acts just like the function you want to evaluate and then evaluate that polynomial instead of evaluating the actual function.
[/quote]
So it's a really long, complex polynomial that calculates the angle? That's why we use calculators to do it, since it does it instantly?

I think it's always good to know how to do it manually.
Like finding the square root of any number. Calculators can do it in an instant with any number, but if we were to do it by hand, it would take a long time and a lot of paper, depending on what the number is.[/quote]

Essentially, yes.  However, it is an infinite polynomial and each term gets progressively smaller and smaller so you can just evaluate a couple terms if you want rather than the entire thing (more terms simply give a better approximation).

In the case of arcsine, we have the polynomial:

$ \sin^{\minus{}1}(x)\equal{}\sum_{n\equal{}0}^{\infty}\left(\frac{(2n)!}{4^n\cdot(n!)^2(2n\plus{}1)}\right)x^{2n\plus{}1}$

This is very nasty to evaluate by hand, so it is best left for computers.

Calculating sine by hand, however is quite easy and can be done to several digits of accuracy using the approximation

$ \sin(x)\approx x\minus{}\frac{x^3}{3!}\plus{}\frac{x^5}{5!}\minus{}\frac{x^7}{7!}$

Thanks. I'm not advanced enough to do any equations with Greek letters.
I can barely do equations with English letters.

You'd think by now someone would have invented a simpler formula for obtaining the angle of a triangle. Aliens probably know how.

The capital sigma ($ \sum$) means to add up that expression, evaluated for $ n\equal{}0,1,2,3,\cdots,\infty$.

For sine, $ \sin(x)\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n\cdot x^{2n\plus{}1}}{(2n\plus{}1)!}\equal{}x\minus{}\frac{x^3}{3!}\plus{}\frac{x^5}{5!}\minus{}\frac{x^7}{7!}\plus{}\cdots$.

See [[Sigma notation]] for more information.  Sigma notation is highly useful in expressing long expressions compactly.


Does there exist a function $f: \mathbb{R}\to \mathbb{R}$ such that $f \circ f(x) = x^{2}-2\,?$

I had posted this problem before,and no body gives a solution,so i decide to post the official french solution :)
Lets put $g(x)=x^{2}-2=f(f(x))$
$g$ have two fixed points $\{-1,2\}$
1) if $y$ is fixed point of $g$ then $f(y)$ too
2)if $x;y$ are fixed points of $g$ and if $f(x)=f(y) \implies f(f(x))=f(f(y)) \implies x=y$ then $\{f(-1),f(2)\}=\{-1,2\}$
3)Put $g(g(x))=x^{4}-4x^{2}+2=h(x)$ then similary as before $f$ inducts a bijection of a fixed points of $h$ which are:$\{-1,2,\frac{-1\pm\sqrt{5}}{2}\}$
Now put $a=\frac{-1+\sqrt{5}}{2}$ & $b=\frac{-1-\sqrt{5}}{2}$
$a$ is not fixed point of $g$ then not of $f$ so we must have $f(a)=b,f(b)=a\implies f(f(a))=f(b)=a$ which means that $a$ is fixed point of $g$ contradiction.

Can anyone explain how the thought process works when coming up with a solution like this? I worked with the problem for about an hour yesterday, and nowhere did it hit me to study fix points.

[quote]
1)if $y$ is fixed point of $g$ then $f(y)$ too .[/quote]

can someone explain why ?

I only know that is $f(y)=y$ $\implies$ $g(y)=ff(y)=y$ .But I dont know the converse is true ?

[quote="Kalle"]He is saying that if $y$ is a fixed point of $g$, then $f(y)$ is also a fixed point of $g$. $f(f(y)) = y\quad \Rightarrow \quad g(f(y)) = f(f(f(y))) = f(y)$[/quote]

that clears my doubt :)


a smooth rod is fixed vertically and infront of  rod there is a disk of mass m and radius R is rolling without slipping on a rough horizontal surface.velocity of center of mass of disc is v.a rod of length L is connected  with disc at A , r/2 distance vertically up from center of disc with pin point (about which it can rotate freely) and other end with a small smooth ring which can move freely in rod OP. at this instant rod makes an angle 30degrees with vertical.find the angular velocity of rod at this instant.

[quote="iit2009"]
a smooth rod is fixed vertically and [b]infront of  rod of mass m and radius R is rolling without slipping on a rough horizontal surface[/b].velocity of center of mass of disc is v.a rod of length L is connected  with disc at A , r/2 distance vertically up from center of disc with pin point (about which it can rotate freely) and other end with a small [b]smooth ring which can freely in rod OP[/b]. at this instant rod makes an angle 30degrees with vertical.[b]find  the angular velocity of rod at this instant[/b].

[/quote]

Please re-phrase your question in a manner that can be easily understood. The [b]bold[/b] part in the quote above is unclear. Is there any relation between $ R$ and $ L$ supplied, or should they be considered independent entities (I dont think we can consider them to be independent (here has to be some relation between $ R$ and $ L$, don't you think so?)?

sorry for my mistake. i have edited my post.


Which of the following would have the smallest remainder?
A. 4,988 ÷ 49
B. 4,993 ÷ 49
C. 4,996 ÷ 49
D. 4,999 ÷ 49
E. 5,002 ÷ 49

which is least ahead but not lower than 4998
it's 4999


I want to change the height and the width of the page, for example:
height = 15 cm
width = 10 cm
How?

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=geometry]geometry package[/url]. The easiest way is to set the margins, for example
\usepackage[margin=2cm]{geometry}
but, if you read the manual you can do much more than that, setting each margin individually or set width and height as you wish.
The geometry package then automatically takes care of other dimensions affected, so is the preferred way to do things.


What is the shape of convex quadrilateral ABCD that known there exist $M\in AB, N\in AC, P\in CD, Q\in DA$ such that AM = BN = CP = DQ and $\vec{AM}+\vec{BN}+\vec{CP}+\vec{DP}=\vec{0}$

[quote]$M\in AB, N\in AC, ...$[/quote]

May be $M\in AB, N\in BC, ...$?

If yes, than we transfer vectors $\vec{BN}=\vec{MN'}, \vec{CP}=\vec{N'P'}, \vec{DQ}=\vec{P'A'}$
It follows from the $sum = \vec{0}$ that $A$ and $A'$ are the same. Hence $AMN'P'$ is a rhombus and $ABCD$ is parallelogramm.

[quote="hien"]Yes, That is $N\in BC$.
What's about points $N', P', A'$, [b]Digger[/b]?[/quote]
Any wrong? When writing "transfer vectors" I've meant their moving preserving directions and lengths.
In other words:
$N'$ is 4th vertex of parallelogram constructed on vectors $\vec{BM}$ and $\vec{BN}$,
$P'$ is 4th vertex of parallelogram constructed on vectors $\vec{CN'}$ and $\vec{CP}$,
$A'$ is 4th vertex of parallelogram constructed on vectors $\vec{DP'}$ and $\vec{DQ}$.


Let $ m,n$ be positive integers such that $ 1\leq n<m$. A box is locked wiht several locks which must all be opened to open the box, and which all have different keys. The keys are distributed among $ m$ people. Suppose that among these people, no $ n$ can open the box, but any $ n\plus{}1$ can open it. Find the smallest possible number $ l$ of locks and then the total number of keys for which this is possible.

[b]Solution[/b]

[u]Lemma[/u] For every group of $ n$ people there is atleast one lock that cannot be opened and the lock is unique.

[u]Proof[/u]  Saythere are two groups of $ n$ people unable to open the same lock. Since the group 2 has a member not in group 1 we can form a group of $ n\plus{}1$ combining group 1 and that member. This contradicts the condition of the problem.

Hence the minimum number of locks=$ m\choose n$. At maximum, a set of $ n$ people may not possess a key. Hence each of the rest $ m\minus{}n$ members have the key. Hence there are $ (m\minus{}n){m\choose n}$ keys.


[quote="r_boris"][quote="NickNafplio"]Πιστευω πως μπορειτε να τη βγαλεται χωρις να κανετε χρηση εξωσχολικων θεωρηματων:

Αν για τους πραγματικους αριθμους $ x,y,z$ ισχυουν:

$ 1/5^x \plus{} 1/5^y \plus{} 1/5^z \equal{} 3$    και    $ x \plus{} y \plus{} z \equal{} 0$

τοτε να αποδειξεται οτι $ x \equal{} y \equal{} z \equal{} 0$[/quote]
Mε Jensen για τρεις αριθμούς? (οταν ισχύει η ισότητα)[/quote]

mallon me am-gm kalytera kai periptosh isotitas.

ps. k.bori ksero oti einai to idio pragma, alla kalytera na mi fovizoume ton kosmo me th jensen.

μπορει να υπαρχουν πολλες λυσεις αλλα πιστευω πως η πιο απλη λυση ειναι η παρακατω:

$ x  \plus{} y \plus{} z \equal{} 0 <\equal{}> x/3 \plus{} y/3 \plus{} z/3 \equal{} 0$

$ 1/5^x \plus{} 1/5^x \plus{} 1/5^z \equal{} 3 <\equal{}>
1/5^x \plus{} 1/5^x \plus{} 1/5^z \equal{} 3*1/5^0 <\equal{}>
1^3/5^{3x/3} \plus{} 1^3/5^{3y/3} \plus{} 1^3/5^{3z/3} \equal{} 3*1/5^{x/3 \plus{} y/3 \plus{} z/3} <\equal{}>
(1/5^{x/3})^3 \plus{} (1/5^{y/3})^3 \plus{} (1/5^{z/3})^3 \equal{} 3*1/5^{x/3}*1/5^{y/3}*1/5^{z/3}$ (1)

Απο την ταυτοτητα $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} c \plus{} b)((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)$
προκειπτει οτι αν $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$ τοτε $ a\equal{}b\equal{}c$ η $ a \plus{} b \plus{} c \equal{} 0$ , οποτε η
(1) γραφεται ισοδυναμα: $ 1/5^{x/3} \plus{} 1/5^{y/3} \plus{} 1/5^{z/3} \equal{} 0$ (αδυνατο αφου αθροισμα θετικων ειναι παντα θετικος), η
$ 1/5^{x/3} \equal{} 1/5^{y/3} \equal{} 1/5^{z/3} <\equal{}>
(1/5)^{x/3} \equal{} (1/5)^{y/3} \equal{} (1/5)^{z/3} <\equal{}>
x/3 \equal{} y/3 \equal{} z/3 <\equal{}>
x \equal{} y \equal{} z$ (2)

αφου ομως $ x \plus{} y \plus{} z \equal{} 0$ απο (2) προκειπτει $ 3x \equal{} 0 <\equal{}> x \equal{} 0$ (3), αρα απο (2) και (3) καταληγουμε τελικα οτι $ x \equal{} y \equal{} z \equal{} 0$

Λογο βιασινης εφαγα ενα 1/2 μπροστα απο το 2ο μελος της ταυτοτητας, αλλα αυτο δεν αλλαζει τιποτα στη λυση της ασκησης


[quote="bpms"][quote="Quevvy"][quote="ckck"]Given that both p and p+1 are prime numbers, what is the least positive composite that is not divisible by p nor p+1?[/quote]
[hide]the only primes that are next to each other are 2 and 3. The least positive composite number will be 5*5 to have the smallest value so then the answer is $\boxed{25}$ unless 1 is composite....... :maybe: [/hide][/quote]
Nah, 1 isn't composite or prime.[/quote]
I wasn't sure so then my answer should be right....


I think you need to know how to construct a unit length. If you know this, then you can multiply, divide, and take square roots. See http://www.egge.net/~savory/maths9.htm .


Let $ X: \{\dfrac{m}{p^n}: m\in\mathbb{Z}, n\in \mathbb{N},  p\in \mathbb{N}, p > 1,\}$

prove that $ \forall a,b\in \mathbb{R}, a < b, \exists x\in X: x\in (a,b)$

:huh:  This true even if you restrict the denominators to be powers of 2, say, or of any specific number. The set $ X$ you describe seems to allow any power in the denominator, which is a serious overkill. Given $ a, b \in \mathbb{R}$ one can always find a number $ n$ such that $ \delta \equal{} 2^{\minus{}n} < (b\minus{}a)/3$ (this is pretty much the meaning of $ 2^{\minus{}n} \to 0$ as $ n \to \infty$). The multiples of $ \delta$ are then equally spaced in $ \mathbb{R}$ with a distance of $ \delta$ between two consecutive points, so one of them must fall strictly inside the interval $ (a,b)$.


This is the problem:
Circle centers O, P, and Q are collinear with circle P passing through O and Q, and circles O and Q are tangent to P. One intersection of Circles O and P is E, and the intersections of circles P and Q are F and G. If FG intersects EQ at X, then find cos(EXO).

I found a bit of information from making a diagram, namely that angle triangle EQO is a 30-60-90 traingle, with the 90 at E and the 30 at Q. Also GXQ is 60 degrees, and therefore EXG is 120. That leaves EOP to be 60 degrees. And if we label the intersection of QP and FG, R, then EXRO is cyclic. Also, FRQ is right.

But Im stumped after this, so I looked in the answers, and it said that QR is 1/2 (if we let the radii in this problem be 1), but I dont see how it is.

I don't quite get it... If circle P passes through points O and Q, then how can circles with centres O and Q be tangent to circle P (internal or otherwise) unless they have radii 0?

Or do you mean that circles O and Q are tangent [b]at[/b] point P?

i figured it out... It was just a little thing with congruent triangles, grr...


calculus
geometry
circumcircle
inradius
analytic geometry
vector
incenter

Given lines f(x) = ax+b, g(x) = cx+d, h(x) = ex+f, find the equation of the circle inscribed in the triangle formed by the vertices of the intersections of f(x), g(x), and h(x).

Are we able to assume that the triangle formed is equilateral?

No.

The triangle need not be equilateral. To convince yourself, draw any three randome lines and see if their intersection points mark off an equilateral triangle.

I have a method of solving the problem, but it is really messy. I guess you can find the center of the circle by using the point-to-line distance formula, and then setting the distacne from center (h,k) equal to f(x) and g(x). Then, you find the distance, which will be the radius, and use (x-h)^2+(y-k)^2=r^2, but it seems too messy to me.

Is there an easier way? Preferably without calculus ( cuz i havne't learned it yet)?

I have already done all of the algebra that you just spoke of... finding the intersection of two of the lines, then finding the mid point and that gives you the X value of the center of the circle, you can then find the middle of the y values which gives your  Y center, then use the distance formula calculate the radius...

this method works fine if and only if the triangle is equilateral... hence the reason i asked the question before.

I don't believe calculus will help with this problem.

That would work for the circumcircle, but we want the incircle. We can find the inradius by using the equation [ABC]=sr, where [t] is the area of triangle t, s is the semiperimeter, and r is the inradius.

i thought about that too, but it looks like too much algebra...

Using barycentric coordinates we can find that OI = (a/a+b+c)*OA+ (b/a+b+c)*OB+ (c/a+b+c)/OC, for any point O and where all these are vectors, noit actual lengths. a, b, c are the sides BC, CA, AB of a triangle ABC and I is its incenter. Maybe this helps. If we calculate the legths of the sides of the triangle formed by the intersections of the lines we can find the coordinates of the incenter by this method. It's messy, but the question involves messy calculations, I don't thnk that can be avoyded.


The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

solutions!

l/w=d/l so l^2=dw=2d. l^2+w^2=d^2, D^2=L^2+4. thus l^2=2d=d^2-4, so we have d^2-2d-4=0.solving this we get (2  :pm:   :sqrt: 4+16)=1 :pm:  :rt5: but since l is positive the answer is 1 +  :rt5:


An artificial satellite  is moving around the earth's surface of th earth .If the magnitude of the universal gravitational constant is decreased at a uniform rate k ,then does the satellite spiral outwards  or inwards?

Logically I feel that the satellite should move outward but look at this

$\frac{GMm}{R^{2}}= \frac{mv^{2}}{R}$

Now $\frac{dG}{dt}=-k$ wherre $k$ is a constant
this says that $\frac{dR}{dt}=-l$ where $l$ is another constant as per  the first equation . This implt=es that the radius should decrease so it should spiral inwards.

Trying to use the equation

$\frac{GMm}{R^{2}}= \frac{mv^{2}}{R}$

assumes that it remains is stable circular orbit with the same velocity. Note that in effect you treated $v$ as constant. If $G$ were to decrease, then the satellite would be going too fast to be in circular orbit. Its inward acceleration would be too small to stay in the same circle, and hence $R$ would increase.


induction
algorithm
number theory
Euclidean algorithm
algebra unsolved
algebra

let $ n \in N$ $ (n>\equal{}2)$
let $ S_n$the sum of all numbers which are writen in this form $ 1/pq$ such that $ p$ and $ q$ are coprime ($ p$^$ q\equal{}1$ )
and $ 0<p<q\equal{}<n$ and $ p\plus{}q>n$
prove that $ S_n\equal{}1/2$

[hide]Proof by induction:

The base case is easily checked: the only possible $ p,q$ for $ n\equal{}2$ are $ p\equal{}1,q\equal{}2$ giving a sum of $ \frac{1}{2}$.

Assume the claim is true for all $ n\le m\minus{}1$.  Then, it is clear that $ S_m$ includes all addends used in $ S_{m\minus{}1}$ accept those for which $ p\plus{}q\equal{}m$.  The only addition made to $ S_m$ are those fractions with $ q\equal{}m$.  Thus, we have

$ S_m\equal{}S_{m\minus{}1}\minus{}\left(\sum_{p\plus{}q\equal{}m,(p,q)\equal{}1}\frac{1}{pq}\right)\plus{}\frac{1}{m}\left(\sum_{(m,k)\equal{}1}\frac{1}{k}\right)$

It is well known that $ (m,k)\equal{}1\Leftrightarrow(m,m\minus{}k)\equal{}1$ so our expression can be rearranged to become

$ S_m\equal{}S_{m\minus{}1}\minus{}\left(\sum_{p\plus{}q\equal{}m,(p,q)\equal{}1}\frac{1}{pq}\right)\plus{}\frac{1}{m}\left(\sum_{(m,k)\equal{}1,k<m/2}\frac{1}{k}\plus{}\frac{1}{m\minus{}k}\right)$

$ \equal{}S_{m\minus{}1}\minus{}\left(\sum_{p\plus{}q\equal{}m,(p,q)\equal{}1}\frac{1}{pq}\right)\plus{}\left(\sum_{(m,k)\equal{}1,k<m/2}\frac{1}{k(m\minus{}k)}\right)$

It is obvious that $ (m,k)\equal{}1\Rightarrow(k,m\minus{}k)\equal{}1$ by the Euclidean Algorithm, and so the two sums are equal and thus cancel leaving $ S_m\equal{}S_{m\minus{}1}$[/hide]


[quote="arqady"][quote="math227"]cauchy [/quote]
The endless cauchy! :D[/quote]

Can you explain what that is? Thanks.

I don't think Cauchy works here...

Since you can't have anything square to a negative and use the standard form of Cauchy - the general form involves compelx conjugates and destroys this argument.

Arqady , does the cauchy inequality holds for infinitely many numbers ? If yes , your solution is amazing.

[quote="arqady"][quote="silouan"]Arqady , does the cauchy inequality holds for infinitely many numbers ? [/quote]
If I don't wrong, for the endless case it's shwartz's inequality.[/quote]
Or we can apply Cauchy for finite number of terms and then take the limit of both sides.

I think that the general inequality holds.
Let $x_{1},...,x_{n}$ positive real numbers and $n$ be a positive integer $n\geq 2$ .Then prove that
$(e^{x_{1}}-1)(e^{x_{2}}-1)....(e^{x_{n}}-1)\geq (e^{\sqrt[n]{x_{1}...x_{n}}}-1)^{n}$


Given an n-sided polygon of perimeter p, not necessarily regular, a circle of radius a is rolled around this polygon.  What is the distance covered by the center of the circle?

[quote="mohd_ahsan"]The distance covered is equal to the perimeter P of the n-sided polygon.[/quote]

Be a little careful now .. the answer is not correct...

Assuming the polygon is convex:

$p+(n-2)r\pi$

To get the answer, think of what the center is doing when the circle is tangent to the polygon at different locations.  When the circle is tangent on a side of a polygon, then the center travels exactly the same distance as the side is long.  When it is tangent on a vertex, the center traces an arc with radius $r$ and degree measure equal to that of the vertex.  Therefore, the arc length at each vertex is $\frac{vr\pi}{180}$, where $v$ is the angle measure of the vertex.  Since the angles in an $n$-sided polygon add up to $180(n-2)$, the total of all the arcs formed when the circle is tangent to a vertex is $(n-2)r\pi$.  Adding this to the perimeter, we get $p+(n-2)r\pi$.

twobomb - again be a little careful.. ..supoose n is very large say a google then the polygon, (assume it is regular)  will almost look like a circle..   Now here a circle is going around a circle ..are you sure that  the answer is correct?  The path can not be  as long as you wish. Can it? ( Hint: [hide]Answer does not depend on n - Or  take a particular case of a square or a triangel and see what you get.[/hide])

his solution was almost correct except that he counted the inner angles, not the outer ones; so we get the answer $P+2r\pi$. with some little effort, one sees that it works for concave ones as well.


hy my freind aziz i see this problem today it was a morrocan tst 2007. the only solution is $f\left(x\right)=\frac{1}{x}$.
1:$f$ is injective:
2: $f\left(1\right)=1$. and we are done. its the first year that morrocan tst was very easy. althoug most of problems are from romanian problem sand short listed problems. see you mon lt :wink:
i hop that who passed the last stage poste the exam for the 3 day


linear algebra
matrix
induction
vector
algebra
polynomial
linear algebra unsolved

Let $A, B$ be third-order matrices with $AB-BA=A.$ Prove that $A^{3}=O_{3}.$ ;)

Matrices are really horrible. I only obtained that $A$ is symmetric  :(  (and not sure about this result either)

Hmm. I'll assume they're real matrices because I know you can also consider those as complex matrices - for other fields, could we similarly extend to the algebraic closure of that field? Anyway, let $v \ne 0$ be an eigenvector of $B$ with eigenvalue $\lambda$. Then $Av = (AB - BA)v = \lambda Av - BAv \Rightarrow B(Av) = (\lambda - 1) Av$, i.e. $Av$ is an eigenvector with eigenvalue $\lambda - 1$. Thus $A^2v, A^3v$ are eigenvectors with eigenvalues $\lambda - 2, \lambda - 3$; note that these are all linearly independent as they have different corresponding eigenvalues. But an order 3 matrix cannot have four indepent (nonzero) eigenvectors! Thus one of the above is $0$; that implies $A^3v = 0$.

Now, if $w$ is is in the root space (right terminology? I mean that thing with the Jordan form, whatever it's called) coressponding to $\lambda$ with height $m$ such that $Bw = \lambda w + v$, where $v$ is thus in the root space with height $m-1$, we have $Aw = A \lambda w + Av - BAw \Rightarrow BAw = (\lambda - 1) Aw + Av$; assuming that $Av$ is in the root space $\lambda - 1$ with height $m-1$, we have that $Aw$ is in the root space with height $m$. By replacing $w$ with $Aw$ and then $A^2 w$, we can get similar results for $A^2w$ and $A^3w$ with $\lambda - 2$ and $\lambda - 3$, respectively. I think this implies, by induction, that these vectors are in the corresponding root spaces and are thus independent, having different eigenvalues. As before, $A^3 w = 0$.

But any matrix $B$ has a basis consisting of vectors in various root spaces (Jordan form), so as $A^3$ is identically $0$ on this basis, it is the $0$ matrix.

Hmmm...so if that was right, I think it generalizes to more than $\dim = 3$?

Hint:
 [hide]$\det A = 0$, $tr A = 0$.[/hide]

Solution:
[hide]
First, note that $tr A = tr(AB - BA) = tr(AB) - tr(BA) = 0$.
Let us prove that $\det A = 0$. Indeed, otherwise, we would have
\[ tr(I) = tr((AB - BA)A^{-1}) = tr(ABA^{-1}) - tr(B) = 0. \]
This contradicts to the fact that $tr(I) = 3$.
So there are 2 possibilities:
1) All eigenvalues of $A$ are equal to 0. Then it is clear that $A^3 = 0$.
2) The eigenvalues of $A$ are equal to 0, $a$, and $-a$ for some $a > 0$. We will show that this is in fact not possible. Assume to the contrary that this is possible.
Note that if $(A, B)$ is a solution then $(\lambda A, B)$ is a solution, and $(CAC^{-1}, CBC^{-1})$ is a solution. Therefore, there is a solution with
\[ A = \begin{pmatrix} 1 & 0 & 0\\ 0& -1 & 0\\ 0 & 0 & 0 \end{pmatrix}. \]
We have
\[ AB - BA = \begin{pmatrix} 1 & 0 & 0\\ 0& -1 & 0\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23}\\ b_{31} & b_{32} & b_{33} \end{pmatrix} - \begin{pmatrix} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23}\\ b_{31} & b_{32} & b_{33} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0\\ 0& -1 & 0\\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} b_{11} & b_{12} & b_{13}\\ -b_{21} & -b_{22} & -b_{23}\\ b_{31} & b_{32} & b_{33} \end{pmatrix} - \begin{pmatrix} b_{11} & -b_{12} & b_{13}\\ b_{21} & -b_{22} & b_{23}\\ b_{31} & -b_{32} & b_{33} \end{pmatrix} = \begin{pmatrix} 0 & 2 b_{12} & 0\\ -2 b_{21} & 0 & -2 b_{23}\\ 0 & 2 b_{32} & 0 \end{pmatrix} \neq A. \]

This finishes the proof.
[/hide]

[quote="mffmartinelli"][quote]1) All eigenvalues of $A$ are equal to 0. Then it is clear that $A^3 = 0$.[/quote]

Why?[/quote]
Here is a technical explanation. $A$ can be represented as ([url=http://en.wikipedia.org/wiki/Jordan_normal_form]Why?[/url])
\[ A = C \begin{pmatrix} 0 & * & 0\\ 0 & 0 & *\\ 0& 0 & 0 \end{pmatrix} C^{-1}. \]
for some $C$, each star equals either 0 or 1.
Now compute
\[ A^3 = C \begin{pmatrix} 0 & * & 0\\ 0 & 0 & *\\ 0& 0 & 0 \end{pmatrix}^3 C^{-1} = 0. \]

Instead of quoting Jordan canonical form, you could have qouted the Cayley-Hamilton theorem instead. If $A$ is a $3\times 3$ matrix, all of whose eigenvalues are zero, then its characteristic polynomial is $p(x)=x^3.$ The Cayley-Hamilton theorem says that if $A$ has characteristic polynomial $p(x),$ then $p(A)=0.$

This is not necessarily true if the field has non-zero characteristic:

Consider $p\times p$ matrices with elements in $\mathbb F_p$ (the field with $p$ elements, where $p$ is a prime). Let $B=\mbox{diag}(0,1,\ldots,p-1)$. $B-I$ will also be a diagonal matrix, with the same diagonal entries as $B$, but in a different order. This means that there is an invertible $A$ such that $B-I=ABA^{-1}\Rightarrow BA-AB=A$, but since $A$ is invertible, it cannot be nilpotent.

Yes, sorry I forgot to add that suplementary hypothesis.

[quote="harazi"]Ok, here goes a little bit more difficult problem: consider $K$ an arbitrary field. Prove that a matrix $A$ verifies $A^n=0$ if and only if there exists $B$ such that $A=BA-AB$. Of course, all matrices have entries in $K$.[/quote]

I'm not exactly sure, but I think $K$ having characteristic $0$ means we can extend to the algebraic closure, so if my proof above was right, it should generalize?

Actually, I think this is slightly different in that we have $BA - AB$ as opposed to $AB - BA$. Well then, the proof above will be almost the same except we'll have eigenvalues $\lambda + 1, \lambda + 2, \lambda + 3$ where before we had minus signs, I think.

Now, to prove the other direction, let $J_k$ be the $k \times k$ matrix with $0$'s everywhere except $1$'s on the superdiagonal (the row just above the diagonal). Let $D_k = (_kd_{ij})$ be given by $_kd_{ij} = (k - i)\delta_{ij}$. Then $D_kJ_k - J_kD_k = J_k$ by direct computation. Now, let $J$ be a block matrix with the blocks being various of these $J_k$. Take $D$ to be the matrix with the blocks being the corresponding $D_k$s. Then by rules of block-matrix multiplication, $DJ - JD = J$. Finally, let $A^n = 0$. Then all eigenvalues of $A$ are $0$, so it has some Jordan form $J = C^{-1}AC$. Then let $B = CDC^{-1}$ where $D$ is the matrix found above corresponding to $J$; we have $BA - AB = CDC^{-1}A - ACDC^{-1}$ $= C(DC^{-1}AC - C^{-1}ACD)C^{-1}$ $= C(DJ - JD)C^{-1} = CJC^{-1} = A$, as desired.

Here's another argument for the original result, over fields of characteristic zero, of any size.

That is, if $A$ is an $n\times n$ matrix over a field of characteristic zero, and there exists $B$ such that $AB-BA=A,$ then $A$ is nilpotent.

Lemma: for all $k\in\mathbb{N},\,A^kB-BA^k=kA^k.$

Proof of lemma: by induction on $k.$ Assume true for some $k$.

Then $A(A^kB-BA^k)=kA^{k+1}$

$A^{k+1}B-ABA^k=kA^{k+1}.$ Now substitute $AB=BA+A.$

$A^{k+1}B-(BA+A)A^k=kA^{k+1}$

$A^{k+1}B-BA^{k+1}-A^{k+1}=kA^{k+1}.$ Add $A^{k+1}$ to both sides and we are done.

So we have that $kA^k$ has trace equal to zero for all $k.$ Given the assumption of characteristic zero, this means that $A^k$ has trace zero for all $k.$

Now all we have to do is to prove that a matrix such that the trace of $A^k$ is zero for all $k$ is nilpotent. I have a proof, but am looking for ways to improve its presentation before posting it.

[quote="Kent Merryfield"]Now all we have to do is to prove that a matrix such that the trace of $A^k$ is zero for all $k$ is nilpotent.[/quote]

By all means, please post your (presentation-improved :D) proof, but I'll try my hand at this ...

Let $\{\lambda_i\}_{i=1}^n$ be the eigenvalues of $A$ with multiplicities (i.e. we can have $\lambda_i = \lambda_j$). Then we know that $A^k$ has eigenvalues $\{\lambda_i^k\}_{i=1}^n$ and since the trace is the sum of the eigenvalues, with multiplicities (Viète on the characteristic polynomial), we have $\sum_{i=1}^n \lambda_i^k = 0 \,\forall k \in \mathbb{Z}^+$. However, we know (Newton's sums or whatever) that each such $\sum_{i=1}^n \lambda_i^k$ can be written in terms of the first $k$ elementary symmetric polynomials (possibly with fractions and whatnot, but that shouldn't be a problem as the field has characteristic $0$). Thus the first such sum above yields that the first elementary symmetric polynomial is $0$, the second then yields that the second such elementary symmetric polynomial is $0$, etc., so all elementary symmetric polynomials are $0$. Since the characteristic polynomial is the polynomial with the eigenvalues as roots, its coefficients are $\pm 1$ times these symmetric polynomials, i.e. the characteristic polynomial is simply $t^n$ so we apply Cayley-Hamilton.

[quote="Kent Merryfield"]Now all we have to do is to prove that a matrix such that the trace of $A^k$ is zero for all $k$ is nilpotent. I have a proof, but am looking for ways to improve its presentation before posting it.[/quote]
(This is only true in characteristic zero)

Why reinvent the wheel? See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=64870]here[/url] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=33753]here[/url] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=17748]here[/url] .

Hmm.. I thought there were more versions of this out there. Note also that these don't all contain proofs.

MysticTerminator's argument is approximately what I would use- make sure to extend to an algebraically closed field first, or similar trickery.

The argument I had was essentially harazi's from the second of your three references and Mystic Terminator's, above - sums of power of eigenvalues and Newton's formulas.

The easiest solution, in my opinion: we have by induction $BA^k-A^kB=kA^k$ and so if $A$ is not nilpotent, then the endomorphism $f(A)=BA-AB$ has infinitely many eigenvalues, false.

Harazi can you please post your solution in detail. :)

:wacko:  I don't get it: do you want me to show how to proceed in the induction step? Or to prove that a non-zero endomorphism of a finite dimensional vectorial space has finitely many eigenvalues? Sorry, I won't do that.

I'm sorry Harazi, I didn't know this theorem. :blush:

I don't think we should call it theorem, as long as we accept that a non-zero polynomial has finitely many roots.


inequalities
LaTeX
inequalities unsolved

let a1, a2, ....an be any positive numbers.
let Si=a1+a2+...+ai
prove 1/S1+2/S2+...+n/Sn<4(1/a1+1/a2+...1/an)

also see http://www.kalva.demon.co.uk/soviet/sov86.html  No.10

Since I just know about the Latex code, let me retype it. :P

Prove for any positive reals $ a_1,a_2,...,a_n,$
We have $ \frac{1}{a_1}\plus{}\frac{2}{a_1\plus{}a_2}\plus{}...\plus{}\frac{n}{a_1\plus{}a_2\plus{}...\plus{}a_n}<\frac{4}{a_1}\plus{}\frac{4}{a_2}\plus{}...\plus{}\frac{4}{a_n}$

This is ASU 1986 Grade 10 Problem 2, should not be that hard. Unfortunately, I got stuck on it for a long time :oops:


inequalities
linear algebra
matrix
algebra
polynomial
linear algebra unsolved

[sorry if it was already discussed, for me is new :)]
Let $A \in M_n(C), \det (A) = 0.$ Then $\exists a \in C$ such that $(A^{*})^2 = a \cdot A^{*}$

But isn't it really obvious? I mean, Sylvester inequality gives that $rank(A^{*})$ is at most 1 and any matrix of rank 1 is annulated by a polynomial of degree 2.


We draw a circle with radius 5 on a gridded paper where the grid consists of squares with sides of length 1. The center of the circle is placed in the middle of one of the squares. All the squares through which the circle passes are colored. How many squares are colored? (The figure illustrates this for a smaller circle.)

[img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1995Number9.jpg[/img]

A. 24
B. 32
C. 40
D. 64
E. None of these

[hide="Not so beautiful solution"]

This problem can be solved by putting the diagram in the coordinate system. Normally, I don't recommend coordinate system because it often requires some brute forcing at the end (same here) but in this case, it makes problem bit easier to approach.

Let the center of the innermost square be at the origin. Then, rest of the squares have their centers with integer coordinates as well. Furthermore, for any square with center $ (h,k)$:

The maximum $ x$ distance is $ h \plus{} 1/2$ and the minimum $ x$ distance is $ h \minus{} 1/2$
The maximum $ y$ distance is $ k \plus{} 1/2$ and the minimum $ y$ distance is $ k \minus{} 1/2$
(All within one square)

Now, since we're looking for squares that the circle touches, we can find this by finding the squares that [i]contain[/i] points $ (a,b)$ such that $ a^2 \plus{} b^2 \equal{} 25$ (from the circle formula of $ (x\minus{}h)^2 \plus{} (y\minus{}k)^2 \equal{} 25$ with the center $ (h,k)$).

From here, all we need is a calculator to save time and rather careful choice of picking the points. We obviously know that there are four points ---- (0,5), (5,0), (0,-5), and (-5,0). As for the rest of them, we can just find number of the squares in one quadrant and multiply by 4.

I focused mine in Quadrant I because the numbers are all positive. You can simply brute force but if you test the centers, points near the boundary, and use symmetry (you'll see why), you'll get that there are 9 such squares with following centers:

(1,5),(2,4),(2,5),(3,4),(4,4),(4,2),(4,3),(5,1),(5,2)

So, the answer is C, or 40.

Note that there is a symmetry in this with the line $ y \equal{} x$. If you noticed this early, it could've saved you time so this isn't [i]bad[/i]. But normally, it's better to stay off from coordinates unless you're willing to do these calculations. [/hide]


There are $ 2\frac{1}{2}$ $ a$'s in $ 4\frac{1}{6}$. What is the value of $ a$? Express your
answer as a mixed number.


I've read the information on the USAMTS site but I'm still confused about this.

For some of my solutions, I'd prefer to format them using HTML/CSS rather than using PDF. Will answers in HTML format be accepted? If not, can I print out the pages and send them by mail? Or is there some way to convert an HTML page to PDF after making it?

We cannot accept HTML documents by email.  If you want to write them in HTML, then print them out and mail or fax them.  You can also "virtually" print them to PDF and then send the PDF file by email.

[quote="TheAmazingOne"]How would I go about virtually printing them as a PDF? It's something I've never done before.

I'd like to send the solutions electronically if at all possible because... well, I heard about the USAMTS by mail just today so I'm having to work on these problems at the last second. I don't want to risk that the solutions might not be mailed in time.[/quote]

The deadline is a postmark deadline, so if your post office gets them by Monday Oct 3 at the latest, you should be fine.  You could also fax them in, or scan & email them.  For converting your work to PDF, try either Go2PDF or CutePDF.  (Google either one and you'll hit 'em.)  They're free, and work with most programs.


summer program
Mathcamp
Stanford
college
email
ratio
AMC

Has anyone here received a Mathcamp brochure in the mail (at home), accompanied by a letter congratulating you on being an AMC top scorer?   (Don't bother posting if the answer is no, but please do post if the answer is yes.)

[quote="joml88"]Both my friend from school and I have gotten one.  It looks nice :) .  .[/quote]

Great!  Our mailers are located in SC so it's not surprising that you would have gotten them first.

[quote]Btw, just out of curiosity what did one have to get on the AMC to get a packet sent to them?[/quote]

It varies depending on grade, gender, and whether you took the AMC10/12.  We sent around 6500 packets, so you can work backwards from that to figure out roughly what the cutoffs were.  I looked at your profile and saw the list of your high school math team, and I think we sent packets to five people on the team.

Math department at my school also recieved one.

I just got one in the mail. It looks terrific. I wish I could go, but my family is already going on a vacation while it takes place.

thanks, all.   i'm glad the brochures are finally starting to arrive -- they should have been mailed two weeks ago, so we were getting nervous!

there's no more need for anyone to continue posting responses to my original question.  thanks!

I got one today as well, and I'm interested in going, but it'll have to wait until after my Stanford High School Summer College application comes back. My fingers are crossed for Stanford, but it'd definitely be fun to attend Mathcamp, especially since I'm probably not doing any traveling at all this summer if I don't.

Just got mine today!  It sounds like a lot of fun; hopefully my parents will (a) let me go, and (b) pay for it :D .
[quote]Btw, just out of curiosity what did one have to get on the AMC to get a packet sent to them?[/quote]
I got 118 on the AMC 12.

Wait, how could this not have been pointed out already... unless I'm missing something.  I figure its already been brought to your attention, but just in case, did you know that the congratulating letter says its at Colby College?  But I'm sure its fine since the brochure and website make it pretty clear.

EDIT: oh sorry, I just read the correction on the website.

Me neither... :(  not that I scored high on the AMC's :blush:
But I don't trust international mail, especially those that come through Mexican customs

[quote="tetrahedr0n"]Aww, I didn't get one. (and my school hasn't either.) I assume I will eventually, and I'm looking forward to this year's.[/quote]

You won't get one automatically -- we delete alums from the list (it's a better use of the brochures to send them to people who don't know about the camp) -- but if you want one send me email and I'll make sure you get one.

[quote="djshowdown2"]Wait, how could this not have been pointed out already... unless I'm missing something.  I figure its already been brought to your attention, but just in case, did you know that the congratulating letter says its at Colby College?[/quote]

Yeah, the staffer (who shall remain nameless) who updated the letter from last year forgot to ask anyone to proofread it before sending it to the printers....  (It wasn't me!)  We've had four phone calls about it so far, so at most 6496 to go!

[quote="xxreddevilzxx"]On the letter it said that you accept 100 talented students. Of those, how many are alums?? And what is the acceptance ratio of the new applicants?[/quote]

The total number of students will be more like 110.  The number of alums is variable, but recently has been in the ballpark of 40, and the acceptance ratio of the new applicants is around 50%.  But (barring a dramatic change in the number of applicants) I bet the acceptance ratio of new applicants who have accounts on AoPS is higher than 50%

[quote="ztbb"]The total number of students will be more like 110.  The number of alums is variable, but recently has been in the ballpark of 40, and the acceptance ratio of the new applicants is around 50%.  But (barring a dramatic change in the number of applicants) I bet the acceptance ratio of new applicants who have accounts on AoPS is higher than 50%[/quote]

Is that necessarily true?? Is that because the AoPS users are usually better than non-users???

[quote="xxreddevilzxx"]Is that necessarily true?? Is that because the AoPS users are usually better than non-users???[/quote]

That's my guess -- but I admit I don't have quantitative evidence to back that up.

I just got one, it looks like a lot of fun

However, I think that if you wanted to get new users that have not heard of MathCamp, then it might have been good to send them earlier (Less than 3 weeks until deadline).

[quote="mcalderbank"]IHowever, I think that if you wanted to get new users that have not heard of MathCamp, then it might have been good to send them earlier (Less than 3 weeks until deadline).[/quote]

Indeed!  We're going to talk to the mailers on Monday to find out what happened: we gave them the addresses on March 18, but the brochures aren't postmarked until April 5.  We are very annoyed!

I received a brochure too, despite my embarrassingly low AMC 12 score: 104.5 However, I cannot attend as I will be attending a free summer math program.

Was this the first time that MathCamp has been mailing out advertisements? 'Cuz sometimes applicant rates increase enormously.

I know that SSP mailed out advertisements this year: the number of applicants jumped up from 115 to 300 in one year. I didn't receive spam from SSP though despite my high PSAT score - because my name was already ont he SSP mailing list.

[quote="Simfish"]Was this the first time that MathCamp has been mailing out advertisements?[/quote]

No, we do it every year.

I got one! And last year, too! But I don't think I will be attending because I'll be busy.

[quote="picounts"]Just got my brochure today! From the descriptions I've heard, it sounds like so much fun! I hope I'm accepted.[/quote]

Haha...knowing you, Tim, I have absolutely no doubt that you will be. If I luck out and get in too, this may be somewhat like the old MOP days (well, with the red group at least...since Yasha will be there too apparently).

[quote="MithsApprentice"]If I luck out and get in too, this may be somewhat like the old MOP days (well, with the red group at least...since Yasha will be there too apparently).[/quote]
I've heard that Andrew G. is applying, and I think Josh will be there, so it really would be like a frosh MOP reunion. Now, the trick will be not to procrastinate and to get the application in on time.

[quote="picounts"]Now, the trick will be not to procrastinate and to get the application in on time.[/quote]

Well said. So we know there's 5 of us applying or alum, and there'll probably be more we don't know about. ^_^

Are they guaranteed???(the MOPpers) Then I have even less chance of making it.

[quote="joml88"]Both my friend from school and I have gotten one.  It looks nice :) .  Btw, just out of curiosity what did one have to get on the AMC to get a packet sent to them?  I'm guessing it was about 120 or so.[/quote]
I suspect it was simply AIME qualifying - I got 104 on the AMC 12, and I got one (unfortunately, I can't apply, as my summer is already booked, but it looks like a lot of fun).

[quote="Susah"]dannngg people who've made MOP go to mathcamp?
i dunno how im gonna get in now  :([/quote]

First of all, contrary to popular opinion, MOPpers are normal people. MOPpers going to MathCamp should not stop you from going at all.

Second of all, you're a girl. MathCamp says they give girls an easier time with the application process.

Reddevil, I don't think anyone except MathCamp vets are "automatically" accepted. We'll be competing for those spots just like you are.

Moppers shouldnt scare you away. The joy and beauty of mathcamp is the range of their classes. At the same time someone can be teaching a really hard(four star as its reffered to) class which Mopers and other talented people will find fun and exiting, while someone else will be teaching a realavly easy(one star) class which anyone there will find accesable. Don't be scared off simply because there will be people smater than you... there will ALWAYS be people smater than you, no matter who when and where you are. As long as you have a good qualifying quiz score... you still have a decent shot at getting in; regardless if youve been to mop 5 years in a row now or never even qualifyed for the AIME. Remember... problem solving is only one of the many branches of math  ;)

[quote="salmononpi"]there will ALWAYS be people smater than you[/quote]

Doesn't that contradict the well-ordering principle?  Or, maybe we are talking about different areas of smartness.  If so, I agree.  Mathematics is now inexhaustible, and for any member X of the human race, there will always be another member that knows a mathematical fact that X doesn't.

[quote="juggle7"]Doesn't that contradict the well-ordering principle?[/quote]

...the signs of a real, live math nerd. Props to you. :-)

Hehe.  The way I think of it is that even the best mathematician alive is not as good as Gauss was.

Hmm...yeah, Mathcamp has math on a lot of levels and a lot of subjects. For almost all of the people at the camp there will be a class they would get completely lost in and another that would seem fairly easy. Also if someone got into MOP that just means they are good at problem solving, and problem solving is only two of the many many classes that Mathcamp offers. That's why the qualifying quiz is not all of the application. Obviously if you get only one question on the quiz you almost certainly won't get in, but a lot of applying depends on what you write about yourself. The reason Mathcamp sets a limit for the number of people accepted is not because the people who didn't get accepted would be lost in even the easiest classes. It is because there is simply a limit to how many campers Mathcamp can accommodate.

And yay for the freshman MOPper reunion. Adam Hesterberg I think will be there too, though technically he wasn't a freshman MOPper though he was a freshman and a MOPper.


How can we solve

Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is

$ \mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 12 } \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 }$

using combinatorics? Rather than just counting

I don't know using combinatorics in this one, but  [hide="the answer is "]16[/hide] done by counting.

The easiest way to approach this is to use restrictions on combinatorics. Basically, we have 9 points on this figure, and we must choose 3 of them to form a triangle. However, we find that each triangle must consist of either two corner points and the one center point, or one center point, one corner point, and one midpoint, or three corner points.

[geogebra]b23caecc9e3ab1293662885b743e921e57f7779a[/geogebra]

Sorry, my geogebra skills suck  :oops: .


The sequence is clearly bounded within $ [1,2]$. To find the possible $ \ell \equal{} \lim_{n \to \infty} x_n$ is again easy; by passing to the limit in the recurrence relation we get $ \ell \equal{} \sqrt {2}$.

Now we are left with the tedious task to prove the sequence is convergent (monotonous it clearly is not, since no more than two consecutive terms can lie on the same side of $ \sqrt {2}$. The following derived relation may be useful.
$ \sqrt {2} \minus{} x_{n \plus{} 2} \equal{} \frac {x_{n \plus{} 1} \minus{} \sqrt {2}} {\sqrt {2}x_{n \plus{} 1}} \plus{} \frac {x_{n} \minus{} \sqrt {2}} {\sqrt {2}x_{n}}$.

[quote="mavropnevma"]The sequence is clearly bounded within $ [1,2]$. To find the possible $ \ell \equal{} \lim_{n \to \infty} x_n$ is again easy; by passing to the limit in the recurrence relation we get $ \ell \equal{} \sqrt {2}$.

Now we are left with the tedious task to prove the sequence is convergent ...[/quote]

I think a rather simple method would be to show that the sequence $ u_n\equal{}\max(\frac{x_n}{\sqrt 2},\frac{x_{n\plus{}1}}{\sqrt 2},$ $ \frac{\sqrt 2}{x_n},\frac{\sqrt 2}{x_{n\plus{}1}})$ is a decreasing sequence whose limit is $ 1$

I received a complicated solution from my friend.He tried to prove for all $ x_n \ge \sqrt 2$ then the sequence $ |x_n\minus{}\sqrt 2|$ is decreasing, and it will be increasing if $ x_n \le \sqrt 2$
However, I hope there will be a nice solution for it.
@pco. I thiink your idea is very nice. Could you continue? :)

[quote="Allnames"]Let $ x_n$ be a sequence such that $ x_1 = x_2 = 1$ and $ x_{n + 2} = \frac {1}{x_{n + 1}} + \frac {1}{x_n}$.Now Find ${ lim_{n\to \infty} x_n}$[/quote]

Let $ u_n: =\max\{x_n,x_{n+1},x_{n+2},x_{n+3}\}$ for every $ n\in\mathbb{N}^*$. We have:
$ x_{n+2} = \frac{1}{x_{n+1}} + \frac{1}{x_n} = \frac{1}{\frac{1}{x_{n}+x_{n-1}}} + \frac{1}{\frac{1}{x_{n-1}+x_{n-2}}} \leq \frac{1}{\frac{1}{u_{n-2}+u_{n-2}}} + \frac{1}{\frac{1}{u_{n-2}+u_{n-2}}} = u_{n-2}$

On the other hand, by the definition of $ \{u_n\}$ we have $ x_{n+1},x_{n},x_{n-1} \leq u_{n-2}$ so $ \max\{x_{n-1},x_{n},x_{n+1},x_{n+2}\} = u_{n-1} \leq u_{n-2},\forall n$ such that those terms exist.

Therefore $ \{u_n\}$ is a non-increasing and under-bounded sequence so there exists $ \lim{u_n} = L$. Hence, $ \forall \varepsilon >0,\exists N, \forall n \geq N, L-\varepsilon < u_n < L + \varepsilon$. Thus, $ \forall n\geq N$ we have $ x_n \leq u_n < L + \varepsilon$. This yields that $ \{x_n\}$ is bounded.

By Bolzano-Weierstrass's theorem, there exists converging subsequences of $ \{x_n\}$, namely $ \{x{}_n{}_k\}$. Let $ L: =\sup\{\lim_{k\to\infty} x{}_n{}_k\}=\overline{\lim x_n}$ and $ l: =\inf\{\lim_{k\to\infty} x{}_n{}_k\}=\underbar{\lim x_n}$.

There exists a subsequence $ \{x{}_n{}_k\}$ converges to $ L$ when $ k\to\infty$. Now $ \{x{}_{n{}_k - 1}\}$ is bounded so by Bolzano-Weierstrass's theorem again, there exists a converging subsequence of $ \{x{}_{n{}_k -1}\}$. We will just take the terms of $ \{x{}_{n{}_k - 1}\}$ with $ \{x{}_n{}_k\}$, so there exists the two converging sequence $ \{x{}_n{}_k\}$ and $ \{x{}_{n{}_k - 1}\}$. Repeat the argument above, we obtain that there exist four subsequences of $ \{x_n\}$ satisfies the following conditions:
$ \lim_{k\to\infty}{x{}_n{}_k}=L \\ \lim_{k\to\infty}{x{}_{n{}_k-1}}=l_1 \\ \lim_{k\to\infty}{x{}_{n{}_k-2}}=l_2 \\ \lim_{k\to\infty}{x{}_{n{}_k-3}}=l_3$

Since $ x_{n+2} = \frac{1}{x_{n+1}} + \frac{1}{x_n}$, put $ k\to\infty$ we have: $ L=\frac{1}{l_1}+\frac{1}{l_2} \leq \frac{2}{l}$ so $ Ll\leq 2$. Analogously (exists subsequence $ \{x{}_{n{}_{k'}}\}$ converges to $ l$ etc) we also have $ Ll \geq 2$. Therefore $ Ll = 2$.

By $ Ll = 2$ so $ L = \frac{1}{l_1} + \frac{1}{l_2} \leq \frac{2}{l} = L$, equality holds, so $ l_1 = l_2 = l$.

Finally:
$ l = \frac{1}{l} + \frac{1}{l_3} \geq \frac{1}{l} + \frac{1}{L}\Rightarrow 2l \geq L + l$

This means $ L\geq l \geq L$, so $ L=l=\sqrt{2}$.

It is followed that $ \lim_{n\to\infty} x_n =\sqrt{2}$, QED.


integration
limit
calculus
real analysis
real analysis unsolved

Let $x(t)$ be a solution of the second order differential equation $x''(t) + a(t)x'(t) + x(t) = 0$ for $t \ge 0$, where $a: [0 ,\infty) \to \mathbb{R}^+$ is non-increasing and $\int_0^\infty a(s)ds = \infty$. Show that $\lim_{t \to \infty} x(t) = 0$.

Seems like it's time to post a solution to this one. Note first of all that $\frac{d}{dt}(x(t)^2+x'(t)^2)=2xx'+2x'x''=-a(t)x'(t)^2$, so the sum $x(t)^2+x'(t)^2$ is non-increasing and $\int_0^\infty a(t)x'(t)^2\,dt<+\infty$ . Thus it suffices to show that $x(t)^2+x'(t)^2$  takes arbitrarily small values. Now note that $x(t)x'(t)$ is bounded (it does not exceed $x(t)^2+x'(t)^2$) and $(xx')'=x'(t)^2-a(t)x(t)x'(t)-x(t)^2$. Thus the integrals $\int_0^T [x'(t)^2-a(t)x(t)x'(t)-x(t)^2]\,dt$ are bounded and, by the Abel-Dirichlet test, so are the integrals $\int_0^T a(t)[x'(t)^2-a(t)x(t)x'(t)-x(t)^2]\,dt$. By Cauchy-Schwartz, $\left|\int_0^T a(t)^2 x(t)x'(t)\,dt\right|\le a(0)\sqrt{\int_0^T a(t)x'(t)^2\,dt}\,\sqrt{\int_0^T a(t)x(t)^2\,dt}\le \text{Const}\sqrt{\int_0^T a(t)x(t)^2\,dt}$. Hence
\[ \begin{aligned} &\int_0^T [x'(t)^2-a(t)x(t)x'(t)-x(t)^2]\,dt \\ &\le -\int_0^T a(t)x(t)^2\,dt+\text{Const}\sqrt{\int_0^T a(t)x(t)^2\,dt}+\text{Const} \end{aligned} \].
It immediately follows from here that the integrals $\int_0^T a(t)x(t)^2\,dt$ are bounded, so $\int_0^\infty a(t)x(t)^2\,dt<+\infty$. Thus $\int_0^\infty a(t)[x(t)^2+x'(t)^2]<+\infty$. The rest should be clear.