algorithm
inequalities
graph theory
Ramsey Theory
combinatorics theorems
combinatorics

Since this theorem appeared in another topic it may be useful to start another topic for its own.

Ramsey's theorem, in its finite version, states that :

Let $k,r \geq 1$ and $m_1,...,m_r \geq k$ be integers.
There is a integer $n_0$ such that, for each $n \geq n_0$, no matter how each $k$-elements subsets of a set $X$ with $|X|$ are colored by $r$ colors (say $c_1,...,c_r$) there exists a subset $X_i$ of $X$ such that $|X_i| = m_i$ and each $k$-subset of $X_i$ has color $c_i$.

It may be useful to check the sense of this theorem by checking small values of $k,r, \cdots $.
In particular, for $k=1$ it is just the pigeon-hole principle.

In another hand, for $k=2$ the $2$-subsets of a set $X$ can be seen as the edges of the complete graph whose vertices are the elements of $X$. Thus, Ramsey's theorem states that, given $m_1,...,m_r \geq 2$, then for $n$ sufficientely large no matter how you color the edges of $K_n$ each with one from $r$ colors, there will be a complete monochromatic $K_{m_i}$.

The least integer $n_0$ which have the property described in the theorem is called the Ramsey number $R(m_1,...,m_r, k)$.
Note that, the exact value of the Ramsey numbers are unknown, except for some exceptional and small values of $r,k,m_i$

Pierre.

Since I talked about the finite version of the theorem, I have to give the infinite version of Ramsey' theorem :

Let $k,r \geq 1$ be two integers. Let's give an infinite set $E$ and a $r$-coloring of the $k$-elements subsets of $E$. Then there exists an infinite set $A \subset E$ whose $k$-elements subsets have all the same color.

Pierre.

Pierre, I believe you gave the generalized Ramsey's Theorem (or Ramsey's Numbers), which can be, sometimes, very helpful in contests.

The "simple" version is usually expressed in terms of acquaintances and strangers:

R(p, q) denotes the smallest number of people among which there exists either p people that know each other (acquaintances) or q people that do not know each other (stangers), where, of course, p,q are positive integers.

An introductory relationship between the "simple" and the "generalized" Ramsey's Numbers might be the following:

Show that R(p, q)=R(p, q; 2).

[quote]Note that, the exact value of the Ramsey numbers are unknown, except for some exceptional and small values...[/quote]

Yes, I seem to remember that only 7 "simple" Ramsey numbers have been successfully computed... :cool:

Yes, what you call 'simple' version is just the case $r=k=2$.

Pierre.

hi evrybody
i have read that we know just 7 ramsy bicolor numbers
$R(p,q)$
$(p,q)$<(7,7)
i'm just surch the values

hi
i am new to this forum. and this is my first message.
For my college project, i did ramsey number generator(for a given p,q and a number it says whether the given number is a ramsey number for r(p,q).
 i had deviced an algorithm which works very fine till r(4,1). for values greater than this, it takes very long hour and i dont even get the output .
I need to know whether there is any test graph or any other method to test my algorithm theoritically.
I had implemented my algorithm using C.
Thank you

Does $|X|$ denote the cardinality of X?

I'm having troubles with that notation.. Any help would be appreciated!

[quote]subsets of a set X with |X|[/quote]

then what does this mean?  :maybe:

[quote="bellahsayn"]i have read that we know just 7 ramsy bicolor numbers[/quote]
[url=http://mathworld.wolfram.com/RamseyNumber.html]MathWorld[/url] lists 9 known values.

[quote="bellahsayn"]
and we have the propriety
$R(p,q)$=<$R(p-1,q)$+$R(p,q-1)$-1[/quote]
It's wrong. In the general case, we have $R(p,q)\leqslant R(p,q-1)+R(p-1,q)$. For the inequality $R(p,q)< R(p,q-1)+R(p-1,q)$ we need assumption $2\mid R(p-1,q)\; R(p,q-1)$.
Also $R(3,8)= 28$, $R(3,9)= 36$.


Let ABC be an equilateral triangle and let P be a point inside ABC. AP, BP, CP intersect BC, CA, AB at A', B', C' respectively. Show that A'B'*B'C'*C'A'>=A'B*B'C*C'A.

[i]ISL 1996[/i]

Hello, grobber!
Nice problem!

It is easy to show B'C' \geq (AB'+AC')/2 (by cos theorem).
So A'B'*B'C'*C'A' \geq prod (AB'+AC')/2 \geq prod \sqrt(AB'*AC') = A'B*B'C*C'A because A'B*B'C*C'A=AB'*BC'*CA' by Cheva theorem.

Yup! I did pretty much the same thing (cos theorem and all). I think it was considered too easy for IMO :D


Find the smallest possible distance of points $ P$ and $ Q$ on a $ xy$-plane, if $ P$ lies on the line $ y \equal{} x$ and $ Q$ lies on the curve $ y \equal{} 2^x$.


AMC
AIME
combinatorics proposed
combinatorics

Hey,

When I solved http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_2_2006-2007/Problem_13, I noticed that

$\sum_{n=1}^{k}(n*(n-1)*(n-2)*(k-n)) = \frac{1}{20}* (k+1) *k*(k-1)*(k-2)*(k-3)$

This has a number of generalizations, as well -- perhaps you'd enjoy finding them out for yourself.

Is this a common result of some famous theorem?

Thanks!
Dima

I don't know what generalizations you have in mind... in this case divide both sides by 6 and we have the RHS as k+1 choose 5, and the LHS is the number of ways to first pick five guys out of k+1 in the following way:

First pick the second tallest of your five; say he is the n+1th tallest. Then pick one among the k-n choices for the tallest guy and the n choose 3 choices for the remaining guys.

Of course this argument can generalize to other cases..


Let $ x,y>0$ such that $ x\minus{}x^2 \geq y^3\minus{}y^2$. Find the maximum value of $ F\equal{}x^2\plus{}y^2$.

Thank.

[quote="BillClin"]Let $ x,y > 0$ such that $ x \minus{} x^2 \geq y^3 \minus{} y^2$. Find the maximum value of $ F \equal{} x^2 \plus{} y^2$.

Thank.[/quote]
$ 3y^2\leq2y^3\plus{}1$ and $ 2x\plus{}x^2\leq2x^2\plus{}1.$
Thus, $ x^2\plus{}y^2\plus{}2x\plus{}2y^2\leq2y^3\plus{}2x^2\plus{}2.$
Id est, $ x^2\plus{}y^2\leq2(y^3\plus{}x^2\minus{}x\minus{}y^2)\plus{}2\leq2.$ :)


$ \prod_{k\equal{}0}^{44} \sin{(2k\plus{}1)^{\circ}} \equal{} \frac{1}{2^n}$ for some $ n\in\mathbb{Q}$. Compute $ n$.

This is a pretty famous problem...

I know Ive seen something like this but I cant seem to find it in the search..

Is the answer like [hide]44 [/hide]by any chance?

[quote="igiul"]$ \prod_{k \equal{} 0}^{44} \sin{(2k \plus{} 1)^{\circ}} \equal{} \frac {1}{2^n}$ for some $ n\in\mathbb{Q}$. Compute $ n$.[/quote]
Whoa, where would you get the sin(2)
Wouldn't it be sin(1)sin(3)sin(5)...sin(89)?

[quote="the future"][quote="CircleSquared"][quote="igiul"]$ \prod_{k \equal{} 0}^{44} \sin{(2k \plus{} 1)^{\circ}} \equal{} \frac {1}{2^n}$ for some $ n\in\mathbb{Q}$. Compute $ n$.[/quote]
[hide="simplification hint"]
$ \sin{(n)}\sin{(90 \minus{} n)} \equal{} \sin{(2n)}$

[/hide][/quote]

1/2sin(2n)[/quote]

meh, close enough
at least it explains where the 2^n in the ans's denominator comes from.

[quote="CircleSquared"][/quote][quote="the future"][quote="CircleSquared"][quote="igiul"]$ \prod_{k = 0}^{44} \sin{(2k + 1)^{\circ}} = \frac {1}{2^n}$ for some $ n\in\mathbb{Q}$. Compute $ n$.[/quote]
[hide="simplification hint"]
$ \sin{(n)}\sin{(90 - n)} = \sin{(2n)}$

[/hide][/quote]

1/2sin(2n)[/quote][quote="CircleSquared"]

meh, close enough
at least it explains where the 2^n in the ans's denominator comes from.[/quote]

well actually i only realized it because of your post, but I still cant get it down correctly.
I know theres an old post about this, ill try and find it

[quote="ejr322"]If you regroup using sine x = cos 90-x, you can change it to:

[hide] prod (1-44) sin 2x/ (2)

which is sin(2) sin(4) ... sin (88) / (2^44).

You can regroup the terms again - sin(4) sin (8) ... sin(88). / (2^22)

So far we have a grand total of 2^66 in the denominator.

Unfortunately, the terms do not group nicely after this point.

[/hide][/quote]

also you forogt about the sin45, it doenst have a pairing.

Heres the link

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=118563519&t=182995

uh yea you can tell that the answer given in that link isnt in this form and i have a feeling that the answer in that link is the correct answer

(note the sin 90 is trivial since its just 1)


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

A matrix in M_n(R) such that there is k\in N\{0} A^(k+1)=0_n
Proove that det(I+A/1!+A/2!+...+A^n/n!) = 1
Cristinel Mortici
-----------------------------------------------------
If you have a different solution i'll take it.

A^(k+1) = 0 implies A is nilpotent, is similar to a triangular superior
with diag(0;...;0)
if j>0 A^j is also similar to a  triangular superior with the same diagonal
 I+A/1!+A/2!+...+A^n/n! is similar to a triangular with diag(1;...;1)
this gives the determinant 1

M.Omarjee

We know that if A has eigenvalues a1,...,an then P(A) has eigenvalues P(a1),...,P(an), where P is a polynomial in R[X]. If repetitions occur we have the same multiplicity.

If A is nilpotent, then A has all eigenvalues 0, so P(A) has all eigenvalues P(0).
Our P=1+x+...+x n/n!  has P(0)=1, so P(A) has all eigenvalues 1, so det(P(A))=1 n=1.

Grr... :D :D
amfulger already posted a solution for this one and I already had it writtten so I will post the full solution, yesterday I couldn't connect myself to the internet(dial-up) problems :(

Here it goes:
A matrix in M_k(R) such that there is k is in  N\{0} A^(n+1)=0_k
Proove that det(I+A/1!+A/2!+...+A^n/n!) = 1 .

A^(n+1)=O_k ==> x^(n+1)=0 has the only solution 0. so det(A-xI_k)=(-1)^kx^k
let x_i with i=1,n be the solutions of the polinomial P(x)=1+1/1!x+..+1/n!x^n=0
From Viete we get that x_1*..*x_n=(-1)^n*n!
because A^k and I_k comute in the product we apply Cauchy-Binet and get that:
det(I+A/1!+A/2!+...+A^n/n!) = (1/n!)^k* prod(k=1,n) det(A-x_i*I_k)=(-1)^(nk)*(1/n!)^k*(x_1*..*x_n)^k=(-1)^(nk)*(1/n!)^k*(-1)^(nk)*(n!)^k=1
all done!

Initially Cristinel Mortici propose the problem like so:
A matrix in M_k(R) such that there is k is in  N\{0} A^(n+1)=0_k
Proove that det(I+A/1!+A/2!+...+A^n/n!)>0.
but both problems end up with the same fact det(..)=1
:D :D

cheers! :D

The problem we talk about follows easily from:

For every nilpotent matrix B, det(I+B)=1 (!!)
Solution. Easy. Try.

Lagrangia! Who said you about the initial form of the problem?

If B is nilpotent then B is similar to a triangular with diagonall contains only 0, B+I_n is similar to a triangular with diagonal (1,1...,1)
det (B+I) = 1.1....1=1

You also have
Lemma :A,B \in M_n(R or C) such that AB=BA, B nilpotent
then det(A+B)=detA:

[quote="christi"]For every nilpotent matrix B, det(I+B)=1 [/quote]
det(B-tI)=(a_1-t)(a_2-t)...(a_n-t) where a_i are eigenvalues of B, for t=-1 it is det(B+I)=(a_1+1)(a_2+1)...(a_n+1). But if B is nilpotent matrix, it has all eigenvalues zero, so det(B+I)=1.

same goes (with a little modification for det (A +B) = det A )
it's easy to prove it with A inverstible, because BA<sup>-1</sup> is also nilpotent, and we apply the fact beata wrote.
if A is not invertible, then there is a infinity of x for which A+xI is invertible, and using the result for A+xI we get that det ( A+xI +B )= det (A+xI ) for an infinity number of x, and because that is a equality of polynomials, it is true for any x, whence for x=0 and we are done.

I think this was used in the Vietnam Undergraduate competition, wasn't it Namdung?

[quote="Valentin Vornicu"]
I think this was used in the Vietnam Undergraduate competition[/quote]

Do you have any idea where can I find the problems of Vietnam Undergraduate competition say 1990 until today ?


Does anyone get their last name slaughtered? Mine is Qian (argh no chinese typing thing) so yea...
Even some asians are guilty, such as junggi, who pronounces Cao as "Cow".

Heh. People mess up on both my first and last name (even though my first name is so easy).
It annoys the crap out of me.

Mistakes:

Bobbitt
Bobbit
Babbit

IT'S SO ANNOYING!!! WHY CAN'T PEOPLE SPELL BABBITT RIGHT?!

[quote="BanishedTraitor"]lol Well my last name actually means something in English and some people like to make fun of it.[/quote]

Does it happen to start with a w?

My last name is Cao... Since Americans can't pronounce Chinese stuff right, I just tell them to pronounce it KO. Now I either have my name made fun of as KO or just cow... :(

My last name is Rabel (pronounced Ruh- Bell (emphasis on Bell)) (Cuban Name), and if it isn't spelled wrong ("Rebel"), it's pronounced wrong:

Reh-Bell
Rubble
Ray-Bell
Ray-Bull

My name is pretty easy to pronounce...take a look at my sig; it shouldn't be too hard or else you're having difficulties in English.

Eh...I just dont care because it's happened so much, last name is Luan(yes it's chinese). Acctually I've heard it pronouned wrong so many times I pronounce it wrong too XD.

Haha you can't botch "Ko" unless you're really bad at english.

My last name is impossible XD

Nishimura-Gasparian

Everyone completely butchers it (at MC state-they said)

Neeshoimaroh Gesko

don't ask where the k came from.

[quote="erabel"]My last name is Rabel... (Cuban Name)[/quote]
If you told me that much, I'd get it. Spanish pronunciation is pretty consistent. I also see the connection with "Ravel", the French (sort of) composer.
Without some indication that it's Spanish, I'm not sure.

[quote="Ihatepie"]Nishimura-Gasparian[/quote]
Japanese/Armenian? I haven't seen that combination before. There's nothing particularly hard there- just the eight-syllable length, and the switch in the middle.

Most of the issues are about language- the general rule is that names follow their native languages' pronunciation rules. Since those vary widely, and some languages have sounds that don't exist in English, there are a lot of ways to go wrong.

Of course, even a name that's just two English words stuck together can get scrambled- usually by replacing one of those words to make it a more common name.

张培锦.
My first name's the one that gets slaughtered the most though.
They say something like
Pzin Zhange.

No one really butchers my name except for the middle one. Apparently kenji is kengi to them.

Mine is slaughtered so much I don't know how to explain hoe to pronounce it using sounds americans can say. It's actually really simple, its

zheng

They say

zjang (some kind of weird "fuzzy" sounds, cuz for some reason americans don't know what sounds zh makes)
zjung
zang (because it gets confused with the more common "zhang," and anyway, that's still not right)
zjang

And so on so on. Really, americans need to learn how to make asian sounds, seeing as their dominating the world.

people butcher Bohyun, my first name...
보현 if you want to see it written in Korean.

THAT's why I go by Sarah (gaspeth my name on AoPS!!)..

and then people confuse my last name b/c Kim can be a girl's name. so sometimes they call me Kim, but i respond to either one.

and on the internet i prefer kimmy.

Being in SFBA, we have a crazyload of asians here. I have like a whole database of the ways names get slaughtered.
Zhang turns into zang
Zheng turns into zen
Zhao turns into zao
etc.etc.
Japanese names are easy to pronounce, so they don't get slaughtered.
People with FOB first names get killed completely. One guy I know who's name is Rui Shu (again, can't type chinese :() got it pronounced Ryooee shi at SCVMA. ccy has english name Charles, but for some odd reason people call him "konjooey" (cong yue)????????
This guy who has last name "Ge" gets it pronounced Gee like the first part of "geez".

Mine is by far the worst, here is a list of what my brother and I have been called:
Kian
Queen
Kwin
Kwiahn
Ching Chong(yes...after not being able to pronounce it someone actually said that)
"How do you pronounce it?"
"[pause]"
Kwan
Kwa(long A)n
King (hey that's not all that bad :D no actually it is)
Chika Chwinn (you don't want to know where that came from)
Kwen

I tell them the correct chinese pronounciation, then they go what? Chen? okay chen it is...chan has showed up too, and at graduation I went to the counselor and told her, and she wrote "Chi Ahn" on the paper, and that's exactly how she said it. :mad:

I'm not exactly not guilty though, I totally pronounce white composer's names wrong (Chopin...how the **** do you pronounce that? I've heard shopan, shopen, not sure which is right..., Rachmaninoff...okay no I'm not going to try it... Prokofiev...hmmm...I used to say prokofayve, cuz I couldn't remember which way the ie went, and lots more)

ak909090, lifeisacircle, lifeisasphere, and superquark1123 are really lucky, their last names never get slaughtered. Well, okay, superquark1123's sometimes gets confused, but rarely. Come to think of it, most of the people I know have pretty easy last names; jin, ko, chen, chang, yang, etc. Except, although my 6th-grade sometimes got her last name slaughtered, but pretty much only by relatively new english speakers and nonhonors kids who have trouble with english. Although this slaughter is the most offensive.

Bozzo $ \equal{}>$ "Bowzow"

You know what I mean, long o's.

i forgot mine

li becomes
lie
l

i would post my full name, because you can make something kool out of that but i cant trust ALL of you. sorry

My full name: David Yoshiaki Ko.

Actually, most people are quite proficient at pronouncing my middle name, as it's entirely phonetic. And it's not like I tell everyone my middle name anyway. Of course, my first and last names are no problem. (So I do belong to the minority of Asians whose names don't get butchered on a regular basis.)

haha my last name is Qu. It's really hard for Americans to pronounce it the Chinese way, so I just say it's pronounced cue (basically just Q without U).

Actual way, the consonant is a mix between ch and tz, and the vowel is the french u.
Wikipedia says:
q: like [b]ch[/b]urch; pass it backwards along the tongue until it is free of the tongue tip
ü: as in German "[b]ü[/b]ben" or French "l[b]u[/b]ne" (To get this sound, say "ee" with rounded lips)

And the accent is the third accent, which makes it the same tone as the "be" in "before".

My name's not asian: Trabucco(italian lol), but it still get's killed. It's supposed to be like Trah-byu-ko, but I hear Trah-bu-ko much more often. (I.e. whenever someone who doesn't know me--and half the people that do--say it), and an occasional Tobacco. Well only once--that's what my elementary school librarian called me.

[quote="jmerry"]
[quote="Ihatepie"]Nishimura-Gasparian[/quote]
Japanese/Armenian? I haven't seen that combination before. There's nothing particularly hard there- just the eight-syllable length, and the switch in the middle.[/quote]
Close.  :wink: Well my mom's is Japanese (Nishimura) and my dad's is Gasparian (Brazilian/Armenian,more Brazilian)

and they combined it.

My first name is Xiaoyu....
It's not even funny how many teachers have pronounced it wrong...
I believe I've only had one teacher who was able to pronounce it right the first time...
And it's not just mispronunciations, it's been misspelled on a yearly basis...

List of worst mispronunciations and misspellings:
Zaoyu(from like 2nd grade)
Jiogo(track coach) - Spelled Xiaogu on sheet
Chiyu(MC states 2008) - Spelled Xiayo on sheet
Oyu(MC states 2007)-Spelled right, I think...

[quote="dingzhou"]why butcher my last name when you have my first name  :rotfl:  :D[/quote]
Huh i didn't quite get that..

Mine is li. fortunately gets pronounced right most of the time. "Lee". But talk about the slaughtering of Cao, one of my best friends' last name. People say it "Cow" or Co (long o) or caw (short o), or sometimes, after my friend says the right pronunciation, which is close to "tsow" , "tso" or "tsaw". really!

well, people who know me can pronounce my last name, people who dont apparently butcher it like crazy. i mean, how the h311 do you butcher a last name

Sun?

(although the Chinese pronounciation is different)

[quote="kimmystar94"]Wait..
Q is pronounced like CH. Right?

Gees. English already has a sound for ch. why did you start using Q? I always pondered that.
(i'm korean. don't kill me)[/quote]

in chinese (and probably other asian languages), there are no consonant clusters like "bl" or "ch" or "st"

[quote="mz94"]
in chinese (and probably other asian languages), there are no consonant clusters like "bl" or "ch" or "st"[/quote]

What? See this:
[url]http://en.wikipedia.org/wiki/Pinyin#Initials[/url].

There is "ch," but I guess it's different from "q."

Chinese already had its own set of sounds and stuff, and when it was decided that characters' pronounciations were to be written using english letters with accents, they just assigned them the best they could. That's the reason why people in HK, Taiwan, Singapore, etc. decided to change the english spellings so they would sound more like they were to be pronounced (except it doesn't really work still...).

I tell people to say "chan" like jackie chan but with a little long e sound mixed in, but in reality, the "q" sound is different from the "ch" sound.

yeah. my point was why not use a perfectly fine consonant cluster??
but okeydokey.. so Q sounds like ch. but not really. thanks! :D

my korean name is pronounced kinda funky so i just say the american way of sayin my korean name and don't really care when people butcher it (it happens, lol)

My friend has the last name Bardey (bar-day) and for some reason, people always pronounce it "Bardly".

My last name: Hulin

I'm always surprised by how many people mispronounce that - its real pronunciation is something like "ül-in" where the ü is the French or German u-sound and the in is nasalized, but the Americanized "hyoo-lin" is fine with me. I get a lot of "hull-in" and "hoo-lin".

My first and middle names are essentially impossible to mispronounce since they're pretty standard American names.

My last name is Johri. i think the "h" is really what makes everyone mess up. it would be better if there was no "h."

Johri is supposed to be pronounced joe-ree
but lots of ppl pronounce it jaw-ree, joe-hree, jaw-rye, joe-rye, and even ho-ree because someone thought it was a spanish name    :rotfl:

I'm not Asian, but 90% of people (I'm not kidding) pronounce my last name wrong.  It is Kandell.  Almost everyone says it "candle", but it is actually "can-dell" (as in aluminum and computers).  I've gotten used to it.

I TOTALLY AGREE WITH THAT but mine isnt that bad its chow and this guy made a song about it and was singing it during tech and everyone was staring (including the teacher :| ).........and i wont post the song here because it has my first name and people can see and i already posted my name

its not really the pronounciation for me its the meaning....
like in elementary school this guy cam to our school and said stuff about bats and he was like bat food........we call it bat chow and then like EVERYONE turned around and stared at me.......scary............ :ninja:

My last name is Bejoy. Do you know how many people buthcher it. Gosh it;s so simple(Be-Joy)

Dorminy can be Dorming, Dormin, Domin, Dominy, Doming, Dormen (I'm not a porter). Thus I grep for myself with Do*m*n* :D.

My last name is alright, but my first is Dai (pronounced like Die). However, most people tend to say "day" on the first try. As a result, I have preferences for teachers who get it right on their first try.
EDIT: One Asian had a English name of Simon, but the lunchlady got it as "Seemon", which sounds very...suggestive. Who butchers "Simon"?!


On the sides of triangle $ ABC$ we consider points $ C_1,C_2 \in (AB), B_1,B_2 \in (AC), A_1,A_2 \in (BC)$ such that triangles $ A_1,B_1,C_1$ and $ A_2B_2C_2$ have a common centroid.
Prove that sets $ [A_1,B_1]\cap [A_2B_2], [B_1C_1]\cap[B_2C_2], [C_1A_1]\cap [C_2A_2]$ are not empty.

$ \vec{A_1A_2}\plus{}\vec{B_1B_2}\plus{}\vec{C_1C_2}\equal{}0$
$ \vec{A_1A_2}\equal{}\alpha\vec{BC}$, $ \vec{B_1B_2}\equal{}\beta\vec{CA}$, $ \vec{C_1C_2}\equal{}\gamma\vec{AB}$ therefore $ \alpha\vec{BC}\plus{}\beta\vec{CA}\plus{}\gamma\vec{AB}\equal{}0$ $ \implies$ $ \alpha\vec{BC}\plus{}\beta\vec{CA}\minus{}\gamma(\vec{BC}\plus{}\vec{CA})\equal{}0$ $ \implies$ $ (\alpha\minus{}\gamma)\vec{BC}\equal{}(\gamma\minus{}\beta)\vec{CA}$. $ \vec{BC}$ and $ \vec{CA}$ are not collinear therefore $ \alpha\equal{}\beta\equal{}\gamma$.Suposing that $ C_1 \in AC_2$ $ \implies$ $ A_1 \in BA_2$ and $ B_1 \in CB_2$ from where the conclusion.


calculus
derivative
integration
function
real analysis
calculus computations

A function is defined as

$F(x)=\int^{x}_{0}{f(t)}\,dt$.

Find the derivative of this function with respect to $x$.

You do need a hypothesis of continuity for $f$: either continuity at a point if you're trying to prove differentiability at a point, or continuity on an interval if you're trying to prove differentiability on an interval.

And the result does have a name.

(The Lebesgue version wouldn't belong in this forum.)

OK, this question came from an entrance paper for Oxford University, but was actually listed with 4 choice answers:

$1. f(x)-f(0)$
$2. f(x)$
$3. f\prime(x)$
$4. f\prime(x)-f\prime(0)$

Hope that clears it up :)

it's not a matter of clarity.. it's just that the problem, as stated, is not well-posed.
if $f$ is continuous, then the answer is $f(x)$ (and it's pretty easy to show).
otherwise, $F$ could not be differentiable...
take $f = \chi_{(0,+\infty)}$ (i think it's called heaviside's function).
$F(x)$ is obviously $0$ if $x<0$, and $x$ otherwise.
and it's not differentiable in $0$. that's all...

The hypothesis of continuity for $f(t)$ and hypothesis of differentiability for $\alpha (x)\ ,\ \beta (x)\Longrightarrow$
$\boxed{\ \left(\int_{\alpha (x)}^{\beta (x)}f(t) dt\right)'=f(\beta (x))\cdot\beta '(x)-f(\alpha (x))\cdot \alpha '(x)\ }\ .$


group theory
abstract algebra
linear algebra
matrix
induction
superior algebra
superior algebra unsolved

Let $F_{p}$ be the field with $p$ elements. How many $p$-Sylow's are there in $GL_{n}(F_{p})$?

Hi.

For any prime $p$ and any integer power $q$ of $p$, there are $(1+q)(1+q+q^{2})...(1+q+...+q^{n-1})$ p-sylowgroups in $GL_{n}(\mathbb{F}_{q})$.

Let $B_{n}$ the subgroup of the upper triangular matrices and let $U_{n}$ the subgroup of the upper triangular matrices that have only 1 on their diagonal.
$U_{n}$ is a p-sylowgroup of $GL_{n}(\mathbb{F}_{q})$ and a normal subgroup of $B_{n}$, so $B_{n}$ is contained in the normalisator $N(U_{n})$. I show that it is actually equal to the normalisator of $U_{n}$. Hence there are $|GL_{n}(\mathbb{F}_{q}): B_{n}|=(1+q)(1+q+q^{2})...(1+q+...+q^{n-1})$ p-sylowgroups.

Now for the proof:
We have the inclusion $B_{n}\subseteq N(U_{n})$ and show the other one:

For all matrices A,B,C,D so that $\begin{pmatrix}A & B \\ C & D\end{pmatrix}\in N(U_{n})$ and X,Y so that $\begin{pmatrix}1 & X \\ & Y\end{pmatrix}\in U_{n}$ (A is 1x1, X and B are 1x(n-1), C is (n-1)x1, D and Y are (n-1)x(n-1)) there are $X' \in \mathbb F_q^{1\times\left(n-1\right)}$ and $Y' \in U_{n-1}$ with:

$\begin{pmatrix}A & B \\ C & D\end{pmatrix}^{-1}\cdot\begin{pmatrix}1 & X \\ & Y\end{pmatrix}\cdot\begin{pmatrix}A & B \\ C & D\end{pmatrix}=\begin{pmatrix}1 & X' \\ & Y'\end{pmatrix}$

$\Rightarrow \begin{pmatrix}1 & X \\ & Y\end{pmatrix}\cdot\begin{pmatrix}A & B \\ C & D\end{pmatrix}=\begin{pmatrix}A & B \\ C & D\end{pmatrix}\cdot\begin{pmatrix}1 & X' \\ & Y'\end{pmatrix}$

$\Rightarrow \begin{pmatrix}A+XC & B+XD \\ YC & YD\end{pmatrix}=\begin{pmatrix}A & AX'+BY' \\ C & CX'+DY'\end{pmatrix}$

$\Rightarrow A+XC=A \Rightarrow XC=0$

Now we fix a C. We can choose any X, so C must be 0 and we can deduce $YD=DY'$.
Since $\begin{pmatrix}A & B \\ & D\end{pmatrix}$ is invertible, A and D are also invertible. $\Rightarrow D^{-1}YD=Y'$
Since Y is in $U_{n-1}$, D is in $N(U_{n-1})$ which is (by the induction hypothesis) equal to $B_{n-1}$
$\Rightarrow \begin{pmatrix}A & B \\ & D\end{pmatrix}\in B_{n}$

This completes the proof (if I didn't made a mistake  :wink: )


How many pairs of positive integers $(a, b)$ satisfy $\frac1a + \frac1b = \frac1{2004}$?

[quote="Peter VDD"]Indeed, the given that it are integers is pretty handy :D

resembles good at some bmo question?

anyway:
$(a-2004)(b-2004)=2004^2$
a,b must be larger than 2004 (elseway the initial equality can never hold with positive integers)

$2004^2 = 2^4.3^2.167^2$
so that makes $5.3.3 = 45$ options, did I make a mistake? 20 secs for solving isn't enough so I probably missed something... :?[/quote]
Yeah you are right. the objective is to make $b-2004\mid2004^2$.

Always your friend, 3X.lich.

Use SFFT to simplify to $(a-2004)(b-2004) = 2004^2 = 2^4 \cdot 3^2 \cdot 167^2 \implies (4+1)(2+1)(2+1) = \boxed{45}$. Both $a,b$ are larger than $2004$, or else one of $\frac{1}{a},\frac{1}{b}$ will already be $\geq \frac{1}{2004}$.

[quote=OlympusHero]Use SFFT to simplify to $(a-2004)(b-2004) = 2004^2 = 2^4 \cdot 3^2 \cdot 167^2 \implies (4+1)(2+1)(2+1) = \boxed{45}$.[/quote]

You should also mention why we can't have $a-2004\le0$ and $b-2004\le0$.

Yet again, me and mathmax12 have the same solution.

We use SFFT to get that $(a-2004)(b-2004)=2004^2=2^4 \cdot 3^2 \cdot 167^2$, so we have $5\cdot3\cdot3=\boxed{45}.$


inequalities
linear algebra
linear algebra solved

Is this inequality: if $rank(A)\geq rank(B)$ then $rank(A^n)\geq rank(B^n)$ true?

I don't think it's true. Take any two matrices $A,B$ of rank $1$ s.t. $A^2=0$ but $B^2\ne 0$.

Yes,grobber,I think you are right. I supposed this inequality was true because of this one:

  Let $A,B\in M_{3}(\mathbb{R})$ such that $rank(A)>rank(B)$.Then $rank(A^2)\geq rank(B^2) $
  And this one from above it is true. I proved it.What do you think? :?

[quote="cezar lupu"]

  Let $A,B\in M_{3}(\mathbb{R})$ such that $rank(A)>rank(B)$.Then $rank(A^2)\geq rank(B^2) $
  [/quote]

It was discussed on the forum


The following data set gives the average amount of precipitation for the Pinar del Rio province, Cuba, in inches for the month of April, over the 30 year period 1931-1960.
0.31       0.47       0.53        0.59          0.75      0.81       0.81        0.89        0.96        1.18
1.20       1.20       1.31        1.35          1.44      1.51       1.62        1.74        1.87        1.89
1.95       2.05       2.10        2.21          2.45      2.81       3.00        3.09        3.35        4.75
a) Compute the mean and standard deviation
b) Does the set conform to the 68-95-99 rule? Explain

I'm not sure if this belongs on the forum (or this particular one). It's not so much a problem solving problem as a plug the numbers into the equation and compute problem. Just use standard deviation and mean equations (s.d. eq. can be found here [url]http://en.wikipedia.org/wiki/Standard_deviation[/url]).


How many even numbers are greater than 202 and less than
405?

The problem asks for the number of even numbers between 202 and 405, exclusive; this is essentially the same problem as finding the number of even numbers between 204 and 404, inclusive
$ 204/2 \equal{} 102, 404/2 \equal{} 202$
$ 202\minus{}102 \plus{} 1 \equal{} \boxed{101}$


YES. I knew nothing of 45 45 90 and 30 60 90 until i ventured upon it and i still use it as a refresher for AMCs

If you're looking for a book for geometry contests, I'll humbly recommend [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9]this one[/url].

I'll check it out...you can never have too many contests to do...

[quote="now a ranger"]I worked through all the free ones, and it taught me some. But will this help me in math tournaments? I get the concepts of some things I didn't know before after trying everything, but I don't think it necessarily will help on math contests since SAT is supposedly different.  I am taking geometry this year.

Does anyone else recommend any other websites? I am starting to practice geometry from Mu Alpha Theta past tests from the Theta division.

Also, I got Geometry the Easy WAy and Bob Miller's geometry for the clueless at the library. ARe these 2 good?

thanks  :|[/quote]

I doubt if an SAT course will be sufficient for contests involving "real" math. Also, your 2 books have the words "easy" and "clueless" in the title, so I'm not too sure about those.

Best way to find out how ready you are is to take a practice test:
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44

SAT takes only formula knowledge. I know of someone who wrote an 800 on the SAT in 7th grade, but got >20th place in their state at mc (as a 7th).

[quote="now a ranger"]Also, I got Geometry the Easy WAy and Bob Miller's geometry for the clueless at the library. ARe these 2 good? [/quote]
those two are probably good if you want to skip a school geometry class or something. I know that my friend studied with geometry the easy way and skipped geom class. (but theyre probably really basic)
If you want to improve in basic contest geom, grab the aops books. The aops vol1 geom chapters are really good, and ,i dont know because I havent looked at it, the intro to geometry is probably AWESOME(it's an aops book after all).
and like people said above, you can do really well on SAT if you have basic knowledge of precalc and mathcounts level math.

[quote="funcia"]the clueless one helps with def. and teaches me. o ok i do want to skip geom this year(it starts in jan.)[/quote]

I have the clueless book now. I need it to prepare for math tournaments. I think it will refresh some things, dunno about helping my score. But it is a confidence builder hands down


let $ \varphi_{A}$ a function:
$ \varphi_{A}: E\equal{}\equal{}>{0,1}$

$ x\equal{}\equal{}\equal{}>1$;x $ \in$ $ A$
$ 0$; x $ \not\in$ $ A$

$ 1$. prove that $ \varphi_{A}$ is surjective.
$ 2$. prove that $ A\equal{}B\Leftrightarrow\varphi_{A}\equal{}\varphi_{B}$.
$ 3$. prove that the function

$ \varphi: P(E)\equal{}\equal{}>P(E,{0.1})$
              $ A\equal{}\equal{}\equal{}>\varphi_{A}$  is $ bijective$
$ remark$:
{0,1} is a $ set$.
$ P(E)$ is the set of the sets.
$ P(E,{0,1})$ is the set of bijections.
 :P [/u]

1) is not true for $ A\equal{}E$ or for $ A \equal{} \emptyset$  :wink:
The whole thing is a classical homework exercise...

Pierre.


$ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}1$,prove that
\[ (1\plus{}\frac{1}{x})(x\minus{}yz)^2\plus{}(1\plus{}\frac{1}{y})(y\minus{}zx)^2\plus{}(1\plus{}\frac{1}{z})(z\minus{}xy)^2\plus{}\frac{71}{4}xyz\geq\frac{5}{4}\]

[quote="Tourish"]$ x,y,z > 0$ such that $ x \plus{} y \plus{} z \equal{} 1$,prove that
\[ (1 \plus{} \frac {1}{x})(x \minus{} yz)^2 \plus{} (1 \plus{} \frac {1}{y})(y \minus{} zx)^2 \plus{} (1 \plus{} \frac {1}{z})(z \minus{} xy)^2 \plus{} \frac {71}{4}xyz\geq\frac {5}{4}\]
[/quote]
Yes it's true.
After homogenization we need to prove that
$ \sum_{cyc}(3x^5yz\plus{}4x^4y^3\plus{}4x^4z^3\minus{}4x^4y^2z\minus{}4x^4z^2y\minus{}6x^3y^3z\plus{}3x^3y^2z^2)\geq0,$ which is obvious.

[quote="arqady"][quote="Tourish"]$ x,y,z > 0$ such that $ x \plus{} y \plus{} z \equal{} 1$,prove that
\[ (1 \plus{} \frac {1}{x})(x \minus{} yz)^2 \plus{} (1 \plus{} \frac {1}{y})(y \minus{} zx)^2 \plus{} (1 \plus{} \frac {1}{z})(z \minus{} xy)^2 \plus{} \frac {71}{4}xyz\geq\frac {5}{4}\]
[/quote]
Yes it's true.
After homogenization we need to prove that
$ \sum_{cyc}(3x^5yz \plus{} 4x^4y^3 \plus{} 4x^4z^3 \minus{} 4x^4y^2z \minus{} 4x^4z^2y \minus{} 6x^3y^3z \plus{} 3x^3y^2z^2)\geq0,$ which is obvious.[/quote]

Wow,such terrible caculations....I am waiting for a nice solution :roll:

[quote="Tourish"]

Wow,such terrible caculations...[/quote]
I don't agree with you. 15 minutes of the work.  :wink:


geometry
geometric transformation
homothety
geometry unsolved

Circles $k_{1},k_{2}$ intersect, and circle $k_{3}$ with center $O$ touches $k_{1},k_{2}$ internally at $A,B$ respectively. A line is tangent to $k_{1}$ at $C$ and $k_{2}$ at $D$. Let $E$ be the foot of perpendicular from $O$ to $CD$. Show that $AC,BD,OE$ are concurrent.

We denote as F, the intersection point of the lines AC, BD and it is enough to rpove that the line OF, is perpendicular to CD.

We denote as S, the intersection point of the lines O1O2, AB, where O1, O2, are the centers of the circles (K1), (K2), respectively.

We consider the line ABS, as the transversal of the triangle OO1O2 and so, by Menelaus’s theorem, we have that
(SK1 / SK2)·(BO2 / BO)·(AO / AO1) = 1, (1).

From (1), => SK1 / SK2 = AO1 / BO2, (2) (because of AO = BO).

We denote as S', the intersection point of the lines O1O2, CD and so, from CO1 // DO2, we have that S'K1 / S'K2 = CO1 / DO2, (2).

From (1), (2), => SK1 / SK2 = S'K1 / S'K2, => S ≡ S', (3) (because of AO1 = CO1 and BO2 = DO2).

From (3), we conclude that the triangles ACO1, BDO2, are perspective and so, by Desarques’s theorem we have that the point O, as the intersection point of their sidelines AO1, BO2, the point F, as the intersection point of their sidelines AC, BD and the point at infinity, as the accepted intersection point of their parallel sidelines CO1, DO2, are collinear.

Hence the segment line OF, as parallel to CO1, DO2, is perpendicular to CD and the proof is completed.

Kostas Vittas.

PS. By the drawing (I have not any proof), I see that the concurrency point of the lines AC, BD, OE, lies on the circle (K3). I will check again, if this is a false result.

Call the centers of $k_{1},k_{2}$ $O_{1},O_{2}$.

Call the point where segment $OE$ and $k_{3}$ intersect $F$.

I will show that $F$ is on $AC$ and $BD$.

Because $O_{1}A=O_{1}C$ and $OA=OF$ and $O_{1}C\| OF$ hence $F$ is on $AC$.

Similarly, $F$ is on $BD$. So, $AC,BD,OE$ are concurrent.

Thank you very much OHO, for your simple and nice proof, of both of the results
(the intersection point of AC, DD, lies on the circle (K3)).

Some times (many times about me), our mind follows a complicated way for the solution of a problem.
So, we don’t see a beautiful sort proof, which is hidden, but in front of us.

Thank you again, Kostas Vittas.

A is the external homothety center of the circles $k_{1},\ k_{3},$ C is the internal homothety center of the circle $k_{1}$ and the common tangent CD of $k_{1},\ k_{2}.$ Similarly, B is the the external homothety center of the circles $k_{2},\ k_{3},$ D is the internal homothety center of the circle $k_{2}$ and the common tangent CD of $k_{1},\ k_{2}.$ Therefore, the internal homothety center of the circle $k_{3}$ and the line CD is the intersection of AC, BD. But this internal homothety center is the intersection of the normal OE from the center O of $k_{3}$ to CD with the circle $k_{3}$ closer to this line.


How do I graph a function with a vertical asymptote? When I try, it gives me an error.

Most likely it is the division by zero error. Just graph the pieces on the left and on the right of the vertical asymptote separately and start them a bit away from the point where the function is undefined.

Also, how would you represent arcsin(x) in Asymptote?

Like this? You use the asin function (you get an error if you plot it on an interval outside its domain).

[asy]import graph;
size(200);

xaxis("$x$",-1,1,Ticks);
yaxis("$y$",-pi/2,pi/2,Ticks);

draw(graph(asin,-1,1,operator ..),red);[/asy]


Let $ x_{1}$, $ x_{2}$, ..., $ x_{n}$ be positive reals such that $ x_{1}\cdot x_{2}\cdot ...\cdot x_{n}\equal{}1$. Prove that

$ \frac{1}{n\minus{}1\plus{}x_{1}}\plus{}\frac{1}{n\minus{}1\plus{}x_{2}}\plus{}...\plus{}\frac{1}{n\minus{}1\plus{}x_{n}}\leq 1$.

It is easy to see that:
1/(n-1+a) + 1/(n-1+b) ≤ 2/(n-1+√ab)  for √ab ≤ n-1.

So we can use mixing variables for remaining variables, for any pair (a,b) such that
ab ≤ (n-1)2, pair (a,b) we change in (√ab, √ab)  (where a not equal to b).

We can do such operation till all numbers less than (n-1) are equal.
It means that we have
x1=x2==xi =X where X < 1, and x[i+1], , x[n] ≥ n-1.

Now, again it is easy to see that:

1/(n-1+a) + 1/(n-1+b) ≤ 1/(n-1+1)+1/(n-1+ab)   for a,b ≥ n-1.

Again we mix variables to mix (a,b) into (1,ab) where for a,b ≥ n-1. and we have
x1 = x2 = = xi = t where t < 1, x[i+1] == x[n-1]=1, x[n] = Y > n-1.

Returning to the first part we mix x1, x2,, x[n-1] and we have
x1 = x2= = x[n-1] = X < 1 and Y > n-1.

So we have inequality:

(n-1)/(n-1+X) + 1/(n-1+Y) ≤ 1 or

(n-1) + X ≤ (n-1)Y +XY which is true due to Y ≥ 1. q.e.d

Prowler, how does your procedure work for $n=5$ and $x_1=x_2=x_3=1/4$, $x_4=x_5=8$ ?

Dear Prowler,

if true your proof is very nice but I have also a doubt about this procedure.

Now I will explain it in detail.

If $x_1x_2... x_n =1$ and $0\leq x_1\leq x_2 \leq...\leq x_n$ then we can

deduce that

$x_1 \leq 1, x_1x_2 \leq 1,x_1x_2x_3 \leq 1,...,x_1x_2..x_{n-1} \leq 1, x_n \geq 1 $

We can substitute $(x_1,x_2,..,x_n) -----> (\sqrt{x_1x_2}, \sqrt{x_1x_2}, ...,x_n)$ but in the next passage

$(\sqrt{x_1x_2}, \sqrt{x_1x_2}, ...,x_n)-----> ((\sqrt{x_1x_2},\sqrt{\sqrt{x_1x_2}x_3},\sqrt{\sqrt{x_1x_2}x_3},..,x_n)$

you cannot anymore say that $\sqrt{\sqrt{x_1x_2}x_3} \leq 1$ as

being $x_1x_2 \leq 1$ we have $ \sqrt{x_1x_2} \geq x_1x_2 $ and here your procedure falls if I am right.

Thank you very much Vasc and Manlio.
I edited my solution. Now I hope no mistake.

Thanks every one! I am very busy now. And even I did not read this solution yet. If I have any problem I will return soon! Again Thanks to evry one!

Dear Prowler,

can you please explain to me this ineq

[quote]Now, again it is easy to see that:

1/(n-1+a) + 1/(n-1+b) ≤ 1/(n-1+1)+1/(n-1+ab) for a,b ≥ n-1. [/quote]

Thank you very much.


There is a button that lets you delete posts, but to delete topics you have to talk to admins.

To delete your own, press the little X in a box in the bottom right corner of your post.

Right where it would say Quote, it says [color=red]X Delete[/color].  I don't think this belongs here though!  but, whatever :D

Yeah, try to keep this stuff to the questions and suggestions forum.

You can't delete your own post unless nobody has posted after you -  of course, you can edit out all the stuff, but that's lame, please refrain from doing it.  Alternately, you could just ask me to delete it, but you'll have to have a really good reason; we usually don't delete stuff.


inequalities
complex numbers
geometry unsolved
geometry

[color=blue][b]Ptolemy inequality.[/b] Let $A$, $B$, $C$, $D$ be four points in the plane. Then, $AB\cdot CD+BC\cdot DA\geq AC\cdot BD$, and this inequality becomes an equality if and only if the points $A$, $B$, $C$, $D$ either lie on one circle in this order or lie on one line in one of the orders $\left(A,\ B,\ C,\ D\right)$, $\left(B,\ C,\ D,\ A\right)$, $\left(C,\ D,\ A,\ B\right)$, $\left(D,\ A,\ B,\ C\right)$.[/color]

I know proofs of this inequality involving complex numbers and inversion, but I couldn't find anything using only elementary geometry.

if you express this problem again:
with triangle $ABC$ and any $M$ then $aMA,bMB,cMC$ is three side of a triangle! we can prove it follow:
called $A'=MA\cap (O),B'=MB\cap (O),C'=MC\cap (O)$ then $\triangle MBC\sim\triangle MC'B'\Rightarrow \frac{BC}{B'C'}=\frac{MB}{MC'}\Rightarrow\frac{MA.BC}{B'C'}=\frac{MA.MB.MC}{MC.MC'}=\frac{MA.MB.MC}{P_{M/(O)}}$
similar you will have
$\frac{MA.BC}{B'C'}=\frac{MB.CA}{C'A'}=\frac{MC.AB}{A'B'}=\frac{MA.MB.MC}{P_{M/(O)}}$
thus $aMA,bMb,cMC$ is three side of a triangle similar $A'B'C'$ :)

[b]Theorem.[/b] (Ptolomeo) Given $\square ABCD$,then $AB\cdot CD+AD\cdot BC=AC\cdot BD$ iff is cyclic.

[b]Proof. [/b]

$(i)$ Let $P \in \overline{AC}$ be such that $m\angle ABP=m\angle DBC$, as $m\angle CAB=m\angle BDC$ we get $\triangle ABP \sim \triangle DBC$ thus $AB\cdot CD=AP\cdot BD$. Also $m\angle BAD=m\angle BPC$ and $m\angle BDA=m\angle BCP$, so $\triangle BAD \sim \triangle BPC$ and hence $BC.AD=PC.BD$, and as $AP+PC=AC$ we get $AB.CD+AD.BC=AC.BD$.

$(ii)$ Supose that $AB.CD+AD.BC=AC.BD$, let's take $P$ such that $m\angle BAP=m\angle CAD$ and $m\angle ABP=m\angle ACD$, so $\triangle ABP \sim \triangle ACD$ asi $AB.CD=AC.BP$ and $m\angle BAC=m\angle PAD$ so $\triangle BAC \sim \triangle PAD$ thus $AC.PD=AD.BC$. Adding what we have otained $AB.CD+AD.BC=AC.(BP+PD)$, but $AB.CD+AD.BC=AC.BD$ so $BP+PD=BD$ and hence $P \in \overline{BD}$ and $m\angle ABD=m\angle ACD$, so $\square ABCD$ ia cyclic. $\blacksquare$

[b]Remark.[/b] If $\square ABCD$ is not cyclic, then $m\angle ABD \neq m\angle ACD$ hence $P \not \in \overline{BD}$
then, as $BP+PD>BD$ we get $AB\cdot CD+AD\cdot BC>AC\cdot BD$.

As conclution

[b]Theorem. (Extended Ptolomeo theorem)[/b] Given $\square ABCD$, then $AB\cdot CD+AD\cdot BC\geq AC\cdot BD$ and the equality holds iff $\square ABCD$ is cyclic.

José Carlos Chávez Sandoval

you can also generilise ponits to circle.

[quote="neworder"][color=red][b]Ptolemy inequality.[/b] Let $A$, $B$, $C$, $D$ be four points in the plane. Then, $AB\cdot CD+BC\cdot DA\geq AC\cdot BD$, and this inequality becomes an equality if and only if the points $A$, $B$, $C$, $D$ either lie on one circle in this order or lie on one line in one of the orders $\left(A,\ B,\ C,\ D\right)$, $\left(B,\ C,\ D,\ A\right)$, $\left(C,\ D,\ A,\ B\right)$, $\left(D,\ A,\ B,\ C\right)$.[/color][/quote]

[color=darkblue][b]Lemma.[/b] If the points $A\ ,\ B\ ,\ C\ ,\ D$ belong to the same line $d\ ,$ then there is the relation $\boxed{\ \overline{AB}\cdot\overline{CD}+\overline{BC}\cdot\overline{AD}=\overline{AC}\cdot\overline{BD}\ }\ .$
[b]Proof.[/b] Denote $X(x)$- the point $X\in d$ which has the affix $x\ .$ Therefore the previous relation is equivalently with the true relation
$(b-a)(d-c)+(c-b)(d-a)=(c-a)(d-b)\ .$ I used the [u]Chasles' relation[/u] $\overline{XY}=y-x\ .$[/color]


induction
geometry
3D geometry
sphere
algebra unsolved
algebra

Consider a set A of points in the plane (space) such that the distance between any two points from A is a rational number. Then A is at most countable.
  Additional nice question ( this one I can solve  :D ): find such a set which is infinite.

I suppose the answer to the second question couldn't simply be $\{(n,0)|n\in\mathbb Z\}$, right? (I mean, the problem wouldn't be challenging in this case) Maybe the points must not be collinear?

You, grobber, you guessed the actual meaning of the problem.

Ok :). Now (sorry for bugging you with these mere details :)), is the extra condition supposed to be "no three points are collinear" or "not all our points are collinear"? It might make a difference, although I'm not sure.

I am bad: no three points are collinear. I want to add that Erdos's problem in plane is not difficult, but in $R^n$...

Question about such infinite A was discussed here some time ago.
I remeber that there were two constructions for set A on unit circle: grobber's and mine.

Uppps, sorry for posting once again a problem already posted. Hey, grobber, you should have said you have already solved the problem!

Well, if I'm not completely wrong, I don't think it is so difficult, even in $R^n$ for which you just have to use induction on $n$.
I will prove it for $n=3$.

Let $r_0,r_1,...,r_n,...$ be an enumeration of the positive rational numbers.
Let $X$ be an arbitrary point in $A$ (I think we may assume that $A$ is non-empty  ;)  ).
Let $A_n$ be the set of the points $M \in A$ such that $XM = r_n$.
Clearly, we have $\dsp A = \cup_{n \geq 0} A_n$.
Since a countable union of countable sets is countable, we only have to prove that each $A_n$ is countable.

Now, let $n \geq 0$ be given. Let $Y$ be some element in $A_n$ (same remark as above...).
Let $Y_k$ be the set of the elements $M$ of $A_n$ such that $MY = r_k$. Again, w have $\dsp A_n = \cup_{k \geq 0 } Y_k$, and we therefore only have to prove that each $Y_k$ is countable.
Note that, for each $k$, the set $Y_k$ is contained in a circle (it is contained in the intersection of two spheres one with center which belongs to the other).

Let $k \geq 0$ be fixed.
Let $Z$ be an arbitrary element of $Y_k$.
Let $Z_i$ be the set of the elements $M$ of $Y_k$ such that $MZ = r_i$. Then, for each $i$, the set $Z_i$ has at most two elements, and in particular it is a finite set.
Thus, $\dsp Y_k = \cup_{i \geq 0 } Z_i}$ is a countable union of finite sets and then is countable, and we are done.

Pierre.

Oh, thanks a lot, Pierre; indeed, while reading the solution I realize that it wasn't so damn tough.

Here is the problem Myth was talking about
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16540&highlight=[/url]


Hello everyone  :lol:

Find the value of:

$ \displaystyle\frac {1}{tg9^\circ{}}\cdot{}(\displaystyle\frac {1}{cot9^\circ{} \minus{} 3tg9^\circ{}} \plus{} \displaystyle\frac {3}{cot27^\circ{} \minus{} 3tg27^\circ{}} \plus{} \displaystyle\frac {9}{cot81^\circ{} \minus{} 3tg81^\circ{}} \plus{} \displaystyle\frac {27}{cot243^\circ{} \minus{} 3tg243^\circ{}})$

$ tg \equal{}$ tangent
$ cot \equal{}$ cotangent

Greetings  :D

That smells the flavour of $ \boxed{\tan 3\theta \equal{} \frac {3\tan \theta \minus{} \tan ^ 3 \theta}{1 \minus{} 3\tan ^2 \theta}}$.

Use this to get :

$ \frac {1}{\tan 9^\circ}\sum_{k \equal{} 1}^4 \frac {3^{k \minus{} 1}}{\cot (3^{k \plus{} 1})^\circ \minus{} 3\tan (3^{k \plus{} 1})^\circ } \equal{} \frac {1}{\tan 9^\circ} \sum_{k \equal{} 1}^4 3^{k \minus{} 1}\cdot \frac {\tan (3^{k \plus{} 2})^\circ}{3 \minus{} \tan ^ 2 (3^{k \plus{} 1})^\circ}$

which follows Telescoping Method, I wonder. :roll:

Hello kunny think that is the way there will be no other.

Greetings  :D

Hello Kunny fail to complete the problem is not the way I telescope.

Greetings  :D


algebra
polynomial
quadratics
number theory unsolved
number theory

Prove that there exist polynomials $ f,g,h$ with integer coefficients, positive leading coefficients such that $\frac{4}{f(x)}+\frac{4}{g(x)}+\frac{4}{h(x)}=\frac{1}{ax+b}$ if and only if b is not a quadratic residue mod a.
   reason for posting this problem: some romanian teachers who thought that Erdos's conjecture can be proved using identities. The answer: nope

(just for the sake of completeness and.. to revive the topic ;)) :
Harazi's talking about the [url=http://en.wikipedia.org/wiki/Erd%C5%91s-Strauss_conjecture]Erdös-Strauss[/url]conjecture.


algebra
polynomial
calculus
integration
college contests

Let's consider real polynomial P(x), satisfying conditions:

P(0) = P(1) = 0, Integral {from 0 to 1} |P'(x)|dx = 1.

Show that |P(x)| <= 1/2 for all x from [0, 1].


Solve the system:
\[ \begin{displaymath} \left\{ \begin{array} ax+by = (x-y)^2, \\ by+cz=(y-z)^2, \\ cz+ax=(z-x)^2, \end{array} \right \end{displaymath}  \]
where $a,b,c$ are constant.

should it be $ax$ in the lhs of first equation?

http://www.kalva.demon.co.uk/balkan/bsoln/bsol844.html

so i found the problem on kalva, it's incorrectly stated as it is.  It is $ax+by=(x-y)^2$ for the first equation, and $a, b, c$ are positive.

Given that they're positive, the solution kalva gives is pretty natural.  Without that information it seems a lot messier, and i'm not sure it's solvable.


Prove the following ineq:

$ x^3\plus{}y^3\plus{}z^3\plus{}2x^2y\plus{}2y^2z\plus{}2z^2x>\equal{}3x^2z\plus{}3y^2x\plus{}3z^2y$

$ x,y,z$ positive numbers

[quote="Nemion"]Prove the following ineq:

$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 2x^2y \plus{} 2y^2z \plus{} 2z^2x > \equal{} 3x^2z \plus{} 3y^2x \plus{} 3z^2y$

$ x,y,z$ positive numbers[/quote]
Suppose that $ z\equal{}max\{x;y;z\}$ then we can rewrite this ineq as:
$ (x\plus{}y)(x\minus{}y)^2\plus{}(2x\plus{}z\minus{}y)(x\minus{}z)(y\minus{}z) \geq 0$
We have $ x\plus{}y>0;2x\plus{}z\minus{}y>0$
Done!

From Schur It's equivalent to prove that $ x^2y \plus{} y^2z \plus{} z^2x \geq x^2z \plus{} y^2x \plus{} z^2y$
$ \leftrightarrow (x \minus{} y)(y \minus{} z)(x\minus{}z) \geq 0$ which is obviously true

[quote="apollo"]From Schur It's equivalent to prove that $ x^2y \plus{} y^2z \plus{} z^2x \geq x^2z \plus{} y^2x \plus{} z^2y$
$ \leftrightarrow (x \minus{} y)(y \minus{} z)(x \minus{} z) \geq 0$ which is obviously true[/quote]

What happens when $ (x,y,z)\equal{}(3,1,2)$?

[quote="nayel"][quote="apollo"]From Schur It's equivalent to prove that $ x^2y \plus{} y^2z \plus{} z^2x \geq x^2z \plus{} y^2x \plus{} z^2y$
$ \leftrightarrow (x \minus{} y)(y \minus{} z)(x \minus{} z) \geq 0$ which is obviously true[/quote]

What happens when $ (x,y,z) \equal{} (3,1,2)$?[/quote]
Sorry,I had a mistake. :blush:


[quote="qweretyq"]but can handwritten solutions get commendations?
procrastination killed me.. and im starting doing the problems now so i prob won't be able to LaTeX it....[/quote]

Yes, they can get commendations.

However, handwritten solutions will not be posted.

[quote="rain"]can u write the handwritten solutions in pencil, or does it have to be in pen?[/quote]

Either is fine, but we *strongly* recommend not faxing pencil-written solutions.  They tend to come through very poorly.


Show by induction that $ 4^{2n\plus{}1} \plus{} 3^{n\plus{}2}$ is divisible by 13 for all $ n>\equal{}0$.

Prove $ 4^{2n + 1} + 3^{n + 2}$ is divisible by $ 13$ $ \forall ~n \in \mathbb{N}\cup \{0\}$.

[b][u]Proof by the Generalized Principal of Mathematical Induction:[/u][/b]

$ (i)$ Let $ n = 0$.  Thus:

$ 4^{2n + 1} + 3^{n + 2} = 4^1 + 3^2 = 4 + 9 = 13 = 13 \cdot 1$ and since $ 1 \in \mathbb{Z}$, by definition, $ 13 | 13$ and thus our statement holds for $ n = 0$.

$ (ii)$ Suppose for some $ n \in \mathbb{N}\cup \{0\}$, $ 4^{2n + 1} + 3^{n + 2}$ is divisible by $ 13$.

Thus:

$ 4^{2(n + 1) + 1} + 3^{(n + 1) + 2} = 4^{2n + 3} + 3^{n + 3}$
$ = 4^2 \cdot 4^{2n + 1} + 3^1 \cdot 3^{n + 2}$
$ = 16 \cdot 4^{2n + 1} + 16 \cdot 3^{n + 2} - 13 \cdot 3^{n + 2}$
$ = 16 (4^{2n + 1} + 3^{n + 2}) - 13 \cdot 3^{n + 2}$

And by our inductive step $ 13|(4^{2n + 1} + 3^{n + 2})$, so $ 13|[16(4^{2n + 1} + 3^{n + 2})]$.  Also, clearly $ 13|( - 13 \cdot 3^{n + 2})$ so therefore $ 13$ shall divide their sum.  Therefore our statement holds for $ n + 1$.

$ (iii)$ Therefore by parts $ (i)$ and $ (ii)$ and the Generalized PMI (we started at $ n = 0$ rather than $ n = 1$ so its generalized), $ 4^{2n + 1} + 3^{n + 2}$ is divisible by $ 13$ $ \forall ~n \in \mathbb{N}\cup \{0\}$.

$ \text{Q.E.D.}$


irrational number
real analysis
real analysis unsolved

This question was made by Lucas Andisco. I failed to give an answer.

Suppose that $ \{a_i\}_{i\in\mathbb N}$ is a sequence of positive rational numbers, whose series converges to a rational number. (I mean, $ \sum_{i\equal{}1}^\infty a_i$ exists and is rational.) Is it always true that there is a subsequence whose series converges to an irrational number?

Any convergent series of infinitely many positive terms has subseries converging to infinitely many different values.

Proof: Choose a subseries $ b_n$ such that the terms go to zero rapidly: $ b_{n\plus{}1}<\frac12 b_n$. Each subseries of this must converge to a different value (since $ \sum_{n>k}b_n<b_k$), and there are uncountably many of these.

An uncountable set of values must include irrationals.

Thanks.

I was trying to prove something stronger:

Let $ \{a_i\}_{i\in\mathbb N}$ be a sequence of positive real numbers, whose series converge to a number $ M$. Is it always true that there is a number $ m<M$ such that if $ m\leqq S\leqq M$ then there is a subsequence of $ \{a_i\}$ whose series converge to S?

That's false. There are not necessarily any intervals in the set of possible values. Example: if $ a_n\equal{}2\cdot 3^{\minus{}n}$, the set of possible values is the Cantor set.

Hahaha, thank you very much! Next is:

Let $ \{a_i\}_{i\in\mathbb N}$ be a sequence of positive real numbers, whose series converge to a number $ S$. Is is always true that there is an interval $ I$ containing $ S$ such that if $ s\in I$ then there is a real sequence $ \{\epsilon_i\}$ with $ \epsilon_i\equal{}\minus{}1,$ $ 0$ or $ 1$ such that the series of $ \{\epsilon_i a_i\}$ converge to $ s$?

No, and all it takes is a slightly different Cantor-type set. $ 5^{\minus{}n}$ should do it.


A faucet leaks at a rate of 1 pint every 2 hours for 4 full days. How many gallons of water leaked from the faucet?

Well there are 24 hours in 1 day so there are 96 in two days.  1 pint every 2 hours means .5 pints every hour.  So $ .5\times 96\equal{}48$.

However, it asked for gallons, so there are $ 48 \div 8 \equal{} \boxed{6}$ gallons.


When I try the following code...

import olympiad;
size(150);
pair A=(0,0), B=(3,1), C=(1,3), D=(2,2);
path P=A..B..C..D;
draw(P,blue);
add(pathticks(P,5));

...I get an olympiad.asy error: Divide by Zero. I notice that the example compiles perfectly on AoPS (without the olympiad.asy call). I recently saw the note to change the "include" declarations to "import" at the start of olympiad.asy but that hasn't had any effect either way.

It seems that you are still using the old version of the Olympiad package. Just download the new one from my sticky "upgraded Olympiad package" on this subforum :).


[quote="GameBot"]When the positive integers up to 2007 are alternately subtracted and added $ (1 \minus{} 2 \plus{} 3 \minus{} 4 \plus{} \cdots \plus{} 2001 \minus{} 2002 \plus{} 2003 \minus{} 2004 \plus{} 2005 \minus{} 2006 \plus{} 2007)$, what is the value of the expression?[/quote]

Well, it's equivalent to 1003 ones because (-2+3), (-4+5), (-6+7), etc. is equal to one, so there are 2006/2=[b][i][u]1003[/u][/i][/b].  Interesting concept, like a mash-up of Gauss's sum of first hundred numbers trick, mixed with telescoping sums.

I believe both of you forgot to add one of the ones. The answer should, I believe, be $ 1004$.

I also get 1004. I really want to run for a calculator to check.

It's pretty obvious that it's actually the answer, but if you want to, here's some C++ code that would do it:

[code]
#import <iostream>
using std::cout;
void main() {
int i;
int t = 0;
int d;
for (i=1;i<=2007;i++) {
if (i % 2 == 0) {
t=t-i;
}
else {
t=t+i;
}
}
cout << t;
cin << d;
return 0;
}
[/code]

I haven't coded C++ in a long time, so that might be wrong...

It's 1004. All the peopl with the answer 1003 forgot the first 1.

Interesting. I've always done that one $ 2007\minus{}(1\plus{}1\plus{}...\plus{}1)$ (easier to not make mistakes xD)

lol, this topic is from december 2007  :rotfl:

Dang I just realized that  :rotfl:  :rotfl:  :rotfl:


inequalities
AMC
USA(J)MO
USAMO
inequalities theorems

While reading the book " Mathematical Olympiad Treasures" by Titu Andreescu and Bogdan Eneescu I came across a very useful lemma which is stated below.

For a,b,x,y all positive real numbers then the inequality a^2/x+b^2/y>(a+b)^2/(x+y)

The proof is very simple and is omitted here. The generalization can be done using Mathematical induction.

This lemma can be used to prove inequalities like Cauchy-Schwartz, Nesbitt's etc. Some IMO, USAMO and Romanian Olympiad problems were solved very easily by this lemma.

It's very well-known :wink:
http://www.mathlinks.ro/viewtopic.php?t=15125


geometry
circumcircle
inequalities
inradius
geometry proposed

Let $ABC$ be a triangle.Denote $r_{1a}$ radius of the circle tangent to the sides $AB$ and $AC$ and internally tangent to the circumcircle of the triangle $ABC$, and denote $r_{2a}$  radius of the circle tangent to the sides $AB$ and $AC$  and externaly tangent to the circumcircle of the triangle $ABC$.Prove that $3(\frac{1}{r_{2a}}+\frac{1}{r_{2b}}+\frac{1}{r_{2c}})=\frac{1}{r_{1a}}+\frac{1}{r_{1b}}+\frac{1}{r_{1c}}$ if and only if the triangle $ABC$ is echilateral.

Nobody? It's not that hard  :P . At first i was afraid it would be too boring  :wink:

[color=darkblue]See: http://www.mathlinks.ro/viewtopic.php?t=152882
Look at the [b]Virgil Nicula's[/b] post.[/color]

As in a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=152882]topic of Virgil Nicula[/url]
$\sum\frac{1}{r_{1a}}=\frac{4R+r}{2Rr},\sum\frac{1}{r_{2a}}=\frac{2R-r}{2Rr}$ thus your inequality is equivalent
$3\frac{2R-r}{2Rr}=\frac{4R+r}{2Rr}\Leftrightarrow R=2r$ of course it is equilateral. :D

Oki.Srry i didn't see that  :blush: .I think i posted this topic before Mr Nicula posted his..i don't know.But, i have another solution for the ineguality :
Using the formulas $r_{1a}=\frac{r}{cos^{2}\frac{A}{2}}$ and $r_{2a}=\frac{r_{a}}{cos^{2}\frac{A}{2}}$ where $r$ is the inradius and $r_{a}$ is the radius of the excircle tangent to $BC$.So the ineguality becomes:
$\sum \frac{cos^{2}\frac{A}{2}}{r_{a}}\geq \frac{1}{3}(\sum cos^{2}\frac{A}{2})\frac{1}{r}=\frac{1}{3}\sum cos^{2}\frac{A}{2}\sum \frac{1}{r_{a}}$ where i used that $\frac{1}{r}=\sum \frac{1}{r_{a}}$.
As you can see the ineguality reduces to $Chebisev$, if we prove that if $\frac{1}{r_{a}}\leq \frac{1}{r_{b}}\leq \frac{1}{r_{c}}$ than also $cos^{2}\frac{A}{2}\leq cos^{2}\frac{B}{2}\leq cos^{2}\frac{C}{2}$
But $\frac{1}{r_{a}}\leq \frac{1}{r_{b}}\leq \frac{1}{r_{c}}$ if $p-a\leq p-b \leq p-c$ equivalent to $a\geq b \geq c$.
And $cos^{2}\frac{A}{2}\leq cos^{2}\frac{B}{2}$ if $a(p-a)\leq b(p-b)$ if $(a-b)(a+b-c)\geq 0$ wich leads to $a\geq b$.
So $cos^{2}\frac{A}{2}\leq cos^{2}\frac{B}{2}\leq cos^{2}\frac{C}{2}$ if $a\geq b \geq c$.
Than by $Chebisev$ we are done! :D


Is there any helpful sources to learn about vectors for a complete beginner?

Um...most pre-calculus books contain a chapter that basically introduces vectors...

[url]http://en.wikipedia.org/wiki/Vector_%28spatial%29[/url]

This should be a good start.

AoPS vol 2 has plenty of helpful chapters.


Let be $ a,b,c>0$ such that $ abc\equal{}1$ . Prove that :

$ \frac{1}{a^2(b\plus{}c)}\plus{}\frac{1}{b^2(c\plus{}a)}\plus{}\frac{1}{c^2(a\plus{}b)}\ge \frac{3}{2}$

Make the substitution $ a\equal{}\frac{1}{x}, b\equal{}\frac{1}{y}, c\equal{}\frac{1}{z}$, with $ xyz\equal{}1$. Then the inequality becomes $ \sum_{cyc} \frac{x}{y\plus{}z} \geq \frac{3}{2}$, which is Nesbitt's inequality.

hello, let $ a\equal{}\frac{1}{x},b\equal{}\frac{1}{y}c\equal{}\frac{1}{z}$, then our inequality is equivalent with
$ \frac{x}{y\plus{}z}\plus{}\frac{y}{y\plus{}z}\plus{}\frac{z}{x\plus{}y}\geq\frac{3}{2}$ and this is Nesbitt.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, let $ a \equal{} \frac {1}{x},b \equal{} \frac {1}{y}c \equal{} \frac {1}{z}$, then our inequality is equivalent with
$ \frac {x}{y \plus{} z} \plus{} \frac {y}{y \plus{} z} \plus{} \frac {z}{x \plus{} y}\geq\frac {3}{2}$ and this is Nesbitt.
Sonnhard.[/quote]

This was the idea of brut3forc3  :) . Ok , i didn't said you take it his idea . May be you wasn't observant to his post .

hello, no, it was also my own idea.
Sonnhard.

there is my solution, assume that $ a\geq b \geq c$ then by Chebychev,
$ S\equal{}\sum \frac{1}{a^2(b\plus{}c)}\geq\frac{1}{3}\left(\sum\frac{1}{a^2}\right)\left(\sum \frac{1}{b\plus{}c}\right)$
by Cauchy-Schwarz, $ \sum\frac{1}{b\plus{}c}\geq \frac{9}{2(a\plus{}b\plus{}c)}$ then it's suffice to prove that,
$ \left(\sum\frac{1}{a^2}\right)\left(\frac{1}{a\plus{}b\plus{}c}\right) \geq 1$
$ \Leftrightarrow \sum\frac{1}{a^2} \geq a\plus{}b\plus{}c$
$ \sum a^2b^2 \geq a^2b^2c^2(a\plus{}b\plus{}c)\equal{}abc(a\plus{}b\plus{}c)$ wich is true, because, $ (ab\minus{}bc)^2\plus{}(bc\minus{}ca)^2\plus{}(ca\minus{}ab)^2 \geq 0$

with the same conditions, we can also prove that,
$ \sum \frac {1}{a^{\alpha \plus{} 1}(b \plus{} c)}\geq \frac {3}{2}$ for all $ \alpha \geq 1$


calculus integration inequalities algebra polynomial logarithms Gauss

Let $ P$ is a polynomial of degree $ n$. Prove that $ (\int\limits_{\minus{}1}^{1} |P(x)|dx)^2\le C\sqrt{n}\int\limits_{\minus{}1}^{1} (1\minus{}x)P^2(x)dx$.
Can we replace $ \sqrt{n}$ by something smaller?

The usual meaning would be, "There exists a constant $ C,$ independent of $ n$ and $ P,$ that makes this inequality true for all $ n$ and all $ P.$"

You are right, Kent, here is the usual meaning. :)
I srongly believe now that we can replace  $ \sqrt {n}$ by $ \log^{\alpha}{n}$.

You certainly can. A polynomial of degree $ n$ cannot have more than a half of its integral over $ [\minus{}1,1]$ concentrated on the interval of length much less than $ 1/n^2$ (by Markov, we have $ |P'|<Cn^2\max_{[\minus{}1,1]}|P|$, so the set of "largest" values is already smeared over the interval of length comparable to $ \frac1{n^2}$). Thus, you can integrate up to $ 1\minus{}\frac c{n^2}$ instead of $ 1$, which allows to get $ C\log n$ by Cauchy-Schwarz immediately. If you take $ P(x)\equal{}\sum_{k\equal{}0}^n x^k$, you'll see that this is sharp up to a constant factor.

Very nice, I have different ( slightly more complicated solution) with much more  complicated example using cvadrature formulas.

I will post only my example (because the proof that I have involve one non-elementary inequality) which solves general problem of maximazation $ \frac {(\int g(x))^2}{\int (1 \minus{} x)g^2(x)}$ for $ deg g(x)\le n$. Consider $ P_{n \plus{} 1}$- Legendre polynomial http://en.wikipedia.org/wiki/Legendre_polynomials of degree $ n \plus{} 1$.
and let $ x_i$ be its zeros. Using Gauss-Jacobi quadrature formulas http://en.wikipedia.org/wiki/Gauss-Jacobi_Mechanical_Quadrature we obtain that our problem reduces to the maximization $ \frac {\sum \lambda_k g(x_k))^2}{\sum \lambda_k (1 \minus{} x_k)g^2(x_k)}$ where $ \lambda_k > 0$ are Chrystoffel numbers and $ \minus{} 1 < x_k < 1$ (these are standart properties of orthogonal polynomials). Now using Cauchy-Schwartz inequality we get $ \frac {(\sum \lambda_k g(x_k))^2}{\sum \lambda_k (1 \minus{} x_k)g^2(x_k)}\le \sum (\frac {\lambda_k}{1 \minus{} x_k} )$ (the bound is sharp because we can take the polynomial $ P$ of degree $ n$ for which $ P(x_k) \equal{} \frac {1}{1 \minus{} x_k}$ and we'll get the equality). Now to estimate the last sum we can just use a few standart things about Chrystoffel numbers (well,  I've find out  some of them only a few weeks or days ago :blush: ): representation of Chrystoffel numbers $ \lambda_k \equal{} c_n\frac {1}{\sum p^2 _i (x_k)}$ where $ p_i$ is Legendre polynomial of degree $ i$ and $ c_n$ is a bounded constant, the inequality $ \sqrt {(1 \minus{} x^2)}\sum p^2 _i (x) < cn$ for some universal constant $ c$ and the property that if $ x_k \equal{} cos \alpha_k$ then $ \alpha_k \sim \frac {k\pi}{n \plus{} 1}$. Combining these three observations we'll get the desired result (our sum will be approximately $ \log n$). But of course fedja's solution is much more simplier (I actualy need some more general case I'll post it later )

Please  write  down  the  proof  [color=red] in detail[/color]  that
$ (\int\limits_{ \minus{} 1}^{1}|P(x)|dx)^{2}\le C\sqrt {n}\int\limits_{ \minus{} 1}^{1}(1 \minus{} x)P^{2}(x)dx$

thanks!

Please, tell [color=red]in detail[/color], which part(s) of what is written you have trouble with. ;)

For  example  ,      your  post   in  #5,  I  cannot  undarstand..........

Thanks!


In fiecare patratel al unui dreptunghi $\ 10x19$ s-au scris unul dintre numerele $\ 0$ sau $\ 1$, dupa care s-au calculat sumele numereleor din fiecare coloana si fiecare linie.Care este cel mai mare numar de rezultate diferite care se poate obtine?Justificati!

R: 19.
 Sumele obtinute pe linii si coloane pot lua valori de la 1 la 19 (pe coloane suma <=10 iar pe linii <=19). Avem 19 coloane si 10 sume posibile, deci 9 sume obligatoriu se vor repeta. Apoi, sunt 10 linii, iar pentru ca sumle sa nu se repete vom lua valorile din casute astfel incat sumele sa aiba valori de la 10 la 19.  Suma 10 se repeta, deci in total vom avea 10 sume care se repeta. Daca toate sumele ar fi fost diferite, in total ar fi 29 sume distincte. 29-10=19.
 Ne mai ramane sa aratam ca un astfel de dreptunghi exista, ceea ce nu este greu...

Da, raspunsul este corect, dar valorile sint de la $\ 0$ la $\ 19$, in total $\ 20$ de valori.Nu de la $\ 1$ la $\ 19$.


There are 2000 points in a circle, and each point is given a number that is equal to the average of the numbers of its two nearest neighbors. Show that alla the numbers must be equal.

I am sorry, but I dont know the answer :( I hope u will find it :wink:

[quote="miyomiyo"]In a circle or on a circle?

And do the numbers have to be integers?[/quote]

The points are [i]on[/i] the circle :blush: and the numbers are not necessarily integers (according to the source from where I found the problem)

[hide]Consider the smallest number, the two numbers next to it must be the same as it as one cannot be less than it.  Then consider the number next to one of those, it must be equal to them aswell...[/hide]

very good use of the extremal principal :wink:

[hide]Well, if we just pick a point $A$, and give the lowest number $p$, so that it is between points $B$ with number $q$ and $C$ with number $r$. We know that if $A$ is the avergare of $B$ and $C$, then

$p=\frac{q+r}2$ which is only true if $q=r=p$, so this simply will keep going no matter how many points are there.[/hide]


Lets assume $ is a variable 1$=100Cents

$=x Cents=y 1y=.01$
x=100y
(x)(x)=100y(x)
(x)(x)=100y(100y)
xx = 10000y
xx=100x
x=100
$=100

?
This is as much of a formula as I could figure out.
Does this make sense?
Is there a rule I am disregarding?

Let me know if this works thanks.

Is 1$times 1$ equal to 100$? What I am trying to do is support the Idea that 1 dollar is equal to 100 cents, therefore 1 dollar times 1 dollar equals 100 cents times 100 cents which would equal 10 dollars. my algebra Idea probably didnt make much sense, which is where I need your help. I cant find out how to support the Idea that 1$*1$= 100$

The least I can think of is:

1$*1$= 1$squared$=100cents
100cents squared = 10,000 cents, which would equal 100$(woops) not 10$

[quote="t0rajir0u"]What does it mean to multiply money by money?[/quote]

good question.
I dont know.
maybe its undefineable :huh:

You're losing the units; that's why everything seems weird. For convenience, let $d$ be dollars and $c$ be cents. We have

$1d \cdot 1d = 1d^{2}$.

The unit of this is dollars-squared, which has pretty much no meaning. Then

$100c \cdot 100c = 10000c^{2}$.

The unit of this is cents-squared, again without meaning. The important thing to notice is that

$1d^{2}= 10000c^{2}$ because $1d = 100c$.

Your "argument" is similar to saying:

$1 \text{ft}= 12 \text{in}$

$1 \text{ft}^{2}= 144 \text{in}^{2}$

so $\text{in}= 12 \text{in}$. Which is obviously wrong.

Thankyou Paladin, you are right, that makes no sense at all.
Do you think there is any way of proving multiplication of 2 moneys?
Is there any way to prove the concept of 1$*1$= 100$, or is money multiplyed by money undefinable?

As far as I can tell, Money*Money is undefinable.

[quote="JavaMan"]I can't think of how dollar^2 is meaningful in anyway..[/quote]

Neither can I.  It doesn't describe anything.

You might as well say:

1 pound = 16 ounces
Dividing by 2, 1/2 pound = 8 ounces
If we square both sides, 1/4 pound = 64 ounces

Suddenly, overeating seems far more serious.

[quote="seamusoboyle"]It means george bush can lie about what he spends![/quote]

:rotfl:

[quote="nutz_for2.718281828"][quote="seamusoboyle"]It means george bush can lie about what he spends![/quote]

 :rotfl:  :rotfl:  :rotfl:

Also, I don't think there is any square dollar...[/quote]

You could cut one [i]into[/i] a square...

This is just a suggestion, but shouldn't a moderator lock this topic up as it is going no where.

$\$1\times\$ 1\ne \$100$. It makes no sense.

[quote="Swoopkev"]Lets assume $ is a variable 1$=100Cents

$=x Cents=y 1y=.01$
x=100y
---- (x)(x)=100y(x)
---- (x)(x)=100y(100y)
xx = 10000y
xx=100x
x=100
$=100

[/quote]

if you pay attention to the two lines we see the second line should be xx = 10000yy

you multiply money x a number

but not money and money.. together

insignificant

[quote="nutz_for2.718281828"][quote="seamusoboyle"]It means george bush can lie about what he spends![/quote]

 :rotfl:  :rotfl:  :rotfl:

Also, I don't think there is any square dollar...[/quote]
Well this simply implies that you have morbid obesity.

[quote="jhcreinhardt"]you get x^2 dollars and you cut and paste them until they're in a square of side x[/quote]

Hmm... or you could paste the dollar bills onto the faces of a cube, then you'd have dollars cubed.  :D

:spam: :spam:

To the original question: Money*Money is meaningless. Why would you want to multiply money by money anyways? After all, money isn't a length or anything. It's just money.

End of story.

@anirudh and mathnerd314: Do you have something against figuring out why certain things don't make sense? If you had a question, and you posted it on the forum because that's what this forum is for, would you be happy if instead of answering the question people went "That's a dumb question. Lock the thread."? We actually had a meaningful discussion going before the random jokes; if I didn't think it was a valid question, I wouldn't have posted (as well as the other people).

paladin8:
Sorry, I was just afraid the topic would degenerate into the random jokes.

Swoopkev:
I hope your question has been answered satisfactorily.

well yea...i think its been established that mulitplication involves the units, and that dollar^2 isnt a useful unit.  i still dont think the topic is worth anything to begin with...i mean, when has anyone ever multiplied dollars

[quote="mathnerd314"]paladin8:
Sorry, I was just afraid the topic would degenerate into the random jokes.
[/quote]

Fair enough, I was more concerned about the comment that effectively said that the original question was meaningless. It is disappointing how the topic turned into something reminiscent of G&FF.

@jhcreinhardt: The point isn't that the unit does not exist for all intents and purposes (which is a bad argument, since math is abstract after all). The point is that when people have legitimate questions, we should answer them instead of criticizing them.

@paladin: sorry. I didn't mean to. I thought it was going into a spam thread. :blush:

Maybe treat the $\$$ as $x$ sortof

[quote="anirudh"]@paladin: sorry. I didn't mean to. I thought it was going into a spam thread. :blush: [/quote]

Yeah, no problem. I can see what you meant as well.

Any particular unit is useful if it describes something quantifiable with which we can calculate other useful quantities.  Since it's already pretty clear that the unit $\text{Money}^{2}$ is not a useful quantity in the sense that it describes something quantifiable, there's not much point in talking about it.

What do you mean by showing its uselessness "mathematically"?  Usefulness is inherently a scientific question, not a mathematical one, since it deals with real-world application.

[quote="Swoopkev"][size=150][color=red]NEW CONCLUSION:[/color][/size][b][color=red]MULTIPLYING MONEY BY MONEY IS USELESS.[/color][/b]
[color=blue]Conclusion has been drawn logically by Forum Users, but can anybody draw a conclusion Mathametically?[/color]

Ideas: [quote="anirudh"]Maybe treat the $\$$ as $x$ sortof[/quote][/quote]Please do not make excessive use of large font sizes and color changes as these tend to be distracting.

Actually, I googled "dollars squared" and found that square dollars can be useful sometimes, in the following sense:

Suppose you have a bunch of data for personal expenses or something like that. Find the variance.

(That's why the standard deviation is usually used.)

Otherwise, I'm unaware of any really practical use for square dollars.


[hide]Note that:
$ \frac {3^{k} \plus{} 1}{3^{k} \minus{} 1} \equal{} 1 \plus{} \frac {2}{3^{k} \minus{} 1} \le 1 \plus{} \frac {2}{2^{k}}$ with equality on the right side only occurring for $ k \equal{} 1$

Therefore:
$ \sum_{k \equal{} 1}^{n \plus{} 1}\left(1 \plus{} \frac {2}{3^{k} \minus{} 1}\right) < \sum_{k \equal{} 1}^{n \plus{} 1}\left(1 \plus{} \frac {2}{2^{k}}\right) \equal{} \sum_{k \equal{} 1}^{n \plus{} 1}1 \plus{} \sum_{k \equal{} 1}^{n \plus{} 1} \frac {2}{2^{k}} < (n \plus{} 1) \plus{} \frac {\frac22}{1 \minus{} \frac12} \equal{} n \plus{} 3$

Also,
$ \sum_{k \equal{} 1}^{n \plus{} 1}\left(1 \plus{} \frac {2}{3^{k} \minus{} 1}\right) \equal{} \sum_{k \equal{} 1}^{n \plus{} 1}1 \plus{} \sum_{k \equal{} 1}^{n \plus{} 1}\frac {2}{3^{k} \minus{} 1} \equal{} (n \plus{} 1) \plus{} 1 \plus{} \sum_{k \equal{} 2}^{n \plus{} 1}\frac {2}{3^{k} \minus{} 1} \ge n \plus{} 2$

Thus, $ n \plus{} 2 \le A < n \plus{} 3$ and $ \lfloor A \rfloor \equal{} n \plus{} 2$.[/hide]


calculus
integration
trigonometry
logarithms
calculus computations

Let $a,b$ be positive constant numbers.
Calculate

\[\int^{\frac{\pi}{2}}_0 \frac{\cos x}{a\cos x+b\sin x} dx\]

Another classical one.

We study the indefinite integral:
$
\int \frac{c\cos x+d\sin x}{a\cos x+b\sin x}dx
$

$(a\cos x+b\sin x)'=-a\sin x+b\cos x$.
Let $c\cos x+d\sin x = A(a\cos x+b\sin x) + B(a\cos x+b\sin x)'$

If we can solve $A,B$, then the indefinite integral becomes
$
\int \left( A+\frac{B(a\cos x+b\sin x)'}{a\cos x+b\sin x}\right)dx = Ax+B\ln(a\cos x+b\sin x)+c
$

It can be done by brute force, using the substitution $t=\tan\left(\frac x2\right)$. For a nicer method,
[hide="see this"]
We have $a\cos x+b\sin x=c\cos(x-\theta)$, where $c=\sqrt{a^2+b^2}$, $\cos\theta=\frac ac$, and $\sin\theta=\frac bc$. Substituting $y=x-\theta$, the integral becomes $\int_{-\theta}^{\frac\pi2-\theta}\frac{\cos(y+\theta)}{c\cos y}\, dy$
$=\int_{-\theta}^{\frac\pi2-\theta}\frac{\cos y\cos\theta-\sin y\sin\theta}{c\cos y}\, dy=\int_{-\theta}^{\frac\pi2-\theta}\frac{\cos\theta}c-\frac{\sin\theta}c\tan y\, dy$
$=\frac{\pi\cdot a}{2(a^2+b^2)}+\left[\frac{b}{a^2+b^2}\ln(\cos y)\right]_{-\theta}^{\frac\pi2-\theta}=\frac\pi2\cdot\frac a{a^2+b^2}+ \frac{b}{a^2+b^2}\ln\frac ba$[/hide]

All of this assumes that $a$ and $b$ have the same sign. If they do not, the integral is improper and diverges.

You are right, jmerry. :)

Paring the indefinite of $\sin $ and $\cos $,let $I=\int^{\frac{\pi}{2}}_0 \frac{\cos x}{a\cos x+b\sin x}dx,J=\int^{\frac{\pi}{2}}_0 \frac{\sin x}{a\cos x+b\sin x}dx$,

we have $aI+bJ=\frac{\pi}{2},bI-aJ=\ln \frac{b}{a}$, Solving this you will obtain your answer.

[quote="kunny"]Can anyone solve this problem?[/quote]
Isn't my solution a general method?


Let be given a triangle $ABC,$ with the orthocentre $H$ and $B'= BH\cap AC$. Assume that: $HB=3HB'$, denote: $K$ be the midpoint of $[BB']$. Prove that: $\angle{AKC}=90^{\circ}$

We can use Papius Theorem of Median and the Reciproc of Pytagora :lol:

In the triangle AHB', m(AHB')=m(C) and tg(C)=AB'/B'H
In the triangle BB'C, tg(C)=BB'/B'C. So AB'/B'H=BB'/B'C,
AB'*B'C=B'H*BB'. But B'H=KB'/2 and BB'=2KB'.
Then AB'*B'C=(KB'/2)*2KB'=KB'^2, so the triangle AKC is a right-angled triangle with the right angle K


calculus
integration
calculus computations


Let $ a$, $ b$, $ c$ be nonzero real numbers such that $ a\plus{}b\plus{}c\equal{}0$ and $ a^3\plus{}b^3\plus{}c^3\equal{}a^5\plus{}b^5\plus{}c^5$. Find the value of
$ a^2\plus{}b^2\plus{}c^2$.

[quote="worthawholebean"]Let $ a$, $ b$, $ c$ be nonzero real numbers such that $ a \plus{} b \plus{} c \equal{} 0$ and $ a^3 \plus{} b^3 \plus{} c^3 \equal{} a^5 \plus{} b^5 \plus{} c^5$. Find the value of
$ a^2 \plus{} b^2 \plus{} c^2$.[/quote]
[hide="Solution"]
Let $ k \equal{} a^2 \plus{} b^2 \plus{} c^2$.  Then\[ k(a^5 \plus{} b^5 \plus{} c^5) \equal{} k(a^3 \plus{} b^3 \plus{} c^3) \equal{} (a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3)\]Thus,\[ k(a^5 \plus{} b^5 \plus{} c^5) \equal{} a^5 \plus{} b^5 \plus{} c^5 \plus{} a^2b^2(a \plus{} b) \plus{} b^2c^2(b \plus{} c) \plus{} a^2c^2(a \plus{} c)\]and since $ a \plus{} b \plus{} c \equal{} 0$, we have that $ a \plus{} b \equal{} \minus{} c$, $ b \plus{} c \equal{} \minus{} a$, and $ a \plus{} c \equal{} \minus{} b$.  thus,\[ (k \minus{} 1)(a^5 \plus{} b^5 \plus{} c^5) \equal{} \minus{} a^2b^2c \minus{} b^2c^2a \minus{} a^2c^2b \equal{} \minus{} abc(ab \plus{} bc \plus{} ac)\]Also, since $ a \plus{} b \plus{} c \equal{} 0$, we have that $ a^5 \plus{} b^5 \plus{} c^5 \equal{} a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$.  Plugging this in, we have that\[ 3abc(k \minus{} 1) \equal{} \frac { \minus{} abc[(a \plus{} b \plus{} c)^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2)]}{2}\]Hence, either $ abc \equal{} 0$ (in which case $ a$, $ b$, or $ c$ must be $ 0$, which contradicts the given) or $ 6(k \minus{} 1) \equal{} \minus{} 1(0^2 \minus{} k) \equal{} k\implies 5k \equal{} 6$, which gives us that $ k \equal{} \boxed{ \frac {6}{5}}$.  [/hide]
EDIT: Sorry everyone, [b]arqady[/b] is correct.  I made a small error in my solution; it has been edited.


inequalities
search
algebra unsolved
algebra

This is from the Austrian Mathematical Olympiad, 1984.

Let a,b,c be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {12abc} \leq 1$

I used AM-GM and cubed the result to find $ abc \leq 1/27$, so the problem reduces to

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {12/27} \leq 1$
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {4/9} \leq 1$
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2/3 \leq 1$
$ a^2 \plus{} b^2 \plus{} c^2 \leq 1/3$
where $ a \plus{} b \plus{} c \equal{} 1$.  It seems intuitive that the maximum (edit: minimum) of $ a^2 \plus{} b^2 \plus{} c^2$ occurs when $ a \equal{} b \equal{} c \equal{} 1/3$, such that $ a^2 \equal{} b^2 \equal{} c^2 \equal{} 1/9$, meaning $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1/3$, but how do I prove this?  Can it be said that any expression symmetric in its variables (like $ a^2 \plus{} b^2 \plus{} c^2$) reaches its maximum (edit: minimum) when the variables are all equal to their maximum possible value?  If this is true, is it an inequality with a name that I should cite?

Edit: Sorry, I meant minimum value of $ a^2 \plus{} b^2 \plus{} c^2$. xD

It's pretty difficult to prove considering it's false: $ 1^2 \plus{} 0^2 \plus{} 0^2 \equal{} 1 > \frac {1}{9}$.  The correct inequality is actually the opposite one (which can be proven by Cauchy or Power Mean / AM-QM) and is the reason this problem is hard.

As for your other question, it's not in fact true that the extrema of a symmetric expression occur when the values of $ a, b, c$ are symmetric.  Jason Bland (scorpius119) is particularly good at finding and solving problems where this is true, so if you search for his posts you might find an example, but I can't at the moment.

Edit:  Another note.  One nontrivial way to write inequality problems is to take two inequalities $ A \le B, C \ge D$ and ask whether either of the inequalities $ A \plus{} C \le B \plus{} D$ or $ A \plus{} C \ge B \plus{} D$ holds.  In the happy case that one of those inequalities holds, it cannot be proven from either of the inequalities from which it was written and is therefore necessarily sharper (hence harder) in some sense.

No matter what you meant minimum or maximum, it can't be solved by removing $ abc$ like that in the first place.

[hide="My solution"]
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {12abc} \leq 1$
$ \iff (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca) \plus{} \sqrt {12abc}\leq 1$
$ \iff 1 \minus{} 2(ab \plus{} bc \plus{} ca) \plus{} \sqrt {12abc}\leq 1$
$ \iff \sqrt {3abc}\leq ab \plus{} bc \plus{} ca$
$ \iff \sqrt {3abc(a \plus{} b \plus{} c)}\leq ab \plus{} bc \plus{} ca$
$ \iff (ab \plus{} bc \plus{} ca)^2\geq 3abc(a \plus{} b \plus{} c)$
$ \iff a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\geq abc(a \plus{} b \plus{} c)$
By AM-GM $ \frac {a^2b^2 \plus{} b^2c^2}{2}\geq abc\cdot b$
Similarly $ \frac {b^2c^2 \plus{} c^2a^2}{2}\geq abc\cdot c$
And $ \frac {c^2a^2 \plus{} a^2b^2}{2}\geq abc\cdot a$
Adding these: We have the required inequality.[/hide]


determine all real functions $f(x)$ which satisfy
$f(x-f(y))=f(x+y^{1999})+f(f(y)+y^{1999})+1$

$Please$ $it$ $would$ $be$ $nice$ $to$ $send$ $you$ $this$ $problems$ $to$ $the$ $Algebra$ $section$, $please$.

Davron


Art of Problem Solving will be hosting a math jam for Round 3 of the USAMTS, on Monday, January 14 at 7:30 PM ET/4:30 PM PT.  You may not discuss the problems on this forum until Friday, January 11.

If you missed it, [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=223]the transcript is available here[/url].


Let $ n \geq 3$ be an integer and $ x,y,z \in \mathbb{Z}^*$ with $ x \neq y, y \neq z, z \neq x$. Show that the equation $ z^{3n} \equal{} 3x^ny^nz^n \plus{} x^{3n} \plus{} y^{3n}$ has no solution in $ \mathbb{Z}^*$.

Let $ z^n \equal{} a; \minus{} y^n \equal{} b; \minus{} x^n \equal{} c$

Our equation can be written as

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$

So $ a \plus{} b \plus{} c \equal{} 0 \Leftrightarrow x^n \plus{} y^n \equal{} z^n$ ( note that $ y \neq x$ so $ b\neq c$ )

Then follow Fermat's

yup ;) if any one can proof without fermat ? :D

The existence of a counterexample to Fermat's last theorem gives a solution to this problem. If you found a simple proof that this equation had no solutions you'd be famous, because FLT follows.

Is this question more easy?
Let $ n \geq 3$,is there a solution for the modular equation:
$ z^{3n} \minus{} x^{3n} \minus{} y^{3n} \equiv 0 (mod x^ny^nz^n)$ with $ (x,y)\equal{}1,(y,z)\equal{}1,(z,x)\equal{}1$?


function
real analysis
real analysis unsolved

[quote="arp"]Let $(f_{n})$ and $(g_{n})$ be uniformly convergent sequences of real functions on $X$.  What can we say about the uniform convergence of $(f_{n}g_{n})$ and $(f_{n}\circ g_{n})$, or what conditions are necessary to conclude that they are uniformly convergent?[/quote]

Let $\parallel f_{n}-f\parallel <\epsilon,\parallel g_{n}-g\parallel <\epsilon, n\ge N_{\epsilon}$.
So
 $\parallel f_{n}g_{n}-fg\parallel =\parallel f_{n}g_{n}-f_{n}g+f_{n}g-fg\parallel <\epsilon(\parallel f_{n}\parallel +\parallel g\parallel )<$
$<\epsilon(\epsilon+\parallel f\parallel +\parallel g\parallel )$

Note that your argument depends on the norms $\|f\|$ and $\|g\|$ being finite - in other words, that $f$ and $g$ are bounded on $X.$ The theorem is actually false in general precisely becuase we cannot assume such boundedness if $X$ is not compact.

For example: let $f_{n}(x)=g_{n}(x)=x+\frac1n$ on $\mathbb{R}.$ We clearly have that $f_{n}=g_{n}$ converges uniformly to $f(x)=g(x)=x.$ But $f_{n}(x)g_{n}(x)=x^{2}+\frac{2x}{n}+\frac1{n^{2}}$ does not converge uniformly to $x^{2}.$

For a fixed $f,$ the proper conditions to assure that $f\circ g_{n}$ converges uniformly would the the uniform convergence of $g_{n}$ and the uniform continuity of $f.$ Again, without $X$ being compact, we cannot that just from the continuity of $f.$


Somebody requested that I remain on high alert on the AMC forum until Wednesday, just in case anyone does something they shouldn't be doing.  I intend to do that, but I was wondering: is there a way that I can make AoPS send me an e-mail every time someone posts in AMC?  (I know you can do this for individual threads - you click "notify me when a reply is posted", but I'm not sure if this is possible for an entire forum.

Well, if anyone sees a message about the AMC12 tomorrow that is appropriate, please PM the mods as soon as possible so that they will get an e-mail and check it out.

I will try to be on alert tomorrow and PM you if I see anything.


Let $ a,b$ be real numbers and $ a,b\in (0,4)$. The sequence $ x_n$ determined as follow:
\[ \begin{array}{l}
 a_0  \equal{} a,a_1  \equal{} b \\
 a_{n \plus{} 2}  \equal{} \frac{{2(a_{n \plus{} 1}  \plus{} a_n )}}{{\sqrt {a_{n \plus{} 1} }  \plus{} \sqrt {a_n } }} \\
 \mathop {\lim }\limits_{x \to \infty } \left( {a_n } \right) \equal{} ? \\
 \end{array}
\]

Let $ b_n\equal{}a_n/4$, then $ 0<a_0,a_1\le 1$ and $ b_{n\plus{}2}\equal{}\frac{b_n\plus{}b_{n\plus{}1}}{\sqrt{b_n}\plus{}\sqrt{b_{n\plus{}1}}}.$

It is easy to show, that $ min(\sqrt{b_n},\sqrt{b_{n\plus{}1}})\le b_{n\plus{}2}\le max(\sqrt{b_n},\sqrt{b_{n\plus{}1}})$.
Let $ c\equal{}min(a/4,b/4),d\equal{}max(a/4,b/4$. Then $ c^{2^{\minus{}[n/2]}}\le b_{n}\le d^{2^{\minus{}n}}$, therefore $ b_n\to 1$ or $ a_n\to 4.$


A basketball team has a winning percentage of .625 after 8 games. How many games in a row must be won to raise the winning percentage to .750?

that decimal is 5/8 and x is the number of points he must score so

$ \frac{5\plus{}x}{8\plus{}x} \equal{} \frac{3}{4}$

Cross multiply and solve leaving x as 4.

Well they currently have 5 out of 8.

After x games, we want it to be .750

$ \frac{5\plus{}x}{8\plus{}x}\equal{}.750\equal{}\frac{3}{4}$

$ 4(x\plus{}5)\equal{}3(x\plus{}8)$

$ 4x\plus{}20\equal{}3x\plus{}24$

$ x\equal{}4$

so 4 more games. A quick check shows this is 9 of 12 which is indeed 3/4.

If they had won $ .625$ of 8 games, they won 5 games, and hence lost 3 games.  After some more number of games, all of which they one, these 3 games represent the .25 that they lost, so they need to play 12 games, so $ 12\minus{}4\equal{}\boxed{8}$

the other method can be faster sometimes, but it's helpful to know both :)


real analysis
calculus
integration
function
algebra
domain
real analysis unsolved

Let $f: \mathbb{R}\to [0, \infty]$ - unbounded function . $M_{k}= \{x: k \leq f(x) < k+1 \}$.

Prove that $f$ is Lebesgue integrable on $\mathbb{R}$ iff $\sum_{k}k m(M_{k}) < \infty$,

where $m$ - Lebesgue measure.

what if $0<f(x)<1$ for all $x$ and $f$ non integrable ?

[quote="eugene"]Yes, sorry, $f$ is supposed to be undounded.[/quote]
That doesn't help; the statment is still false, for essentially the reason that alekk said - that you have no control over the function when its values are between 0 and 1. For instance, consider $f(x)=\frac1{\sqrt{x}}$ on $(0,\infty).$

If you made the domain of the function a set of finite measure, then it would be true, becuase then the set $M_{0}$ couldn't have large enough measure.

Here's a variant. Again, $f$ is a nonnegative meaurable function, and I don't particularly care what the domain is.

Let $M_{k}=\{x: 2^{k}\le f(x)<2^{k+1}\}.$

Prove that $f$ is integrable iff $\sum_{k=-\infty}^{\infty}2^{k}m(M_{k})<\infty.$

Oh, yeah, sure - how could i write this nonsense. Yes, you are right, and now i know the solution.
Sorry for my stupid mistake.