algebra
polynomial
algorithm
quadratics
calculus
integration
number theory

Find the polynomial with least degree that leaves a remainder of $ 2x$ when divided by $ (x\minus{}1)^{2}$, and a remainder of $ 3x$ when divided by $ (x\minus{}2)^{3}$

This is a very interesting question.
[hide="hint"]It will help to expand the divisors. $ (x\minus{}1)^{2}\equal{}x^{2}\minus{}2x\plus{}1$ and $ (x\minus{}2)^{3}\equal{}x^{3}\minus{}6x^{2}\plus{}12x\minus{}8$

Now consider that $ \frac{a}{x^{2}\minus{}2x\plus{}1}\equal{}b\plus{}\frac{2x}{x^{2}\minus{}2x\plus{}1}$ and $ \frac{a}{x^{3}\minus{}6x^{2}\plus{}12x\minus{}8}\equal{}c\plus{}\frac{3x}{x^{3}\minus{}6x^{2}\plus{}12x\minus{}8}$

Now solve those for $ a$, something good might happen.

[/hide]

[quote="Kyron"]It will help to expand the divisors.[/quote]

It usually doesn't.

[hide="Solution"] Let the polynomial we are looking for be $ P(x)\plus{}3x$.  $ P(x)$ is uniquely defined $ \bmod (x\minus{}1)^{2}(x\minus{}2)^{3}$, so it has degree at most $ 4$.  Therefore

$ P(x) \equal{} (x\minus{}2)^{3}(ax\plus{}b) \equal{} (x\minus{}1)^{2}Q(x)\minus{}x$

Now, $ P(1) \equal{}\minus{}a\minus{}b \equal{}\minus{}1$ and $ P'(1) \equal{} 2a\plus{}3b \equal{}\minus{}1$.  Hence $ a \equal{} 4, b \equal{}\minus{}3$ and we have

$ P(x)\plus{}3x \equal{}\boxed{ (x\minus{}2)^{3}(4x\minus{}3)\plus{}3x }$ [/hide]

[quote="t0rajir0u"]

 Let the polynomial we are looking for be $ P(x)\plus{}3x$.  $ P(x)$ is uniquely defined $ \bmod (x\minus{}1)^{2}(x\minus{}2)^{3}$, so it has degree at most $ 4$.

 [/quote]

Why degree at most $ 4$? If it's uniquely defined that modulo, shouldn't it have degree at most $ 5$?

Also, did you need to check for a degree-3 one?

>_> if it is uniquely defined that modulus, then everything of the form $ k\cdot(x\minus{}1)^{2}(x\minus{}3)^{3}\plus{}\text{solution polynomial}$ will work.

If the solution polynomial is of the form $ ax^{5}\plus{}bx^{4}\plus{}cx^{3}\plus{}dx^{2}\plus{}ex\plus{}f$, then simply let $ k\equal{}\minus{}a$.

It is still not proven that polynomials of degree 1,2,3 don't work.

I also think that we could do like a euclidean algorithm thing for this also. I am not too good with the reverse euclidean algorithm, but i think that it should essentially be the same. [Instead of integer congruences, we are given polynomial congruences...since the euclidean algorithm holds with that, we should be able to apply the same process...].

[quote="Altheman"]It is still not proven that polynomials of degree 1,2,3 don't work.[/quote]
As the highest degree divisor is of the third degree and leaves a residual $ 3x$ this means that there must at [i]least[/i] be an $ x^{4}$ in the dividend.

It's unique modulo $ (x\minus{}1)^{2}(x\minus{}2)^{3}$. If a polynomial of degree $ 3$ or less satisfied it, then it's not unique modulo $ (x\minus{}1)^{2}(x\minus{}2)^{3}$.

Okay what if it was $ p(x)\equiv x\mod (x\minus{}1)^{2}$ and $ p(x)\equiv x\mod (x\minus{}2)^{3}$?

Both of your 'proofs' would follow the same as this example, but the minimal degree polynomial is $ p(x)\equal{}x$. So clearly, those are not valid explainations as they stand.

It is a trivial point, but I am just pointing out that we did not actually prove that we found the minimal degree polynomial.

[quote="Altheman"]I am just pointing out that we did not actually prove that we found the minimal degree polynomial.[/quote]

As I've stated repeatedly, the Chinese Remainder Theorem is valid over polynomials, which means the solution is [b]unique[/b] $ \bmod (x\minus{}1)^{2}(x\minus{}2)^{3}$.  :)

Edit:  By the way, if you read my solution carefully, it should be pretty clear that if a cubic polynomial had worked we would've discovered that $ a \equal{} 0$.  And a polynomial of degree $ 2$ or less would leave itself as a remainder upon division by $ (x\minus{}2)^{3}$, and $ 3x$ is clearly not a suitable solution.  ;)

[quote="Altheman"]Okay what if it was $ p(x)\equiv x\mod (x\minus{}1)^{2}$ and $ p(x)\equiv x\mod (x\minus{}2)^{3}$?

Both of your 'proofs' would follow the same as this example, but the minimal degree polynomial is $ p(x) \equal{} x$. So clearly, those are not valid explainations as they stand.

It is a trivial point, but I am just pointing out that we did not actually prove that we found the minimal degree polynomial.[/quote]

In that case the unique polynomial modulo $ (x\minus{}1)^{2}(x\minus{}2)^{3}$ would be $ x$. No quadratic, cubic, or quartic would work.

[quote="cancer"][quote="Lazarus"]>_> if it is uniquely defined that modulus, then everything of the form $ k\cdot(x\minus{}1)^{2}(x\minus{}3)^{3}\plus{}\text{solution polynomial}$ will work.

If the solution polynomial is of the form $ ax^{5}\plus{}bx^{4}\plus{}cx^{3}\plus{}dx^{2}\plus{}ex\plus{}f$, then simply let $ k \equal{}\minus{}a$.[/quote]

Hmm... actually, can $ a$ be a non-integer?[/quote]
In theory $ a$ can be any number (integer, real, imaginary, rational, irrational, fractional, etc.). Polynomials do not [i]tend[/i] to have imaginary or irrational coefficients. In this case, $ a$ cannot be irrational because an irrational $ a$ would cause the division to return the dividend. $ a$ is also not imaginary because that would (probably) return an imaginary quotient. Rational (non-integer) returns rational (non-integer) quotients. Fractional is almost the same as rational, except it allows certain irrational to be expressed (like $ \frac{5}{7}$).

This leaves real, integer solutions.

First off, $ \frac{5}{7}$ is not irrational.

Second, it doesn't matter whether or not $ a$ is integral or not, since we can still reduce it.

Third, $ a$ will be integral, because everything we are given are integer polynomials, and we are applying an algorithm that gives us another integer polynomial. If we were given non-integer polynomials, then there is a chance that $ a$ is not integral.

[quote="Hamster1800"]First off, $ \frac{5}{7}$ is not irrational.[/quote]
Oh yeah  :wallbash:  :blush: just repeating.

[quote="Hamster1800"]$ a$ will be integral, because everything we are given are integer polynomials, and we are applying an algorithm that gives us another integer polynomial. If we were given non-integer polynomials, then there is a chance that $ a$ is not integral.[/quote]

Elaboration:  We are working over the ring of integer polynomials, that is, $ \mathbb{Z}[x]$, so the division algorithm (for monic polynomials) will only produce integer polynomials.  The Chinese Remainder Theorem holds here (for monic polynomials).  Generally, it holds over $ F[x]$ with $ F$ a field, so if you like we can work over $ \mathbb{Q}[x]$ (which is what we would do if the polynomials were not monic), but when the polynomials we are dealing with are monic it doesn't matter.

Edit:  The point I am trying to make is that it is not necessary that the given polynomials be non-integer as long as they are not monic.


let $ a,b,c \geq 0$ prove that :

\[ a^ab^bc^c \geq (abc)^\frac {a \plus{} b \plus{} c}{3}
\]
2)

consider $ a\geq b \geq c \Longrightarrow A \equal{} a^ab^cc^b \plus{} b^bc^aa^c \plus{} c^cb^aa^b \leq B \equal{} 3a^ab^bc^c$
AM-GM:
\[ (abc)^\frac {a \plus{} b \plus{} c}{3} \leq A \leq B
\]

let $ a\geq b\geq c$
use  Chebyshev     inequality
$ a\ln(a) \plus{} b\ln(b) \plus{} c\ln(c)\geq\frac {a \plus{} b \plus{} c}{3}(a\ln(a) \plus{} b\ln(b) \plus{} c\ln(c))$
that is !
done~


Hey, i just wanna know some good music to get on my ipod, so if anyone could post their favorite artist or song, that'd be good. Thanks. I don't care what genre either.

I strongly recommend Queen, They Might Be Giants, Spock's Beard, and, of course, the Beatles.

Songwise, I also recommend 99 Red Balloons (Goldfinger,) Mr. Blue Sky (Electric Light Orchestra,) and A Whiter Shade of Pale (Procul Harum.)

pretty girl by sugarcult

the freshman by the verve pipe

perfect situation by weezer

only one by yellowcard

maria maria by santana

Jack johnson's In between dreams CD

Maroon 5's new cd

coldplay

right here by staind

[b]In the shadows by The Rasmus[/b]

Bring me to life by Evanescence

Unforgiven i,ii and nothing else matters by Metallica

Highway star by Deep Purple

Sweet child o'mine and November rain by Guns n Roses

[b]Without me,lose yourself,cleanin out my closet,Like toy soldiers by Eminem[/b]

With arms wide open , Higher by Creed

[b]Welcome home by Coheed and cambria[/b]

In da club,Candy shop by 50cents

Somewhere I belong,breaking the habit by Linkin Park

I just wanna live by Good Charlotte

[b]Its my life by Bonjovi[/b]

[b]Lets get it started by The Black eyed peas[/b]

green day
[hide]basket case *                                                 american idiot
warning*(more pop-rock though)
holiday [/hide]

  blink-182 (even tho they r not a band anymore)
     all the small things, i miss u, adam's song*, wats my age again?*

[hide] *= fav songs[/hide]

[quote="furious"]led zeppelin (particularly stairway to heaven)
pink floyd
radiohead[/quote]

oh yeah,how could i forget them

Asian Kung-fu generation is pretty good....try their song "24hrs"

Dude, techno\trance will get you so pumped-its what they play in all the night clubs across the world.

Sandstorm-Darude (World Best Seller, any genre ?1998?)
Elements of Trance (Midnight Club II Soundtrack)
Summer Breeze CD-DJ Tiesto
Rush CD-Darude
DJ Quicksilver
Trance Theory

a lot more (some hard to remember), but this is a good start. One of the top radio stations in the world- http://www.di.fm . You can listen to this to get a feel for the music.

OMG YELLOWCARD FTW best band in the world

[quote="Chocoboom"][quote="shadysaysurspammed"]I just liked the drummer of yellowcard,he's awesome,the rest seemed nothing so special[/quote]

rest not so special?!? what about crazy violin guy????!#$!@%
$&*[/quote]

Well Actually i've just heard lights and sound,not any other.In that the drummer goes crazy  :D

[quote="shadysaysurspammed"][quote="h_s_potter2002"]Has anyone mentioned the Alkaline Trio yet?[/quote]

only you :D[/quote]

Ok.  I just wanted to make sure someone mentioned the Alkaline Trio.

[quote="B-tone_Party"]99 Red Balloons (Goldfinger)[/quote]I prefer the original version by Nena, not as hard-rocky.  :P

Anyways, I recommend:

matchbox twenty (basically anything on their first CD is good, with Real World and 3 AM being my favorites)

Counting Crows (Mrs. Potters Lullaby, A Long December are pretty cool)

John Mellencamp

.38 Special

[quote="furious"]led zeppelin (particularly stairway to heaven)
pink floyd
radiohead[/quote]

I know the Summerville crowd loves Pink Floyd.  My favorite Pink Floyd song is [b]Another Brick in the Wall[/b].  I love that song!  There are like a billion versions though, so choose whichever one you like; I suggest one of the longer ones!  But this song is an absolute must-have for anyone who has ever gone to school.

Green day, definitely.
Holiday
Boulevard of Broken Dreams
Jesus of Suburbia

[quote="mysmartmouth"][hide="Disclaimer"] I don't actually [i]love[/i] some of these songs, and some are just on my playlist because they're popular.  But some are little known hits that I really enjoy.  And I don't  dislike any of these; they all range from decent to excellent, mostly very good.  Furthermore, some songs may contain explicit content, just note the labels on the iTunes downloads warning of explicitness.[/hide]

[u]My iTunes Favelist[/u]
[i]Song - Artist[/i]
[b]Favorites listed in bold type[/b]
Beverly Hills - Weezer
Candy Shop - 50 Cent
Carolina On My Mind - James Taylor
Crack Music - Kanye West
Did You Ever Have to Make Up Your Mind - Lovin' Spoonful (an oldie LP!!!)
Disco Inferno - 50 Cent
[b]East Bound and Down - Jerry Reed[/b]
Fired Up - MC Hush
[b]Forgot About Dre - Dr. Dre[/b]
[b]Gold Digger - Kanye West[/b]
[b]Hotel California - The Eagles[/b]
I'll Make A Man Out of You - Mulan Soundtrack (hahaha...my friend's fave song, still on my CPU...)
[b]I'm In Love (Wit A Stripper) - T-Pain[/b]
Lazy Generation - F-Ups
My Band - D-12
[b]My Humps - Black Eyed Peas[/b]
[b]Papa Was a Rollin' Stone - The Temptations[/b]
[b]Photograph - Nickelback
Rollin' on a River - CCR
The Watcher - Dr. Dre
These Boots Are Made For Walkin' - Nancy Sinatra (NOT THE JESSICA SIMPSON VERSION!)
Westside - TQ[/b]
You Are My Sunshine - 100 Songs for Kids (I love this one!)

Please comment![/quote]

Oh yeah,you reminded me of my humps(oh yeah ma hump,ma hump,ma...) and with that i remember another anooying (but awesome) song,CRAZY FROG(axle f)
Fergot 'bout dre's fine,prety fast and nice.

I think you guys should really try my fav song of all time ,[b]IN THE SHADOWS[/b] by  THE Rasmus.Its an awesome song,who ever Listens to this falls in love with it,whether he likes rock or not. [url=http://www.mp3shits.com/download_mp3/The_Rasmus_In_The_Shadows_mp3/33461/]heres the free download[/url]

Take a look at some jazz too, if you like it.

Miles Davis, Chick Corea, Thelonius Monk, Weather Report, Diana Krall, and the Modern Jazz Quartet are all pretty good, in addition to standards like Armstrong and Fitzgerald.

kanye west - late registration awesome totally, everyone go get it
yellowcard - lights and sounds, awesome album
eminem - curtain call, all your favourites rolled into one cd. OMG SHAKE THAT FT. NATE DOGG IS THE BEST SONG IN THE WORLD

I can't believe only one person has said the Alkaline Trio.

Yeah.  Kanye West is a beast, but he's also a philanthropist.  I'm not that into "crack music...that black music" a.k.a. rap. (That's a Kanye West quote.)  However, Kanye West brings something new to rap that is worth enjoying.  At least check him out.

Green Day.

I also like some random songs with a good beat but really bad lyrics that I don't pay attention to.

The Empire Strikes Back- G-UNIT

so many good songs from their new album

I know what you're thinking -

"Ewwwww, the Alkaline Trio?  Who wants to listen to them?  They probably smell or something."

But seriously, they're a really good band.  you should go listen to them.

who doesn't love Mariah Carey cmon, she can sing like $7^{1645747}!$ octaves

even though shes a little crazy ¬_¬

The Doors:
Waiting for the Sun
The Crystal Ship
Break on Through
House of the Rising Sun
People are Strange
Riders on the Storm
Touch Me
Light my Fire

Hendrix:
I Don't Live Today
Highway Chile
Changes
If 6 was 9
Red House
Crosstown Traffic
Bold as Love
Vodoo Chile
Vodoo Chile (Slight Return)
The Wind Cries Mary
Hey Joe
All Along the Watchtower
Manic Depression
Castles Made of Sand
Let Me Stand Next to Your Fire
Little Wing
Purple Haze

Clapton  (Derek and the Dominos/Diserali Gears/Cream/Eric Clapton):
Layla
Bell Bottom Blues
Key to the Highway
Cocaine
Crossroads

Marley:
Redemption Song
No Woman, No Cry
Could You Be Loved
War
Fussing and Fighting
Buffalo Soldier
Jammin

Dylan:
The Hurricane
Don't Think Twice it's Alright
The Times They are a'Changing
Shelter From the Storm
All Along the Watchtower
Subterranean Homesick Blues (greatest lyrics of all time)
The Ballad of Hollis Brown
Oxford Town
Masters of War

The Grateful Dead:
Saint Steven
Dark Star
The Golden Road (to unlimited devotion)
Casey Jones
Friend of the Devil
Uncle John's Band
Turn on Your Love Light
Mexicali Blues
Sugar Magnolia

Allman Brothers:
Midnight Rider
Ramblin Man
Melissa
Blue Sky
Jessica
Revival

I haven't heard any good NEW song.Modern rock bands usually play the same tune,jhan jhan jhan jhan.........No one different is coming up.I haven't heard the afmous arctic monkeys yet.Even the new INXS song bites. :|

I heard this song of thegame,dreams or something.It was O.K.

Everyone loves kanye west but the way he boasts around these days and says that he is the onle DIFFERENT rapper these days makes me sick.Eminem's is daddy if he talks about different.

And yeah,any favourite lines you guys have from your fav songs?

Best band in the world => [b]Incubus[/b]

I really like dance dance by fallout boy for some reason ... and just basically all songs by fallout boy, yellow card, and... yeah, groups like them.

and I'm a girl, so I like all of these hip hop/girl rock singers... but I doubt you'd enjoy them.  ;)

Green Day is one of my favorites, and U2. basically, any rock group.

[quote="shadysaysurspammed"]And yeah,any favourite lines you guys have from your fav songs?[/quote]

"Drive yourself insane tonight it's not that far away and I just filled up your tank earlier today."

"Hey,here's a concept that works,20 million other white rappers emerge,but no matter how many fishes in the sea it'll feel so empty without me"-Eminem(Without me)

If you are not obsessed with English, I can suggest Debabrata Biswas - an awesome singer. Gimme your mail address if you want some of his smashers.

black eyed peas - gone going - ft. jack johnson
black eyed peas - the apl song

check these songs out :D awesome

[quote="oooooh ya"]The Doors:
Waiting for the Sun
The Crystal Ship
Break on Through
House of the Rising Sun
People are Strange
Riders on the Storm
Touch Me
Light my Fire
[/quote]

Whoa, freaky!  Those are [i]exactly[/i] the same Doors songs I like (although you forgot "Strange Days" and "Love Her Madly" :) )
I also listen to The Who, the Beatles, David Bowie, Depeche Mode, Boston, Emerson/Lake/Palmer, Jim Croce, U2, the Rolling Stones (just a little.  I like exactly 3 of their songs), and Pink Floyd.

Current favorite line(s):
"You run and you run to catch up to the sun, but it's sinking
 Racing around to come up behind you again
 The sun is the same in a relative way, but you're older
 Shorter of breath, and one day closer to death"
          -Pink Floyd, "Time"
It reminds me of  [i]The Great Gatsby[/i]: "tomorrow we will run faster, stretch out our arms further…And on fine morning - So we beat on, boats against the current, borne back ceaselessly into the past."

And:
"You can rent your blues and photograph your soul
You can even dig some diamonds out of rock-n-roll
You can change the world, but if you lose control
They will take away your t-shirt..."     -Emerson, Lake, and Palmer, "Nobody Loves You Like I Do"
Just because it's random.

And one more (for this one, imagine Mick Jagger speaking with a southern drawl):
"I was drivin' home, early Sunday mornin', through Baker's Field, listening to gospel music on the college radio station.  And the preacher said, 'You know you always have the Lord by your side.'  And I was so pleased to be informed of this, that I ran twenty red lights in his honor.  Thank you, Jesus.  Thank you, Lord!"  - Rolling Stones, "Far Away Eyes"
 :rotfl:
If you don't think that's funny, I guess you just have to hear it :)


Prove that :

$ tg\left(\frac{\pi}{4n}\right)\le \frac{1}{n}\ ,\ (\forall)n\in \mathbb{N}^*$

[hide]Since half of the secant is shorter than half of the arc, we have $ \tan{\pi\over 4n}<{\pi\over 4n}\cdot{1\over\cos{\pi\over 4n}}$, and for $ n\geqslant 2$, we have $ \cos{\pi\over 4n}\geqslant \cos{\pi\over 8}\equal{}\frac{\sqrt{2\plus{}\sqrt{2}}}{2}>{\pi\over 4}$. Therefore $ \tan{\pi\over 4n}<{\pi\over 4n}\cdot{4\over\pi}\equal{}{1\over n}$ for $ n\geqslant 2$. For $ n\equal{}1$ we have equality.[/hide]


Let $n$ and $k$ be given relatively prime natural numbers, $k<n.$ Each number in the set $M=\{1,2,...,n-1\}$ is colored either blue or white. It is given that [list] [*] for each $i\in M,$ both $i$ and $n-i$ have the same color, [*] for each $i\in M,i\ne k,$ both $i$ and $\left \vert i-k \right \vert $ have the same color. [/list] Prove that all numbers in $M$ have the same color.

We write $a \sim b$ iff we can prove that $a$ and $b$ must have the same color. This clearly is an equivalence relation.

Observation 1: if $a \equiv b \mod k$, then $a \sim b$.
Proof: W.l.o.g. $a<b$ and write $b=a+mk$, $m>0$. Then we have $b \sim |b-k| = b-k \sim |b-k-k| = b-2k \sim ... \sim |b-mk| = b-mk =a$.

Observation 2: if $a \equiv -b \not \equiv 0 \mod k$, then $a \sim b$.
Proof:
by observation 1, we may assume $0<a<k$ ($a=k$ is excluded by $a \not \equiv 0 \mod k$). Then $a \sim |a-k| = k-a \sim b$, the latter by observation $1$ again.

Observation 3: if $b \equiv a-n  \not \equiv 0 \mod k$ then $a \sim b$.
Proof:
we have $a \sim n-a \sim b$, the latter by observation 2.

We finish the problem by remembering that $\gcd(n,k)=1$, thus $-n, -2n , -3n , ... -(k-1)n , -kn \mod k$ are all residue classes. And if $a_1,...,a_k$ from the allowed interval represent them in that order, we always have $a_i \sim a_{i+1}$ by the last observation. Thus all have the same color.

I think it should be
[quote="ZetaX"]

Observation 2: if $a \equiv -b \equiv 0 \mod k$, then $a \sim b$.

Observation 3: if $b \equiv a-n   \equiv 0 \mod k$ then $a \sim b$.
Proof:
we have $a \sim n-a \sim b$, the latter by observation 2[/quote]

And why would you think that¿ This special cases would be useless.


calculus
integration
trigonometry
limit
calculus computations

Please, prove or disprove that:
$ \int_{1}^{\infty}\frac {\cos x}{x^2}dx$ converges.

You could use:

∫Cos(x)/x^2 dx = -Cos(x)/x - ∫Sin(x)/x dx

It would yield:

∞
∫Cos(x)/x^2 dx =
1

 ∞
-∫Sin(x)/x dx - limit(Cos(x)/x,x→∞) + Cos(1)/1 =
 1

-limit(SinIntegral(x),x→∞)+SinIntegral(1) + Cos(1) =

-π/2+SinIntegral(1)+Cos(1)

Why is that necessary?  $ \minus{} \int_{1}^{\infty} \frac{1}{x^2} \, dx \le \int_{1}^{\infty} \frac{\cos x}{x^2}  \, dx \le \int_{1}^{\infty} \frac{1}{x^2} \, dx$

A more interesting integral:  $ \int_{1}^{\infty} \frac{ \cos x}{x} \, dx$, which I believe also converges by an extension of the Alternating Series test.

that is right $ \int_1^\infty\;\frac{\cos\,x}{x^2}\;dx$ is [i]absolutely[/i] convergent.

we have, $ \frac{|\cos\,x|}{x^2}\;\leq\;\frac{1}{x^2}$, and we know by the p-test, $ \int_1^\infty\;\frac{1}{x^2}\;dx$ is convergent. the basic comparison test then implies that

$ \int_1^\infty\;\frac{|\cos\,x|}{x^2}\;dx$ is convergent, i.e. $ \int_1^\infty\;\frac{\cos\,x}{x^2}\;dx$ is [i]absolutely[/i] convergent.

for $ \int_1^\infty\;\frac{\cos\,x}{x}\;dx$, let us look at $ \lim_{a\to\infty}\;\int_1^a\;\frac{\cos\,x}{x}\;dx$

integrate by parts to get $ \lim_{a\to\infty}\;\left\{\; \frac{\sin\,x}{x}\;\bigg|_1^a \;\; \plus{} \;\; \int_1^a \frac{\sin\,x}{x^2}\;dx\; \right\}$

now $ \lim_{a\to\infty}\;\left\{\; \frac{\sin\,x}{x}\;\bigg|_1^a\;\right\}\;\equal{}\;\minus{}\sin\,(1)$

and $ \lim_{a\to\infty}\;\left\{\;\int_1^a \frac{\sin\,x}{x^2}\;dx\; \right\} \;\equal{}\; \int_1^\infty \frac{\sin\,x}{x^2}\;dx$ which can be shown to converge [i]absolutely[/i] just like $ \int_1^\infty\;\frac{\cos\,x}{x^2}\;dx$  above.

hence, $ \int_1^\infty\;\frac{\cos\,x}{x}\;dx\;\equal{}\;\;\minus{}\sin\,(1) \;\; \plus{} \;\; \int_1^\infty \frac{\sin\,x}{x^2}\;dx$

converges.

now, let me ask this, show $ \int_0^\infty\;\frac{\sin\,x}{x}\;dx$ is [i]conditionally[/i] convergent. :D


Assume that a,b,c,d is positive numbers.Prove that:
[tex]\frac{a^2}{3a^2+bc}+\frac{b^2}{3b^2+cd}+\frac{c^2}{3c^2+da}+\frac{d^2}{3d^2+ab}\leq\frac{ab}{3c^2+ab}+\frac{bc}{3d^2+bc}+\frac{cd}{3a^2+cd}+\frac{da}{3b^2+da}[/tex]
[i] modified by Megus [/i]

sorry, can you express your question more clearly?

nobody interested in this problem?I think it is difficult and interesting.

first, transform it to an equivalent inequality,
a,b,c E R+ such that abc=1, then
\[
\frac{1}{{3a^3  + 1}} + \frac{1}{{3b^3  + 1}} + \frac{1}{{3c^3  + 1}} \ge \frac{3}{4}
\]
because
\[
\begin{array}{l}
 \sum {\frac{{a^2  - bc}}{{3a^2  + bc}} \le 0}  \\
  \Leftrightarrow \sum {\frac{{4a^2 }}{{3a^2  + bc}} \le 3}  \\
  \Leftrightarrow \sum {\frac{{a^2 }}{{3a^2  + bc}} \le \frac{3}{4}}  \\
  \Leftrightarrow \sum {\frac{{3a^2 }}{{3a^2  + bc}} \le \frac{9}{4}}  \\
  \Leftrightarrow \sum {\frac{{bc}}{{3a^2  + bc}} \ge \frac{3}{4}}  \\
 \end{array}
\]

to prove the equivalent inequality
expand all the things and it is equivalent to
\[
3\left( {a^3 b^3  + b^3 c^3  + c^3 a^3 } \right) + 5\left( {a^3  + b^3  + c^3 } \right) \ge 24
\]
which is true by amgm

sorry,siuhochung,maybe you don't understand my question.I will rewrite it:
[tex]\frac{a^2}{3a^2+bc}+\frac{b^2}{3b^2+cd}+\frac{c^2}{3c^2+da}+\frac{d^2}{3d^2+ab}\leq\frac{ab}{3c^2+ab}+\frac{bc}{3d^2+bc}+\frac{cd}{3a^2+cd}+\frac{da}{3b^2+da}[/tex]

Ok,It's a problem of mine in Vietnamese forum.
I think it's quite easy because:

$\sum\frac{a^2}{3a^2+bc} \le 1 \le \sum\frac{ab}{3c^2+ab}$


geometry
cyclic quadrilateral
geometry open

Let [b]ABCD[/b] be a cyclic quadrilateral with diameter [b]AB[/b], S is the intersection point of [b]AC[/b] and [b]BD[/b], T is the foot of perpendicular from S to [b]AB[/b]. Show that [b]TS[/b],[b]AD[/b] and [b]BC[/b] are concurrent.

It's trivial!
Let $R=AD\cap BC$ thus we have: $S$ is the orthocenter of triangle $RAB,$ so $RS\perp AB$
Then $T\in RS$ and hence $AD,\,BC,\,ST$ are concurrent.


I am studying electromagnetic theory. And I thought I would put
the theory to the test. I got a 22 gage copper wire and a 9 volt battery. I wraped the coil around a screw and hooked up the battery to the wire. I thought it would make the metal screw
into a temporary magnet. I have tried it with a spoon and other metal objects but I see no noticable change in attractiveness.

Does anybody know how much voltage I need to do such a thing.
The funny thing is they do it on youtube with less voltage I believe.

I just want to pratice making small fans and temorary magnet's.
With the knowledge I have.

Question I have all different kinds of wire and nothing noticeable happens. But what is the difference between magnetic wire , and
just regular wire.

I have different gage wire from 18 up to 22. Got it all at radioshack.

The only thing I can think of is that the wire must be providing to much resistance and the current is moving so slow that it induces a magnetic field so weak. I = V/R

But I thought copper was one of the best conductors.
Anyway any help would be great.

My copper wire only gives the info of rating = 300 volts and the gage number.
What exactly is the rating for?
What happens if I used 400volts on a wire rated at 300volt?
Would it burn or not work ...

I know for resistors their are ratings and tolerences (usually 1%,5%,10% 25% )

And if you go way over the tolerences the resistor can just burn out. Kind of like a led if to much voltage is applied.

But would the wire catch on fire?

How many times did you twist the wire around the screw?  The more turns, the stronger the magnet - several hundred turns would not be unreasonable.

If you exceed the rated voltage of the wire, the heat it produces (given by $ P \equal{} I\Delta V$) will cause the wire to melt, or catch fire, or do something bad.  It is possible that you can exceed the rated conditions for short periods of time without damage, but do so only with caution and at your own risk.  Speaking as one who has caused explosions in the school physics lab under similar conditions, please don't do it.

[quote]How many times did you twist the wire around the screw? The more turns, the stronger the magnet - several hundred turns would not be unreasonable.[/quote]

So how many times would you have to twist the wire in loops before you get a considerable magnetic force?

I twisted it about 20 times. I could have sworn I saw this done on youtube with a 6 volt battery and it looked like
nothing over 20 coils in the wire.

I could be wrong though.

Does insulated or none insulated wire matter?
And is their some formula to calculate the strength of the magnetic field given the wire, number of coils and current/volts/resistance?

Also does anybody know what the difference in radio shacks magnetic wire and just plain wire is?

Thanks for any help.


Example:

Morning of Day 1

You have one dollar bill.  You divide it into equal fifths.  Each fifth may be connected to other fifths to form larger fractional fifths of a dollar bill.  If you have a majority of a bill  (3/5, 4/5) then you get to cash those in for a WHOLE dollar bill.  But a 1/5 or a 2/5 piece is worthless on its own, but may be carried over to the next day.  However, a 1/5 or a 2/5 piece, for example, can be attached to like smaller pieces to make them larger to be cashed in for a whole dollar bill.

So you decide to put three - 1/5 pieces together and cash them in for a dollar bill while having two separate 1/5 pieces tagging along.
This arrangement is what you will start off with on Day 2.
-------------------------------------------------------------

Question:  Going back to the morning of Day 1, how much total money CAN you have (maximum) if you pick one of any way(s) of strategy if you do this "fifths method" for 10 days?

I don't understand. We receive 1 dollar note everyday?

[hide]You will end up getting 10 dollars, or 50 fifths of a dollar. From this, you can make 16 sets of 3/5 dollars, with 2/5ths left over, so you will get 16 dollars.
[/hide]
This is a true scenario isn't it? I took the SAT a few years ago, and I remember that there was an essay about money and how if you could get over half a bill, you could redeem it....

You could make some pretty good money (for a high schooler) with 100 dollar bills, although I think that they might check to make sure that your 3 3/5ths of a bill are from the same bill somehow.

mustafa,

hold on till this evening about the confirmation (or not) of your
candidate of the answer of the number of total dollars.

I must work a bit later at a job . . .

But I think you need the one in the middle in every new bill we make, and there are 10 of them.

1 dollar, because the serial numbers won't match :D

That is funny to me, but seriously, we will IGNORE the serial numbers of all bills.  We are not treating them as real/legal bills
in that sense.

<><><> more on this...update on this solution tomorrow

I won't post a solution YET because a few of you don't/didn't get what my problem represented.

Day 1 -- 1  one dollar bill

Split it into 3/5 & 2/5

They become...

Day 2 -- 1  one dollar bill & 2/5

Split them into 3/5 & 2/5 & 1/5 & 1/5

Rearrange them into 3/5 & 3/5 & 1/5

They become...

Day 3 -- 2  one dollar bills & 1/5

Split them into 3(3/5) + 1/5 + 1/5

Rearrange into 3(3/5) + 2/5

They become...

Day 4 -- 3  one dollar bills & 2/5

Split them into 5(3/5) + 2/5

They become...

Day 5 -- 5  one dollar bills & 2/5

Split them into 8(3/5) + 1/5 + 2/5

Rearrange them into 8(3/5) + 3/5

Rearrange these into 9(3/5)

They become...

Day 6 -- 9 one dollar bills

#
..#
....#
......#

 This continues on to Day 10.

Do you notice a pattern with the dollar amounts?  Take the amount of dollars (including any leftover fraction) from Day (n-2)
and add it to the amount of dollars (and fractions) from
Day (n-1). Then subtract (1/2) from that sum.  The final result will be the amount of dollars (and any fractions that go with them) for Day (n).

This can be used to shorten up the process instead of how I was showing all the above steps for up to Day 6.

Someone can/should else try to finish it now.

[quote="Summation"]Day 6 ... $9 dollars. Day 7 \to 9/(3/5) \to$15 dollars.

Day 8 -> 15/(3/5) -> $25 dollars. Day 9 \to 25/(3/5) \to 125/3 \to 123/3+(2/3 of 3/5) \to .........$41 dollars & 2/5.

Day 10 -> 41/(3/5) + 2/5 -> 205/3 + 2/5 -> 204/3 +
.........(1/3 of 3/5) + 2/5 -> $68 dollars+1/5+2/5 \to$68 dollars + 3/5 -> $68 dollars+$1 -> $69 dollars. ======================. I got a final answer of:$69 dollars.

======================.

I am NOT trying to do Latex; this jumble above is NOT what I intended.

Note:  That pattern of dollar amounts, which I mentioned in a prior post, did not hold up for larger numbered days.[/quote]

The $\$$ symbols causes the things in the middle of them to be rendered in Latex. To get just the $\$$ symbol, use
[code]$\$$[/code]

[quote="mustafa"]This is a true scenario isn't it? I took the SAT a few years ago, and I remember that there was an essay about money and how if you could get over half a bill, you could redeem it....

You could make some pretty good money (for a high schooler) with 100 dollar bills, although I think that they might check to make sure that your 3 3/5ths of a bill are from the same bill somehow.[/quote]

yeah if u have over 50% of a bill then u can get a while dollar bill. But a highschooler couldn't make money of it because of the serial numbers and u cant combine 2 different bills together...


[quote="Virgil Nicula"]$\boxed {\ \cos \left(x+20^{\circ}\right)+\cos \left(2x-20^{\circ}\right)=\frac 12+\sin \left(30^{\circ}-x\right)\ ,\ 40^{\circ}<x<140^{\circ}\ }\ .$
[hide="Raspuns"]$x=60^{\circ}\ .$[/hide][/quote]
    The answer is x=60.It isn't so difficult. :roll:


function
calculus
integration
real analysis
real analysis solved

a classical one, but i would like to see different approach:
let $f:R->C^*$ be a continuous function. Prove that there exists functions $g$ and $h$ (from $R$ towards $R$) such that:
$f(t)=h(t)e^{ig(t)}$

First of all, we can assume $h(z)=|f(z)|$, so we may assume WLOG that $f:\mathbb R\to \mathbb S^1$ (the unit circle in the plane) and that we must show that there is a continuous $g:\mathbb R\to\mathbb R$ s.t. $f(z)=e^{ig(z)}$.

Assume we have $\mathbb R=\bigcup_{n\in\mathbb Z}[a_n,a_{n+1}]$, and we have, for each $n$, continuous $g_n:\mathbb R\to\mathbb R$ s.t. $f(x)=e^{ig_n(x)},\ \forall x\in[a_n,a_{n+1}]$. Then, we can glue the $g_n$ by taking $g(x)=g_n(x)+k_n\cdot 2\pi$ whenever $x\in[a_n,a_{n+1}]$ for some suitable integral constants $k_n$.

The above implies that we can take $f:[0,1]\to\mathbb S^1$. We can divide $[0,1]$ in intervals small enough so that $f$ takes at most one of the values $\pm 1$ in each such interval (because of the uniform continuity of $f$ on $[0,1]$). If we lift $f$ on each of these intervals, then we can lift $f$ on the entire $[0,1]$, by the above observation.

This means that it suffices to prove the following lemma (which we also apply by replacing $1$ and $(0,2\pi)$ with $-1$ and $(-\pi,\pi)$):

Let $f:[0,1]\to \mathbb S^1\setminus\{1\}$ be continuous. Show that there is a continuous $g:[0,1]\to\mathbb (0,2\pi)$ s.t. $f(x)=e^{ig(x)},\ \forall x\in[0,1]$.

Take $g(x)=\frac 1i\cdot \exp^{-1}(f(x))\cap (0,2\pi)$. It's not hard to check that this is indeed continuous.


I was wondering, and I didn't know what forum to post this in, so a moderator may move it if they want to.

Why does $e^{\pi\imath}+1=0$?

It comes from the fact that any complex number in the form
$\rho (\cos \theta + i \sin \theta)$ can be written as
$\rho e^{i \theta}$. So $-1= 1 \cdot (\cos \pi + i \sin \pi) =e^{i \pi}$.

If you want to know how we arrive to the fact that

$\rho (\cos \theta + i \sin \theta)=\rho e^{i \theta}$, just consider the Taylor series of
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...$
$\sin x= x -\frac{x^3}{3!}+\frac{x^5}{5!}-...+(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}+...$
$\cos x=$ like sinx but with even coefficients

Thanks for the explanation, even though I haven't heard of a lot of it...

I can understand it now.


Hi there !

Two questions about footnotes in math mode:
1.- I saw that \footnote works well in text mode. However it seems that it isn't allowed in math mode ? I attach you a simple example: I'd like to substitute the equation under the equal sign by a simple star and then display it as a footnote at the bottom of the page.

\begin{eqnarray}
        H^a  & \underset{{\uparrow \atop (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc}}{=} &     ((P_1+P_2)+(R_1+R_2)+(Y_1+Y_2))^2  \\
                & = & (P_1+P_2)^2+ (R_1+R_2)^2 + \hdots
\end{eqnarray}

2.- Under the equal sign I set a reference to a couple of equation numbers. However latex leaves too much space between the equation at the left side of the equal sign and right side. Somebody knows how to reduce this space? Example attached.

\begin{eqnarray}
        H^a  & \underset{{\uparrow \atop ~\eqref{eq:BLUE}   \text{,}~\eqref{c1c2:relation}}}{=} & ((P_1+P_2)+(R_1+R_2))^2  \\
                & = & (P_1+P_2)^2+ (R_1+R_2)^2 + \hdots
\end{eqnarray}

Thanks a lot !

1. Use \footnotemark{} where you want the asterisk to appear, having previously used the line \renewcommand{\thefootnote}{\fnsymbol{footnote}} to ensure you get the asterisk.
Then to actually generate the footnote, outside of math mode write \footnotetext{footnote text goes here}

2. Use negative space such as \hspace{-1pc} but use whatever units you please

Got it! Now I understood how \footnotes works  :oops:
Thanks again Stevem


geometry
incenter
trigonometry
inradius
geometry proposed

I suppose this problem is well known and quite easy but anyway:
In Triangle ABC let C1 be the circle tangent to AB, AC and the incircle of ABC,and
let r1 be its radius. Similary define r2,r3. Prove that r1+r2+r3>=3r.

Of course, I'm sorry, it's my mistake, it should be:
r1+r2+r3>=r

I didn't know that. Have you got a nice solution, because mine isn't so?

I'd love to have a geometric solution, but I don't..

Let $D,E,F$ be the points where the incenter touches $BC,CA,AB$ respectively, and let $X_d,Y_d$ be the midpoints of the small and large arcs $EF$ on the incircle. Now let $M$ be the midpoint of $EF$. It's not hard to show that $\frac{r_1}r=\frac{X_dM}{Y_dM}=\frac{1-\cos D}{1+\cos D}$, where $D$ is $\angle EDF$.

The problem is thus reduced to this: Show that in an acute triangle $ABC$ (changed notation) we have $\sum\frac{1-\cos A}{1+\cos A}\ge 1$, which, in turn, means $\sum \tan^2\left(\frac A2\right)\ge 1$. Since $\tan^2$ is convex on $(0,\frac \pi 4)$, that's that.

Something small to notice:
    If you draw the internal common tangent of (I_1) and (I) and it intersects AB and AC at P and Q, then APQ is an issoceles triangle having (I_a) the incenter and (I) excenter. Hence $\frac{r_!}{r} = tg^2{\frac{A}{2}}$.
    This reminds me a very cool problem :
  LeT $B,A_1, A_2, ... A_{n-1}, C$ are collinear points in that order and A is an abritary point not lying on that line. Let $r_1, r_2,.... r_n, r$ be repectively the inradius of triangles $ABA_1, ..... AA_nC$ . Let $R_1, R_2,......, R_n, R$ be respectively the exradius of trinagle $ABA_1, ..... AA_nC$. Then
    $\frac{r_1r_2....r_n}{R_1R_2....R_n}=\frac{r}{R}$
 :P

What can I say, it's cool for sure!!!
Did you solve it by yourself?

Well it is easy to obtain this solution I also solved it with this solution.


quadratics
algebra
polynomial
sum of roots
product of roots

umm... I'm looking for an equation.

like a quadratic equation

Chinaboy wrote:Create an equation whose two roots are:

x_1 = (a+b)/(a-b)
x_2 = (a-b)/(a+b)

Spoiler:

[hide]An equation with roots p and q is (x-p)(x-q) = x^2 - (p+q)x + pq, so the equation we want is

x^2 - [(a+b)/(a-b) + (a-b)/(a+b)]x + (a+b)/(a-b)*(a-b)/(a+b)

= x^2 - [(a+b)^2 + (a-b)^2]/[(a+b)(a-b)]*x + 1

= x^2 - [a^2 + 2ab + b^2 + a^2 - 2ab + b^2]/[(a+b)(a-b)]*x + 1

= x^2 - (2a^2 + 2b^2)/(a^2-b^2)*x + 1

Or you could multiply by a constant.[/hide]

Edited: Oops - why did I use the difference of the roots before?

Not positive why some of the exponents are showing up funny (as in not emoticon form), but here you go:

[hide]

Let quadratic be cx:^2: + dx + e = 0

Sum of roots: -d/c = ((a+b):^2: + (a-b):^2:)/ (a+b)(a-b)

                              = [2a:^2: + 2b:^2:]/ [a:^2: - b:^2:]

Product of roots: e/c = 1

Therefore, a general quadratic equation would be:

k(x:^2: [-[2a:^2: + 2b:^2:]/ [a:^2: - b:^2:]]x  + 1) = 0

I guess the k in front is kind of silly, but I like it.

Of course there might be fourth-degree equations that also have these roots as double roots (maybe). Somebody want to check stuff like that?

Also, my answer seems to be different from mathfanatic's, but I can't find the flaw in my reasoning. Is he wrong? Am I wrong? Mathfanatic's solution solves for a quadratic of x:^2: + bx + c, where the solution might be ax:^2: + bx + c.

[/hide]

rlee - you're right.  I accidentally subtracted the roots instead of adding them.


Just mutate the sentence above yours slightly, just like in a real-life telephone game.

They epic-failed on the USAMO.

[quote=happyhippos]A one year, 23 day bump. Impressive.

Rain scars my proof.[/quote]
not very impressive, some people do 6 year bumps. [size=50]and get reported for doing so[/size]

Ran scars by hoof.

The "impressive" was mostly just for show. But show isn't quite the right word. You know what I mean though.


inequalities
induction
inequalities solved

USA 1994 inequality
  xi is a infinite sequence of positive reals such that for all n,
x1 + x2 + ... + xn >= n1/2. Show that x12 +
x22 + ... + xn2 > (1 + 1/2 + 1/3 + ... + 1/n) / 4 for all n.
 I read solution of this problem in john scholes, but i don't like.
 Who can solv?

Lemma : Suppose a_1,a_2,...,a_n and b_1,b_2,...,b_n are positive
number such that

(*) b_1>b_2>....>b_n

(**) b_1+...b_k  \leq  a_1 +....+a_k for k= 1;2;....;n

Then

b_1 2 +...+b_n 2  \leq a_1 2 +...+a_n 2

Proof: We suppose b_(n+1) = 0, (*) and (**) give

 \sum _{k=1 to n}(b_k-b_(k+1)). \sum_{j=1 to k}b_j  \leq  \sum _{k=1 to n}(b_k-b_(k+1)) \sum _{j=1 to k}a_ j

 \sum _{j=1 to n}b_j \sum _{k=j to n}(b_k-b_(k+1)) \leq  \sum _{j=1 to n}a_j \sum _{k=j to n}(b_k-b_(k+1))

and

 \sum _{j=1 to n}b_j 2  \leq  \sum _{j=1 to n}a_jb_j

Now we are use Cauchy-Schwarz inequality

( \sum _{j=1 to n}b_j 2 ) 2  \leq ( \sum_{j=1 to n}a_jb_j) 2  \leq ( \sum _{j=1 to n}a_j 2 )( \sum _{j=1 to n}bj 2 )

we simplify and we get

b_1 2 +....+b_n 2  \leq a_1 2 +...+a_n 2

For our problem

b_ j = sqrt(j) - sqrt(j-1)= 1/(sqrt(j) + sqrt(j-1))

then (*), (**) are satisfied

a_1 2 +...+a_n 2  \geq  \sum 1/(sqrt(j)+sqrt(j-1)) > \sum _{j=1 to n}1/(2sqrt(j)) 2=1/4(1+1/2+...1/n)

 :D 2  Probably  the official solution :D 2

 [quote] Can someone help me understand this inequality? I used the search function to find this problem. I am not here to bump this thread.[/quote]

I think what the question is asking is this:

$x_i$ is a infinite sequence of positive reals such that for all $n$,
$\sum_{i=1}^n x_i \ge \sqrt n $.

Show that $\sum_{i=1}^n  x_i ^2 > \sum_{i=1}^n
\frac{1}{4n}  \forall n$.

Oh. Thank you for clearing it up. What do I do next?

Let $\, a_1, a_2, a_3, \ldots \,$ be a sequence of positive real numbers satisfying $\, \sum_{j=1}^n a_j \geq \sqrt{n} \,$ for all $\, n \geq 1$. [url=https://artofproblemsolving.com/community/c6h57393p353102]Prove that[/url], for all $\, n \geq 1, \,$  \[ \sum_{j=1}^n a_j^2 > \frac{1}{4} \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right).  \]


So are how many USAMO qualifiers are there from MA?

I did horrible on the AIME :( , got a 9, still have a chance of qualifying though since I'm in 9th grade...

Massachusetts probably has more than MI...

Congrats to the following Massachusetts USAMO qualifiers:

Rahul Banerjee 11 Saint Johns High School Shrewsbury
Sergei Bernstein 8 Belmont High School Belmont
Sway Chen 9 Lexington High School Lexington
Yuga Cohler 10 Lexington High School Lexington
Nickolas Fortino 12 Phillips Academy Andover
Andrew Geng 10 Westford Academy Westford
Feiqi Jiang 9 Acton-Boxborough Regional High School Acton
Pavel Kamyshev 12 Deerfield Academy Deerfield
Boping Lai 9 Roxbury Latin School West Roxbury
Cary Malkiewich 12 Doherty Memorial High School Worcester
Natee Pitiwan 12 Brooks School North Andover
Jarupon Sathirapongsasu 13 Deerfield Academy Deerfield
Jacob Sharpe 10 Canton High School Canton
Pawat Techapongtada 12 Middlesex School Concord
Stephen Xu 9 Lexington High School Lexington
Tianyuan Zheng 10 Algonquin Regional High School Northborough

Note how the MI forum has 100% of its qualifiers posting on it, and how the MA forum has what, like 20%?

I didn't qualify, but some of the USAMOers this year are from the Western Mass ARML team. Rahul doesn't have a username on AoPS, but Tim Zheng and Cary have usernames here...so totally we have 3 USAMO qualifiers on Westen Mass...

[quote="filletwho"]But you forget that Feiqi is really from Troy High...[/quote]

And I'm not on Western MA ARML team.

Maybe he's from Troy High School, in CA.

But we didn't forget you.  NO really!

I remember that Michigan has 17 and Massachusetts has like 30.

[quote="jclarkemathL314159"]i think you're forgetting many people.

off the top of my head, dan suo, noah arbesfeld, and sterling chu.[/quote]

Umm...this post is from last year?

I'm a USAMO qualifier from MA too  :P

It probably is, since Stephen Xu is in 10th grade now.


So I've taken AP Chemistry last year, this year I'm preparing for the USNCO using the textbook "Chemical Principles" by Peter Atkins, but this book covers too much stuff as followed:

    1. Atoms: The Quantum World
    2. Chemical Bonds
    Major Technique 1 Infrared Spectroscopy
    3. Molecular Shape and Structure
    Major Technique 2 Ultraviolet and Visible Spectroscopy
    4. The Properties of Gases
    5. Liquids and Solids
    Major Technique 3 X-Ray Diffraction
    6. Thermodynamics: The First Law
    7. Thermodynamics: The Second and Third Laws
    8. Physical Equilibria
    Major Technique 4 Chromatography
    9. Chemical Equilibria
    10. Acids and Bases
    11. Aqueous Equilibria
    12. Electrochemistry
    13. Chemical Kinetics
    14. The Elements: The First Four Main Groups
    15. The Elements: The Last Four Main Groups
    16. The Elements: The d Block
    17. Nuclear Chemistry
    18. Organic Chemistry I: The Hydrocarbons
    Major Technique 5 Mass Spectrometry
    19. Organic Chemistry II: Polymers and Biological Compounds

I was wondering what chapters are helpful to the first round USNCO exam in the upcoming April?  Also, do i have to do some additional readings/problems beside using the "Chemical Principles"?

All of them are potential sources for questions....so my suggestion is to read and study all of them :)

I'd say all chapters are extremely important. If one is to read Atkins's book, then one should read everything about the book. For instance, one might think that chapters 14-16 are boring and useless. However, there are at least two reasons why one should read them. First, there is so much information there that it would help in one's knowledge of chemistry trivia. Second, Atkins does a good job explaining material in inorganic chemistry, and it is bound to be useful later. For instance, on the national exam, there is always an inorganic question or two, and it would be helpful to have a general idea of what inorganic chemistry is.

The first chapter might be tricky in that there's a lot of calculus, but it is extremely helpful to understand what's going on in that chapter. Plus, it's fun and enlightening!...

I have this book too, while the calculus in the first chapter is interesting, it's a little hard to fully grasp.  Is there any chance that this kind of stuff could come up on the national exam?

I've seen questions about the energy of electrons which I'm pretty sure is in the first chapter.  Also from the first chapter: electron structures,m periodic trends, quantum numbers.  Wavefunctions is useful and knowledge of the electromagnetic spectrum also comes in handy.  The entire book is handy with the trivia found in captions.

From what I've done so far, I find that even if I read Atkins, the information does not sink in completely. You can have flawed understanding or not know what to pay attention to while reading and things don't go very smoothly.

Working on actual USNCO problems helped me a lot more. If you have the AP Chem material down, you should know enough to begin on the problems, using Atkins as a supplementary reference when you need it for a problem.

My current plan is doing the 10 practice tests on the website and read Atkins at least once while also doing problems from it for more important chapters.

What really gets me on these chemistry competition tests are the first couple of questions.  They usually deal with some sort of chemistry trivia, especially colors.  I always screw up on colors, but I've been working on it so hopefully if a color question comes up I'll have seen it before.

hah ye lol. seems like a lot of its just random stuff spread throughout atkins, esp some in the chpts 14-16.

adchia I know exactly what you mean by when you read atkins and it doesn't stick.  I'm hoping to read it twice and also do plenty of problems so I'll have a complete understanding.


MATHCOUNTS
ARML
trigonometry
summer program
MathPath

AoPSer = a person who has an AoPS account
met = talked to in person

 :)

*sigh* another terribly ranged poll of mine

About 20 but I do not know what to vote for.... exponents left my mind long ago...

GAH!

[color=blue] * ernie is confuzzled[/color]

but i've met none, so i just put none!

yay!

ok hold on...........do they have to come on regularly? i guess not... ok 1-2-3-4-5 yeah about 5 :D school math club...

At least 20... wow AoPS is getting poopular.

Um.... did you misspell popular on purpose trying to insult AoPS or...

i have met 1 (i was the one who directed him to AoPs   :rotfl:  )

well, at mathcounts nats, i met a bunch of AoPSers, we all wrote our accts on our nametags.  I was the only 1 from my state, but like mass and tex had all four ppl on AoPS

sadly, also none.  :( i tried to get my friend from school to start but she made an account and then quit AoPS.  :(

If I made Nats then it'd be a lot.

But I only made states so that is why I only met 3 (and maybe, partially perhaps RussianRocket too, sry if i didn't include u in my previous post)

exactly 1.
and the reason really is just cause my parents found aops and then recommended it to my friend

I know like 10*counts again* oh, actually 12, yay! But, 8 of them is from my school and 1 was in my school.

[quote="gauss1181"][quote="Math Geek"]Hmm, a bunch of people on my ARML team and some people from nats (xpmath, dnkywin, PI-Dimension, etc.)=~10[/quote]
You forgot the tx ppl: stevenmeow, ra5249, (^_^), and budi713.
Oh, and andersonw, kymathman, uldivad9, someperson01, etc.[/quote]

False, I never talked to those people.

PS: Is budi713 Bobby?

i can think of like 20 off the top of my head but probly more.....

Actually, 6!

JoshD
whitesoxfan
jgjgjmath
Stargirl240
RussianRocket
avec_une_h

WOAH SOMEONE KNOWS ME

Ok like there are like 5 that live in my state (mgao ternary dnkywin bisbmard not_trig Teki_teki b_flat jb05 ok this list is getting a lot bigger than 5, lets say 10 :D)

Ok there was like 50 at nats (no point at listing them)

I met like 2 at ARML (I went to U of G so not that many)

so thats like 70 so I just 10^1.8 =/

I wish I was bobby.

I know about 20-30 or more but most of them not very well.

Wow thats alot.

I wanna go to Nats to meet that many :D

umm well a bunch at math camps and contests... maybe around 100? idk

[quote="PI-Dimension"]
I met like 2 at ARML (I went to U of G so not that many)
[/quote]

There was probably more than you think.  A great deal of the people there probably at least have an account even though they don't post.  Anyway, I was there along with a couple others from my team (I'd say about 3 AoPS regulars from SC).  MCrawford was there (definitely a big name for AoPS...).  I also remember Billy from GA.  I dunno...I didn't seek out AoPSers or anything while I was there but I know I came across 5-10 that I know of.

Overall I've probably met more than 50 but probably not more than 100.

I went to mathpath last year and arml this year... probably at lest 30-40...

hmm... I've talked to a lot of AoPSers but the only people I've had a "real" conversation with are ra5249, PI-Dimension and dnkywin

Probably like 50 from MOP alone...then I've been to AMP 07, MathPath 06, MathCounts 06, 07, I'd like to say around 100 but probably not that many.

From ARML and AMP, I'd say about 20-35, given that almost all the people at AMP have AoPS accounts.

Edit: Oh yeah, that includes the tc.


quadratics
modular arithmetic
number theory
number theory solved

If $n$ is a positive integer, let $A = \{n,n+1,...,n+17 \}$.
Does there exist some values of $n$ for which we can divide $A$ into two disjoints subsets $B$ and $C$ such that the product of the elements of $B$ is equal to the product of the elements of $C$?

the answer is no. you might wonder why? :P
if any of the numbers is divisible by 19, then exactly one is divisible by 19, so we cannot have $\displaystyle \prod_{x\in B} x = \prod_{x\in C} x$.
so none of the numbers is divisible by 19.
However if none of the numbers is divisible by 19, let us denote by $a$ the residue of the product of the whole numbers modulo 19, and $b$, $c$ the residues of $\displaystyle \prod_{x\in B} x,\ \prod_{x\in C} x$ modulo 19 as well.
Then we must have $b=c$, and also $a=bc$, so $a=b^2$, so in fact $a$ is a quadratic residue.
Using Wilson's theorem, we obtain that $18! \equiv -1 \pmod {19}$, but it is well known that if $p\equiv 3 \pmod 4$ then $-1$ cannot be a square residue and we are done.

Indeed this is quite easy.

Yes that is exactly what I've done with it.

Pierre.

i have been studying the quadratic residues and the legender symbol from Naoki Sato 's note on Number Theory and solved the problem using them in a same way.


modular arithmetic
number theory unsolved
number theory

Let n be positive integer number such that $ 3^n\minus{}1$ divisible by $ 2^{2009}$. Prove that $ n\ge 2^{2007}$

we know that $ 3^{\phi(2^{2009})}\equal{}3^{2^{2008}}\equiv 1\pmod{p}$. so we know that the order, $ k$, of $ n \pmod{2^{2009}}$, must divide $ 2^{2008}$, so $ k\equal{}2^{r}$ for some $ r$. so the claim is equivalent to showing that $ 3^{2^{2006}}\not\equiv 1\pmod{2^{2009}}$. if we did have $ 3^{2^{2006}}\equiv 1\pmod{2^{2009}}$, then repeatedly using the fact $ a^p\equiv b^p \pmod{p^k} \Leftrightarrow a\equiv b \pmod{p^{k\minus{}1}}$ would give $ 3^{2^{1}}\equiv 1 \pmod{2^{4}}$, which is false -- contradiction.

Much better  \[ \,{2^{2007}}/n\]
We can use the fact that the order of 3 (mod 2^2009) is a power of 2 and his power noted L>=2007
Use the Lemma:
\[ \forall k \ne 0\,\,,\,\,{V_2}({3^{{2^k}}} \minus{} 1) \equal{} k \plus{} 2\]


linear algebra
matrix
calculus
derivative
calculus computations

need help on those two.   I sort of got the first question in problem#3

the matrixs are the same

[img]http://farm4.static.flickr.com/3232/3000661214_70cde3ef94_o.jpg[/img]

$ \frac{\partial u}{\partial t}\equal{}\frac{1}{2}(cf'(x\plus{}ct)\minus{}cf'(x\minus{}ct))\Rightarrow\frac{\partial^{2}u}{\partial t^{2}}\equal{}\frac{c^{2}}{2}(f''(x\plus{}ct)\minus{}f''(x\minus{}ct))$
$ \frac{\partial u}{\partial x}\equal{}\frac{1}{2}(f'(x\plus{}ct)\minus{}f'(x\minus{}ct))\Rightarrow\frac{\partial^{2}u}{\partial x^{2}}\equal{}\frac{1}{2}(f''(x\plus{}ct)\minus{}f''(x\minus{}ct))$
Then note that by multiplying the second partial derivative by $ c^{2}$, we get the first one, satisfying the equality.

[quote="yzc717"][quote="JRav"]$ \frac {\partial u}{\partial t} \equal{} \frac {1}{2}(cf'(x \plus{} ct) \minus{} cf'(x \minus{} ct))\Rightarrow\frac {\partial^{2}u}{\partial t^{2}} \equal{} \frac {c^{2}}{2}(f''(x \plus{} ct) \minus{} f''(x \minus{} ct))$
$ \frac {\partial u}{\partial x} \equal{} \frac {1}{2}(f'(x \plus{} ct) \minus{} f'(x \minus{} ct))\Rightarrow\frac {\partial^{2}u}{\partial x^{2}} \equal{} \frac {1}{2}(f''(x \plus{} ct) \minus{} f''(x \minus{} ct))$
Then note that by multiplying the second partial derivative by $ c^{2}$, we get the first one, satisfying the equality.[/quote]

thank you, I kinda figured out myself. thanks anyway.

how about the 3 and 4[/quote]

I mean part 2 in problem#3

I got #4

$ f(\mathbf{x}) \equal{} \| \mathbf{x} \| \equal{} \sum x_i ^2$

$ \frac {\partial f}{\partial x_k} \equal{} \frac {\partial}{\partial x_k} \left( x_k ^2 \plus{} \sum_{i \neq k} x_i ^2 \right) \equal{} 2 x_k$

$ \therefore \nabla f \equal{} (2 x_1, 2x_2 \ldots 2x_n) \equal{} 2 (x_1, x_2, \ldots x_n) \equal{} 2 \mathbf{x}$


when you have the distance around a shape, the largest shape (in area) that can be formed is a circle, right?

For a fixed perimeter, a circle gives the maximum area, yes.

but only if there are no restrictions on the shape you can have. if it has to be a rectangle, the highest area is obtained with a square.

For $n$ sides, the maximum area will be a regular $n$-gon.  As the number of sides increases, so does the area.  The maximum number of sides is $\infty$, or a circle.


geometry
calculus
MATHCOUNTS
Harvard
college
real analysis

Next year my school will be offering the IB diploma and I will be a junior. But the math and science courses are only Standard Level, there are some High Level English and Social Studies courses. And of course, like most people here, I would like to do something in engineering or science. So is it better to do the IB diploma or stick with AP classes? The AP science courses have chem, physics B/C, biology. Math include Stats, Calc BC, and there is honors multivariable calculus.

Thanks in advance for your input.

It's not clear that the IB diploma will be any more advantageous for college admission than several AP courses, if that is your question.

It seems you already know what would you would like to do, and usually that means you will work harder and succeed better if you do what you like to do. Also, college admissions officers get tired of seeing people doing what they have been told is good for getting into college rather than doing what they really care about. So I think you already know what your best choice is at your school.

Yeah, I don't think IB is going to be better for you than those other classes - IB is just a ton of work, helping you to get knowledgeable about a good number of subjects, and since Math HL and Futher SL is not offered at your school while you could just do BC and honors multivariable, i think that you will get a lot more out of the non-IB route. If I had that decision I would have chosen the BC and multivariable of IB (I'm currently enrolled in IB and its a good amount of work - but if I got out of IB I'd have to leave the school I was at and no other school nearby offer anything near to the opportunities you've written about.)

I heard that IB is more liberal arts oriented, is that true? And what is so special about the IB program that makes it different and even better than AP in certain areas?

Thanks for the previous responses, they have been very helpful.  :)

Well, the main difference between IB and AP is that IB is more of a program in the US - you must take 6 classes (there are certain restrictions on what combinations of classes you can take - check ibo.org and surf around it a bit to see the requirements) and for two semesters you must take the Theory of Knowledge course as well as write a 3-4,000 essay.

Unlike IB, AP doesn't even require you to take the class, all you have to do is take the test. There is no minimum for AP classes, so you can make your schedule a lot more flexible. Therefore the IB can be seen as a much more comprehensive "program" than the AP and students who go through IB tend to have more challenging workloads because of the various things that you need to do in order to complete the diploma (aside from the two things i mentioned earlier, you have to things for specific classes that count into your IB grade ,i.e. lab/lab writeups, orals, essays for history/english, math portfolios, etc.)

Now, as for IB being more liberal arts, it seems to be much more geared towards liberal arts in the United States (where many schools do not offer Physics,Math HL, Math Further SL and courses that are geared to a more mathematical mindset). So although IB may show that you put more work into it, in your situation I do not think that the colleges will matter that you didn't do the IB when you chose instead to pursue a more mathematical approach (since your school doesn't offer the highest level of Math that IB offers).

IB definitely not worth it IMO, research, TOK, all bs classes. However in my school, all the "smart people" are in IB and the "dumb people" in honors/ ap classes so the teachers are better for IB and for honors they make you take life manangement or somehting

I  think in the long run IB will help you learn how to manage your time and it requires you to develop deeper analytical skills than the short-term memory that is required for AP courses.

However, I take both AP and IB courses, and thus far, it is true that my IB courses have bombarded me with a much larger workload relative to the AP/Honors (I don't take Honors anymore, but it is much easier than AP from my previous experiences)

If you don't think the extra workload is worth it, then IB is definitely not right for you. IB tests (as opposed to AP's multiple choice/essay) begins with much more comprehensive and straight-to-the-point essays followed by oral presentations (eww)

And lets not forget the World Literature Papers, CAS essays, portfolios, TOK (which may be bs in practical applications but I got a glimpse of it and I found it very interesting), Extended Essays, Internal Assessments, etc.etc.etc.  :(

I still think it's worth it though.  :)

Go Rickards!!!

Hey Chinyu, Rickards has an IB program? Ha, your not that far away from the best IB program in the world at Stanton in Jacksonville. lol, jk u guys own us in math and we suck but our english program is good. What is your six subject?

Haha, Stanton.... and Eastside was ranked 4th in the US or something.

I'm personally doing:
Math Analysis
Pre-IB Chemistry I
Pre-IB Latin II
AP Physics B
PIB Precalculus
AP European History
PIB English II (Oh my god, she is terrible)

Chen takes AP Psych instead of Math Analysis.

EDIT: Yeah we do own in math  :P , jk.  Maybe this will be our year, haha.

lol, Newsweek rated stanton #1 in country for 2002 and like #3 for 2004. We're slippin cause the admission lets too many ghetto people in. What grade are u guys in? Im a sophmore. Im taking same classes ur taking cept math analysis (what the heck is that? is that like IB math studies cause it sounds like real analysis and thats college level) and how are u in ap physics? im only in physics 1 which sux but yea...btw if u ever go to state or national latin, ill see u there

[quote="xxazurewrathxx"]I think the Newsweek rankings were very deceptive, and I honestly don't believe that Chiles is that far behind Rickards.

First off, they decided to exclude schools like Exeter, TJ, AAST, etc. (because, as they [newsweek guys] claim, they have too many smart kids there. I'm not even kidding, thats what they said. Something about the schools had an above average student body? Read it yourself, my memory fails me.)

Second, they rank the schools by the NUMBER of AP/IB tests taken. So theoretically, if I attended a school that forced all its students (1000) to take the tests and all of them failed, and another school had 500 students take the test and pass, then "my school" wins. That seems like a lousy way to rank high schools, by participation (which seems to be valued much more here in the States) as opposed to actual "skill", for lack of better terms. This may explain why schools like Eastside rank so high up (no offense).

I'm a sophmore and I go to Rickards and yes it does have an IB program. They messed up my schedule early in the year (forcing me to drop French V) so I was planning to switch schools into Lincoln. In the end, however, the administration told my parents they would allow "exceptions" for me and let me take Psychology (I'm technically in IB Psychology even though I'm not in the IB program yet, I'm in Pre-IB) and AP Physics B. Then of course, Saiem (pakman) and Tushar (another guy) had to whine to the teacher and follow me into AP Physics from their Physics H class  :P

My schedule, as Saiem already graciously revealed to the world, is as follows:

Pre-IB Pre-Calculus
Pre-IB English II
Pre-IB Latin II
AP Physics B
AP European History
IB Psychology HL (it's supposed to be AP for me, but they haven't fixed my schedule yet, we're all in the same class though)
Pre-IB Chemistry I

See, no French V  :mad:

As far as academics go, Rickards does have a pretty good math team relative to other places in FL (in the amc competitions anyway, I'm sorry to say that the same couldn't be said for our MAO), and of course JCL and Brainbowl are both pretty good (which I'm sure Saiem could tell you more about). If you ever decide to move to Tallahassee, come to Rickards!  :)[/quote]

haha, I moved from tallahasse when i was 10. I think i lived in alumni village with a lot of other chinese people (sadly a lot of them moved away). My dad's has good friends there including that guy who is building his own house. Btw chinyu, i met u at state mathcounts where u completely owned everyone (including us).

Sadly, you are right about the newsweek ratings and they are completely bs. Stanton isn't what its all said to be and the IB program does have many drawbacks. For ie, I could not take IB psyche and IB other class because i am a sophmore and haven't been inducted.  One thing to remember in IB is that you have to take a sixth subject which basically leaves your junior year with no free classes. (TOK, Research all bs classes) However, the IB literature department is not bad if you like poetry.

2 more things- I take it that you didn't skip precal cause of mu alpha theta and other competitions. A lot of people in my school are in calc cause we suck at math.

And how were you going to take French V? So are u going to do latin as language or french then?

peace
                        :)

yep, Yang's right in this case. Stanton doesnt live up to it's hype... they just overwork you. That.. and math ppl here are an extreme minority... i used to be the only one until i showed Yang (rain) this site.  :D

[quote="rain"]
haha, I moved from tallahasse when i was 10. I think i lived in alumni village with a lot of other chinese people (sadly a lot of them moved away). My dad's has good friends there including that guy who is building his own house. Btw chinyu, i met u at state mathcounts where u completely owned everyone (including us).

Sadly, you are right about the newsweek ratings and they are completely bs. Stanton isn't what its all said to be and the IB program does have many drawbacks. For ie, I could not take IB psyche and IB other class because i am a sophmore and haven't been inducted.  One thing to remember in IB is that you have to take a sixth subject which basically leaves your junior year with no free classes. (TOK, Research all bs classes) However, the IB literature department is not bad if you like poetry.

2 more things- I take it that you didn't skip precal cause of mu alpha theta and other competitions. A lot of people in my school are in calc cause we suck at math.

And how were you going to take French V? So are u going to do latin as language or french then?

peace  :)[/quote]

Oh so you lived in Alumini village? I lived there before I moved over to the Lincoln school zone but ended up attending the IB schools on the other side of town.

Well I'm not "technically" in IB Psychology. I'm in the class but I should be in AP Psychology (I'm the only sophmore so they didn't want to start a new class just for me, everyone else takes the IB course). What do you mean by the "research" class, I don't recall that as being a necessity for the IB diploma. I've heard of the inquiry class but Rickards doesn't require it. Man, I can't wait for TOK... that looks like so much fun!

As for the pre-calculus thing you mentioned, no I didn't skip it even though I was hoping to take it over the summer and be done with it. I recall not being able to get permission from my math teacher (he was also my MAO sponsor), I suppose I wouldn't be much help to the Calculus team and it would jeopardize our Alpha team.

Additionally, I'm not going to take French V (I started in French I in 7th grade at Fairview, a Pre-IB middle school program. Yeah, that's right, I've been in this IB program practically all my life.) I was supposed to skip to V this year and then maybe be done with French VI as a junior, but oh well.

Pre-IB middle schools? That is just scary but might explain the fact how many people skip bio/chem/physics honors to go to AP science. (is that what Fairview does?). Research is a quarter class (assuming you have block scheduling) mainly for your extended essay. I think all IB schools have to have that.

Back to the foreign language, wow, Chenyu, does that mean you already took AP French (IV)? A master of math and language? hahaha I personally suck with all foreign languages except for latin so yea. Do you know this guy Abner or Justin Bird? They go to FAMU (or something) and I'm going to compete against them at states.

Nah I don't know whom you're talking about. Anyway, I dropped French because Rickards forced me to, made me choose between French and Latin since they were on the same period. That almost forced me to Lincoln  :mad:

Yea, I'm planning to be an IB freshman next year, but the recent events: At my future school (let's call this SchA), there were some kids in the IB-4.5GPA, WON THE NATIONAL SCIENCE BOWL, top #4 in the whole school (1600 people), perfect score on SAT. Now, In another school (not IB) (SchB), SchB's students got like 2000s in SAT, and they're not really famous. But guess what:
Guys in SchA made it into John Hopkins (I think this was the highest college acceptance), but in SchB, people made into Harvard, MIT.
So: IB=not so impressive on college acceptances-extra work, yea, and you become better at "life", but you kinda "ruin" your life because your acceptances weren't so good >.<


linear algebra
matrix
algebra
polynomial

I have a question, can somebody help?

Let U, V be two mxn full rank matrices (m<=n). It is known that there exists a diagonal matrix (mxm) D=diag(1,a,a^2,..) that decreases the rank of (U-DV) by one. How can I find D?

regards,

tiffi

Cut out the upper left $ m\times m$ block of $ U$ and $ V$, and ignore the other columns. Call these reduced matrices $ U'$ and $ V'$. We want $ \det(U'\minus{}DV')\equal{}0$; this is a polynomial equation in $ a$. Solve that equation.

Choosing any other $ m$ columns of $ U$ and $ V$ would also work; we get a different polynomial for each choice.


$ rank(BCA) \leq 2$

$ (ABC)^3\equal{}576ABC$ so $ 2\leq rank(BCA)$

$ BCA$ is invertible

$ (BCA)^4\equal{}576(BCA)^2$

$ (BCA)^2\equal{}576I$

We have
 $ BCA$ is diagonalisable since $ X^2\minus{}576\equal{}(X\minus{}24)(X\plus{}24)$ splitt with simple roots

$ trace(BCA)=trace(ABC)=0$;$ trace((BCA)^2)=trace((BC)ABCA)=trace((ABC)^2)=1152$. Let $ spectrum(BCA)=\{\lambda,\mu\}$; then ${ \lambda=24,\mu=-24}$ and $ BCA$ is diagonalizable. Thus $ (BCA)^2$ is diagonalizable and $ (BCA)^2=576I$.

Thanks you !!!
Thanks you !!!
Thanks you !!!
 :roll:  :roll:  :roll:


inequalities
geometry
inequalities proposed

[quote="N.T.TUAN"]Assume that $ a,b,c$ are positive reals that cannot be the sides of a triangle. Prove that $ (abc)^{2}(a\plus{}b\plus{}c)^{2}(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\geq (a^{2}\plus{}b^{2}\plus{}c^{2})^{3}(a^{2}\plus{}b^{2}\minus{}c^{2})(a^{2}\minus{}b^{2}\plus{}c^{2})(\minus{}a^{2}\plus{}b^{2}\plus{}c^{2}).$[/quote]
Ok it is false  :D . If the above clause is true then negative of it is true if
\[ (abc)^{2}(a\plus{}b\plus{}c)^{2}(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\leq (a^{2}\plus{}b^{2}\plus{}c^{2})^{3}(a^{2}\plus{}b^{2}\minus{}c^{2})(a^{2}\minus{}b^{2}\plus{}c^{2})(\minus{}a^{2}\plus{}b^{2}\plus{}c^{2}).\]
then $ a,b,c$ are three side of triangle, but with $ a,b,c$ are three side of triangle we can prove
\[ (abc)^{2}(a\plus{}b\plus{}c)^{2}(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\geq (a^{2}\plus{}b^{2}\plus{}c^{2})^{3}(a^{2}\plus{}b^{2}\minus{}c^{2})(a^{2}\minus{}b^{2}\plus{}c^{2})(\minus{}a^{2}\plus{}b^{2}\plus{}c^{2}).\]
thus combine them we get
\[ (abc)^{2}(a\plus{}b\plus{}c)^{2}(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\equal{}(a^{2}\plus{}b^{2}\plus{}c^{2})^{3}(a^{2}\plus{}b^{2}\minus{}c^{2})(a^{2}\minus{}b^{2}\plus{}c^{2})(\minus{}a^{2}\plus{}b^{2}\plus{}c^{2}).\]
for all $ a,b,c$ are three side of triangle of course that clause is false.
However two equivalent following clause is true
If $ a,b,c$ are three sides of triangle then
\[ (abc)^{2}(a\plus{}b\plus{}c)^{2}(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\ge (a^{2}\plus{}b^{2}\plus{}c^{2})^{3}(a^{2}\plus{}b^{2}\minus{}c^{2})(a^{2}\minus{}b^{2}\plus{}c^{2})(\minus{}a^{2}\plus{}b^{2}\plus{}c^{2}).\]
Or negative of it
\[ (abc)^{2}(a\plus{}b\plus{}c)^{2}(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\le (a^{2}\plus{}b^{2}\plus{}c^{2})^{3}(a^{2}\plus{}b^{2}\minus{}c^{2})(a^{2}\minus{}b^{2}\plus{}c^{2})(\minus{}a^{2}\plus{}b^{2}\plus{}c^{2}).\]
then a,b,c aren't three sides of triangle.
By the way I can prove the stronger for $ a,b,c$ are three sides of triangle
\[ (abc)^{2}(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\ge (a^{4}\plus{}b^{4}\plus{}c^{4})(a^{2}\plus{}b^{2}\minus{}c^{2})(a^{2}\minus{}b^{2}\plus{}c^{2})(\minus{}a^{2}\plus{}b^{2}\plus{}c^{2}).\]
:D

[hide="Hint for it"]
Ok here is hint
If triangle $ ABC$ is right or obtuse then of course the inequality is true.
If triangle $ ABC$ is acute then $ a^{2},b^{2},c^{2}$ are three side of a triangle $ A'B'C'$ that area of $ ABC,S \equal{} 4r'(4R'\plus{}r')$ thus we need to prove
$ a'b'c'(16S)\ge (a'^{2}\plus{}b'^{2}\plus{}c'^{2})\frac{16S'^{2}}{2p'}\Leftrightarrow a'^{2}\plus{}b'^{2}\plus{}c'^{2}\le 8R'^{2}\plus{}2R'r'$
which is true.

[/hide]
[hide="Some word for 979"]
Maybe I think this post is the last post of me, I want to keep my memories with mathlinks like the number $ 979$ the wonderful number of me.
[/hide]


geometry
rectangle
combinatorics unsolved
combinatorics

Please help me prove: If one can tile a rectangle with finitely many squares, then one side of the rectangle is a rational multiple of the other.
Thanks alot !

I think, [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=20599&highlight=#20599]here[/url] is a solution for this problem.  It is really difficult.


geometry
rectangle
calculus
rotation
analytic geometry
geometric transformation
reflection

Prove that any curve of length 1 can be covered by a rectangle of area less than or equal to 1/4.

(Edited) 1/8 changed to 1/4

Note: Thanks to JBL for pointing out the mistake.

Regrettably, this is not true: consider as our curve two segments, each of length 1/2, perpendicular to each other at their endpoints.  They require a rectangle of area at least 1/4 to cover them.

Yeah, I was thinking of that exact same counterexample.  Is it then true for 1/4 instead of 1/8?

Maybe somebody wants to make this more rigorous, but I think I have the idea.

For a given curve, draw the smallest area-wise rectangle around it with width w and length L.  Clearly since the curve is of length 1 we can say that 0 :le: w :le: 1.  For a given w, the largest value L can take is [tex]\displaystyle \frac{1}{2} \sqrt{1-w^2}[/tex] which occurs when the curve forms an isosceles triangle with base w.  Therefore the area of this rectangle, L*w is bounded by: 0 :le: L*w :le: [tex]\displaystyle \frac{1}{2} w \sqrt{1-w^2}[/tex] and a bit of calculus tells us that [tex]\displaystyle \frac{1}{2} w \sqrt{1-w^2}[/tex] is at most 1/4 on [0, 1].

I might be doing something wrong, but I think I have a counterexample to what you just said, zscool:

Let each of those segments have length 1/2 and be at a 135 degree angle.  The width will be [tex]\displaystyle \frac{1}{2\sqrt{2}}[/tex], while the height will be [tex]\displaystyle \frac{1}{2} + \frac{1}{2\sqrt{2}}[/tex], which is more than [tex]\displaystyle \frac{\sqrt{1-(\frac{1}{2\sqrt{2}}})^2}{2}[/tex].  The problem with your proof is that you assumed that any old rectangle will do it, which is not necessarily the case.

you're right.

I'm thinking that maybe I can fix up that argument if we rotate the curve until its projection onto the x-axis is maximal, and call that length w.  Then move the curve until it touches the y axis and line x = w, then project it onto the y-axis with a length of L, and the object fits into the rectangle bounded by x = 0, x = w, y = 0, y = L.  Maybe then we can put a maximum bound on L using the method I orininally thought of.

I'm inclined to believe that what you said is true; however, I would still need you to convince me that what I showed can't happen, which in particular would require you to figure out why it didn't work in my counterexample and why that can't happen when you take the maximal projection.

I know, I just posted my ideas so that somebody might be able to expand on them and prove it correctly, i.e. brainstorming.

I'm really curious about what went wrong with your proof in the case of it that I gave, because the statement you made seemed valid on its face but clearly wasn't true.  But I don't know why it wasn't.  Any ideas?

[quote="zscool"]what if the curve is closed? In that case it has no well defined end points.[/quote]
Then it fits into a rectangle with both side length <= 1/2 because the distance between any two points on the curve is <=1/2.

We have already seen that closed curves of length 1 can be enclosed by a rectangle of area 1/4.

Notation - Suppose that some open curve C terminates at A and B.  We will call the set of points contained in the closed region formed by connecting A and B with a straight line the "region" of C.  We call C a "nice curve" if all of C lies on one side of the line AB and bewteen the perpendiculars to AB at A and B.  zscool has shown that for any nice curve, the "largest minimum enclosing rectangle" has area of 1/4.  We will call any curve C that requires a rectangle of area [tex]$A_{max}$[/tex] to be enclosed a "worst case curve."

All that is left is to show that we can transform any open curve C into a nice curve C' while not removing any points from its region.

The key to this problem appears to be reflection.  Let S be a worst case curve that terminates at A and B.  We will transform S into a nice curve by two reflections.  First, let H be the convex hull of S.  (Correct me if I'm wrong, a convex hull is the shape you would get if you pulled a rope taut around a shape; in any case, H is the shape of a taut rope around S.)  If there is more than one line segment of H that is not part of S, then in at least one instance we can flatten S up against H, appending the region of S while decreasing the length of S (continuous form of the triangle inequality.)  If this is the case, the entire shape can be dialated to have length of 1, but this increases the minimum enclosing rectangle - this is a contradiction.  By a similar argument, if A or B is not part of the convex hull of S, we can arrive at a contradiction.  So there is one line segment [tex]$CD$[/tex] such that [tex]$CD \in H$[/tex] and [tex]$CD \notin S$[/tex]. Specifically, this must be [tex]$AB$[/tex].  Now, overlay coordinates such that A is the origin and that B lies on the positive x-axis, with the remainder of S above the x-axis.  Suppose WLOG that S rotates clockwise as it is traversed from A to B.  Let D and E be points on S such that their x-coordinates are minimum and maximum respectively.  We may now reflect the part of S between D and A over the perpendicular from D to the [tex]$AB$[/tex] and the part  of S between E and B over the perpendicular from E to AB, forming nice curve S' that terminates at A' and B'. Note that A' and B' now have the smallest and largest x-coordinates of all points on S', which still has length of 1, yet the region of S is a subset of the region of S', so we did not decrease the minimum enclosing rectangle.  Returning to zscool's original argument, the proof appears to be complete.

* Sorry I do not have graphics to make this easier to follow and verify/disprove.

fuzzylogic wrote:Make two sides of the rectangle to be parallel to the line segment connecting the two end points of the curve. The touching point of one side together with two vertices of the opposite side forms a triangle of area <= 1/8 by showing the sum of two sides is at most the length of the curve (which is 1).

Finally, I managed to draw a picture using Windows Paint.

[hide]

It's not hard to see AC + BC  :le: (the length of the curve) = 1. So the area of the triangle  :tri: ABC = 1/2(AC)(BC) sin( :ang: ACB)  :le: 1/2(AC)(BC) :le: 1/8(AC+BC) :^2: :le: 1/8. Therefore the rectangle area  :le: 1/4.

[/hide]


algebra
polynomial
calculus
function
analytic geometry
number theory
prime factorization

[quote="Xixas"]Show that the polynomial $f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}$ is irreducible in $\mathbb{Q}[x]$.[/quote]

Here's an ugly way involving calculus:

We show that this equation has no real roots.

Case 1: x >= 0:

In this case there are obviously no roots.

Case 2: x < 0:

We see that this is the nth degree approximation of $e^x$. Since for x < 0, the series is alternating, the error $|e^x - f(x)| \leq \frac{x^{n+1}}{(n+1)!}$.

If the approximation is an overestimate, there are clearly no roots as $e^x$ has no roots. Thus we assume an underestimate:

$e^x - f(x) \leq \frac{x^{n+1}}{(n+1)!}$
$e^x - \frac{x^{n+1}}{(n+1)!} \leq f(x)$

We show that $e^x - \frac{x^{n+1}}{(n+1)!} > 0$

$e^x > \frac{x^{n+1}}{(n+1)!}$

With the implication that n must be odd for there to be an underestimate.

Except this doesn't look like a very true statement.

In fact, for n = 3, f(x) = 1 + x + x^2/2 + x^3/6 definitely has a root. Proof:

f(0) = 1

f(-6) = 1 - 6 + 18 - 36 = -5 - 18 = -23

The function is obviously continuous, so by the Intermediate Value Theorem there is a zero on (-6,0).

Perhaps I misunderstand the meaning of reducible?

Yes, because he is asking about irreducibility in $\mathbb{Q}[x]$.

The polynomial obviously has real roots if $n$ is odd, that's not the issue.

I can prove irreducibility for prime $n$ by Eisenstein's criterion, but beyond that... :(

This problem was posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=64794
Perhaps you are interested in a gerneralisation: http://www.mathlinks.ro/Forum/viewtopic.php?t=36701

[quote="Xixas"]I have a very concise proof on my textbook but I can not understand it completely. It goes as follows:

It suffices to show that $n!f(n)=x^n+nx^{n-1}+n(n-1)x^{n-2}+\cdots+n(n-1)\cdots3\cdot 2\cdot x+n!$ is irreducible in $\mathbb{Z}[x]$. But this would follow if we proved that there exists a prime $p$ such that $p\le n <2p$. However using Bertrand's postulate we easily see that such a prime $p$ exists.

Why is it sufficient to prove the existence of such $p$? I can not see how the result follows once we have shown it (in my opinion author is refering to Eisenstein's criterion or to its generalization but how do they imply the required?)...[/quote]

I was actually thinking along those lines, but I thought Eisenstein's criterion required that $p$ divide every coefficient but the first and the last, and such a $p$ clearly does not divide $n$...

it's eisenstein's generalisation:
bertrand's postulate shows that all posible irreducible factors of $p$ have degree $>n/2$

Can you please state this generalisation? I know the following one:

Let $P(x)=H(x)G(x),\ P,H,G\in\mathbb{Z}[x]$ and $P(x)=\sum_{i=0}^{n}a_i x^i$. Suppose there exists a prime $p$ such that $p|a_0,\ a_1,\ldots,\ a_{n-m}$ but $p$ does not divide $a_{n-m+1}$ and $p^2$ does not divide $a_0$. Then $m<min(\deg{H},\ \deg{G})$.

There is also this criterion of Dumas' which is also a generalization of Eisenstein's criterion but I don't think the author is refering to it (and in my opinion it does not imply anything stronger in this case).

First a few conventions: a factor of $f(x)$ means a polynomial factor with rational coefficients; a lattice point in $\mathbb R^{2}$ is a point with integer coordinates; the width of a segment in $\mathbb R^{2}$ is the difference of the $x$-coordinates of its endpoints (not the length of the segment).

Now go to business.

Let $a_{0}, a_{1}, \dots , a_{k}$ be the digits of $n$ in base $p$, $p$ a prime, so that $0\leq a_{i}<p$, $n=\sum_{i=0}^{k}a_{i}p^{i}$, then we have the following easy lemma, whose proof is left to the reader.

LEMMA. The lattice points split the $p$-adic Newton polygon of $f(x)$ into $a_{0}+a_{1}+\dots+a_{k}$ segments: $a_{k}$ segments of width $p^{k}$, followed by $a_{k-1}$ segments of width $p^{k-1}$, ..., followed by $a_{1}$ segments of width $p$, followed by $a_{0}$ horizontal segments of width $1$.
NOTE: Of course, this segments are free of inner lattice points, I mean.

Now let $d$ be the degree of a factor of $f(x)$, and let $n=p_{1}^{e_{1}}\dots p_{s}^{e_{s}}$ be the prime factorization of $n$. We just apply the lemma above to the primes $p_{1},\dots, p_{s}$ dividing $n$:

The $p_{1}$-adic Newton polygon of $f(x)$ proves that $p_{1}^{e_{1}}|d$.
...
The $p_{s}$-adic Newton polygon of $f(x)$ proves that $p_{s}^{e_{s}}|d$.

This yields $n|d$ so that $d$ equals $0$ or $n$.


calculus
integration
LaTeX
calculus computations

This is one hard question...hoepfully you guys know your ODE's.

F(w) = integral(-inf,inf) e^(iwx) * e^-x^2 dx

by getting the ODE of F(w) in the form:

dF/dw + h(w)F, F(0) = Fo,

Where Fo is the constant which you have to define explicitly.

Note: Sorry..i tried to put this in latex for the longest time but kept failing.

Note: This  is not a calculus problem..thats why i put it here.  It is a problem solving type prolbem...

-----

Hint: Solving this, you could easily see that F(0) = root(pi)

MOD: Moved to the calculus forum.

i dont think calculus is allowed in this forum... it says clearly at the top

$F(\omega)=\int_{-\infty}^{\infty}e^{i\omega x}e^{-x^{2}}\,dx$

Find the first-order linear ODE that $F$ satisfies, then solve that equation. Use a separate explicit calculation of $F(0)$ as an initial condition for the ODE.

(Click on the above to see the $\LaTeX$ code.)

I know three ways to compute the Fourier transform of a Gaussian. This is one of them.


probability
combinatorics unsolved
combinatorics

Suppose we mark $n$ points on a rod so that it is divided into $n + 1$ parts. If $L$ is the total length of the rod, What is the probability that the length of each part is $\leq \frac{L}{n}$

[hide]There is some connection between the answer and representation on n-dimensional graphs.[/hide]

I hope everyone has understood the question. It's not often that you don't get a solution in 2 days on this forum

Come on people has no one even thought of n=2 or n=3 cases

[hide]The general answer is $ \frac{1}{n^n} $ . Now anything????[/hide]

I'm sorry I'm bringing back an OLDDD forum. But please post your ideas/thoughts for this problem!


If you haven't already, I would suggest you to consider converting the forums to IPB.  Here is an example of an IPB powered board:  http://forums.invisionpower.com/.  It's more secure, customizable, and has alot of mods available.. while phpBB has alot of bugs/glitches that allow hackers to screw the board up.  I've seen alot of boards on the web which have been hacked because they were powered by phpBB!  Even the website of phpBB itself was hacked once.

[quote="Algebra 1 ©"]If you haven't already, I would suggest you to consider converting the forums to IPB. It's more secure, customizable, and has alot of mods available.. while phpBB has alot of bugs/glitches that allow hackers to screw the board up.  I've seen alot of boards on the web which have been hacked because they were powered by phpBB!  Even the website of phpBB itself was hacked once.[/quote]Thanks but no thanks. Please take more time to read and understand what each system does better. phpBB was not hacked because of phpBB, but because of AwStats v. 6.1 which has nothing to do with the forum software.

Good luck convincing other people to pay for something worse than phpBB :) Topic closed.


[quote="stancioiu sorin"]Cine credeti ca va castiga O.N.M.2006? :)[/quote]
     Comentati raspunsul.

La a 11-a, Bacaul...  :D No comment.

Ce sens are topicul asta?

nici eu nu inteleg ce sens are, dar ma rog. Valentin ar fi o idee sa inchizi topicul asta. :)

Evident Cezar. Domnu' Sorin o sa ia in curand o pauza binemeritata daca o tine tot asa ;) :)


abstract algebra
superior algebra
superior algebra solved

Let $ \left(G,\cdot \right)$ be a group with $ p^3$ elements, where $ p$ is a prime. Show that $ x^p\in Z(G)$ for all $ x\in G$.

Hmm, is it that easy :?
$G$ is a $p$-group, so it has non-trivial center $|Z(G)|\in\{p,p^2,p^3\}$. In case $|Z(G)|=p^3$ or $|Z(G)|=p^2$ we're done, since then $G/Z(G)$ is cyclic, so $G$ is abelian. If $|Z(G)|=p$ then $G/Z(G)\sim \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ or $G/Z(G)\sim\mathbb{Z}/p^2\mathbb{Z}$. In the first case we're done since any element has order dividing $p$. In the second case $G/Z(G)$ is cyclic, so $G$ must be abelian.

I have a few questions
1) How do you know that $Z(G)$ is non-trivial? Is there a theorem or something?
2) Why is $G/Z(G)$ cyclic for $|Z(G)|=p^2$ ?
3)
[quote="Christian Hirsch"]If $|Z(G)|=p$ then $G/Z(G)\sim \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ or $G/Z(G)\sim\mathbb{Z}/p^2\mathbb{Z}$.[/quote]
Why?

1. Yes there is a theorem stating that the center of a p-group is non-trivial. One can prove this by group action. See for example [url=http://www.math.harvard.edu/~elkies/M250.01/center.html]www.math.harvard.edu/~elkies/M250.01/center.html[/url]
2. if $|Z(G)|=p^2$ then $|G/Z(G)|=p$. But it follows from Lagrange that if the order of a group is prime, then it must be cyclic. A group $G$ is abelian iff $G/Z(G)$ is cyclic
3. The statement amounts to saying that every group $H$ of order $p^2$ is of one of the types mentioned. We have $|H|=p$ or $|H|=p^2$. As $H/Z(H)$ is cyclic in both cases, we know that $H$ must be abelian. If there is an element of order $p^2$ than $H$ is isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$. Otherwise take an nontrivial element $a\in H$ and a nontrivial element $b\not\in\langle a\rangle$. Then one can check that we must have $\langle a\rangle \cap\langle b \rangle=\{e\}$. Thus the map $\langle a\rangle \times \langle b\rangle\to G$, $(x,y)\to xy$ is a monomorphism ($H$ is abelian) and must be an isomorphism by equality of the cardinalities $|G|$ and   $\langle a\rangle \times \langle b\rangle$

Did it help?


Older television screens have an aspect ratio of $ 4: 3$. That is, the ratio of the width to the height is $ 4: 3$. The aspect ratio of many movies is not $ 4: 3$, so they are sometimes shown on a television screen by 'letterboxing' - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $ 2: 1$ and is shown on an older television screen with a $ 27$-inch diagonal. What is the height, in inches, of each darkened strip?
[asy]unitsize(1mm);
defaultpen(linewidth(.8pt));
filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black);
filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black);
draw((0,2.7)--(0,13.5));
draw((21.6,2.7)--(21.6,13.5));[/asy]$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.25 \qquad \textbf{(C)}\ 2.5 \qquad \textbf{(D)}\ 2.7 \qquad \textbf{(E)}\ 3$

Let the longer side be 4x and the shorter side be 3x.  Thus the original aspect ratio is 4:3.  However, for the movie, the aspect ratio is 2:1. Therefore, the dimensions of the "white strip" are 2x * 4x.

$ 9x^2\plus{}16x^2\equal{}27^2$
$ 25x^2\equal{}27^2$
$ x^2\equal{}\frac{27^2}{5^2}$
$ x\equal{}\frac{27}{5}\equal{}5.4$

Therefore, each darkened strip is $ 2.7\equal{}\boxed{D}$

A simpler method would be to pretend the actual width and length were 4 and 3.  To change that to a 2:1 ratio, you have to take 1 away from the height, or .5 away from the top and .5 from the bottom.  Using 4 and 3, your diagonal is 5 long, so you multiply .5 by 27/5 to scale it back up, getting 27/10 or 2.7.

No Algebra.. mostly

I completely brute-forced with no method at all.

[quote="worthawholebean"]Older television screens have an aspect ratio of $ 4: 3$. That is, the ratio of the width to the height is $ 4: 3$. The aspect ratio of many movies is not $ 4: 3$, so they are sometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $ 2: 1$ and is shown on an older television screen with a $ 27$-inch diagonal. What is the height, in inches, of each darkened strip?

[asy]unitsize(1mm);
filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black);
filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black);
draw((0,0)--(21.6,0)--(21.6,16.2)--(0,16.2)--cycle);[/asy]

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.25 \qquad \textbf{(C)}\ 2.5 \qquad \textbf{(D)}\ 2.7 \qquad \textbf{(E)}\ 3$[/quote]

Method 2:

Let the diagonal be 5x, the legs by 3x and 4x. Since the diagonal is 27, $ x\equal{}\frac{27}{5}$. The one with 3x has the strips. The non-strip part is half of 4x. Therefore the stripped part is x and since it is symmetrical, each strip is x/2. Each strip is $ 2.7 D$

[hide=solution]By $3\text{-}4\text{-}5$ right triangles, the width of an older television screen is $27\cdot\cfrac{4}{5}=21.6$ and the height is $27\cdot\cfrac{3}{5}=16.2$. Let $x$ be the height of each darkened strip. Then \[\cfrac{21.6}{16.2-2x}=2\rightarrow 16.2-2x=10.8\rightarrow 2x=5.4.\] Therefore, $x=\boxed{\textbf{(D)} ~2.7}$.[/hide]


In the game "two in a row" players flip a coin twice and they win if they get heads both times. Four contestants play in the game and they take turns flipping their two coins. The first one to get two heads wins. Mike goes first. What is the probability that no one wins on their first turn and that Mike wins on his second turn?

[hide]The probability of winning is $\frac14$, so the probability of not winning is $\frac34$. So the probability that 4 people don't win, then the 5th wins is $\frac34*\frac34*\frac34*\frac34*\frac14$ or $\frac{81}{1024}$[/hide]


geometry
3D geometry
calculus
integration

Prove: A cube cannot be decomposed into finitely many smaller cubes, all of different sizes.

If you can't decompose a square into smaller squares of different sizes, then you can't decomnpose a cube into smaller cubes of different sizes. This is because the bottom of a cube is a square, and as a result of decomposing the cube, the bottom of the cube would be decomposed into squares.

mathdool wrote:Works for square too, unless maybe you say it decomposes into one object: [hide]itself.[/hide]

Can you prove this? If you can, then you have just proved the cube problem also, for the reason Yasha gave. I know a proof to the cube problem, but it is much different than this method. I'd like to see if this method also works.

[quote]Works for square too, unless maybe you say it decomposes into one object: itself.[/quote]

But you can dissect a square into squares of different sizes, see here:

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

[hide]Consider all small cubes that touch the bottom face of the big cube. There must be a smallest one.

In a dissection of a large square into uniquely sized smaller squares, the smallest square can't touch the edge of the large square.

Using this, the smallest cube touching the bottom face can't touch an edge of the face. On top of this cube, we have a square surrounded by walls on which we must place more cubes.

There must be a smallest cube in this set, which doesn't touch the edge of the square so the square on top of this cube is surrounded by walls.

We can continue this process to produce smaller and smaller cubes, contradicting the fact that the dissection contains a finite number of squares.[/hide]

JS1527 wrote:This may be a stupid question but does this have anything to do w/:[hide]fermat's last theorem[/hide]?

Not really. Fermat's Last Theorem would only help in that if you add two cubes you can't get a third one, but we're talking about dissecting a cube into many cubes, so it doesn't apply.

JS1527 wrote:This may be a stupid question but does this have anything to do w/:[hide]fermat's last theorem[/hide]?

What Yasha said, and in addition, the case a^3+b^3=c^3 is impossible in a,b,c E N is not particularly easy to prove, but nevertheless well known.

Also, Fermat's Last Theorem states that there are no [b]INTEGRAL[/b] solution to [tex]a^3+b^3=c^3[/tex]


Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\equal{}4.$ Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}\geq4
\]
Remark. This inequality is not true for the condition $ a^5\plus{}b^5\plus{}c^5\plus{}d^5\equal{}4.$

I'm happy! I could with this one! (I know that is old and everything, but it's not sooo trivial and I'm an idiot at maths.)

Suppose that $ \{a,b,c,d\} \equal{} \{x,y,z,t\}$ with $ x\ge y\ge z\ge t$. We will see that

$ \frac{a^2}b\plus{}\frac{b^2}c\plus{}\frac{c^2}d\plus{}\frac{d^2}a\ge \frac{x^2}y\plus{}\frac{y^2}x\plus{}\frac{z^2}t\plus{}\frac{t^2}z$.

Doing the Abel Summation trick, we see that the difference is

$ (x^2\minus{}y^2)\left(\frac1p\minus{}\frac1y\right) \plus{} (y^2\minus{}z^2)\left(\frac1p\plus{}\frac1q\minus{}\frac1x\minus{}\frac1y\right)\plus{}$
$ (z^2\minus{}t^2)\left(\frac1p\plus{}\frac1q\plus{}\frac1r\minus{}\frac1x\minus{}\frac1y\minus{}\frac1t\right)$,

where $ p$ is the denominator of $ x$ in the lhs and $ q$, $ r$ and $ s$ are defined similarly for $ y$, $ z$ and $ t$. We have that $ x^2\ge y^2$ and also $ \frac1p \ge \frac1b$ since $ p$ cannot be $ x$ and then $ p\le y$. We have that $ y^2\ge z^2$ and that $ \frac1p\plus{}\frac1q\minus{}\frac1x\minus{}\frac1y$ obviously. Finally we see that $ z^2\ge t^2$ and the only way of $ \frac1p\plus{}\frac1q\plus{}\frac1r\minus{}\frac1x\minus{}\frac1y\minus{}\frac1t$ to be negative is that $ \{p,q,r\}$ are defined to be $ \{x,y,z\}$ but that would imply that $ s$ is defined to be $ t$, which cannot be true since $ s$ is the denominator of $ t$ in the lhs.

In conclusion, every factor of every denominator is non negative, which proves our assertion.

So what is left to prove is that, for $ a\ge b\ge c\ge d\ge 0$ such that $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\equal{}4$ we have that $ \frac{a^2}b\plus{}\frac{b^2}a\plus{}\frac{c^2}d\plus{}\frac{d^2}c\ge 4$. In fact, $ \frac{a^2}b\plus{}\frac{b^2}a\ge 2\sqrt[4]{\frac{a^4\plus{}b^4}2}$ (just put $ b\equal{}1$, square two times and divide by $ (a\minus{}1)^2$) and $ \frac{c^2}d\plus{}\frac{d^2}c\ge 2\sqrt[4]{\frac{c^4\plus{}d^4}2}$, so it's left to prove $ 2\sqrt[4]{\frac{a^4\plus{}b^4}2}\plus{}2\sqrt[4]{\frac{c^4\plus{}d^4}2}\ge 4\sqrt[4]{\frac{a^4\plus{}b^4\plus{}c^4\plus{}d^4}4}\equal{}4$, which is nothing more than fourth power mean. That's it.


Let g(x) be a function such that g(a+b)=g(a)g(b) for all real numbers a and b.  If 0 is not an element of the range of g, find g(-x) in terms of g(x).

[hide]

Let , and we get , and since  is not in the range of the function, then .

Then, let , and , and we get , so .

[/hide]