Euler
integration
calculus
analytic geometry
trigonometry
function
logarithms

Mathematica says $\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2}$
Euler says $\Gamma(\frac{3}{2}) = \int_0^\infty t^{\frac{1}{2}}e^{-t}\ dt$

Is there an elementary (equivalent of calc 1 & 2) method of integrating $\int_0^\infty t^{\frac{1}{2}}e^{-t}\ dt = \frac{\sqrt{\pi}}{2}$?

Integration by parts establishes the recursion $\Gamma(x+1)=x\Gamma(x).$

Given that, it's easier to calculate

$\Gamma\left(\frac12\right)=\int_0^{\infty}t^{-1/2}e^{-t}\,dt=\sqrt{\pi}.$

Make the substitution $t=u^2,\,dt=2u\,du$ to get

$2\int_0^{\infty}e^{-u^2}\,du=\int_{-\infty}^{\infty}e^{-u^2}\,du$

Now, have you seen the standard trick (doing a two-dimensional integral in both rectangular and polar coordinates) for that?

Look here

 [url]http://www.maths.bris.ac.uk/~maxyl/Probability/Gamma.pdf[/url]

and here

 [url]http://www.phys-astro.sonoma.edu/people/faculty/tenn/GammaFunction.pdf[/url]

I am not sure what you mean by elementary...

[quote="djshowdown2"]Is there an elementary (equivalent of calc 1 & 2) method of integrating $\int_0^\infty t^{\frac{1}{2}}e^{-t}\ dt = \frac{\sqrt{\pi}}{2}$?[/quote]

I think there is one and it involves trig substitution...

swapnillium, I will guarantee you that  the antiderivative
$\int t^{1/2}e^{-t}\,dt$ is not an elementary function. Note that finding an antiderivative for this would be equivalent to finding an antiderivative for the Gaussian, and we know we can't do that in elementary terms.

Hence any method of exact computation must be something that works for the particular definite integral without relying on antiderivatives. Such methods do not fall within the "Calculus I and II" definition that djshowdown2 was going for, if you consider "Caculus II" to be the same as "BC" and to not contain any double integrals. The standard method I spoke of involves double integrals in both rectangular and polar coordinates, and thus falls into "Calculus III" - and into "Calc. II" if the course moves fast enough.

[quote="djshowdown2"]Is there an elementary (equivalent of calc 1 & 2) method of integrating $\int_0^\infty t^{\frac{1}{2}}e^{-t}\ dt = \frac{\sqrt{\pi}}{2}$?[/quote]
Assume as definition of $\Gamma: (0,\infty)\to (0,\infty)$ following  (H.Bohr and J.Mollerup): the restriction of Gamma function at $(0,\infty)$ is the unique function $G: (0,\infty)\to (0,\infty)$ which satisfies:
1) $G(x+1)=xG(x) \; ,\; \forall x>0$,
2) $\ln{G(x)} \; ,\; x \in (0,\infty) ,$ defines a convex function on $(0,\infty)$,
3) $G(1)=1$ .
Note that 1)-3) implies $G(x)\equiv \Gamma(x)=\int\limits_{0}^{\infty}e^{-t}t^{x-1}\; dt \; , \; x>0 .$
Supposing $x>0,y>0,$ try to use $\Gamma(x+1)=x\Gamma(x)$ and /or     $\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\equiv B(x,y): =\int\limits_{0}^{1}t^{x-1}(1-t)^{y-1}\; dt ,\; x>0,y>0.$

I would add that historically the first to claim that $\left(\frac12\right)!=\frac{\sqrt{\pi}}2$ was John Wallis, in his [i]Arithmetica Infinitorum[/i] (a book dismissed by Thomas Hobbes as a "scab of symbols.")

This was a half-generation or so before there was such a thing as calculus. Wallis recognized the following pattern that holds for nonnegative integers $n:$

$\int_0^1(x(1-x))^n\,dx=\frac{(n!)^2}{(2n+1)!}$

Note that this is a special case of the Beta function in flip2004's last post. Of course, Wallis was using other notation - the concept of "integral" didn't really exist yet, although mathematicians did know how to compute what amounted to the integrals of polynomials.

Wallis then boldly set $n=\frac12$ in this equation (without any particularly convincing justification) and noted that

$\int_0^1\sqrt{x(1-x)}\,dx=\frac{\pi}8$ because it is the area of a semicircle.

Wallis's boldness in dealing with fractional powers probably helped inspire Newton in his own development of the binomial series.

[quote="Kent Merryfield"]I would add that historically the first to claim that $\left(\frac12\right)!=\frac{\sqrt{\pi}}2$ was John Wallis, in his [i]Arithmetica Infinitorum[/i] (a book dismissed by Thomas Hobbes as a "scab of symbols.")
[/quote]Very interesting,Alex


complex analysis
complex analysis unsolved

Prove that if $f(z)$ is meromorphic on $U \setminus \{z_0\}$, then $e^{f(z)}$ has an essential singularity at $z_0$.


Eto, sutra počinje IMO 2009 i stvorih ovaj topic [size=59]kad ga već nitko od vas 34 koji idete tamo nije stvorio[/size] iz jednog čisto egoističnog razloga - kako bih poželio svima puno sreće i, posebno, puno zabave u Bremenu. Možda ne bacamo stolice, baklje, tupe i oštre predmete po fakultetima i školama, ali znajte da ima ljudi koji navijaju za vas.  :clap:  :clap2:  :clap:

[size=59](Ujedno se ispričavam onima kojima sam na, kako bi rekao Harun, "najpopularnijem siteu za društveno umrežavanje" zatrpao home svojim višestrukim lijepim željama. Bude još toga popodne. :))[/size]

Evo od mene sticky za ovu temu, zaslužila je. Zavidim onima koji idu (čovječe, slušat ćete Terryja Taoa i Lovasza...)

Puno sreće svima, uživajte, iskoristite svaki trenutak IMO no. 50 i pokažite šta znate  :wink:

Ja svečano obećavam da neću ljudima pisati good luck poruke po LicoKnjizi (najpopularnijem siteu za društveno umrežavanje  :lol: ), dosta su vam Melkiorove nadahnute riječi.

Još jednom, sretno!

Vidim, zainteresiranost za Olimpijadu je ogromna na ovom podforumu... gotovo da bi čovjek pomislio da neće pasti barem 8 zlata.  :P

Je li se javio ko iz slobodarskog Bremena sa novim informacijama from the face of the place?  :)

Komentari na zadatke? Možda sam malo subjektivan, ali doista mi se čine laganim. Ne prelaganim, naravno, ali očekivao sam nešto strašnije za jubilej  :lol:

1. Lak relativno (ma lak je ja sam ga resio :D)
2. Mozda naginje malo na stranu lakseg za IMO ipak
3. Ne gledam ja trece zadatke :D

Cini mi se da su ovi zadaci lepi, ali previse laki. Ja licno ne bih voleo da ih vidim na takmicenju, jer ovakav prvi dan verovatno povlaci visoke granice, a 6. ce sigurno biti tezak (mozda neka 'lepa' kombinatorika, koja je lepa sve dok ne moras da je resavas). Najtezi zadatak prvog dana je naravno bio onaj niz, ali kako ja volim takve stvari, meni je on bas legao tako da sam ga vrlo brzo resio. Prvi zadatak je ispod nivoa IMO-a i verovatno najlaksi zadatak u poslednjih desetak godina. Drugi zadatak je lep zadatak, ali da bude prvi (tezi), jer u resavanju ne postoji nijedan nelogican korak. U poslednjih 15-ak godina padaju mi na pamet barem tri geometrije koje su bile prvi zadatak, a koje nisu bile lakse od ovoga (2005/1, 2000/1, 1998/1).
U svakom slucaju srecno nasima i nadam se dobrim vestima!

Ah, dakle opet je došlo vrijeme kada se mi koji se ne natječemo hvalimo zlatima koja bismo zasigurno osvojili...  :lol: Ja ću biti iskren i priznati da sam lijen kao svi vi ostali zajedno i da nisam sam rješavao niti jedan od ovih zadataka - počeo sam 1., ali se nakon minutu pojavilo rješenje, pa sam odustao. :blush: Stoga je moje mišljenje bazirano uglavnom na izgledu zadataka i na prezentiranim rješenjima (u svoju obranu, zbilja sam ih temeljito prolazio...), pa možda i nije toliko mjerodavno.

[b]1. zadatak[/b] je lagan, ali ga ne bih proglasio uvjerljivo najlakšim do sada. Da je lagan, jest, slažem se. Ali bilo je i drugih laganih zadataka - 2006/1. je bio posebno poznat kao lagan, a i po mom skromnom mišljenju 2007/1. je jako lagan kad ga nacrtaš (2008/1. neću komentirati - njega sam analitički  :lol: ). U svakom slučaju, zadatak je lijep, tu se slažem s Goblinom, a mislim da treba uzeti u obzir još jednu stvar - teorija brojeva nije bila 1. zadatak od 1992., ako se ne varam, tako da je to meni bilo dosta neočekivano i zapravo osvježavajuće.

[b]2. zadatak[/b] je, vidim, isto naišao na neodobravanje mathlinkerske publike - "That makes the problem actually useless infact. There is nothing invovled int he problem. I shall be surprised if some contestant doesnt solve this problem.". Da je nešto ultratežak, nije, ali zahtijeva, bar u meni najljepšem rješenju, korištenje potencije točke i kuta između tetive i tangente, što nisu teške stvari, ali ipak nije čisti angle chasing. Lijep zadatak, u svakom slučaju - da je možda mogao biti teži, mogao je, ali nije loše ponekad ni riješiti ponešto na Olimpijadi.  :lol:

[b]3. zadatak[/b] je zbilja zbilja lijep. (Jesam li spomenuo da mislim je ovo valjda najljepši prvi dan olimpijade ikada?). Naravno, šanse da bih ga riješio su jednake onima da me pogodi grom, ali moja je procjena da je ipak bilo i težih trećih zadataka. Ako ništa, onda je ovaj zadatak barem dovoljno zanimljiv da se njime baviš kada riješiš prva dva (ili nedajbože odustaneš od njih). Ne koristi (bar u rješenju po kojem sam ja išao) nikakvo teoretsko znanje, nego čisto razmišljanje, što mislim da je idealno za težak zadatak na IMO.

Općenito, jako sam zadovoljan time kako prvi dan izgleda. Da je možda iznadprosječno lagan, to je moguće, ali svakako nadomješta to ljepotom i zanimljivošću. Preostaje da vidimo drugi dan.  :)

A dobrih vijesti će naravno biti, pogotovo ako se naši natjecatelji slažu s Goblinovim procjenama težine.  :D

2. zadatak: Po meni je sasvim OK zadatak, solidan, ne vidim zasto ga ne odobravaju, ipak ima 600 takmicara tamo i ne mogu svi da rese tako nesto.

[quote="jmerry"]

If you're tuned into geometry and see the right idea, its simple and quick. If you're not, you can flounder for hours getting nowhere.

[/quote]

Tacno, i zato su i dali ovakav zadatak, ipak nece svi odmah primetiti kako se resava i gotov posao...

Ali, mozda postoji jos jedan razlog za ipak posle svega laksi 2. zadatak...
Citajuci malo, dosao sam do nekog zakljucka da se problemi 2. i 5. biraju poslednji, i da ima vise dobrih kandidata za ta mesta iz svake oblasti... A onda, mozda je neka druga teza geometrija izbacena za 2. zadatak zbog tezine 2. dana... Ocekujem da ce biti tezak... A ostali?

Izvinite zbog duplog posta, ali evo i drugog dana!!!

Komentari?

4. Nije tezak, elementarno resenje je dodavanjem jedne tacke, zatim cisto jurenje uglova, dok se moze uraditi trigonometrijom... bez ideje, sa puno racuna (ja sam prvobitno pogresio, sad se ispravio sa takvim resenjem)... Tezi od prvog juce

5. OK zadatak, funkcionalna jednacina (koje se protezu kroz BMO i IMO vec dugo) po resenju user-a Soarer (ako je tacno) nije ni pretezak zadatak, mada mnogo tezi od drugog juce.

6. ... Ne znam ni kako da pocnem :D

pozdrav svim matematicarima! jednom godisnje 'svratim' ovde dok cekam prvo nezvanicne, a onda zvanicne rezultate. drzimo palceve svima kod koordinacije i odredjivanja granica. nego, nije mi jasna poruka na bremenskom sajtu za imo: 'pictures will be posted SOON' - i ta se poruka 'dinsta' DVA dana, a deca su tamo vec 4 dana. kad planiraju da stave te slike? kad se IMO zavrsi?  :oops:
teova mama

Sta ocekuju takmicari? Meni su javili posle prvog dana da je Teodor jedini uradio sva 3 (svaka cast)...

pitanje: sta ocekuju takmicari? je bilo upuceno meni? ja sam u sms kontaktu samo sa teom. takmicari se valjda standardno nadaju najboljem (u odnosu na to sta su uradili). izvinjavam se za pogresan citat u prethodnom postu. usli smo u PETI dan olimpijade, a na nemackom sajtu i dalje stoji obavestenje: 'Pictures will be released as soon as possible.' izgleda da ni nemci nisu sto su nekad bili. na proslogodisnjoj naucnoj olimpijadi u juznoj koreji, svakodnevno su na sajtu bile i nove slike i vesti. nadam se da mi niko ne zamera sto se ovde bavim ne-matematickom temom. BTW, ko je bugi? imam neprijatna iskustva komuniciranja sa nepoznatim osobama, a ja pripadam starijim generacijama - prvo se predstavim, pa onda ... srecno na koordinaciji!!!

Pa ajde da završim drugi dio svojeg mišljenja o zadacima na Olimpijadi. I dalje vrijede isti disclaimeri kao za prvi dan. :)

[b]Zadatak 4.[/b] mi se ne čini strašno težak (naravno, mislim da je težinom usporediv s drugim zadatkom iz prvog dana, ako nije i teži) - vidio sam to planimetrijsko rješenje koje se čini dosta razumno i intuitivno, pogotovo ako natjecatelju geometrija nije najslabija strana (kao nažalost meni). Ali, čak i ako je, mislim da se dosta može izvući trigonometrijom (kao što je pokazao Bugi). Ja sam jučer popodne probao analitičko rješenje - znam da koordinate ne vole kuteve jer upliću tangense i nisam ga dovršio do kraja jer je opet u igru ušla lijenost, ali nalazim poprilično vjerojatnim, nemojte me držati za riječ :), da bi se i tako moglo izvući.

[b]Zadatak 5.[/b], usprkos svojem netipičnom izgledu, zapravo i nije toliko čudan zadatak. Primijetit ćete da se nejednakost trokuta zapravo koristi u rješenju samo dva puta ako točno brojim, jednom od toga u sasvim trivijalnom slučaju (kad je jedna stranica jednaka jedan). Zapravo, dobro je pitanje može li se tvrdnja zadatka nekako jednostavno transformirati, a da iz toga izbacimo cijelu priču s trokutima koja bi mene, da sam ja bio tamo, dosta zbunila.

[b]Zadatak 6.[/b] ima to rješenje od probability1.01a koje se čini dosta jednostavno, ali ne znam je li postignut konsenzus da je to rješenje zaista točno (čini mi se da su komentari uglavnom nešto u stilu "Da, rješenje je zbilja lagano, pa valjda mora imati grešku, ali nigdje ne mogu naći grešku..."). Naravno, ni ja nisam našao grešku, ali to je sasvim očekivano. Ako ispadne da rješenje jest točno, a da sam ga ja dobro shvatio, onda je zadatak po mom mišljenju nešto lakši od šestih zadataka inače, makar je i to diskutabilno - pitanje je koliko vremena izgubiš da se sjetiš ideje, a vremena nemaš na pretek ako si prije toga riješio/rješavao prva dva zadatka.

Moje generalno mišljenje o drugom danu je lošije nego nego o prvom. Mislim da zadaci jesu nešto teži, makar se opet ne može reći da se radi o nekoj posebnoj težini. Ali to nije ono što me smeta, nego mislim da su zadaci iznadprosječno ružni: u četvrtom zadatku uvjet da je neki kut između ne pretjerano povezanih točaka $ 45^{\circ}$ ne čini mi se previše intuitivan, u petom zadatku, kažem, ta priča o nejednakosti trokuta zbunjuje malo i nije prelijepa, možda bi se zadatak mogao malo modificirati, a na šesti zadatak zapravo i nemam nekih prevelikih primjedbi, jedino, kažem, ne znam je li ono rješenje točno...

No, kažem po stoti put, lako je nama odavde pričati, na natjecanju je ipak puno drugačije - uz tradicionalnu nervozu, postoji i vremensko ograničenje koje ti možda i ne igra neku ulogu ako si (kao što sam ja  :maybe:) prezadovoljan ako ćeš riješiti jedan zadatak svaki dan, ali ipak dosta pritišće ako ideš na svih šest.

Vidjet ćemo završne rezultate - ljudi na Mathlinksu uvijek ocjenjuju zadatke lakšima nego što su zapravo većini natjecatelja, pa mislim da sva očekivanja ovdje treba uzimati s određenim podozrenjem.  :ninja:

[quote="mornik"] Ali, čak i ako je, mislim da se dosta može izvući trigonometrijom (kao što je pokazao Bugi). [/quote]

Na kraju nisam ni pokazao na forumu :) Pokusao sam, ali onda sam se izgubio koji trougao da gledam, a jos sam napravio gresku pa je ispalo tako kako je :)
 Posle sam se ispravio, a alexilic je lepo pokazao resenje :)

Mislim da su nam se gotovo sve želje do kraja ispunile.

:winner_first: (32-42)
Teodor von Burg, Srbija

:winner_second: (24-31)
Bodan Arsovski, Makedonija
Mihajlo Cekić, Srbija
Nina Kamčev, Hrvatska
Luka Milićević, Srbija
Dušan Milijančević, Srbija

:winner_third: (14-23)
Matej Aleksandrov, Slovenija
Ivo Božić, Hrvatska
Predrag Gruevski, Makedonija
Bojan Joveski, Makedonija
Adrian Satja Kurdija, Hrvatska
Matko Ljulj, Hrvatska
Jelena Radović, Bosna i Hercegovina
Stefan Stojčevski, Makedonija
Vukašin Stojisavljević, Srbija
Goran Žužić, Hrvatska

[b]Pohvala[/b] (jedan cijeli zadatak)
Adnan Ademović, Bosna i Hercegovina
Admir Beširević, Bosna i Hercegovina
Borna Cicvarić, Hrvatska
Ratko Darda, Bosna i Hercegovina
Gregor Grasselli, Slovenija
Anja Komatar, Slovenija
Matjaž Leonardis, Slovenija
Nikola Milinković, Crna Gora
Filip Talimdžioski, Makedonija

Čestitke apsolutno svima!

izvinite ako se ponavljam, ALI veliki je uspeh stici do IMO-a. zato, cestitke SVIMA, a posebno osvajacima medalja i NARAVNO MOM SINU  :rotfl:

Cestitke svima! (najvise Teodoru :) )

Kako ne volim ono bod do sledece nagrade... Vukasinu i Stefanu falio bod za srebro i bronzu?

Čestitam i ja svima, posebno teu :)
 čestitam i jeleni za medalju osvojenu za BiH team, iako je vidljiva tendencija pada broja medalja i ukupnog plasmana našeg tima...

Svaka cast Tediju. I svima ostalima.
Sta ces Veca bez tebe ne mozemo nista

u teodorovo ime se svima zahvaljujem!
neka i ova IMO bude podstrek onima koji rade sa talentima da podmlade svoje ekipe:
- Raul Arturo (11 god.), Peru - bronza - prvi dan 2 sedmice (drugi dan se siroce umorio :)
- Omer (14 god.), Rumunija - zlato - zahvaljujuci talentu, ali i predanom radu Dana Schwarza
mi smo se silno namucili da teodoru obezbedimo propisnu pripremu u toku CELE godine (a ne dve nedelje pred olimpijadu). zato vas molim kao roditelj da ne zapostavljate MLADE talente.

jos jednom cestitke svim takmicarima i ono sto se obicno propusta cestitke liderima i zamenicima, posebno KRTINICU I DJUKICU!

Omer je sedmak... Bio je sa mnom na JBMO sad... I zlato na IMO... To je pravi rad i talenat (nista bez kombinacije).

bugi, znas li mozda gde mozemo da vidimo ko je dobio specijalne nagrade.
cak ni na zvanicnom sajtu IMO-a, ne mogu to da nadjem za pojedinu godinu.
jedino se javlja u hall of fame.

A ja to hteo da pitam nekoga :) Ne znam...

Zanimljivo, nigdje ne vidim da postoji takav registar... Bilo bi zanimljivo pokrenuti temu u IMO podforumu u svrhu pravljenja takvog registra, jer bi bio useful addition trenutnom imo-official siteu.

prema krtinicu i djukicu teovo resenje 3-eg zadatka je bilo ekstra lepo. no, znamo (ko zna :)) da kod dodeljivanja specijalnih nagrada ima i malo politiziranja. posto mene kao majku obicnu ignorisu, ajte vi matematicari pokrenite akciju da se u rezultatima prikaze ILI ako se one dodeljuju dan posle, na poslednjem sastanku zirija, onda da se kasnije zvanicnim rezultatim DODA i ta kolona - jer je cak i meni kao laiku jasno da ona ukazuje na posebne talente. hvala!

A da li je uopće iko dobio posebnu nagradu u zadnjih 4 godine, nakon Iure 2005? Za 2006 i 2007 sam poprilično siguran da nije bilo posebnih nagrada...

evo, stigli su momci, pa je stigo i odgovor - nije bilo specijalnih nagrada.
inace, za treci zadatak jos jedan momak je dao resenje kao teo, i tako  :maybe:
ima vremena  :)
pozdravi svima do sledeceg IMO-a!

Sa malo većim zakašnjenjem čestitam svima  :)


Some time ago I found such a task:

We can construct bracket expressions, i.e. $[\ [\ ]\ ]\ [\ [\ [\ ]\ ]\ [\ ]\ ]$ and $[\ [\ [\ ]\ [\ ]\ ]\ ]\ [\ ]\ [\ [\ ]\ ]$ are proper bracket expressions. $[\ [\ [\ ]\ [\ ]\ ]\ ]\ [\ ]\ ]\ [\ [\ ]\ ]$ is improper one. Now we are given the $n$ (where $2*n$ is the length of bracket expression) and $k$ numbers $0<s_1<s_2\ldots s_k\leq 2n$ - that means that in bracket expression on positions $s_i$ there are opening brackets. The question is: how many proper expressions of length $2n$ there is, if opening brackets are on posiotions $s_1\ldots s_k$.

I can see that the number of all proper expressions of length $2n$ is Catalan number. So for $n=18$ (so length of bracket expression is $36$), $k=1$ and $s_1=1$ (all proper expressions have opening bracket on position $1$) the answer is $477638700$ (Catalan number).

But for input: $n=10$, $k=3$, $[s_1=5, s_2=10, s_3=15]$ the answer is $1966$. As far as I know - solution should be $O(2n*n)$. Can anyone give me a hint how to solve it?

Any help appreciated...

This problem can be solved using
dynamic programming in O(2n*n)
what you do is create an arran of
a[2*n][n] and here a(i,j) must be numer of possible
expressions of length i such
that there are j open brackets more than number of closed
brackets now initialle a[1][1] is equal to 1 and all others are 0 in the first row .. now you can build the solution bottom up .. ( think about it .. ) .. and at last just pring a[2*n][0] .. I think this should be sufficient .. I think it's clear now ..


inequalities
function
inequalities proposed

If $a,b,c$ are positive numbers such that $a+b+c=3$, then
$\frac{a}{3a+b^{2}}+\frac{b}{3b+c^{2}}+\frac{c}{3c+a^{2}}\le \frac 3{4}$.

The inequality is equivalent to
\[\sum\frac{b^{2}}{b^{2}+3a}\ge \frac{3}{4}\]
By Cauchy Schwarz inequality, we have
\[LHS \ge \frac{(a^{2}+b^{2}+c^{2})^{2}}{\sum a^{4}+(a+b+c)\sum ab^{2}}\]
It suffices to prove
\[4(a^{2}+b^{2}+c^{2})^{2}\ge 3\sum a^{4}+3\sum a^{2}b^{2}+3\sum ab^{3}+3\sum a^{2}bc \]

\[\Leftrightarrow (a^{2}+b^{2}+c^{2})^{2}-3\sum ab^{3}+3(\sum a^{2}b^{2}-\sum a^{2}bc) \ge 0 \]
By VasC's inequality, we have
\[(a^{2}+b^{2}+c^{2})^{2}-3\sum ab^{3}\ge 0 \]
By Am -GM inequality,
\[\sum a^{2}b^{2}-\sum a^{2}bc\ge 0 \]
We are done. :)

I will try the same think like in manlio's ineguality.I still don't know if it's corect, so forgive me if i'm mistaken :D
Take the function $f(x,y)=\frac{x}{3x+y^{2}}$ where we have that $f$ is concave by calculating the second derivate of this function.
So,by aplying Jensen, we obtain:\[\sum_{cyc}f(a,b)\leq3f(\frac{a+b+c}{3},\frac{a+b+c}{3})=3f(1,1)=\frac{3}{4}\] the conclusion!


Not exactly.
It should be $\sum_{i=0}^{\infty}\frac{1}{(4x+1)(4x+2)(4x+3)(4x+4)}$.
Anyway, the way to solve it should be very similar.


Whenever someone hides something, or link to a URL, for some reason, since today I cannot open it. When I click I immediately go to the top of the page and then nothing happens. The link turns purple but nothing shows up.

EDIT- Not just on AoPS, but everywhere! I can't click on the chat button and notifications button in FaceBook either! I use IE8 with Vista Home Premium SP2.

Then it would be a problem with your computer. Try restarting or even reinstalling IE8.

For the Facebook, it happens to me almost always, and I just open a new window with the same page, and then everything normally works. Although I use Google Chrome...


vector
superior algebra
superior algebra solved

A question: In a ring, is every left zero divisor a right zero divisor?

I find in some books, a zero divisor is exactly left zero divisor in other books... So I ask this question.

No.

For an example, let $S$ be a vector space of infinite sequences (any of the standard ones will do) and $R$ the ring of endomorphisms of $S$.

Let $L$ be the left-shift operator: $L(a_0,a_1,a_2,\dots)=(a_1,a_2,a_3,\dots)$ and $P$ be projection onto the first component: $P(a_0,a_1,a_2,\dots)=(a_0,0,0,\dots)$. Then $LP=0$, but since $L$ is surjective $xL\neq0$ for all $x\neq0$.
To get the reverse behavior, consider the right-shift operator $R$: $R(a_0,a_1,a_2,\dots)=(0,a_0,a_1,a_2\dots)$. $PR=0$, but since $R$ is injective, $Rx\neq 0$ for all $x\neq 0$.

This is a very useful example to remember- it's also where you get elements that are left-invertible or right-invertible but not both.


Let $ A,B\in M_n(\mathbb{R})$ with $ n$ odd and $ A^2 \plus{} B^2 \equal{} I_n$. Show that $ det(AB \minus{} BA) \equal{} 0$.

Let $ U\equal{}A\plus{}iB$ and $ n\equal{}2p\plus{}1$ . $ U\bar{U}\minus{}I\equal{}i(BA\minus{}AB)$. $ det(U\bar{U}\minus{}I)\equal{}det(\bar{U}U\minus{}I)$ and thus, is a real number. $ det(i(BA\minus{}AB))\equal{}i(\minus{}1)^pdet(BA\minus{}AB)$ is a real number and $ det(BA\minus{}AB)\equal{}0$.


calculus
integration
rotation
complex analysis

I need to evaluate:

integral dx/(x-a) rotated counterclockwise around abs(z) = 1 where a is any constant such that abs(a) does not equal 1.

I believe that if abs(a) < 1, then I can use the Cauchy-Integral formula and the integral is 2pi*i.
If abs(a) > 1, then Cauchy-Goursat says the integral is zero.

Is this true or am I missing something?


calculus
integration
logarithms
function
calculus computations

substituition $ x \equal{}e^y$ give the definition of incomplete gamma function, (with some manipulations)

http://en.wikipedia.org/wiki/Incomplete_gamma_function

with some manipulation we can get $ \Gamma(a,x)\equal{}(a\minus{}1)!e^{\minus{}x}\sum_{k\equal{}0}^{a\minus{}1}\frac{x^k}{k!}$ and make a substituition on the integral.

[quote="kunny"]Prove that $ \frac {( \minus{} 1)^n}{n!}\int_1^2 (\ln x)^n\ dx \equal{} 2\sum_{k \equal{} 1}^n \frac {( \minus{} \ln 2)^k}{k!} \plus{} 1$.[/quote]
we have, using straightforward integration,
$ \int_1^2 x^p\;dx\equal{}\frac{2^{p\plus{}1}}{p\plus{}1}\minus{}\frac{1}{p\plus{}1}\;\;\;\quad\quad \text{for}\;\;\;p>\minus{}1$

also,
$ \frac{\partial ^n}{\partial p^n}\;\int_1^2 x^p\;dx\equal{}\int_1^2 x^p\cdot(\ln x)^n\;dx\;\;\;\quad\quad\ldots\;\;\;(1)$

and,
$ \frac{\partial ^n}{\partial p^n}\;\left(\frac{2^{p\plus{}1}}{p\plus{}1}\minus{}\frac{1}{p\plus{}1}\right)\equal{}\mho\;\minus{}\;\frac{(\minus{}1)^n\cdot n!}{(p\plus{}1)^{n\plus{}1}}\;\;\;\quad\quad\ldots\;\;\;(2)$

where $ \mho\equal{}\frac{\partial ^n}{\partial p^n}\;\left(\frac{2^{p\plus{}1}}{p\plus{}1}\right)$

now observe
$ \frac{\partial }{\partial p}\;\left(\frac{2^{p\plus{}1}}{p\plus{}1}\right)\equal{}\ln\,2\;\frac{2^{p\plus{}1}}{p\plus{}1}\minus{}(1)\frac{2^{p\plus{}1}}{(p\plus{}1)^2}$

$ \frac{\partial ^2}{\partial p ^2}\;\left(\frac{2^{p\plus{}1}}{p\plus{}1}\right)\equal{}(\ln\,2)^2\frac{2^{p\plus{}1}}{p\plus{}1}\minus{}(2)\ln\,2\frac{2^{p\plus{}1}}{(p\plus{}1)^2}\plus{}(2\times1)\;\frac{2^{p\plus{}1}}{(p\plus{}1)^3}$

$ \frac{\partial ^3}{\partial p ^3}\;\left(\frac{2^{p\plus{}1}}{p\plus{}1}\right)\equal{}(\ln\,2)^3\frac{2^{p\plus{}1}}{p\plus{}1}\minus{}(3)(\ln\,2)^2\frac{2^{p\plus{}1}}{(p\plus{}1)^2}\plus{}(3\times2)\;\ln\,2\frac{2^{p\plus{}1}}{(p\plus{}1)^3}$
$ \minus{}(3\times2\times1)\;\frac{2^{p\plus{}1}}{(p\plus{}1)^4}$

$ \frac{\partial ^4}{\partial p ^4}\;\left(\frac{2^{p\plus{}1}}{p\plus{}1}\right)\equal{}(\ln\,2)^4\frac{2^{p\plus{}1}}{p\plus{}1}\minus{}(4)(\ln\,2)^3\frac{2^{p\plus{}1}}{(p\plus{}1)^2}\plus{}(4\times3)(\ln\,2)^2\frac{2^{p\plus{}1}}{(p\plus{}1)^3}$
$ \minus{}(4\times3\times2)\;\ln\,2\frac{2^{p\plus{}1}}{(p\plus{}1)^4}\plus{}(4\times3\times2\times1)\;\frac{2^{p\plus{}1}}{(p\plus{}1)^5}$

$ \ldots\;\;\;\ldots$

$ \ldots\;\;\;\ldots$

and so,
$ \mho\equal{}\frac{\partial ^n}{\partial p ^n}\;\left(\frac{2^{p\plus{}1}}{p\plus{}1}\right)\equal{}\sum_{k\equal{}1}^{n\plus{}1}\;\left\{\frac{(\minus{}1)^{k\plus{}1}\cdot(\ln\,2)^{n\minus{}k\plus{}1}\cdot 2^{p\plus{}1}}{(p\plus{}1)^k}\;\frac{n!}{(n\minus{}k\plus{}1)!}\right\}$

plug this in $ (2)$, then put $ p\equal{}0$ inside both $ (1)$ and $ (2)$, equate the two and simplify to get,

$ \int_1^2 (\ln x)^n\;dx\equal{}2\sum_{k\equal{}1}^{n\plus{}1}\;\left\{(\minus{}1)^{k\plus{}1}\cdot(\ln\,2)^{n\minus{}k\plus{}1}\;\frac{n!}{(n\minus{}k\plus{}1)!}\right\}\minus{}(\minus{}1)^n\cdot n!$

i.e.
$ \frac{(\minus{}1)^n}{n!}\int_1^2 (\ln x)^n\;dx\equal{}2\sum_{k\equal{}1}^{n\plus{}1}\;\left\{(\minus{}1)^{n\plus{}k\plus{}1}\cdot(\ln\,2)^{n\minus{}k\plus{}1}\;\frac{1}{(n\minus{}k\plus{}1)!}\right\}\minus{}1$

$ \equal{}2\sum_{k\equal{}1}^{n\plus{}1}\;\left\{(\minus{}\ln\,2)^{n\minus{}k\plus{}1}\;\frac{1}{(n\minus{}k\plus{}1)!}\right\}\minus{}1$

$ \equal{}2\sum_{k\equal{}1}^{n}\;\left\{(\minus{}\ln\,2)^{n\minus{}k\plus{}1}\;\frac{1}{(n\minus{}k\plus{}1)!}\right\}\plus{}1$


Let $ABCD$ be a square with $AB=a,AC \cap BD=\{0\}$ and let $M,N,P,Q,R,S,T$ be the middlepoints of $AD,OB,OC,OD,OA,DC$ and
$BC$ respectively. If $\{E\}=QP \cap SR,\{F\}=MN \cap SR,\{G\}=MN \cap TR,\{H\}=PN \cap TR$,$\{I\}=QP \cap SO$,$\{K\}=HO \cap SR,\{L\}=BO \cap TR$ then show that:
a) $HO \bot RS$;
b) $IL \bot RT$;
c) Area of the $EFGHP$ is $\frac{3a^2}{20}$

 :ninja:

Maybe there is a typo:
S, T are both the midpoint of DC?

sorry for my mistake , i've edited my problem. $S \in (DC)$ and $T \in (BC)$ (middlepoints)


Solve in nonnegative integers the equation $ (1\plus{}x!)(1\plus{}y!)\equal{}(x\plus{}y)!$

$ (1 \plus{} x!)(1 \plus{} y!) \equal{} (x \plus{} y)!$

If $ x,y > 2$ then $ xy > x \plus{} y$ then $ LHS > RHS$
$ x \equal{} 1$
$ 2 (1 \plus{} y!) \equal{} (1 \plus{} y)!$
$ 2 \plus{} 2 y! \equal{} y! (1 \plus{} y)$
$ 2 \plus{} y! \equal{} y y!$ then $ y \le 2$
only solution for $ y \equal{} 2$

$ x \equal{} 2$
$ 5 (1 \plus{} y!) \equal{} (2 \plus{} y)!$
$ 5 (1\plus{} y!) \equal{} (1\plus{}y)! (2 \plus{} y)$ then $ y \le 3$
solution for $ y \equal{} 1$

It is quite easy to solve.
Let $ x\leq y$ then
$ 1\plus{}x!\plus{}y!\plus{}x!\plus{}(xy)!\equal{}(x\plus{}y)!$
So $ x!|1$
It mean that $ x\equal{}1$
So $ 2y!\plus{}2\equal{}(y\plus{}1)!$
It mean that $ y|2$
Check for $ y\equal{}1$


Prove that with 4 any real-number we always have 2 real-number x;y satisfying:
       $ 0\leq\frac{x\minus{}y}{1\plus{}xy}\leq 1$

[quote="karis"]Prove that with 4 any real-number we always have 2 real-number x;y satisfying:
       $ 0\leq\frac {x \minus{} y}{1 \plus{} xy}\leq 1$[/quote]

 We can find $ 0 \leq d\leq c\leq b\leq a\leq 90^{0}$  angles satisfying  ---->$ x\equal{}tana;y\equal{}tanb;z\equal{}tanc;t\equal{}tand$---->    $ \frac {x \minus{} y}{1 \plus{} xy}\equal{}\frac{tana\minus{}tanb}{1\plus{}tana*tanb}\equal{}tan(a\minus{}b)$----> $ 0\leq tan(a\minus{}b)\leq 1$ ---> $ 0\leq a\minus{}b \leq 45^{0}$

Let's consider that we can't find 2 angles that their minuse is less than 45^{0}
$ a\minus{}b\geq 45^{0}$
$ b\minus{}c\geq 45^{0}$
$ c\minus{}d\geq45^{0}$----> $ (a\minus{}b)\plus{}(b\minus{}c)\plus{}(c\minus{}d)\geq 45^{0}\plus{}45^{0}\plus{}45^{0}\equal{}135^{0}$---> $ a\minus{}d\geq135^{0}$.   So our consideration is false. .Also x,y is always found satisfying the question


easy qn  , but i am messing up ........ big time !!

the answer shud be $ \frac{2}{3} ms^{\minus{}2}$

since $ 6 \equal{} 3v \plus{} 6v$ :D
but i m sure this can't be ur doubt!!!! :P  :D

[quote="rituraj007"]the answer shud be $ \frac {2}{3} ms^{ \minus{} 2}$
[/quote]

Your answer is certainly wrong!  :huh:

[hide="because"]its dimensionally incorrect  :P :rotfl: [/hide]


function
algebra
functional equation
algebra unsolved

[quote="new_member"]Find all the functions $ f: Z \minus{} > Z$ such that:

           $ f((x^2 \minus{} y^2)^2) \plus{} f((2xy)^2) \equal{} f((x^2 \plus{} y^2)^2)$.

We can easily see that the functions $ f(x) \equal{} cx$ where c is constant satisfy the conditions of the problems but
how do we prove that those are the only ones?[/quote]

There are obviously infinitely many such functions :

Since the functional equation involves only perfect squares, the function may take any values we want for non perfect squares entries.

For example
$ f(x) \equal{} [100\sin(x^2)]$ $ \forall x < 0$
$ f(x) \equal{} x^9$ if $ x\ge 0$ is not a perfect square
$ f(x) \equal{} 2x$ if $ x$ is a perfect square

...


trigonometry
inequalities unsolved
inequalities

If $ a,b,c$ are positive reals then prove that :
$ |\frac{a\minus{}b}{a\plus{}b\plus{}2c} \plus{} \frac{b\minus{}c}{b\plus{}c\plus{}2a}\plus{} \frac{c\minus{}a}{c\plus{}a\plus{}2b} | < \frac{1}{8}$

Let $ \frac{b+c}{a+b}=\tan \alpha ,\ \frac{c+a}{b+c}=\tan \beta ,\ \frac{a+b}{c+a}=\tan \gamma (0<\alpha ,\ \beta ,\ \gamma <\frac{\pi}{2})$,

$ |\frac{a-b}{a+b+2c}+\frac{b-c}{b+c+2a}+\frac{c-a}{c+a+2b}|=|\tan (\alpha +\frac{\pi}{4})+\tan (\beta +\frac{\pi}{4})+\tan (\gamma +\frac{\pi}{4})|$.

[quote="arqady"][quote="IWW"]If $ a,b,c$ are positive reals then prove that :
$ |\frac {a \minus{} b}{a \plus{} b \plus{} 2c} \plus{} \frac {b \minus{} c}{b \plus{} c \plus{} 2a} \plus{} \frac {c \minus{} a}{c \plus{} a \plus{} 2b} | < \frac {1}{8}$[/quote]
$ |\frac {a \minus{} b}{a \plus{} b \plus{} 2c} \plus{} \frac {b \minus{} c}{b \plus{} c \plus{} 2a} \plus{} \frac {c \minus{} a}{c \plus{} a \plus{} 2b} | \leq \frac {1}{22}$ is true too. :wink:[/quote]
Let : $ {a}\ge {b} \ge {c}$ then $ LHR \equal{} \frac {( a \minus{} b)(b \minus{} c)(a \minus{} c)}{(2a \plus{} b \plus{} c)(a \plus{} 2b \plus{} c)(a \plus{} b \plus{} 2c)}$;
$ Max(LHR)$ - then $ c \equal{} 0 ;$   Let : $ a \plus{} b \equal{} 1 , b \equal{} y , a \minus{} b \equal{} z ; \equal{} > 2y \plus{} z \equal{} 1;$
$ max(LHR) \equal{} max(\frac {yz(y \plus{} z)}{(1 \plus{} y)(1 \plus{} y \plus{} z)})$ then $ 2y \plus{} z \equal{} 1 < \equal{} > max(LHR) \equal{} max(\frac {y(1 \minus{} 2y)(1 \minus{} y)}{(1 \plus{} y)(2 \minus{} y)})$    then $ y\le {1/2}$;
$ Max(LHR)$ will be then $ b \equal{} \frac {1 \minus{} \sqrt {13 \minus{} 4\sqrt {10}}}{2} , a \equal{} \frac {1 \plus{} \sqrt {13 \minus{} 4\sqrt {10}}}{2}$;
$ Max(LHR) \equal{} \frac {\sqrt {13 \minus{} 4\sqrt {10}}(\sqrt {10} \minus{} 3)}{\sqrt {10} \minus{} 1} < {1/22}$; :)

[quote="thanhnam2902"]Hi arqady! I think you are wrong. Let $ P \equal{} |\frac {a \minus{} b}{a \plus{} b \plus{} 2c} \plus{} \frac {b \minus{} c}{b \plus{} c \plus{} 2a} \plus{} \frac {c \minus{} a}{c \plus{} a \plus{} 2b} |$ $ \Longrightarrow P \boxed{ < } \frac {1}{22}$. You give us $ P \boxed{\le} \frac {1}{22}$.[/quote]
Why I am wrong? $ P\leq\frac {1}{22}$ is true because $ P < \frac {1}{22}$ is true. :wink: :P

[quote="Sergic Primazon"]
$ Max(LHR) \equal{} \frac {\sqrt {13 \minus{} 4\sqrt {10}}(\sqrt {10} \minus{} 3)}{\sqrt {10} \minus{} 1} < {1/22}$; :)[/quote]
Actually, $ 13 \minus{} 4\sqrt {10} \equal{} (2\sqrt{2}\minus{}\sqrt{5})^2$, and the expression above can be simplified:
$ Max(LHR) \equal{} \frac {\sqrt {13 \minus{} 4\sqrt {10}}(\sqrt {10} \minus{} 3)}{\sqrt {10} \minus{} 1} \equal{}\frac {(2\sqrt{2}\minus{}\sqrt{5})(\sqrt {10} \minus{} 3)}{\sqrt {10} \minus{} 1}\equal{}$
$ \equal{}\frac{7\sqrt{5}\minus{}11\sqrt{2}}{\sqrt {10} \minus{} 1}\equal{}\frac{(7\sqrt{5}\minus{}11\sqrt{2})(\sqrt {10} \plus{} 1)}{9}\equal{}\frac{8\sqrt{2}\minus{}5\sqrt{5}}{3}.$  :P


assuming that the chances of each final score are equal, the answer is [hide]somebody always gets 6, so there are 6 options for the other participant's score $(0-5)$.  the only time that the score is not more than six is if the other participant does not find anything, so $\frac{5}{6}\approx 83\%$[/hide]

Is the answer [hide] $0$? [/hide] :maybe:

I think I'm right if it counts both people's scores, but individually $0$ is right

[quote="perfect628"]Ya, That's what it seems to be. I think the wording is wrong.[/quote]
No, it is impossible to get more than six. He also didn't say anything about the probability of getting 0-5. 0 could be twice as likely as 1.

[quote="Ignite168"]To get the Golden Power of Veto, you must dumpster dive and find 6 veto symbols before anyone else does in the dump. Once someone finds 6 symbols, the game is immediately over. What is the probability that if you pick two competitors at random, one of the number of vetoes they will have found at the end of the competition will be more than 6?[/quote]
I saw that big brother episode.... :D

I loved that episode!

<3 Janelle <3. (I'm a big BB fan).
[hide]
Back on topic: How many vetoes are there? Are the vetoes individually numbered (1 to x), or is 6 the number of vetoes? Will Danielle get Janelle evicted this week?[/hide]

[quote="Fanatic"]I loved that episode!

<3 Janelle <3. (I'm a big BB fan).

Back on topic: How many vetoes are there? Are the vetoes individually numbered (1 to x), or is 6 the number of vetoes? Will Danielle get Janelle evicted this week?[/hide][/quote]

Most likely, I doubt that S6 will get another HOH.

It should say the 2 competitors combined and yes, getting 1 veto is as likely as getting 0 or 3 but getting 6 vetoes has 0 probability if someone else already has 6.

Boogie, Marcellas, Danielle, Janelle, and James will be competiting in the veto competition and George has the Get Out of Slop Pass now.

[quote="Fanatic"]Guess who Danielle nominated Janelle with as a pawn!

[hide]JAMES! LOL. Howie's nom of James + Sarah 2.0?[/hide][/quote]

REALLY!?I haven't gotten the 24 hour thing this season yet. I should go do that now.

I get live feed recaps from a site, and it's pretty frequently updated (like once every minute during lively conversations and such).

Edit: Only five veto competitors now?

[quote="Ignite168"]It should say the 2 competitors combined and yes, getting 1 veto is as likely as getting 0 or 3 but getting 6 vetoes has 0 probability if someone else already has 6.
[/quote]
does that mean my answer is correct? :wow:

[quote="Fanatic"]GUESS WHO WON POV!!!!

[hide]JANEY! YES!!!![/hide]

Wow, this topic has gone so far off topic. I feel guilty :(.[/quote]
you want janey to win? I want dr. will to win. :D

does this really belong in the mathcounts section???

Since this topic has gone so off topic, I'll just reveal the answer and hope it dies.

[hide]$\frac{1}{2}$[/hide]

Why? Could you post a solution?  :maybe:


Se considera numerele pozitive a,b,c,d care au suma egala cu 1.Sa se demonstreze inegalitatea:
bcd/(a+2)+ acd/(b+2) + abd/(c+2) + abc/(d+2)< 1/13

Woohoo.Am gasit solutia:
Din medii avem $bcd\leq(\frac{b+c+d}{3})^{3}=\frac{(1-a)^{3}}{27}$ rezulta ca $\frac{bcd}{a+2}\leq\frac{(1-a)^{3}}{27(a+2)}$.Iar cum $a\geq0 \Rightarrow 27(a+2)\geq 27\cdot 2$ de unde rezulta ca $\frac{(1-a)^{3}}{27(a+2)}\leq\frac{(1-a)^{3}}{27\cdot 2}\leq\frac{1}{27\cdot2}$[Caci $1-a\leq 1$].
Prin urmare am aratat ca $\frac{bcd}{a+2}\leq\frac{(1-a)^{3}}{27(a+2)}\leq \frac1{27\cdot 2}$,rezulta ca si ceilalti termeni vor fi la fel,rezulta ca toata suma noastra va fi$<\frac4{2\cdot 27}<\frac1{13}$
Final.
PS:Am scris cam in graba..sper ca se intelege ideea de baza :)

e mai mic strict acolo ? oare care e cea mai buna estimare, eventual cu $\leq$ ? (acum ma gandesc la asta)

Este mai mic strict pentru ca nu toate numerele a,b,c,d poot fi egale cu 0.
Eu cred ca nu e cea mai buna estimare,dar poate $\frac{2}{27}$ e cea mai buna estimare cu functii elementare.Cred ca maximul expresiei date are o valoare irationala,si nu cred ca se poate exprima intr-o forma elementara.Dar e doar o banuiala :wink:


Two particles are projected upwards from a cliff with velocities v and 2v

The graph relative motion of the second particle with respect to the first

particle as time increases until they reach the ground is

a)parabolically       b)parabola and linearly  c)linearly   d)linearly and para

bolically

ans is leniar. because each partical posses a constnt VELOCITY in the other's frame

Ground reference frame:
$v_{1}(t) = v-gt$
$v_{2}(t) = 2v-gt$

Moving into reference frame of (1) and using relative velocities

$v_{2\rightarrow1}= v_{2\rightarrow ground}-v_{1\rightarrow ground}$
$\Rightarrow v_{2\rightarrow1}= 2v-gt-(v-gt)=v$


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ \mathbb{Z}[\alpha]$ be an extension ring of $ \mathbb{Z}$ and let $ f$ be the monic irreducible polynomial for $ \alpha$ over $ \mathbb{Z}$. $ \overline{g}\in \mathbb{Z}_{p}[X]$ denotes the image of $ g \in \mathbb{Z}[X]$ in $ \mathbb{Z}_{p}[X]$ with $ p$ prime.
Show that $ p \mid g(\alpha)$ in $ \mathbb{Z}[\alpha]$ iff $ \overline{f} \mid \overline{g}$ in $ \mathbb{Z}_{p}[X]$.

$ \mathbb{Z}_p$ : do you mean the ring of p-adic integers? for the field of p elements the claim is wrong.

As the finite field is a quotient of the $ p$-adics, it's wrong in $ p$-adics, too  :wink:

By the way (@all): Do [u]not[/u] use $ \mathbb Z_p$ for the field with $ p$ elements. That notations is ridiculous, useless and overloaded. Use $ \mathbb F_p$ or $ \mathbb Z / (p)$ instead, both being much more meaningful and they are unique. $ \mathbb Z_p$ are the $ p$-adic integers, nothing else (well, it could be some localisation at $ p$, but then again this collision is easily avoided and especially not really harmful)!

Sorry, I mean $ \mathbb{F}_{p}$. What about if $ \alpha$ is an algebraic integer of a number field?

It was already ba definition an algebraic integer (you assumed that it has a monic minimal polynomial over the integers). The problem you face is that $ f$ may not be irreducible $ \mod p$. For an example, take $ x^4\plus{}1$.

Do you claim to have a counterexample? If so, could you please elaborate.


Ring Theory
LaTeX
number theory
superior algebra
superior algebra solved

How do we prove that $\displaystyle \mathbb Z \left[ \sqrt{d} \right]$ is an UFD for a particular $d$ like $5$ or $-1$?

I'm especially interested in the case $\mathbb Z \left[ i \right]$, because I'm trying to understand a solution by grobber...

Generally I think this is an open problem. Again generally we work with the ring of integers of $\mathbb Q[\sqrt d]$ for $d$ a square free integer. This ring is \[ A=\left\{\begin{array}{cc}\mathbb Z[\sqrt d],&\mbox{if }d\equiv 2,3(mod\ 4)\\\mathbb Z\left[\frac{1+\sqrt d}2\right],&\mbox{if }d\equiv 1(mod\ 4)\end{array}\right..  \]
On the ring $A$ there is a multiplicative norm $\varphi: A\to\mathbb N$ which extends to a multiplicative norm on $\mathbb Q[\sqrt d]$. Let $\omega\in\{\sqrt d,\frac{1+\sqrt d}2\}$ be the generator of $A$ as an extension of $\mathbb Z$. $\omega$ verifies a second degree equation with integer coefficients of the form $x^2+ax+b=0$. Let $\omega^*=-a-\omega=\frac b{\omega}$. Then the norm on $A$ is given by $\varphi(u+\omega\cdot v)=|(u+\omega v)(u+\omega^*v)|$.

With this norm you sometimes can prove that $A$ is euclidean. This is the case for and only for $d\in\{-1, -2, -3, -7, -11,2,3,5,6,7,11,13,17,19,21,29,33,37,41,57,73\}$.

Sometimes you can prove that $A$ is principal. For this we use:
[b]Hasse-Dedekind:[/b] Let $A$ be a subring of $\mathbb C$ and let $\varphi: A\to\mathbb N$ such that:
i)$\varphi(x)=0\Leftrightarrow x=0$.
ii) For all $x,y\in A$ such that $x\not| y$ there exist $a,b\in A$ such that $0<\varphi(ax+by)<\varphi(x)$.
Then $A$ is a principal ideal domain.

In our case we usually take $\varphi$ to be our multiplicative norm and then we can replace (ii) in the theorem by
ii') For all $x,y\in A$ such that $x\no|y$ there exist $\alpha,\beta\in\mathbb Q[\sqrt d]$ such that $0<\varphi(\alpha+\beta\cdot\frac yx)<1$.

If the above doesn't work either, we usually go to Algebraic Number Theory for help. There we can compute the ideal class group, and we have that $A$ is an UFD iff the ideal class group has one element iff $A$ is a principal ideal domain.
The ideal class group is obtained as follows:
-As a set is is $\mathcal I/\sim$, where $\mathcal I$ is the set of nonzero ideals of $A$ and $\sim$ is the equivalence relation given by $I\sim J$ iff there exist $a,b\in A$ such that $aI=bJ$.
-The group operation is given by $\hat I\cdot\hat J=\widehat{I\cdot J}$.
It is easy to prove that $\sim$ is an equivalence relation. It is not as easy to prove that "$\cdot$ " is a group operation.

Good luck with these :)
Fortunately $\mathbb Z[\imath]$ falls in the first category.

[i][Edit] An incomplete $\text{\LaTeX}$ Formula corrected[/i]

[quote="perfect_radio"]You probably meant this:

[quote="amfulger"]Generally I think this is an open problem. Again generally we work with the ring of integers of $\mathbb Q[\sqrt d]$ for $d$ a square free integer. This ring is $A=\left\{\begin{array}{cc} \mathbb Z[\sqrt d] , & \mbox{if } d \equiv 2,3(mod \ 4) \\ \mathbb Z \left[\frac{1+\sqrt d}2\right] , & \mbox{if } d\equiv 1(mod\ 4) \end{array} \right\}$.[/quote]

And, yes, I knew that it was unsolvable generally. (one of the few things i knew about this)[/quote]

In the book I am reading, it shows $Z[i]$ is Euclidean which it follows that Z[i] is PID thus UFD.. I am a very beginner at algebra so I don't know much about any method to determine whether a general ring is UFD, but I think for some famous rings like Z[i], approaches like proving it is Euclidean/PID may work.


What's Wilson's theorem ?
How to prove it?

http://mathworld.wolfram.com/WilsonsTheorem.html
And, of course, learn to use google.com ;)

This webpage http://www.libraryofmath.com/wilson-theorem.html has what you are looking for.

There are some basic properties of Wilson's Theorem with some examples.

[url]http://www.libraryofmath.com/wilson-theorem.html[/url]


induction
algebra
functional equation
algebra solved

Put $m=0$ into (c) and (d), then using (b) we obtain $f(n)+g(0)^2=g(n)+f(0)^2$. Putting $n=0$ we get $f(0)=g(0)$ or $g(0)+f(0)=1$. If $g(0)+f(0)=1$ then $g(0)=0$ and $f(0)=1$ or vice versa. Putting $n=0$ in (b), $g(1)=1$, and putting $n=1$ we arrive at $f(1)=0$. Contradiction to (a). So $f(0)=g(0)$ or $f(n)=g(n)$. If we solve functional equation $f(m^2+f(n))=f(m)^2+f(n)$ we are done. Put $m=0$, so $f(f(n))=f(0)^2+f(n)$ and $f((k-1)f(0)^2+f(n))=kf(0)^2+f(n)$ for all positive integers $k$ (induction). Let $k=f(0)^2+1$. On one hand $f(f(0)^4+f(n))=f(f(0)^2)^2+f(n)$ but on the other $f(f(0)^4+f(n))=f(0)^4+f(0)^2+f(n)$, so $f(f(0)^2)^2=f(0)^4+f(0)^2$ or $(2f(f(0)^2))^2+1=(2f(0)^2+1)^2$ and $f(f(0)^2)$, $f(0)^2$ are non-negative integers. Hence $f(0)=0$. Putting $m=1$ and $n=0$ in $f(m^2+f(n))=f(m)^2+f(n)$ we establish $f(1)=f(1)^2$, therefore $f(1)=1$. Using induction we can go on with $f(n)=n$, just put $m=1$, then $f(1+f(n))=1+f(n)$.


[quote="N.T.TUAN"]Prove that for every couple of positive integers $n,k$ the number $(\sqrt{n}-1)^{k}$ can be written in the form $\sqrt{a_{k}}-\sqrt{a_{k}-(n-1)^{k}}$, with $a_{k}\in\mathbb{N}$.[/quote]
Let $(1+\sqrt n )^{k}=x_{k}+y_{k}\sqrt n$, then $(\sqrt n-1)^{k}=(-1)^{k}(x_{k}-y_{k}\sqrt n )$, therefore
$(n-1)^{k}=(-1)^{k}(x_{k}^{2}-ny_{k}^{2}).$
If k is odd, then $a_{k}=ny_{k}^{2}$ give $\sqrt{a_{k}}-\srt{a_{k}-(n-1)^{k}}=y_{k}\sqrt n-x_{k}=(\sqrt n-1)^{k}$.
If k is even $a_{k}=x_{k}^{2}$ work.


floor function
quadratics
inequalities
number theory
prime numbers
Quadratic Residues

Let $M$ be an integer, and let $p$ be a prime with $p>25$. Show that the set $\{M, M+1, \cdots, M+ 3\lfloor \sqrt{p} \rfloor  -1\}$ contains a quadratic non-residue to modulus $p$.

Ok. I solved it but the second lemma is a bit ugly. maybe someone sees an improvement so we can improve that $ 484$ thing

[b]lemma1[/b]
there exists a quadratic non-residue $ a$ modulo $ p$ which is less or equal to $ \lfloor \sqrt {p} \rfloor \plus{} 1$

Proof:
let $ a$ be the smallest quadratic non-residue mod $ p$. Now define $ b \equal{} \lfloor \frac {p}{a} \rfloor \plus{} 1$. Than $ ab < p \plus{} a$ meaning $ ab \minus{} p < a$ (obviouslly by substituting $ p \equal{} ka \plus{} r$ we easily conclude $ ab > p$). therefore $ ab \minus{} p$ must be a quadratic residue. we have (these are legendre symbols)
\[ 1 \equal{} (\frac {ab \minus{} p}{p}) \equal{} (\frac {ab}{p}) \equal{} (\frac {a}{p})(\frac {b}{p})
\]
and since $ a$ is a non-residue also $ b$ must be.
but than because $ a$ is the smallest $ a\leq b \equal{} \lfloor \frac {p}{a} \rfloor \plus{} 1$ which leads to $ a\leq \lfloor \sqrt {p} \rfloor \plus{} 1$

[b]lemma2[/b]
there exists a quadratic non-residue $ c$ mod $ p$ ($ p > 484$ :blush: ) such that
\[ \frac {p}{3\lfloor \sqrt {p} \rfloor \minus{} 1} \leq c \leq \frac {3\lfloor \sqrt {p} \rfloor \minus{} 1}{2}
\]
Proof:
consider $ a$ (the smallest non-residue mod$ p$). $ a\leq \frac {3\lfloor \sqrt {p} \rfloor \minus{} 1}{2}$ by first lemma (obviously $ \frac {3\lfloor \sqrt {p} \rfloor \minus{} 1}{2}\geq \lfloor \sqrt {p} \rfloor \plus{} 1$ for $ p > 8$)
if $ a\geq \frac {p}{3\lfloor \sqrt {p} \rfloor \minus{} 1}$ than $ a \equal{} c$. But if that is not the case than we consider numbers $ a,4a,4^2a,4^3a...$. these numbers are all quadratic non residues mod $ p$ because legendre's symbol is multiplicative. and more. one of these numbers is between $ \frac {p}{3\lfloor \sqrt {p} \rfloor \minus{} 1}$ and $ \frac {3\lfloor \sqrt {p} \rfloor \minus{} 1}{2}$. that is because $ \frac {3\lfloor \sqrt {p} \rfloor \minus{} 1}{2} > 4\cdot \frac {p}{3\lfloor \sqrt {p} \rfloor \minus{} 1}$ for suficiently large $ p$ and it's therefore imposible for the sequence $ a,4a,4^2a,4^3a...$ to miss this interval.
the inequality is equivalent to $ 8p < 9(\lfloor \sqrt {p} \rfloor)^2 \minus{} 6\lfloor \sqrt {p} \rfloor \plus{} 1$. now we substitute $ p \equal{} x^2 \plus{} d$(where $ d < 2x \plus{} 1$) and we get $ 0 < x^2 \minus{} 22x \plus{} 1$ which is true for all $ x$ grater than $ 21$ mening $ p > 22^2 \equal{} 484$

[b]now for the problem[/b]
assume the statement is false and $ p > 484$. therefore there exist $ 3\lfloor \sqrt {p} \rfloor$ consecutive numbers mod $ p$ that are all quadratic residues (one of them may be zero). now consider the set$ \{cM, c(M \plus{} 1), \cdots, c(M \plus{} 3\lfloor \sqrt {p} \rfloor \minus{} 1) \}$ where $ c$ is such a non-residue as in lemma2.
because the legendre's symbol is multiplicative this now is a set of quadratic non-residues(one of them can be zero) that are at most $ 2c$ appart one from another.(if one is zero). we have $ 2c\leq3\lfloor \sqrt {p}\rfloor \minus{} 1$ by definition of $ c$. therefore it is imposible to pick $ 3\lfloor \sqrt {p}\rfloor$ consecutive numbers such that none of them would be a non-residue starting between $ cM$ and $ c(M \plus{} 3\lfloor \sqrt {p} \rfloor \minus{} 1)$.
however this interval has lenght of $ c(3\lfloor \sqrt {p} \rfloor \minus{} 1)$ which is grater or equal than $ p$ due to definition of $ c$ and lemma2. we have therefore covered the whole modul with non-residues that are at most $ 3\lfloor \sqrt {p} \rfloor \minus{} 1$ appart. it is therefore impossible to choose $ 3\lfloor \sqrt {p} \rfloor$ consecutive numbers which do not contain a non-residue. Q.E.D

all that remains is to check all prime numbers up to $ 484$. which is quite a suferable task. i didn't check them myself but i trust the problem holds and hope someone will now come up with a lower bound than $ 484$

I have proved that the set $ \{M, M\plus{}1, \cdots, M\plus{} 2\lfloor \sqrt{p} \rfloor \plus{}2\}$ contains a quadratic non-residue to modulus $ p$ for all primes $ p$. Is it right?

Yes. if u consider $ 0\mod{p}$ as non-residue that is the best result u can get using the same method as i have done above. i however didn't consider zero as a non-residue.
if u didn't consider $ 0$ as a non-residue than i don't know. u must have used modificated or completely different method. why don't u show us your proof

I know I'm quite late to this question, but since apparently this a better result than the other ones in the thread, here's a proof showing that the set {$M, M+1, ..., M+2\lfloor \sqrt{p} \rfloor$} contains a quadratic non-residue mod p for all odd primes p and integers M, even while considering 0 as neither a quadratic residue or a quadratic non-residue.
To show this, assume $M$ and $k$ are integers such that {$M, M+1, ..., M+k-1$} does not contain a quadratic non-residue. It would therefore be sufficient to show that $k < 2\lfloor \sqrt{p} \rfloor + 1$. To do this we take 2 cases:
Case 1: The set {$M, M+1, ..., M+k-1$} does not contain a value 0 mod p.
In this case, assume we take an integer $t$ such that $\frac{p-k}{k-1} \leq t \leq k$, and consider the arithmetic sequence $Mt, (M+1)t, ..., (M+k-1)t$. By considering the integers split into repeating regions of the residues in the set {$M, M+1, ..., M+k-1$} (length k) and the values not in the set (length p-k), we realize that the arithmetic sequence may not stay within one region of length $p-k$ (since $(M+k-1)t-Mt = t(k-1) \geq p-k$), and it may not "jump" over a region of length k (since the difference between consecutive terms is $t \leq k$), and so it's clear that there exists an element of the sequence that is equivalent to a value in {$M, M+1, ..., M+k-1$} mod p. Therefore there exists 2 elements $a, b$ in {$M, M+1, ..., M+k-1$} such that $at \equiv b \pmod p$, and since both $a$ and $b$ are non-zero quadratic residues mod p by definition, we must have $t$ a quadratic residue as well. This tells us that all integers within the set $[\frac{p-k}{k-1}, k]$ are quadratic residues mod p.
Now for any positive integer $c \leq k$, consider $c, 4c, 9c, ...$. At least one of these terms must exist in the range $[\frac{k+1}{4}, k]$, because if not then there exists a positive integer $n$ such that $cn^2 \leq \frac{k}{4}$ and $c(n+1)^2 \geq k+1$, but then $4 < 4(\frac{k+1}{k}) \leq \frac{c(n+1)^2}{cn^2} = (1+\frac{1}{n})^2 \leq 4$, contradiction. But now if $\frac{p-k}{k-1} \leq \frac{k+1}{4}$, all values in the range $[\frac{k+1}{4}, k]$ are quadratic residues, and since for all $c \leq k$ there exists a positive integer $n$ such that $cn^2$ lies in this range, and is therefore a quadratic residue, c is a quadratic residue, which makes all values within $[1,k]$ quadratic residues. It's been proven in C3 that if this holds then $k < \sqrt{p}$, but $\frac{p-k}{k-1} \leq \frac{k+1}{4}$ implies $k \geq \sqrt{4p+5}-2 \geq \sqrt{p}$, contradiction, and so we must have $k < \sqrt{4p+5}-2 \leq 2\sqrt{p}-1 < 2\lfloor \sqrt{p} \rfloor + 1$, giving the desired result in this case.

Case 2: The set {$M, M+1, ..., M+k-1$} does contain a value 0 mod p.
In this case the set {$M, M+1, ..., M+k-1$} is equivalent to {$-a, -a+1, ..., -1, 0, 1, ..., b-1, b$} mod p for some non-negative integers $a$ and $b$, with $k = a+b+1$.
If $p \equiv 3 \pmod{4}$, $-1$ is a quadratic non-residue, forcing $a = 0$, and since all values within $[1,b]$ being quadratic residues mod p implies $b \leq \lfloor \sqrt{p} \rfloor$, as mentioned earlier, $k = b+1 \leq \lfloor \sqrt{p} \rfloor +1 < 2\lfloor \sqrt{p} \rfloor + 1$, giving the desired result.
If $p \equiv 1 \pmod{4}$, we may improve the lemma to state that if all integers in the range $[1,n]$ are quadratic residues mod p, then $n < \sqrt{p}-1$. To show this, let $t+1$ be the smallest quadratic non-residue mod p, and consider $t+1, 2(t+1), ..., t(t+1)$. These are all clearly quadratic non-residues. As all integers in the range $[p-t,p-1]$ are quadratic residues as -1 is a quadratic residue, and also no term is $p$ as $t+1 < p$ clearly, no term in the range $[p-t,p]$ appears in this arithmetic sequence of difference $t+1$. $t+1$ isn't big enough to jump this region, so clearly the sequence stays below it, forcing $t(t+1) < p-t$, which implies $t < \sqrt{p+1}-1$, and since clearly neither $p$ nor $p+1$ are perfect squares, $t < \sqrt{p}-1$, and since all integers in the range $[1,n]$ are quadratic residues mod p implies $n \leq t$, we must have $n < \sqrt{p}-1$. This therefore implies $n \leq \lfloor \sqrt{p} \rfloor - 1$. And now, since all values from in region $[1,a]$ are quadratic residues mod p (due to multiplying by -1) and all values in region $[1,b]$ are quadratic residues mod p, $a \leq \lfloor \sqrt{p} \rfloor - 1$ and $b \leq \lfloor \sqrt{p} \rfloor - 1$, and so $k = a+b+1 \leq 2\lfloor \sqrt{p} \rfloor - 1 < 2\lfloor \sqrt{p} \rfloor + 1$, giving the desired result.

In every case, {$M, M+1, ..., M+k-1$} consisting of no quadratic non-residues implies $k < 2\lfloor \sqrt{p} \rfloor + 1$, and since {$M, M+1, ..., M+2\lfloor \sqrt{p} \rfloor$} is this set for $k = 2\lfloor \sqrt{p} \rfloor + 1$, it therefore must have a quadratic non-residue mod p, and we're done.


I have been getting only 1-point questions in a row for the past few days, and there were no 2,3, or 4 questions I have encountered. Why is that?

Ive only been getting 2 point questions.
So far, i have answered 153 2-Point questions,
and 1 4 point question, and that's all.
What does that mean???

hmmmmmm i usually get 50% 2 pts, 25% 1 pts, 12.5% 3 pts, and 12.5% 4 pts. (Estimate)

is there a 5 pt question?????

I've been getting approximately 70% 2 pts, 20% 4 pts, >5% 1 pt, and <5% 3 pts.

My points are too unfair. I get 75% 1-point questions, 25% 2-point questions, and NO 3 or 4-point questions.

[quote="AIME15"]It's probably because your Alcumus score is increasing, so the system feels that the questions shouldn't be hard for you.  That happened to me also, even when I encountered AIME problems.[/quote]

Read this.  It's because you are good, so be happy. :lol:

I didn't know you could get 3 points either, but maybe I missed some that were like that, so I put <5% 3 pts in my previous post.  :)

[quote="connaissance"]I didn't even know you could get 3 and 4 points.[/quote]

are there rly any 3 or 4 point questions, because i have answered all of the alcumus problems, and i dont recall seeing any question above 2 points...any word on if they are still offering 3 and 4 point questions?

i've seen three or four point questions before
but all my questions have been worth one point for a long time....

my guess is that alcumus gives the problem points based on how other users rates the problem
if other users rate the problem as easy, it'll get less points
if other users rate the problem as hard, it'll get more points

It is very interesting to note that often there are two or three 4-point questions in a row among a bunch of 2-point questions, suggesting that the system treats those questions as sort of a "midterm evaluation" between those types of questions going from 2-point to 1-point.

oops i always rate questions "very easy"

No 3 pts? Hm.. I swear I saw one before.. What about 5? :D

Um, pit.ike314...your two posts just contradicted each other...?

Before, when I started Alcumus, I got all these 4 and 3 point questions and now I only get 1 point questions and rarely a 2 point...

For instance I keep getting questions like this:

12 balls numbered 1 through 12 are placed in a bin. Joe produces a list of three numbers by performing the following sequence three times: he chooses a ball, records the number, and places the ball back in the bin. How many different lists are possible?

[quote="levans"]Some clarifications:

1. There are no 3 point questions.

2. You are most likely to get 1 and 4 point questions at the very beginning and when you begin to run out of problems.

3. The number of points per problem is based on your projected chance of getting the problem correct. It has nothing to do with how others have rated the problem.[/quote]

based on number 3 in this post

I'm guessing that when you first started, Alcumus didn't really know your ability
therefore, the problems were worth more points
but now that you've gotten better, Alcumus knows that you'll probably be able to get these problems right, so the problems aren't worth as much

I have never got a 1 point questions before
Are they really easy??
I keep on getting 4 point ones
I think if u have been doing bad or good recently, it changes the diffuculty

[quote="MathWhiz2009"]I've never gotten a three or four point question. Is this bad? And what happens when you run out of question?[/quote]

when you run out of questions, Alcumus starts giving you the problems you've missed before
once you finish all those, you've got to wait until the admins add more questions

[quote="MathWhiz2009"]I've never gotten a three or four point question. Is this bad? And what happens when you run out of question?[/quote]
None of the problems are worth 3 points.

ANOUNCEMENT!

Guys, you're repeating the same things over and over.

1) There are NO 3-Point questions! Stop talking about them!!
2) If you get only 2 point questions, it means you are improving.

If you're getting only 2 point questions, you are not improving (at least the Alcumus AI does not think so).  If you're getting mostly 1 point questions, then you are improving.

I'm always getting 4 point and 2 point questions...

Then the Alcumus AI does not think you are improving substantially.

[quote="louiethechewy123"]Does rating problems effect the point amount of problems? And everybody will eventually have all of the same points and questions once they finish alcumus, so what is the point of worrying?[/quote]

first question, no
Alcumus assigns the problem a point amount based on how likely you will get the question right

because of that, i believe the second question is no as well
since Alcumus gives problems different point values based on the person, i believe that if everyone finished all the problems, there still might be a difference of a few points between their total points

And once you get all of the questions correct, you get 1 point questions and keep adding to your point value. The growth of your points never stops.


Let $ S\subseteq X$  where $ X$ Banach space. Prove that $ S$ is sequentially compact if and only if $ \forall \varepsilon >0$ there exists subset $ \{x_1,\dots,x_n\}\subseteq S$ such that $ S\subseteq K(x_1,\varepsilon)\cup\dots\cup K(x_n,\varepsilon)$.

I have proved $ \Leftarrow$, but I can't seem to get the other way around.

Suppose $ S$doesn't satisfy the second property, let's call it "doesn't have finite $ \epsilon$-net".
Take any point $ a_1 \in S$, the there exists $ a_2 \in S$ s.t. $ \parallel{}a_1 \minus{} a_2\parallel{} \geq \epsilon$, in the same way there must exists $ a_3 \in S$ such that $ \parallel{}a_i \minus{} a_3\parallel{} \geq \epsilon$ for $ i\equal{}1,2$, otherwise $ a_1,a_2$ would be the finite $ \epsilon$-net.
Having constructed $ a_1,a_2,...,a_k$ choose $ a_{k\plus{}1}$ such that $ \parallel{}a_i \minus{} a_{k\plus{}1}\parallel{} \geq \epsilon$ for $ i\equal{}1,2,\ldots k$. Then it is clear that the sequence $ (a_n)_n$ doesn't have convergent subsequences, and hence, $ S$ is not sequentially compact.


It's $ (a \minus{} b)\cdot(a \plus{} b)$

It came to me while I was in a car, thinking...
I was stupid enough to forget that... :oops:

[quote="Chinaboy"]Never forget that now![/quote]
I won't... I used that last year, but I forgot it this year..


Find all the integers x and y such that x^2+x^3y = y^3+2y^2.

Set x = da, y = db, where gcd(x, y) = d. d = 0 yields a solution. So consider d != 0.

Then a^2 + d^2 a^3 b = d b^3 + 2 b^2.

But then b | a^2. Suppose b had a prime factor p.

Then p | a^2 -> p | a. Then (a, b) != 1, a contradiction.

So b = 1.

Then a^2 + d^2 a^3 = d + 2.

Then d | a^2 - 2. So d \leq a^2 - 2.
But a^2 | d + 2, so a^2 \leq d + 2.

From above, we see that d = a^2 - 2.

Then d = 0 or a = 0. d != 0, so a = 0.

Finally, we obtain the solutions

(x, y) = (0, 0), (0, -2).

I was rummaging through my olympiads book and I found the question to which you referred, orl - the correct statement is:

Find all integers x, y such that

x^6 + x^3 y = y^3 + 2y^2.

I'm pretty sure the proof is analogous to the one I posted above.


[hide="start here"]
[hide="3.14159"]
[hide="26535"]

[/hide][/hide][/hide]
more digits to come later- mods plz dont delete

Just out of curiousity, did you memorize all of these?  Why?

Er, why are we repeating posts... :?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19128&postorder=asc&highlight=Memorizing+pi&start=30]Pi stuff[/url]

Haha. I know
[hide]
3.14159265 I think that's right.
[/hide]

only because I wonder, what is the reason you're memorizing those all

because we don't know any math at all and still think pi is just that wierd black and white movie by the guy who did requiem for a dream

Website: http://www.freewebs.com/howtomemorizepi

I made it back during spring break a year ago when I was really bored and memorized 80 digits just for ****s and giggles because I had 10,000 digits in a book. It really is a pretty cruddy site.

Last year, when I was in 7th grade i memorized 225 digits of pi for pi day. No joke either. Its the school record. It took me about 5 days though.


Prove that for all $k$ $>$ $3$ there is exist $(a,b,c)$ is the triple Pitagore primitve take $k$ is one of there number $a,b,c$

Is it:

Show that for all $k$ there is a primitive Pythagorean triple $(a,b,c)$ such that $k \in \{a,b,c\}$¿

Then:
Use that all primitive ones are given by $a=m^{2}-n^{2}, b=2mn , c=m^{2}+n^{2}$ for coprime $m,n$, not both odd.


Given $ a,b,c,k\geq0$.
Problem 1:Prove that
$ \frac{a\plus{}b}{\sqrt{ab\plus{}c^2}}\plus{}\frac{b\plus{}c}{\sqrt{bc\plus{}a^2}}\plus{}\frac{c\plus{}a}{\sqrt{ca\plus{}b^2}}\geq4$

Problem 1:Find the min value of
$ P\equal{}\frac{a\plus{}b}{\sqrt{kab\plus{}c^2}}\plus{}\frac{b\plus{}c}{\sqrt{kbc\plus{}a^2}}\plus{}\frac{c\plus{}a}{\sqrt{kca\plus{}b^2}}$

[quote="dduclam"]Given $ a,b,c,k\geq0$.
Problem 1:Prove that
$ \frac {a \plus{} b}{\sqrt {ab \plus{} c^2}} \plus{} \frac {b \plus{} c}{\sqrt {bc \plus{} a^2}} \plus{} \frac {c \plus{} a}{\sqrt {ca \plus{} b^2}}\geq4$

Problem 2:Find the min value of
$ P \equal{} \frac {a \plus{} b}{\sqrt {kab \plus{} c^2}} \plus{} \frac {b \plus{} c}{\sqrt {kbc \plus{} a^2}} \plus{} \frac {c \plus{} a}{\sqrt {kca \plus{} b^2}}$[/quote]

[quote="dduclam"]Given $ a,b,c,k\geq0$.
Problem 1:Prove that
$ \frac {a \plus{} b}{\sqrt {ab \plus{} c^2}} \plus{} \frac {b \plus{} c}{\sqrt {bc \plus{} a^2}} \plus{} \frac {c \plus{} a}{\sqrt {ca \plus{} b^2}}\geq4$
[/quote]
It's the hungkhtn's problem.
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166714

[quote="dduclam"]Ok,arqady! The idea'Hungkhtn chime in with mine :D
And Problem 2?[/quote]
I think, it's very hard problem. See here the private case:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160854
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166714


If $ 1\frac12$ hens lay $ 1\frac12$ eggs in $ 1\frac12$ days, how many eggs will 24 hens lay in 24 days?

besides the fact that 1.5 hens and 1.5 eggs is a somewhat disturbing idea, we just solve: \[ \frac{1.5\cdot 1.5}{1.5}\equal{}\frac{24\cdot 24}{x}\Rightarrow x\equal{}\boxed{384}\]

For reasoning, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253924]here[/url]


FOr the pastr 2 days, except for one time yesterday, when I click on the link to come here the page is shown as lots of random charachters on the blue background... refreshing doesn't do anything either.

Am I the only one that is getting this? :(

eum.. I guess you are?

Cuz I'm online like 24 hours a day (ok not overnight but still...)

really got no clue what you're talking about :?
perhaps can you make a print-screen of what you see? may be useful in helping you...

I got that messeage, several times before, but only using AOL and not IE or Netscape

[quote="Peter VDD"]:o:o:o:o:o:o:o

O_o that's the source code...

anyway I see you're entering through AoPS..

have you tried entering through the normal site (http://www.mathlinks.ro) ?

as far as I know, that one has no errors at present[/quote]

The core of both entries is the same.  I suggest deleting your cookies - we made some changes to cookie handling that might cause this error (from either site).

[quote="rrusczyk"]The core of both entries is the same.  I suggest deleting your cookies - we made some changes to cookie handling that might cause this error (from either site).[/quote]

Hmm no, if you don't even get to the login page it matters in style. SubSilver contains no bugs and is all-round compatible. The AoPS stlye may be more creative, but is certainly not the style to pick if there are problems.

By the way, I'm posting this from artofproblemsolving.com link, since the mathlinks.ro link isn't responding. So I won't tell it's completely the same ;)

[quote="RC-7th"]How would I change my style to subsilver?[/quote]Go to profile, and change style from AoPS to MathLinks. It will switch to SubSilver style.

Thanks for your help, I went through mathslinks and it works fine now
thanks :lol:


function
geometry
3D geometry
number theory proposed
number theory

Find the positive integers $n$ with $n \geq 4$ such that $[\sqrt{n}]+1$ divides $n-1$ and $[\sqrt{n}]-1$ divides $n+1$.
[hide="Remark"]This problem can be solved in a similar way with the one given at [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=97&year=2006]Cono Sur Olympiad 2006[/url], problem 5.[/hide]

Yes, very similar to Cone Sul's one..!

to solve, see the perfect square a² more closer to n, and do the substituition n = a²+r, with r < 2a+1.


induction
superior algebra
superior algebra unsolved

I AM NOT ABLE TO SOLVE THESE QUESTIONS......PLS HELP!!!!!!

1.use the principle of induction to show that the order of Sn is n! for
n1.

2.use induction to prove if A1.....An are n subsets of S then S\(union
of all subsets of S)=intersection of(S\all subsets of A).

3.use the well ordering principle to prove that every ideal of Z is of
the
form nZ, for some non-negative integer n.

--------------------------------------------------------------------------------

IS THE SOLUTION TO 4TH QUESTION CORRECT
4.use the principle of induction to prove that any integer greater than
27 can be written as 5a + 8b where a,b are elements of integers
SOL:
Claim : any integer > 27 = 5a + 8b where a,b are elements of integers
(***)
Note: 1 = 5 *(-3) + 8 * 2
Basic: When n = 28, n = 28(-5 *3 + 8* 2) = 5*(-84)+ 8*56

Induction Hypothesis:
Assume when n = k, n = 5a + 8b for some integers a, b
then n+1 = 5a + 8b + 1 = 5a + 8b + + 5 *(-3) + 8 * 2
= 5(a-3) + 8(b +2)
This shows (***) is true when n = k+1
The complete the inductive proof.

I think, you incorrectly stated problem 4, because each integer number can be written as $5a+8b$. Why we need condition about number 27?
But if we restrict ourselves to nonnegative integers then each integer number greater than 27 can be written as $5a+8b$, $a,b\geq 0$, and there is no such representation for 27.

I don't understood pr.1; and, definitely, pr.3 should be placed to Advanced Fields Section.

Questions about group/ring theory belong in the college section, not the high school section.

1 thing I dont understand about the 1st question why do you go for induction when it can be solved easily.

Basically while counting the permutations,note the permutations are bijective.
Now there are n choices for 1,n-1 choices for 2 & so on.


function
logarithms
calculus
derivative
inequalities
real analysis
real analysis unsolved

What is the infimum of the reals of the form $ \frac{1}{3^t}\plus{}\frac{1}{3^{1/t}}$, where $ t\in (0,\infty)$?

[quote="zhoraster"]Any problems with differentiating it?[/quote]
Well:
$ ( \frac{1}{3^{t}}\plus{}\frac{1}{3^{1/t}} )' \equal{}\ln 3
(\frac{1}{t^2} \frac{1}{3^{1/t}} \minus{} \frac{1}{3^t} )$

but I don't know what I should do next. I can't solve the equation
$ 3^t \equal{} t^2 3^{1/t}$.  :blush:
Could you help me?

Probably you can guess a solution.  Then perhaps you should try proving that it's unique.  (You also should consider the limits of your function as $ t \to 0$ and $ t \to \infty$.)

The answer is $ \frac {2}{3}$. The proof is not nice at all (it's ugly, to be perfectly precise), but here it goes... I hope it's correct, I might have easily made an error or two in the calculation. Anyway:

We can obviously achieve $ \frac {2}{3}$ for $ t \equal{} 1$. What's left to prove is that $ \frac {1}{3^t} \plus{} \frac {1}{3^{\frac {1}{t}}}\geq\frac {2}{3}$. We multiply both sides by $ 3^t > 0$ and get $ 1 \plus{} 3^{t \minus{} \frac {1}{t}}\geq 2\cdot 3^{t \minus{} 1}$.

We will now differentiate this and prove the function $ f(t) \equal{} 1 \plus{} 3^{t \minus{} \frac {1}{t}} \minus{} 2\cdot 3^{t \minus{} 1}$ has a minimum at $ t \equal{} 1$ (i.e. that $ f'$ changes sign at $ 1$) $ f\in C^{\infty}(\langle 0, \plus{} \infty\rangle)$ so we won't have any problems with differentiation or continuity or anything in that manner.

Let's notice that $ f'(t) \equal{} \ln 3\cdot 3^{t \minus{} \frac {1}{t}}\cdot (1 \plus{} \frac {1}{t^2}) \minus{} 2\cdot\ln 3\cdot 3^{t \minus{} 1}$ has the same sign as $ g(t) \equal{} 3^{ \minus{} \frac {1}{t}}\cdot (1 \plus{} \frac {1}{t^2}) \minus{} \frac {2}{3}$ (we merely divided $ f'(t)$ by $ \ln 3\cdot 3^t > 0$).

As obviously $ f'(1) \equal{} g(1) \equal{} 0$, we want to prove that $ g(t) > 0$ for $ t > 1$ and $ g(t) < 0$ for $ t < 1$. That is equivalent to $ 3^{ \minus{} x}\cdot (1 \plus{} x^2) \minus{} \frac {2}{3} < 0$ for $ t > 1$ and $ 3^{ \minus{} x}\cdot (1 \plus{} x^2) > 0$ for $ t < 1$ (we substituted $ x$ for $ \frac {1}{t}$).

Let's define $ h(x): \equal{} (3^{ \minus{} x}\cdot (1 \plus{} x^2) \minus{} \frac {2}{3})3^x \equal{} 1 \plus{} x^2 \minus{} 2\cdot 3^{x \minus{} 1}$. Obviously $ h(1) \equal{} 0$. Since the sign of $ h$ is the same as the sign of $ 3^{ \minus{} x}\cdot (1 \plus{} x^2) \minus{} \frac {2}{3}$, it is enough for us to prove that $ h'(x) < 0$ (then, we will have $ h(x) > 0$ for $ x < 1$ and $ h(x) < 0$ for $ x > 1$).

Now $ h'(x) \equal{} 2x \minus{} 2\ln 3\cdot 3^{x \minus{} 1}$. Proving $ h' < 0$ is equivalent to proving $ j(x) \equal{} x \minus{} \ln 3\cdot 3^{x \minus{} 1} < 0$. Since $ j'(x) \equal{} 1 \minus{} (\ln 3)^2\cdot 3^{x \minus{} 1}$ is a strictly decreasing function, all that's left to find is a maximum of this function, i.e. $ x_0$ such that $ j'(x_0) \equal{} 0$. That'd mean $ 1 \minus{} (\ln 3)^2\cdot 3^{x_0 \minus{} 1} \equal{} 0$, which is equivalent to $ 3^{x_0 \minus{} 1} \equal{} \frac {1}{(\ln 3)^2}$, which means $ x_0 \equal{} 1 \plus{} \log_3(\frac {1}{(\ln 3)^2}) \equal{} 1 \minus{} 2\log_3\ln 3$.

Now, let's just see what $ j(x_0)$ is and hope it's less than $ 0$. It really is (I've used my calculator at this point, I currently have no idea how to check it manually (and I don't think it really has a point), but I guess it would be possible somehow): $ j(x_0) < \minus{} 0.0814 < 0$.

And now we're done. Since $ j(x_0) < 0$ and $ j(x)\leq j(x_0)$ for all $ x$, we have $ h' < 0$. Thus, we have $ h(x) > 0$ for $ x < 1$ and $ h(x) < 0$ for $ x > 1$. So, $ g(t) < 0$ for $ t < 1$ and $ g(t) > 0$ for $ t > 1$. Therefore, $ f'(t) < 0$ for $ t < 1$ and $ f'(t) > 0$ for $ t > 1$ and $ f$ really does have a minimum at $ 1$. So, the answer is $ \frac {2}{3}$.

(Edit: Sorry, made a typo. Nonetheless, it didn't change anything. Anyway, zhoraster's solution is much less complicated. :))

Well, it should not be so complicated.

We have $ f'(1) \equal{} 0$.

For $ t < 1$, $ t \minus{} 1/t < 2\log t$. Indeed, for $ t \equal{} 1$ we have an equality here, for $ t < 1$ the derivative of the first, $ 1 \plus{} 1/t^2 > 2/t$ by Cauchy. So $ 3^{t \minus{} 1/t} < 3^{2\log t} < e^{2\log t} \equal{} t^2$ (the last inequality is because $ \log t$ is negative).
So the derivative $ f'(t) \equal{}\log{3}(3^{ \minus{} 1/t}/t^2 \minus{} 3^{ \minus{} t})$ is negative.

For $ t > 1$ similarly $ t \minus{} 1/t > 2\log t$ and $ 3^{t \minus{} 1/t} > 3^{2\log t} > e^{2\log t} \equal{} t^2$ (now $ \log t$ is positive). So $ f'(t)$ is positive.

Therefore, $ t \equal{} 1$ is the point of global minimum.


I have heard it said that ALL solvable geometry problems can be solved using analytic geometry. Is this true? Could someone provide me a bit insight into what this is all about? Thanks.

Roughly:  any geometry problem can be expressed as an algebra problem, even in many dimensions.  In fact, historically, people did all of their algebra problems as geometry problems -- this was something that may have greatly hindered the advancement of some types of mathematics.  (For example, the Greeks never developed much of an algebra -- when they talked about multiplication, the thing they imagined was the area formed by a rectangle with sides of two given lengths, which makes it impossible essentially to deal with non-homogenous equations.  (If you told an ancient Greek a*b*c + d = e*f, he wouldn't know what you were talking about because you just added a volume to a length and equated it with an area.))

Anyhow, after that long tangent -- and geometric system can be expressed in terms of coordinates, and if the system is determined geometrically, then the systems of equations must be also determined with exactly equivalent solutions.

You still probably don't want to do analytic geometry most of the time, however.

hmm. I think it depends on what you mean by "solvable". There are different senses of that word.

"solvable" meaning it can be solved using Euclidean geometry (ie not analytic).

Well euclidean geometry can be modeled completely by coordinate geometry. So yes, every euclidean geometry problem can be transformed into an algebraic problem. But that doesn't mean the resulting algebraic problem can be solved [i]exactly[/i] in terms of roots and radicals. That's what I mean by "solvable". It depends on what tools and theorems you restrict yourself too in the solution of geometric and algebraic problems. For example, in euclidean geometry you can restrict yourself to just ruler and compass constructions as your tools and then some problems aren't solvable. Likewise some algebra problems aren't solvable in closed forms with roots and radicals.

But since all of the axioms of euclidean geometry can be modeled by algebra - every theorem of geometry can be transformed into an algebraic statement. I believe this is what Hilbert used to prove the consistency of Euclidean Geometry.

Hilbert didn't actually prove the consistency of Euclidean Geometry. He proved that if arithmetic is consistent, then so is Euclidean Geometry. (That's probably iff, but I'm not sure.)

Umm...I used to use analytic geometry a lot when faced with a difficult geometry problem, and actually, I became quite good with it.  It does not even come close to replacing the beauty of the Euclidean style though. ;-)


If the sum of the first 10 terms and the sum of the first 100 ters of a given arithmetic progression are 100 and 10, respectively, then what is the sum of the first 110 terms?

$(A) 90$
$(B)-90$
$(C) 110$
$(D)-110$
$(E)-100$

[quote="Ubemaya"]I got D but I'm not sure...[/quote]
That's correct.

Just use $S_{n}=\frac{n(2a+(n-1)d)}{2}$.

[quote="rnwang2"][quote="Ubemaya"]How did you get that formula, and what do the variables represent?[/quote]
This is the well-known formula for the sum of an arithmetic sequence.

$a=$ first term of the sequence
$n=$ number of terms
$d=$ common difference

It is easily derived using Gaussian pairing.[/quote]
So what would those numbers be in this situation?


What is the remainder when $(999999)^3$ is divided by $5$ ?

The units term in $(999999)^3$ is 9 thus on dividing by 5 we get remainder as $4$.

You can just expand it as $(10^6 - 1)^3$ and it's trivial.

[hide]$999999^3\equiv 4^3=64\equiv 4\pmod{5}$

So the remainder is $4$.[/hide]


Hi ,

For what value of $ n$ sum of digits of $ n$ in decimal presantion is $ 9$.

good luck !

There are infinitely many $ n$:
$ 9, 18, 27, 36, 45, 54, 63, 72, 81, 90$ are just a few, but now stick in $ 0$'s anywhere and everywhere.

Extremly sorry
 It should be $ n!$ with sum of digits $ 9$. Sorry again


Find all the distinct triplet of positive integers such that the sum of the three numbers are  minimized among the pairs in which the sum of any two numbers is square number.Note that the rearranged pairs such as $\"1,2,3\"$ and $\"3,2,1\"$ are regarded as the same pair.


That's how you rank on the forums. Well, maybe rank is too strong of a word, but that just measures how many posts you have. You'll have noticed the people with more posts have more stars next to their name and have a different problem as a rank.

For more details, check out this topic.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13045

[quote="takuanitromars36"]This is wierd, but I think that the AoPS website did some changes to its website.[/quote] We did! :) AoPS and MathLinks merged, and also merged features, so if some things seem weird don't panic, you'll get accustomed with them. There is also an options menu to set up your view of the forum, and you have more settings on the profile menu (including choosing a flag, gender, and another style).

The rank is one of the seven millennium problems.  You can find information about them at http://www.claymath.org/millennium .

1. Moderators don't have any normal millenium problem ranks only when in their moderated forums. If they post in a forum in which they do not moderate, they go back to the normal ranking system. Administrators always have their special rank.

2. Mathlinks is a major math-related website/forum just like AoPS. They merged about a month or so ago, making this the AoPS-MathLinks Forum.

How long did it take to get 2500 post that seems like it would take 1/2 a lifetime.

It had been nearly two years since the last post in this thread.  Please, let ghosts rest in peace. :)

By the way, as an example, Valentin Vornicu has made 4.51 posts per day, so at that rate he would have 2500 posts after 555 days.

sorry didnt even look at the date.
 :blush:


Find a positive integral solution to the equation

$\frac{1+3+5+...+(2n-1)}{2+4+6+...+2n}=\frac{115}{116}$.

[hide]
the sum of n even numbers is $n(n+1)$. the sum of n odd numbers is $n^{2}$. thus, we can make the equation: $\frac{n^{2}}{n(n+1)}\=\frac{115}{116}$.
this becomes $\frac{n^{2}}{n^{2}+n}=115/116$.
then, we find that 115^2/115^2+115 can simplify to 115(115)/115(116), and hence give us our answer. N, of course, is 115.
$\boxed{115}$[/hide]

[quote="perfect628"]Would you like inductive or deductive?[/quote]

You're obviously taking geometry this year  :wink:

Yes, but only until next week. Then I start Alg. 2. anyway,

[hide="solution"]The numerator is the sum of the first n odd #s, the denominator even. So, the numerator, for a value n, is equal to n^2, the denominator n^2+n. Thus, the simplified fraction, for n, is n/(n+1). This =115/116, so n=115.[/hide]


Let $a,b,c$ are length of a triangle then:
$\frac{(b+c)bc}{b+c-a}+\frac{(a+c)ac}{a+c-b}+\frac{(a+b)ab}{a+b-c} \geq 2(ab+bc+ca)$
  It is very easy. For those who want to practice. :D

OK! It's Easy! Use Tchebyseff in basic form!

Hung, can u solve me how to use Chebusev in orginal form?I don't know that. Because when I use Chebusev then:
  Suppose $a \geq b \geq c$ we have
$\frac{1}{b+c-a} \geq \frac{1}{a+c-b} \geq \frac{1}{a+b-c}$
$(b+c)bc \leq (a+c)ac \leq (a+b)ab$
 Then it is wrong!

It is equivalent to $\displaystyle \sum \frac{1}{b+c-a}\geq \sum \frac{1}{a}$,which comes from $\displaystyle\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}$. :)

Vlad, i think it is $\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}.$ ;)

:D .Yeah, it is easy!
  Frankly, I have an inequal that is stronger than this one but it is very ugly.  :lol:
I think you mean write the numerator something likes:
$(b+c-a+a)bc$. Is it your sols?
 So how is about:
$\frac{1}{b+c-a}+\frac{1}{a+b-c}+\frac{1}{c+a-b}\geq \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ba}$

[quote="Gibbenergy"]
 So how is about:
$\frac{1}{b+c-a}+\frac{1}{a+b-c}+\frac{1}{c+a-b}\geq \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ba}$[/quote]
It isn't hard.
It is equivalent to
$\sum abc(a+b-c)(c+a-b)\ge\sum a^2(b+c-a)(c+a-b)(a+b-c) \leftrightarrow \sum a^5 +\sum a^3bc\ge\sum(a^4b+b^4a) \leftrightarrow\sum a^3(a-b)(a-c)\ge0$


Ok in the book it says a=b modm and p=q modm. Then a+b=p+q modm. Well I tried some numbers and say it is mod10. 4=14 and 5=15. Adding those I get 18=20 mod10. That's not right. What did I do wrong?

That's wrong. If $ a \equiv b \mod n$ and $ p \equiv q \mod n$, then

$ a\plus{}p \equiv b\plus{}q \mod n$

Therefore,

$ 4\plus{}5 \equiv 14 \plus{} 15 \mod 10$

It is kind of like adding equations. What page is it on?

Yes the 'newer' book has that incorrect. That mistake has been pointed out by (me) as well as others, I think. It's on page 44 or something like that in chapter 5, in the modulo section.

Yea that is what I thought modularmarc101, but the book said otherwise. Yes it is in Chapter 5 pg 44.


Princeton college geometry probability calculus HMMT ARML

All right. If everything goes right (that is, nothing sudden comes up), I'm going to be competing at Princeton Math Competition this year. But because this one requires team round and other teams, I wonder if anyone else can come. I need 9 people so please post here if you're interested!

Sure. I'll join if you need an extra guy. Hopefully you don't care if you win or lose. I'll be happy to just have fun.

oh dude, we're going too as the SC team though.

This sounds like an excellent time for Quagga to resurface...

I just realized that I can't go.. Sorry.. You guys can still talk about the team though.

[quote="mysmartmouth"]Yes, the SC team is going.  Team size is maximum 10 people, but you don't need 10 (see [url=http://www.artofproblemsolving.com/Wiki/index.php/Princeton_Math_Competition]here[/url]).
I think we will have 10, however, if not more.[/quote]

Yeah, but if you don't have 10 then it seems they will assign you some random person/people (which could be bad or good...).  If you did have less than 10 people wouldn't be able to do all of the relays and you would have less people taking individual (if scoring is done by weighting...then this wouldn't matter).  Additionally, you will of course be at a disadvantage on team/power.

We won't require partial teams to combine if they don't want to. If you want to be the SC team and you only have 8 people, we're not going to force you to take on 2 more that you don't want.

By the way, we won't disqualify anyone from anything for having too few people. If a team has fewer than 5 people for one (or both) of the relays, people will sit in the last few seats and start from the middle. Usually you'll have to be really lucky to get it right.

In response to joml88, scoring will NOT be weighted for partial teams.  So you can see that partial teams will be at a distinct disadvantage.  However, if you still want to do it, we will certainly not stop you.  If you are just short on people, though, and are looking to join up with another partial team, we recommend creating a half team (of 5 people) so that the joining process will be easier.

[quote="Iversonfan2005"][quote="mysmartmouth"]Yes, the SC team is going.  Team size is maximum 10 people, but you don't need 10 (see [url=http://www.artofproblemsolving.com/Wiki/index.php/Princeton_Math_Competition]here[/url]).
I think we will have 10, however, if not more.[/quote]

um, we will most likely have a max of 10 people going anyway. Think about it, it's Christmas vacation time. I doubt more than 12 or so people will want to go anyways. Especially for the SC team, it'll be hard to think about who will take what tests. I mean, I don't know how many people really have a "specialty" in a certain math field.[/quote]

Well, I know you specialize in geometry.  And I think most people have a subject area they find easier than others.  For me it is Counting & Probability, and for Asif it is definitely Number Theory...

[quote="mysmartmouth"]
Well, I know you specialize in geometry.  And I think most people have a subject area they find easier than others.  For me it is Counting & Probability, and for Asif it is definitely Number Theory...[/quote]

There is no Counting and Probability section :( . I think you should do Calculus :rotfl:

If it's like HMMT used to be, Advanced Topics has a lot of combinatorics.

what does 'SC' stand for? Southern California?
(sorry, i'm new at princeton math contest thing..)
if so, how can I join?

Thanks,

Um, no.

Now we need to stop digressing and let the thread go back on topic.

especially when two of the teams are named quagga

[quote="Xevarion"][quote="chess64"]Uh, what's the topic of this thread again? It's kinda become a place where anyone who wants to discuss the princeton competition posts.[/quote]
Well, originally someone was trying to organize an AoPS team. (Hence the name.) But that quickly got distracted, and now that it's too late to register more teams (technically you could but there's a long waiting list), this has just turned into the thread of random discussion about the teams going to the contest. Which is fine with me -- what's the fun of a team contest when you can't have a bit of rivalry with the other teams? ;)[/quote]

Well yeah that's basically what I was trying to say. So there isn't really a topic to get back on ;)

Hey I have an idea.  Let's call our team "Arnav and Friends."

[quote="joml88"]Oh ok whatever.  Just don't mess with the derty south or we'll have some beef.  Nathan knows what I'm talking about.  Him and I have had beef between us before.[/quote]
Yeah lets keep things quiet around here. I don't want any incidents over by the O.K. Corral.

Yeah, that's almost as bad as hair in your yellow chives  :o

[quote="MysticTerminator"]ok now c'mon guys let's keep this calm ok

dude and I strongly believe that both teams should be named quagga

helll some other teams can name theirs quagga too if they want

whatev

it's all good

I envision a math contest one day where each and every team will be named quagga

we would pwn the awards ceremony[/quote]

LOL

can we (the woot team) be quagga for hmmt?

Next year if there's an upper division and a lower division (this year too many ppl signed up for upper), then we will call the upper division "Quagga Division" and the lower division "Regular Division."

Teams can only enter the Quagga Division if they are named "Quagga".

[quote="CatherineAsaro"]Anyone on a team called Quagga has to come clean all the coffee I just spluttered all over my keyboard. :D

Best,
Dr. Asaro, whose teams have nowhere near such innovative names.

Unless you all come up with one in the hour before I mail the check! :wink:[/quote]

I thought we would be the Fighting 42's, 'cause we win whenever we're the Fighting 42's, and besides, 42 is awesome. :D

Maybe we should be the Quagga42's.  :P


logarithms
real analysis
real analysis unsolved

[quote="harazi"]Let $ \ x_1,x_2 > 0$  and define $ \ x_{n \plus{} 1} \equal{} x_1^{1/n} \plus{} x_2^{1/n} \plus{} ... \plus{} x_n^{1/n}.$ Prove that $ \frac {x_n \minus{} n}{\ln n}$ tends to $ \ 1$.

[/quote]

$ \ x_{n \plus{} 1} \equal{} x_1^{1/n} \plus{} x_2^{1/n} \plus{} ... \plus{} x_n^{1/n}\implies \frac {x_{n} \minus{} n}{\ln n} \equal{} \sum _{r \equal{} 1}^{n \minus{} 1}\frac {x_r^{1/(n \minus{} 1)} \minus{} 1}{\ln n} \minus{} \frac {1}{\ln n}$

Am i on right path? can someone help me?

as a first step you could try to show that $ \frac{x_n}{n} \to 1$. It may be helpful to prove that
\[ \sum_1^n k^{\frac{1}{n}} \equal{} n \plus{} \ln(n) \plus{} o(\ln(n))\]


analytic geometry
geometry
circumcircle
incenter
trigonometry
graphing lines
slope

$ \boxed{1}$

Given $ O(0,\ 0),\ A(10,\ 0),\ B(8,\ 6)$ on the $ xy$ plane.

(1) Find the equation of the line by which $ \angle{AOB}$ is equally divided.
(2) Find the coordinates of the circumcenter $ O'$ and incenter $ I$ of the $ \triangle{AOB}.$

$ \boxed{2}$

Given three lines on the $ xy$ plane:

$ \l_1: \ 8x \plus{} y \equal{} 53$
$ \l_2: \ 2x \minus{} y \equal{} \minus{} 3$
$ \l_3: \ x \minus{} 3y \equal{} \minus{} 9$

(1) Find the area of the triangle formed by these lines.
(2) Determin the shape of the triangle.

[quote="kunny"](2) Find the coordinates of the excenter $ O'$ and incenter $ I$ of the $ \triangle{AOB}.$
(2) Determin the shape of the triangle.[/quote]

Which excenter? What do you mean "shape"? As in acute, obtuse, right, scalene, etc.?

[quote="1=2"]What do you mean "shape"? As in acute, obtuse, right, scalene, etc.?[/quote]
Right.

$ \boxed{1}$ (2) I have just edited, sorry.

I don't really have a lot of time to do it now [i]and[/i] this would be ugly BUT

[hide="one"]Take a circle of radius $ r: 0<r<d$ where $ d$ is the smaller of the two sides AO and BO... Then, find the intercept of this circle with AO and BO each. Take the midpoint of the two intersection points and then construct a line then runs through that point and the origin.[/hide]

[hide="two"]taking the equations two at a time we can quickly find their points of intersection using matrices and reduced row echelon form. The type of the triangle should follow accordingly.[/hide]

[hide="1"]
Uh, if we label $ \angle AOB \equal{} 2 \theta$, then all we need to do is find $ \tan \theta$, then we have the slope of the line we want.

So..$ \tan 2\theta \equal{} \frac {3}{4} \rightarrow \frac {2 \tan \theta}{1 \minus{} \tan^2 \theta} \equal{} \frac {3}{4} \rightarrow 3 \tan^2 \theta \plus{} 8 \tan \theta \minus{} 3 \equal{} 0$

So, $ \tan \theta \equal{} \frac {1}{3}, \minus{} 3$, but the slope is positive, so the answer is $ \frac {1}{3}$, so the answer is $ \boxed{y \equal{} \frac {1}{3} x }$

part (2)

For the incenter, bisect the angle (like above), and find the intersection point.  I did it and got $ ( \minus{} \sqrt {10} \plus{} 10, \frac{\minus{} \sqrt {10} \plus{} 10}{3}$.  It's ugly, and I didn't check my work :(.

For the circumcenter, all we do is label the center $ (x,y)$, and we apply the distance formula three times to the vertices to get $ \sqrt {x^2 \plus{} y^2}, \sqrt {(x \minus{} 10)^2 \plus{} y^2}, \sqrt {(x \minus{} 8)^2 \plus{} (y \minus{} 6)^3}.$  The first two give us $ x \equal{} 5$, and plugging it into the last two gives us $ y \equal{} \frac {5}{3}$
[/hide]

[hide="2"]
Uh, solving for all three points of intersection using elimination, we get $ (6,5), (5,13), (0,3)$

Using the distance formula $ 3$ times, we get that this is a right triangle.[/hide]

1's answer is correct,  :) ,but for 2, you seemed to have forgotten find the area of the triangle, regrettably the shape of the triangle is incorrect.

[quote="kunny"]$ \boxed{2}$

Given three lines on the $ xy$ plane:

$ \l_1: \ 8x + y = 53$
$ \l_2: \ 2x - y = - 3$
$ \l_3: \ x - 3y = - 9$

(1) Find the area of the triangle formed by these lines.
(2) Determine the shape of the triangle.[/quote]

[hide="Matrices"]
$ \begin{pmatrix}
8 & 1 & 53 \\
2 & -1 & -3 \\
\end{pmatrix}\to
\begin{pmatrix}
1 & 0 & 5 \\
0 & 1 & 13 \\
\end{pmatrix}$
$ \begin{pmatrix}
8 & 1 & 53 \\
1 & -3 & -9\\
\end{pmatrix}\to
\begin{pmatrix}
1 & 0 & 6  \\
0 & 1 & 5 \\
\end{pmatrix}$
$ \begin{pmatrix}
2 & -1 & -3\\
1 & -3 & -9\\
\end{pmatrix}\to
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 3 \\
\end{pmatrix}$[/hide][hide="Triangle"]
$ \begin{array}[b]{ccc}
(5,13) & (6,5) & (0,3)\\
f(x)=53-8x & g(x)=2x+3 & h(x)=\frac{1}{3}(x+9)
\end{array}$

$ \displaystyle
\begin{eqnarray*}
\therefore\text{[ABC]}&=&\int^5_0 (g(x)-h(x))\,dx +\int^6_5 (f(x)-h(x))\,dx\\
&=&\frac{5}{3}\int^5_0 x\,dx+\frac{25}{3}\int^6_5 (6-x)\,dx\\
&=&\frac{125}{6}+\frac{25}{6}=\boxed{25}\\
\end{eqnarray*}$

A simple sketch shows that $ \triangle ABC$ is obtuse scalene.

I conveniently rewrote the system of linear equations binding our triangle so we can see the slopes. All we'll do is take negative reciprocals, reconstruct altitudes and utilize a matrix to show that the orthocenter is outside the bounded region. This yields the following altitudes:

$ \begin{array}[b]{ccc}
x-8y=-24 & x+2y=16 & 3x+y=28
\end{array}$

$ \begin{pmatrix}
1 & -8 & -24\\
1 & 2 & 16\\
3 & 1 & 28\\
\end{pmatrix}\to
\begin{pmatrix}
1 & 0 & 8\\
0 & 1 & 4\\
\end{pmatrix}\Longrightarrow H=(8,4)$

This point clearly lies outside of our bounds, which makes $ \triangle ABC$ obtuse.
As for scalene, (I don't feel like doing this) all you need show is that the altitudes' height
by ratio to the base is never the same for all three sides.[/hide]

Sorry about the calculus but it seemed the fastest way to find the area; Herons formula yields nasty radicals that need simplification even after you factor things out; Solving for the proper lengths to implement the formula $ \frac{1}{2}bh$ was giving me the wrong values for some reason (Prolly screwed the algebra somewhere).

Now you get the correct answer.  :)

Generally for $ O(0,\ 0),\ A(a,\ b),\ B(c,\ d)$, we can have $ [OAB]\equal{}\boxed{\frac{1}{2}|ad\minus{}bc|}$

[hide="2"]We do b first, since that is the simplest. We find the vertices of the triangle: (5,13), (0,3) and (6,5). Pythagorean bashing, we get that the side lengths of the sides are $ 5\sqrt{5}$, $ 2\sqrt{10}$, and $ \sqrt{65}$. Since $ (5\sqrt{5})^2>(2\sqrt{10})^2\plus{}(\sqrt{65})^2$, the triangle is [b]obtuse and scalene[/b], with obtuse angle at (6,5).

For a, we plot (6,3), (0,13), and (6,13). We now have a rectangle with the original triangle inscribed in it. We find the area of the other triangles: 6, 4, and 25. We subtract those from the area of the rectangle, which is 60: $ 6\plus{}4\plus{}25\equal{}35$, $ 60\minus{}35\equal{}\boxed{25}$[/hide]

The simplest method without calculus


$ x\bullet y$

This black circle is kinda big. How can I get a nicer black circle?

[size=75]Edited by stevem to show the missing text after bullet[/size]

\bullet is a binary operation so expects a symbol before and after it. It must think that dollar is is the symbol after it so there is no dollar ending the formula and  hence things go wrong after that.

If you are using a dot for scalar multiplication of vector then you could use \cdot $ x\cdot y$. Otherwise is there anything in [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LATEX Symbol List[/url] you could use?

Of course in extreme you could define your own dot where you have complete control over its size and placement eg
[code]\newcommand{\mydot}%
{\begin{picture}(3,3)
\put(1.2,3){\circle*{0.5}}
\end{picture}}
This is text and $x \mydot y$ includes a dot[/code]

Okay, thanks.

As for makin' a white circle?

I make it with $ \circ.$ Is there another way to make it nicer?

What do you mean by nicer?
Using \circle rather than \circle* in the above code gives an open circle.

For example, I've made this

[img]http://img292.imageshack.us/img292/8508/dfgdfun9.png[/img]

I don't like the \circ. Do you know another way to make another white circle?

Basically you need to change the bullet in the temize environment. Try this, based on the code I pasted above
[code]\newcommand{\mydot}%
{\begin{picture}(3,3)
\put(1.2,3){\circle{3}}
\end{picture}}
\renewcommand{\labelitemi}{\mydot}
\begin{itemize}
\begin{itemize}
  \item $\{a,b\}$
  \item $\{a,c\}$
  \item $\{a,d\}$
  \item $\{b,c\}$
  \item $\{b,d\}$
  \item $\{c.d\}$
\end{itemize}\end{itemize}[/code]You can play with the sizes and positions in the picture environment to get the circle you want.


Find two natural numbers . Sum is 168 , LCM is 1001 .  :oops:

I'm pretty sure you mean LCM as in Least Common Multiple...

[hide="solution"]
First factor $ 1001$ to get $ 7\times11\times13.$ The two numbers can only have these numbers as factors and each factor that appears in each number can only have a power of 0 or 1. Now it's basically easy guess and check. \[ 143\plus{}13\equal{}158\newline143\plus{}11\equal{}154\newline143\plus{}77\equal{}230\newline\boxed{77\plus{}91}\equal{}168.\][/hide]


calculus
integration
function
logarithms
calculus computations

What is the indefinite integral of ln(x)/(1+x)?  Ive tried everything i know but nothing works so far.

The short answer is that it's not an elementary function. There's a related function that goes by the name Li.

I can give you a power series for it.
Substitute $y=x+1$; this makes the integral $\int\frac{\ln(y-1)}y=\int\frac{y-\frac{y^2}2+\frac{y^3}3-\cdots}y =\int\sum\frac{(-y)^n}{n+1}=\sum_{n=1}^\infty (-1)^{n-1}\frac{y^n}{n^2}$

It is possible to evaluate some values of this: $\int_0^1\frac{\ln x}{x+1}=-\sum\frac1{n^2}=-\frac{\pi^2}6$.


Consider the recurrence relationship $a_{n+1}=2^{n}-a_{n}-42$ for all natural numbers $n$.

For which values of $a_{0}$ will the sequence:
a. Escape to infinity?
b. Escape to negative infinity?
c. Converge?
d. Enter an infinite loop?


[quote="N.T.TUAN"]A perfect square $ d_1d_2\cdots d_{2n}$ , with $ d_1\not \equal{} 0$ and $ d_k\not \equal{} 0$ for some $ k\geq n \plus{} 1$, is called special if numbers $ d_1d_2\cdots d_n$ and $ d_{n \plus{} 1}d_{n \plus{} 2}\cdots d_{2n}$ are also perfect squares. Prove that for any positive integer $ m$ there exist at least one $ (4m \plus{} 2) \minus{}$ digit special number and at most two $ 4m \minus{}$ digit special numbers.[/quote]
Let
$ x^2 \equal{} d_1d_2...d_n, \ y \equal{} d_{n \plus{} 1}...d_{2n}, \ z^2 \equal{} x^210^n \plus{} y^2, \ 10^{n \minus{} 1}\le x^2 < 10^n, \ y^2 < 10^n.$
If $ n \equal{} 2m$ (is even), then $ z \equal{} (x10^m \plus{} a)^2, 2ax < 10^m,$ therefore only $ a \equal{} 1,x \equal{} \frac {y^2 \minus{} 1}{2*10^m}, \ 10^m|(y \minus{} 1)(y \plus{} 1)$
Therefore one of $ 5^m|y \minus{} 1,2^m|y \plus{} 1$ or $ 5^m|y \plus{} 1,2^m|y \minus{} 1.$
First give $ y \equal{} 1 \minus{} 2r\mod 10^m, r \equal{} 5^{2^m}\mod 10^m$, second $ y \equal{} \minus{} 1 \plus{} 2r\mod 10^m, r \equal{} 5^{2^m}\mod 10^m.$
If $ n \equal{} 2m \plus{} 1$ we can take y=1 and find solution Pell's equation $ z^2 \minus{} 10(10^mx)^2\equal{}1.$