Io sono italiano!!! Grandioso! Pensavo che nessuno si fosse accorto del fatto che nella sezione italiani nessuno avesse postato qualcosa! Sara' per il fatto che arriviamo 40esimi alle olimpiadi...
Sono da Venezia, studio Matematica all'univ di Trieste (2nd anno). Tu?

leggermente piu piccola... :P
ho 15 anni sono di prato..ok se non hai presente dove è chiedimelo pure...ci sono abituata
vado a una normalissima schifosa scuola...in confronto poi a qst americani che vanno a scuole per cervelloni..fanno gare ogni due giorni e roba del genere...mi sento un po come se in italia si fosse indietro di qualche secolo..non ti ha faatto anche a te qst impressione?
comunque cerco di dare il mio contributo qua nel sito con il mio inglese strampalato...tanto perchè mi piacciono questi americani(anche se il forum è rumeno) competitivi. mi fa piacere che ci sia un italiano.. :lol:
ah te come mai sei su questo sito?

[quote="hammond"]Ehi! Da dove è saltato fuori il forum italiano? Mai visto prima d'ora... Beh, ciao a tutti ci sono anch'io ;)[/quote]
è saltato fuori che l'admin ha chiesto se tutti gli stati erano rappresentati e io gli ho detto di no....bene ora siamo in 3 :D

per la tua gioia, so dove si trova Prato.
 :P

Io mi trovo qui per vari motivi; principalmente mi piacciono i problemi matematici e questi evitano la fossilizzazione dei miei poveri neuroni, inevitabile controindicazione delle universita' italiane.... Inoltre faccio parte della giuria delle gare a squadre locali, quindi durante gli allenamenti e' utile avere nuovi problemi (quest'anno mi hanno dato anche una maglietta....yuppie!).
In effetti in italia siamo un "pochettino" indietro, soprattutto per quanto riguarda il problem solving olimpico... comunque qui i "cervelloni" sono tali in quanto molti sono universitari... alcuni sono addirittura professori...

Tu some mai ti trovi qui?

io trovo qui perchè non posso vivere senza matematica!!! ovviamente non vivo DI matematica :D
e riguardo alla matematica che si fa a scuola io sono un po piu avanti perchè me la studio da sola  a casa
quindi mi annoia limitarmi a fare gli esercizi che da il mio prof..che cmq stimo moltissimo..è lui che mi ha fatto appassionare alla materia..

riguardo a quello che dici delle università italiane si l'ho sentito dire anch'io..e proprio per qst pensavo di andare alla normale di Pisa che si dice continui ad essere tosta..dato che è anche vicina a casa..se poi non dovessi superare l'esame pace, del resto è giusto che rimanga, come dire, un po una scuola non accessibile a tutti ma solo a quelli che sono veramente in gamba...dall'ammissione o no capirò anche in che modo fare della matematica una professione...l'idea di andare a insegnare non mi attrae molto...però mi piacerebbe fare la ricercatrice(leggermente ambiziosa eh :lol: ?)

Anch'io non vivo DI matematica (tant'e' che studio composizione al conservatorio).
Se ti prepari per la normale di Pisa ti consiglio di prepararti bene anche la parte di fisica; e' prorpio questo che mi ha fatto desistere dal provare l'esame (mi dichiaro un completo ignorante in ottica), la prova di matematica si e' (purtroppo) un po' semplicificata in questi ultimi anni, il che toglie un po' il sapore di sfida intellettuale...
Anch'io vorrei fare il ricercatore, ma non penso in italia, anche perche' i campi di mio interesse (combinatoria algebrica e strutture discrete) in italia sono molto trascurati. Quindi mi sto preparando ad un possibile trasferimento (per studio) od in Ungheria od a Linz.. Princeton rimane un sogno (per questioni finanziarie). Come vedi anch'io sono poco ambizioso.
Per ora vedo di finire rapidamente (e bene) l'universita'...

ehm ehm e che c'entra la fisica ccn la matematica? :?
voglio dire piuttosto è la fisica che ha a che fare cn la matematica, non il contrario, giusto? spiegati meglio...non vorrai dire che le domande sono anche di fisica? :ninja:  :huh:  :what?:

[quote="Catraga"] quest'anno mi hanno dato anche una maglietta....yuppie! [/quote]
Giura!  :lol:  :lol:
Non vi dispiace se mi unisco al vostro gruppetto di italiani, anche se italiano non sono...

[quote="Catraga"]Il bando per l'ammissione alla classe di scienze prevede due prove scritte (che variano in base alla facolta') per matematica sono previste una prova di matematica (specifica per matematica) ed un di fisica (generale); i quesiti di fisica sono altrettanto difficili quanto quelli di matematica e la prova e' strutturata come quella di matematica.All'orale i professori sono "liberi di verificare la preparazione generale del candidato per stabilirne l'idoneita'", dal punto di vista legale ti possono chiedere qualunque cosa dalla fisica, alla matematica alla chimica, al nome completo di Leibniz (cosa che in realta' non fanno ne' hanno mai fatto, ma, se vogliono, possono)
Se vai sul sito http://www.sns.it puoi scaricare il bando e qualche prova degli anni passati. Se vuoi prepararti veramente bene ci sono due libri che raccolgono tutti i problemi (un libro di fisica ed uno di matematica) di ammissione alla normale dagli albori (o quasi) fino agli anni novanta.
Comunque un matematico non e' esonerato dallo studiare le altre scienze :P ... io mi sono ritrovato a studiare le dinamiche di popolazioni per effettuare simulazioni al computer.[/quote]
grazie...x ora fisica non la studio...almeno a scuola....ma potrei cominciare ad aggiungere qualche libro di fisica ai molti libri di mate

[quote="Catraga"]niente affatto, dal momento che sei loggato e sei seduto sul computer due file piu' avanti a me e sei tutto preso dalla lettura di qualche forum....
Girati   :D  ciao  :D
Oggi hai provato l'esperienza "legovini" per la seconda volta, eh? Hai comprato il calcolatorino? :lol:

Messaggio subliminale: metti nel piano di studi fisica 2.... fisica 2.... fisica 2...[/quote]
a chi era riferito il messaggio subliminale :? ?

benvenuto vasya. korolev

Il messaggio subliminale e' per Vasya.

Per fisica sono sufficienti i libri di testo che ti verranno dati al liceo. Vedi solo di studiarli bene, tutto qui.
Se hai bisogno di altre infos piu' specifiche, scrivimi pure alla mail (che controllo sicuramente piu' spesso del forum) catraga@linuxmail.org e' molto probabile che ti risponda quasi immediatamente ed inoltre non monopolizziamo il forum.

Comunque, oltre che a prepararti per la Normale ti consiglio di dedicarti anche a quella che viene detta matematica ricreativa (Martin Gardner), che sicuramente aiuta, inoltre (secondo me) alcune volte e' molto piu' divertente; in piu' se non passi l'ammissione, almeno hai ottime speranze di ottenere altre borse di studio (io ne ho vinte 2 cosi'  :D )
Comunque c'e' ancora tempo per parlare di queste cose...

[quote="stancioiu sorin"]I'll visit Rome this summer with my parents.I've seen photos with Rome and the other cities in Italy and I think it's a very beautiful country(maybe the most beautiful).But how is the math in Rome?I mean, the children are good,it's alwais a competition between them...that's a real mathematical city.[/quote]
Uh, I think the 99% of italian students doesn't care about maths, and Rome isn't an exception..
By the way, Francesco Cavazzani (frengo, silver medalist al IMO last year) is from Rome  :D

You can visit via panisperna in rome, where Enrico Fermi did his experiments ;)

Con le lacrime agli occhi, chinando la testa, entrando al tempio, mi accosto a questo sacro simposio.

Eccomi l'ultimo degli accoliti...

Speriamo di non far fare figuracce all'Italia :)

Ciao a tutti!

ciao a tutti, mi sono appena registrato! sono di pordenone e studio informatica a milano.

italia rox
ciao ciao

ciao e benvenuti ai nuovi arrivati...

@stancioiu: sprichst du auch deutsch? :)

ciao a tutti, sono anch'io un italian appassionato di matematica. mi chiamo davide e vivo a venezia (il fine settimana,visto che da lun a ven sono all'università di padova)

Ciao, mi presento anch'io, anche se forse qualcuno mi ha già visto bazzicare l'OliForum... Sono Andrea di Torino e ho 18 anni...
Ci sentiamo!  :)

salve a tutti... Mi chiamo Luca e scrivo dalla provincia di pavia.... sono in vacanza dopo il diploma classico e aspetto di cominciare Facoltà di Scienze Matematiche a pavia....

Ciao a tutti.  :)
Mi sono registrato già da un po' di tempo sul forum italiano.
[url=http://olimpiadi.dm.unipi.it/oliForum/viewtopic.php?p=75499&highlight=#75499]Qui[/url] c'è la mia presentazione.
Ciao,  :bye:

Startrek89

Ci sono italiani, ma è anche pieno di russi. Io sono Il_Russo sull'oliforum.

Oh, non avevo mai notato questa discussione... :)
Io vengo qui raramente, di solito mi limito a visitare l'oliForum italiano :roll:

Io vengo qui ormai tutti i giorni ma sono all'università, ora studio Torino...

Ma tutti che si preoccupano delle olimpiadi, in realtà non misurano veramente le tue capacità matematiche. Comunque io ho intenzione di andare all'IMC a luglio.

quanto bei italiani
io sono jordan del forum, che cavolo, pure il nome nn era disponibile  :mad:
un saluto a eucla, mitchan e tutti glialtri del forum

Ehilà sono Carlein sul forum italiano....visto che originalità coi nick...certo che è alquanto rilassante bazzicare il nostro forumino dopo essere passati di qua : ci si disorienta qui  :(
Saluti!

ciao a tutti!!!!

ci sono anch'io!

p.s. sull'oliforum sono gian92  :D


Let $ x,y,z\geq 0$, s.t. $ x^2\plus{}y^2\plus{}z^2\equal{}xyz$. Prove that:

$ xy\plus{}yz\plus{}zx\geq 4(x\plus{}y\plus{}z)\minus{}9$

[quote="Inequalities Master"]Let $ x,y,z\geq 0$, s.t. $ x^2 \plus{} y^2 \plus{} z^2 \equal{} xyz$. Prove that:

$ xy \plus{} yz \plus{} zx\geq 4(x \plus{} y \plus{} z) \minus{} 9$[/quote]
See here a stronger one:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103629

Conditional$ p^2\minus{}2q\equal{}r\Leftrightarrow p^2\equal{}2q\plus{}r\leq\frac{2}{3}p^2\plus{}\frac{1}{27}p^3\Rightarrow p\geq 9$
ineq$ \Leftrightarrow q\geq 4p\minus{}9\Leftrightarrow p^2\minus{}8p\minus{}r\plus{}18\geq 0\Leftrightarrow \frac{\minus{}p^3}{27}\plus{}p^2\minus{}8p\plus{}18\geq 0$
Setting$ f(p)\equal{}\frac{\minus{}p^3}{27}\plus{}p^2\minus{}8p\plus{}18(p\geq 9);f'(p)\equal{}0\Rightarrow p\equal{}12$
$ \Rightarrow f(p)\geq f(9)\equal{}0$


calculus
derivative
algebra
binomial theorem

Nice solution. It seems that binomial theorem is pretty useful while dealing with "abnormal" infinite series.

These are some of my ideas while working on zabelman's problem:

[hide]Let a = 2    + 3/4          + 3/4 *5/8             + 3/4 * 5/8 * 7/12        + ...

      b = 1/4 + 3/4 * 3/8 + 3/4 * 5/8 * 5/12 + 3/4 *5/8 *7/12*7/16 +...

      Evaluate a-b by using the top term subtract the one right below it, we

have a-b=z (1), where z = zabelman's series.

      Also, we have

      a = 2 + 3/4 + 3/4 *5/8  + 3/4 * 5/8 * 7/12 + ...   (3)

      b = 0 + 1/4 + 3/4 * 3/8 + 3/4 * 5/8 * 5/12 + ...    (4)

      Evaluate a+b in a similar manner, we get a+b=2+z (2)

      (2) - (1), we get

      b = 1.

      Comment: Evaluate (3) - (4), we get

      a-b = 2 + 2/4 + 3/4 * 2/8 + ...= T (the series in the previous thread) = z[/hide]

      As I said, I did not figure out z, but find out b instead. Well, I will take

that as a bonus.

Somehow I like to see the series in compact form. This is probably just a restate of what Prof. Merryfield did.

Let [tex]\displaystyle a_n=\frac{(2n)!}{4^n(n!)^2}(2n+1)\left(\frac{1}{\sqrt{2}}\right)^{2n}[/tex] and [tex]\displaystyle b_n=\frac{(2n)!}{4^n(n!)^2}(2n-1)\left(\frac{1}{\sqrt{2}}\right)^{2n}[/tex] for [tex]n\ge0[/tex].

Then [tex]\displaystyle S=\sum_{n=0}^{\infty}a_n[/tex] is [b]zableman[/b]'s sum and [tex]\displaystyle T=\sum_{n=1}^{\infty}b_n[/tex] is [b]your[/b] sum. Notice that [b]your[/b] sum starts with [tex]n=1[/tex].

So [tex]\displaystyle S-(b_0+T) = \sum_{n=0}^{\infty}(a_n-b_n) = 2\sum_{n=0}^{\infty} \frac{(2n)!}{4^n(n!)^2}\left(\frac{1}{\sqrt{2}}\right)^{2n}[/tex].

From what we knew before about the power series:

[tex]\displaystyle \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\arcsin{x})=\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n}[/tex].

we find that the value of the infinte series is:

[tex]\displaystyle S-(b_0+T)= 2 \frac{1}{\sqrt{1-x^2}}\bigg|_{x=\frac{1}{\sqrt{2}}} = 2\sqrt{2} [/tex].

Therefore [tex]T=-b_0=1[/tex] since [tex]S=2\sqrt{2}[/tex] and [tex]b_0=-1[/tex].

it can be done without knowing [b]zableman[/b]'s sum too.

From what we knew before about the power series:

[tex]\displaystyle \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\arcsin{x})=\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n}[/tex] [tex]\displaystyle = 1 + \sum_{n=1}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n}[/tex], ----- (1)

By taking derivatives and multiplying by [tex]x[/tex] on both sides of (1), we have:

[tex]\displaystyle x\frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right) = \sum_{n=1}^{\infty}\frac{(2n)!}{4^n(n!)^2}(2n)x^{2n}[/tex], ----- (2)

Subtracting (1) from (2) and adding 1 to both sides, we get:

[tex]\displaystyle \sum_{n=1}^{\infty}\frac{(2n)!}{4^n(n!)^2}(2n-1)x^{2n}= 1 + x\frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right) - \frac{1}{\sqrt{1-x^2}} [/tex]. ----- (3)

It can also be done as follows,

[tex]\displaystyle \sum_{n=1}^{\infty}\frac{(2n)!}{4^n(n!)^2}(2n-1)x^{2n}= x^2\frac{d}{dx}\left(\left(\frac{1}{\sqrt{1-x^2}}-1\right) \frac{1}{x}\right)[/tex]. ----- (4)

(4) came from subtracting 1 then dividing by [tex]x[/tex], taking the derivatives and then multiplying by [tex]x^2[/tex]  on both sides of (1).

Carrying out the derivatives calculation (I just did two more within the last 2 days without doing any in the last 20 years :-)), both (3) and (4) are equal to (they should be the same):

[tex]\displaystyle 1 + x^2(1-x^2)^{-\frac{3}{2}}-(1-x^2)^{-\frac{1}{2}}[/tex].

The rest is straightforward by plugging [tex]\displaystyle x=\frac{1}{\sqrt{2}}[/tex].


A penny, a nickel, a dime, and a quarter are tossed simultaneously. What is the probability that at least two heads show and one of the heads showing is the dime? Express your answer as a common fraction.

[hide=" :nhl: "]
So you need the dime to be heads and at least one of the others to be heads.
Chance that the dime is heads: 1/2
Chance that none of the other coins are heads: 1/8
Chance that at least one is heads: 7/8

Total chance: $\frac 12 \cdot \frac 78 = \frac{7}{16}$[/hide]

[hide]The probability of the dime being heads is 1/2 (duh). The probability of at least one head in the other three is seven eigths. 1/2*7/8=7/16[/hide]

[hide=" :fool: "]
There are 7 ways to have the dime have a heads with at least one other coin. There are 16 possible arrangements for the number of heads. So, the answer is $\boxed{\boxed{\frac{7}{16}}}$[/hide]


algebra
function
domain
geometric series
partial fractions
complex analysis

Show that when $0<|z-1|<2$
$\frac{z}{(z-1)(z-3)}$= $-3$$\sum_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}}$- $\frac{1}{2(z-1)}$

I know that $\frac{1}{1-z}= \sum_{n=0}^{\infty}z^n$ and the sigularities are at z=1 and z=3, but I am confused that do  I need to do the decomposition of $\frac{z}{(z-1)(z-3)}?$ If so, what will be? and what to do with the domain of $0<|z-1|<2?$ Thx

Partial fractions. You should be able to expand $\frac{k}{(z-1)-2}$ as a geometric series, and the other piece is a single term.

I tried to do like this....

$\frac{z}{(z-1)(z-3)}=\frac{z}{(z-1)(z-1)-2)}$= $(-1)\frac{z}{1-z}$ $(-\frac{1}{2})\frac{1}{1-(\frac{z-1}{2})}$

Am I am on right track?? Cause it did not come with the same result as required.
Need some help, Thx

Don't write it as $\frac{k}{(z-1)-2}$ until after you've done the partial fractions for confusion's sake; use $\frac{k}{z-3}$ to do that algebra. We're expanding a proper fraction, so neither term will have a $z$ in the numerator; the general form is $\frac{a}{z-1}+\frac{b}{z-3}$, with $a$ and $b$ constants.


DO NOT TALK ABOUT CHAPTER SCORES!!!!!!!!!!!!!!!!!!!!!!!!

PLEASE!!!!!!!!

Lock please.  We don't get to discuss Chapter Competitions until they tell us we can.

oops! sorry guys!
ok, yeah can someone lock this please?
thank you


We have two lines: the first one has a points and the second one b points. How many intersections MAXIMUM are formed from the points of each line?
It is in fact a very easy problem but the formula that I figure out is wrong for examples that I try.:( Any help is welcomed!!!

Label the points on the first line $ M_1, \cdots, M_a$ and the points on the second line $ P_1, \cdots, P_b$.
Now let's fix the segment $ M_iP_k$.
It is intersected by $ M_jP_t$ if and only if $ j<i$ and $ t>k$, or $ j>i$ and $ t<k$. This gives $ (i\minus{}1)(b\minus{}k)\plus{}(a\minus{}i)(k\minus{}1)$ intersection points on the segment $ M_iP_k$. But each intersection point is counted twice since it can be obtain for $ (i,k)$ and $ (j,t)$, and for $ (j,t)$ and $ (i,k)$.
Thus, assuming that no three segments are concurrent (*), it follows that the total (maximal) number of intersection points is
$ \frac 1 2 \sum_{i\equal{}1}^a \sum_{k\equal{}1}^b ((i\minus{}1)(b\minus{}k)\plus{}(a\minus{}i)(k\minus{}1)) \equal{} \frac {ab(a\minus{}1)(b\minus{}1)} 4$.

It is easy to find configurations for which (*) is satisfied.

Pierre.

The maximum number of intersections occurs if no two intersection points created are the same, that is, the maximum number of quadrilaterals that can be formed with the points given on the two lines: $ \binom{a}{2}\binom{b}{2}\equal{}\frac{ab(a\minus{}1)(b\minus{}1)}{4}$.

Hopefully, I have understood the problem correctly.


no, where did you get the problems from?...

[quote="kaileyluvsmath93"]anyone have answers or solution to 1995 state sprint and target[/quote]

i probably have them. pm me. i would go look, but this couch is really comfy  :D


geometry
geometric transformation
trigonometry
rectangle
trig identities
Law of Cosines
area of a triangle

[img]http://img396.imageshack.us/img396/9897/hxath3.jpg[/img]

[color=darkred]Translation[/color]

$ ABCDEF$ is a regular hexagon and $ P$  is a fixed point inside .

If Area of $ \triangle PDE \equal{}4$  ,and Area of $ \triangle ABP \equal{}16$ , then

Area of regular hexagon =?

[hide]well, suppose a line segment perpendicular to AB is drawn through P, and let it intersect AB at X and DE at Y
then XY=AE because AB and DE are parallel and we are given $ \frac {AB \cdot XP}{2} \equal{} 16, \frac {AB \cdot PY}{2} \equal{} 4 \implies \frac {AB \cdot (XP \plus{} PY)}{2} \equal{} 20 \implies AB \cdot XY \equal{} 40 \implies AB \cdot AE \equal{} 40 \, (1)$
but by law of cosines (and other methods), $ AE \equal{} \sqrt {3} \cdot AB$
plugging into (1), we have $ AB \cdot AB \cdot \sqrt3 \equal{} 40 \implies (AB)^2 \equal{} \frac {40}{\sqrt3}$
but the area of the hexagon is just $ \frac {6(AB)^2\sqrt3}{4} \equal{} \frac {6 \cdot \frac {40}{\sqrt3} \cdot \sqrt3}{4} \equal{} \frac {6 \cdot 40}{4} \equal{} \boxed{60}$[/hide]

[hide]Since the area of a triangle is half the base times the height, we can see that $ [ABED] \equal{} 40$ . If we call the center of the hexagon point $ O$ we can see the $ [OED] \equal{} \frac{1}{6}[ABCDEF]$ since $ ABCDEF$ is regular. Triangle $ OED$ has the same base and half the height of rectangle $ ABED$ so its area is $ 40 \cdot \frac{1}{4} \equal{} 10$.

$ [ABCDEF] \equal{} 60$[/hide]

[hide="solution"]Let the height of the smaller triangle be $ h_1$, and let the height of the larger triangle be $ h_2$. We thus have $ \frac{sh_1}{2}\plus{}\frac{sh_2}{2}\equal{}\frac{s(h_1\plus{}h_2)}{2}\equal{}\frac{s^2\sqrt{3}}{2}\equal{}20$. Thus $ s^2\equal{}\frac{40\sqrt{3}}{3}$. The area of the hexagon is $ \frac{3s^2\sqrt{3}}{2}$, or $ \boxed{60}$.[/hide]


vector
complex numbers
complex analysis
complex analysis unsolved

Hey, been racking my brains all evening and still can't get an answer to these 2 problems, please help....

1. Show that the points are collinear in the complex plane if and only if:

(Z3-Z1)/(Z2-Z1) = a for some a in R (the reals)

2. Find all complex numbers that satisfy the following equations:
a.) modulo [(z-12)/(z-8i)]=5/3
b.) modulo [(z-4)/(z-8)]=1

Sorry it's not written out properly.... hope you can understand what it means, any help would be greatly appreciated...
Thanks... desperate math student x

Hints:

1. Try to consider $ Z_3 \minus{} Z_1$ and $ Z_2 \minus{} Z_1$ as vectors in the plane. The three points $ Z_1$, $ Z_2$, $ Z_3$ are collinear if and only if these vectors form the same angle with the $ x$-axis. How do you find these angles (hint: what is the argument of a complex number?), and how do you compare them (hint: $ \arg\left(\frac{x}{y}\right)\equal{}\arg x\minus{}\arg y$) ?

2. $ \left|\frac {x}{y}\right| \equal{} \frac {\left|x\right|}{\left|y\right|}$. Now what do $ \left|z \minus{} 12\right|$, $ \left|z \minus{} 8i\right|$, $ \left|z \minus{} 4\right|$, $ \left|z \minus{} 8\right|$ mean geometrically?

  darij


A few years ago I accidently discovered an approximation for pi.

My approximation is the following:   sqrt(987)/10.

It is a better approximation than 22/7.

Should I publish this result ????

Try this:
$\pi=\frac{4}{1}-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\cdots$
This is the Gregory-Leibniz series

[quote="mathnerd314"]Oh yeah? I have an even better approximation for pi:

$\frac{3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034}{10^{87}}$

No, I don't think there's much need for approximations.[/quote]

LOL  :lol:  No we don't really need approximations for mathematics unless you are doing engineering or other subjects that require you to actually to evaluate actual values of expressions.

thats all very well, but it is not a good approximation. You have approximated using another irrational number. if people don't have a calculator, it is easy to remember pi than the square root of 987

LOL at mathnerd's approximation. That is just pi to about 10-150 digits (I'm to lazy to count.)

The ability to read a simple "$87$" suffices...

The rational number $\frac{355}{113}$ is 200 times as accurate as your approximation - and it's been known (at least in China) as a good approximation for over 1500 years.

I use the 355/113 approximation if I ever require $\pi$ as such.

Thomas J. Osler's 'Get Billions and Billions of correct digits of pi from a Wrong Formula', is really interesting. It's not particularly useful by any means, but as I said, interesting.(Google it)

$\pi=\sqrt{6 \cdot \sum^{\infty}_{j=1}{\left(\frac{1}{j}\right)^{2}}}$

How's that for an approximation?

[quote="ZetaX"]All mentioned infinite series are rather bad for approximating $\pi$. Take a search here on the forum for other threads that already discussed this.[/quote]

Why? there are infinite series at equate to exactly pi.

Unless you mean infinte series as pi are very unpragmatic for everyday use, that i agree with.

Approximating numbers means to calculate (real) values that are as close to them as possible. And an approximation is better if for the same exactness we need less calculations.
In that sense, the sums mentioned here are bad as other ways of calculating converge much faster.

[quote="buzzer11"][quote="ZetaX"]All mentioned infinite series are rather bad for approximating $ \pi$. Take a search here on the forum for other threads that already discussed this.[/quote]

Why? there are infinite series at equate to exactly pi.

Unless you mean infinte series as pi are very unpragmatic for everyday use, that i agree with.[/quote]

Well, unless you're actually going to calculate the entire infinite series... :D

\[ \frac{1}{\pi}= \frac{2\sqrt{2}}{9801}\sum_{k=0}^{+\infty}\frac{(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}\]
This series, accredited to Ramanujan, converges rapidly (like 8 decimal places per iteration). Many supercomputers used to approximate $ \pi$ use this or a slightly different form to calculate billions or trillions of digits of $ \pi$.

Just the base case, $ k=0$ gives an approximation for $ 7$ digits of $ \pi$. Try it!

Source: Wikipedia

wow.. i would just use pi the symbol...  :heli:


For $ n\geq 2$, let $ a_n$ be an increasing integer sequence such that $ \sum_{k\equal{}1}^n \frac{1}{a_k}\equal{}1$. Prove that $ a_n$ has more than one even numbers.

[quote="kunny"]For $ n\geq 2$, let $ a_n$ be an increasing integer sequence such that $ \sum_{k \equal{} 1}^n \frac {1}{a_k} \equal{} 1$. Prove that $ a_n$ has more than one even numbers.[/quote]

Suppose that there is only one even number, $ a_i\equal{}2^i \cdot q$ with $ i \ge 1$ and odd $ q$,
Let $ d\equal{}LCM(a_1,\cdots,a_i,\cdots,a_n)\equal{}2^i \cdot LCM(a_1,\cdots,q,\cdots,a_n)\equal{}2^i \cdot o$ with odd $ o$ because all the members of $ \{a_1,\cdots,q,\cdots,a_n\}$ are odd.

then $ d \equal{} 2^i o\equal{}\sum_{k \equal{} 1}^n \frac {2^i \cdot o}{a_k}$ notice that all the terms are even but the $ a_i$ term which is odd. Contradiction.

Still pending to prove that there is no solution with all odd $ a_i$

There are many counter-examples.  For example,
\[ \frac{1}{3} \plus{} \frac{1}{5} \plus{} \frac{1}{7} \plus{} \frac{1}{9} \plus{} \frac{1}{11} \plus{} \frac{1}{15} \plus{} \frac{1}{35} \plus{} \frac{1}{45} \plus{} \frac{1}{231} \equal{} 1.\]


Find the x-and y-intercepts of each polynomial function below.

(1) f(x) = -1.2x^4 + 0.5x^2 -[sqrt{3}}(x) + 2

(2) f(x) = pi(x^5) + pi(x^4) + [sqrt{3}](x) + 1

this really belongs in high school basics.. mods?

Actually, this stuff is taught in middle school I think, but whatever.

[hide="hint"]What is an x-intercept?  What is a y-intercept?[/hide]

[hide="bigger hint"]When a line crosses the x-axis, y=0.  When a line crosses the y-axis, x=0.[/hide]

So, in both functions, I replace x with 0 and then y with 0 to find the x and y intercepts, right?

yeah just set x to 0 and solve for y to get the y-intercept, and then set y to 0 and solve for x to the the x-intercept


The interior of a drinking glass is a cylinder of diameter 8 cm and hight 12 cm. The glass is held at an angle of 45 degrees from the vertical and filled until the bass is just covered. How much water, in milliliters, is in the glass?

i am really not sure, but using my not too accurate intuition, the water has half the volume of the circumscribed cylinder (diameter 8 cm height 8 cm)

going from this...
$ V\equal{}\frac12 \pi \left( \frac82 \right)^2 8 \equal{} 64\pi ml$

a bit more reasoning to the first 2 lines:
seppose we have a cylinder with diamater 8 cm and height 8 cm.
distinguish 1 side as A and the other as B
hold the cylinder at 45 degrees to the vertical with A facing down and pour water so A is just completely covered
now so the same for face B
i think every point in the cylinder will be filled by the water for EXACTLY 1 base, and since the volumes are identical by symmetry, the water's volume should be one half of the cylinder

Your intuition seems accurate enough to me. That answer makes sense and is what I would have said.

Here is my solution,

When it is tilted at $ 45$ degrees we get a "side-on" triangle, this triangle has an angle of $ 45^{\circ}$ degrees as the water is parallel to the ground and by alternate angles.

So we want to know how high the water goes up to, we know one side and just need the other (let it be $ x$). So

$ tan 45^{\circ} = \frac {x}{8}$

$ x = 8 \times tan 45$

now we know that $ tan 45$ equals 1 as we just use the right angled isosceles triangle with sides 1 we get that, So

$ x = 8 \times 1$

$ x = 8$

So we have a bit of a weird object going on, but we can imagine this water space as half of the cylinder of 8cm high and 4cm radius, thus the volume of this cylinder is

$ V = A_{base} \times h$

$ = \pi r^2 \times h$

$ = \pi \times 4^2 \times 8$

$ = 128\pi$

Now this is for the cylinder, and not the water, but the water is half the volume of the cylinder, thus

$ \frac {1}{2} \times 128\pi$

$ = 64 \pi$

[quote="ocha"]The height of the cylinder is 12cm not 8cm
therefore only 1/3 of the glass is full[/quote]

If you are referring to this line

[quote="kurt.math"] ... but we can imagine this water space as half of the cylinder of 8cm high and 4cm radius, thus the volume of this cylinder is...[/quote]

This is the high that the water makes on that angle of 45 degrees in the cup.


We have the numbers$a_{1},a_{2},....,a_{7}$

$a_{1}=\frac{a_{2}+a_{7}}{2}\\ a_{2}=\frac{a_{1}+a_{3}}{2}$
 And so on...(we can arrange the numbers on a circle so each number is the AM of his adiacent numbers).
Prove that their sum is divisible by 7.

I have seem this question before, if I can remember, a[size=59]1[/size]=a[size=59]2[/size]=...=a[size=59]7[/size], therefore the sum of all the numbers equal 7a[size=59]1[/size], which is divisible by 7. :P

Thank you, but i already figured that out, i need to know why?  :)

We can say that the sequence $a_{1},a_{2},a_{3},...,a_{7}$ is an arithmetic sequence becuase every number is the average of its adjacent numbers.  Then it is easy to see that the common difference is $0$ becuase $a_{1}=\frac{a_{2}+a_{7}}{2}$.

We can also work as the following way:

a[size=59]2[/size]=(a[size=59]1[/size]+a[size=59]3[/size])/2
a[size=59]4[/size]=(a[size=59]3[/size]+a[size=59]5[/size])/2

Add them together we will get:
a[size=59]2[/size]+a[size=59]4[/size]=(a[size=59]1[/size]+a[size=59]5[/size])/2+a[size=59]3[/size]
4a[size=59]3[/size]=a[size=59]1[/size]+a[size=59]2[/size]+a[size=59]4[/size]+a[size=59]5[/size]

Hence we can deduce that:
2a[size=59]3[/size]=(a[size=59]1[/size]+a[size=59]5[/size])

Similarly, we can deduce that:
2a[size=59]6[/size]=(a[size=59]4[/size]+a[size=59]1[/size])
2a[size=59]7[/size]=(a[size=59]5[/size]+a[size=59]2[/size])

2(a[size=59]1[/size]+a[size=59]2[/size]+...+a[size=59]6[/size]+a[size=59]7[/size])
=2(a[size=59]1[/size]+a[size=59]2[/size]+a[size=59]4[/size]+a[size=59]5[/size])+2a[size=59]3[/size]+2(a[size=59]6[/size]+a[size=59]7[/size])
=8a[size=59]3[/size]+2a[size=59]3[/size]+2[a[size=59]4[/size]+a[size=59]1[/size]+a[size=59]5[/size]+a[size=59]2[/size]]
=8a[size=59]3[/size]+2a[size=59]3[/size]+4a[size=59]3[/size]
=14a[size=59]3[/size]

Therefore, (a[size=59]1[/size]+a[size=59]2[/size]+...+a[size=59]6[/size]+a[size=59]7[/size]=7a[size=59]3[/size]), which is divisible by 7.

I hope this can help you! :P

[b]Solution.[/b] Assume that there exists a minimal number, say $a_{1}=\frac{a_{2}+a_{7}}{2}$.  If any one of $a_{2},a_{7}>a_{1}$, then $a_{1}<\frac{a_{2}+a_{7}}{2}$.  As a result, $a_{1}=a_{2}=a_{3}=\hdots=a_{7}$.

And the conclusion follows.$\blacksquare$


limit
integration
calculus
trigonometry
calculus computations

How can i find the fallowing limit :
$\lim_{n\to \infty} 2^n \int_0^{\pi/2} cos(x)cos(2x)cos(4x)...cos(2^{n}x) dx$

multiply the inside of the integral by $\frac{ \sin x }{ \sin x}$, then using the 2's and the double angle formula, the integral becomes

$\lim_{n\rightarrow\infty} \int_{0}^{\frac{\pi}{2}} \frac{ \sin 2^{n+1}x }{\sin x} \, dx$

i don't know what to do now, but hopefully, you can get it from here  :P


please tell me what is Fuhrmann triangle?

thank you!

naser u can write it ,in Iranian community.(oonja behtar komaket mikonan)


calculus
integration
logarithms
trigonometry
function
derivative
real analysis

Just as the title suggests, this was an integral sent to me by a friend. He said this was found it off of a site related to yours.
My solution to $ \int_0^1\ln\left(\sqrt {1 - x} + \sqrt {1 + x}\right)\,dx$

[hide]

Let

$ I = \int_0^1\ln\left(\sqrt {1 - x} + \sqrt {1 + x}\right)\,dx$

$ \Rightarrow{2I = 2\int_0^1\ln\left(\sqrt {1 - x} + \sqrt {1 + x}\right)\,dx}$

$ = \int_0^1\ln\left(\left(\sqrt {1 - x} + \sqrt {1 + x}\right)^2\right)\,dx$

$ = \int_0^1\ln\left(2 + 2\sqrt {1 - x^2}\right)\,dx$

$ = \int_0^1\ln(2)\,dx + \int_0^1\ln\left(1 + \sqrt {1 - x^2}\right)\,dx$

Now for the second integral letting

$ z = \sqrt {1 - x^2}$

$ \Rightarrow\forall{x} > 0\quad{x = \sqrt {1 - z^2}}$

$ \Rightarrow{dx = \frac { - z}{\sqrt {1 - z^2}}}$

So we have

$ \int_0^1\frac {z\ln(1 + z)}{\sqrt {1 - z^2}}\,dz$

Now using integration by parts with $ u = \ln(1 + z)$ and $ dv = \frac {z}{\sqrt {1 - z^2}}\,dz$ Gives

$ = - \sqrt {1 - z^2}\ln(1 + z)\bigg|_0^1 + \int_0^1\frac {\sqrt {1 - z^2}}{1 + z}$

Now for this second integral we can calculate it indefinitely

Let

$ z = \sin(\theta)$

$ \Rightarrow{dz = \cos(\theta)d\theta}$

So we get

$ \int\frac {\cos^2(\theta)}{1 + \sin(\theta)}\,d\theta$

$ = \int\frac {1 - \sin^2(\theta)}{1 + \sin(\theta)}\,d\theta$

$ = \int\left\{1 - \sin(\theta)\right\}d\theta$

$ = \theta + \cos(\theta)$

Back subbing gives

$ = \arcsin(z) + \sqrt {1 - z^2}$

So now we see then that

$ \int_0^1\frac {\sqrt {1 - z^2}}{1 + z}\,dz$

$ = \left\{\arcsin(z) + \sqrt {1 - z^2}\right\}\bigg|_0^1$

$ = \frac {\pi}{2} - 1$

So putting this all together we get

$ \int_0^1\ln(2)\,dx + \int_0^1\ln\left(1 + \sqrt {1 - x^2}\right)\,dx$

$ = \ln(2) + \frac {\pi}{2} - 1$

$ = 2I$

$ \Rightarrow{I = \frac {\ln(2) + \frac {\pi}{2} - 1}{2}}$

$ \therefore\quad\boxed{\int_0^1\ln\left(\sqrt {1 - x} + \sqrt {1 + x}\right)\,dx = \frac {\ln(2) + \frac {\pi}{2} - 1}{2}}\quad\blacksquare$

[/hide]

[hide]Primitivating by parts you get

$ \int \ln(\sqrt{1\minus{}x} \plus{} \sqrt{1\plus{}x})\,dx \equal{} x \left[\ln(\sqrt{1\minus{}x} \plus{} \sqrt{1\plus{}x}) \minus{} \frac12\right] \plus{} \frac12 \arcsin x \plus{} C$.

Therefore,

$ \int_0^1 \ln(\sqrt{1\minus{}x} \plus{} \sqrt{1\plus{}x})\,dx \equal{} \left[x \left[\ln(\sqrt{1\minus{}x} \plus{} \sqrt{1\plus{}x}) \minus{} \frac12\right] \plus{} \frac12 \arcsin x \right]_0^1$

$ \equal{} \frac{\ln 2}{2} \minus{} \frac12 \plus{} \frac \pi 4$.[/hide]

[quote="Carcul"][hide]Primitivating by parts you get

$ \int \ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x})\,dx \equal{} x \left[\ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x}) \minus{} \frac12\right] \plus{} \frac12 \arcsin x \plus{} C$.

Therefore,

$ \int_0^1 \ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x})\,dx \equal{} \left[x \left[\ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x}) \minus{} \frac12\right] \plus{} \frac12 \arcsin x \right]_0^1$

$ \equal{} \frac {\ln 2}{2} \minus{} \frac12 \plus{} \frac \pi 4$.[/hide][/quote]

Primitivating? Integrating by parts. Well when I take

$ x\cdot\left[\ln\left(\sqrt{1\minus{}x}\plus{}\sqrt{1\plus{}x}\right)\right]'$ I sure don't get a function that nicely integrates to arcsine?

[quote="Carcul"]I am not understanding your doubt.[/quote]
No doubt, I am sure that you are correct, it is just then when I differentiate $ \ln\left(\sqrt{1\minus{}x}\plus{}\sqrt{1\plus{}x}\right)$ and then multiply it by x I do not get an integral that is anywhere near resembling the derivative of the arcsine function. This is all based upon the notion that you are in fact using integration by parts which may not be the case.

As I stated, I used integration by parts. Try it again.


Maybe this one realizes my goal. :D

Let $a, b, c$ be non-negative reals, and $k$ an arbitrary real. Prove that

\[
\sum_{sym} a^4b^2 + a^4bc + \left(\frac{1}{2}k^2-1\right)a^3b^3 + \left(-k^2-2k-2\right)a^3b^2c + \left(\frac{1}{2}k^2+2k+1\right)a^2b^2c^2
\]

is nonnegative. When is it 0? (Yes, there's a non-trivial equality case.)

----------

Maybe this one is easier. For any reals $a, b, c, k$,

\[
\sum_{sym} \frac{1}{2}k^2a^4 + \left(-k^2-2k\right)a^3b + \left(2k+2\right)a^2b^2 + \left(\frac{1}{2}k^2-2\right) a^2bc \ge 0
\]

[quote="peeta"]sorry telling u but i wouldnt call that a very pretty inequality  :dry: ;)[/quote]

Haha! I never claimed they were pretty, just symmetric in three variables, with interesting equality cases, probably nontrivial (though they both have quick solutions), etc. ;) Essentially they are me looking for truly new ideas, not ones that can be done based on ideas from the already vast Mathlinks lore. I wonder does anyone know where factoring in more than two variables is discussed? In particular, I am curious if there is some general algebraic interpretation of non-trivial equality cases.

For the easier I expand and I make a second degree equation with unknown the k and I want to sovle this in $\geq 0 $ Could anyone help me?
Sorry for my bad english :blush:

If this I wrote above is wrong could Mildorf give a hint?

I think you mean that you are using the discriminant principle. If my computations are correct, that yields the equivalence of this with the inequality

\[
\sum_{sym} a^6b^2-a^6bc-a^4b^4+2a^4b^3c-a^3b^3c^2 \ge 0
\]

I am not sure how to prove this one directly. Perhaps we could try something along the lines of writing it as $\sum_{sym} (a^4-a^2b^2)(a^2b^2-a^2bc) + a^4b^3c-a^3b^3c^2 \ge 0$.

Anyway, my solution is completely different - it uses Vornicu-Schur.

It is Vornicu-Schur, since it rewrites as

\[
\sum_{cyc} (b+c-ka)^2(a-b)(a-c) \ge 0
\]


For natural number $n,$ Let $S_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}.$ Prove the following equation for $n\geq 2.$
\[\sum_{k=2}^{n}\frac{1}{k(k-1)}S_{k}=2-\frac{1}{n}-\frac{1}{n}S_{n}\]

possible hint (I keep screwing up the solution and am not even sure if this would work):
[hide]
Find a closed form for $S_{n}$ and then put it into the summation, then split the summation into two separate summations, both of which telescope?

$S_{n}= \frac{n(n+1)}{2n}$

$\sum^{n}_{k=2}\frac{1}{k(k-1)}\times \sum^{n}_{k=2}\frac{k(k+1)}{2k}$[/hide]

EDIT: WOW, that $S_{n}$ is sooooo wrong, I also distributed addition over multiplication, please disregard.

[quote="max_tm"]possible hint (I keep screwing up the solution and am not even sure if this would work):
[hide]
Find a closed form for $S_{n}$ and then put it into the summation, then split the summation into two separate summations, both of which telescope?

$S_{n}= \frac{n(n+1)}{2n}$[/hide][/quote]

Hum, maybe you calculated as follows,

$S_{n}=\sum_{k=1}^{n}\frac{1}{k}= \frac{\sum_{k=1}^{n}1}{\sum_{k=1}^{n}k}=\frac{n}{\frac{n(n+1)}{2}},$ which is incorrect. :(

[quote="N.T.TUAN"][hide="hint"]Induction on $n$  :wink: [/hide][/quote]

It is also easy enough to do non-inductively: just use [hide]$\frac{1}{k(k-1)}= \frac{1}{k-1}-\frac{1}{k}$ twice and $S_{k}= S_{k-1}+\frac{1}{k}$ once.[/hide]

Yes, we can prove the equation directly. :lol:

Could someone who has a full direct solution please post it? Would be much appreciated!

Well, since you asked so nicely ;)

[hide]$X = \sum_{k}\frac{S_{k}}{k(k-1)}= \sum \left(\frac{S_{k}}{k-1}-\frac{S_{k}}{k}\right) = \sum \left(\frac{S_{k-1}}{k-1}-\frac{S_{k}}{k}+\frac{1}{k(k-1)}\right)$.

Now, the $\frac{S_{k}}{k}$ terms all telescope, leaving me with $X = \frac{S_{1}}{1}-\frac{S_{n}}{n}+\sum\frac{1}{k(k-1)}= 1-\frac{S_{n}}{n}+\sum\left(\frac{1}{k-1}-\frac{1}{k}\right) = 1-\frac{S_{n}}{n}+\frac{1}{1}-\frac{1}{n}$, where we used another telescoping sum, and that's exactly what we wanted.[/hide]


quadratics
Vieta
algebra
polynomial
quadratic formula

The quadratic $ 8x^2\plus{}12x\minus{}14$ has two real roots. What is the sum of the squares of these roots? Express your answer as a common fraction in lowest terms.

Remember that $ a^2 \plus{} b^2 \equal{} (a \plus{} b)^2 \minus{} 2ab$, so all we need to do is find the sum of the roots and the product of the roots and substitute into that factorization. By Vieta's the sum is $ \minus{}\frac{12}{8} \equal{} \minus{}\frac{3}{2}$, and the product is $ \minus{}\frac{14}{8} \equal{} \minus{}\frac{7}{4}$. Substituting gives us $ \frac{9}{4} \plus{} \frac{28}{4} \equal{} \frac{37}{4}$.

Hmm... I agree we have to use Vieta but the $-2ab$ part is equal to $\frac{7}{2}$ because the product multiplied by $-2$ is $\frac{-14}{8}\times(-2)=\frac{7}{2}$ So the answer is $\frac94+\frac72=\frac{23}4$

[hide="Solution"]
Q: The quadratic $8x^2+12x-14$ has two real roots. What is the sum of the squares of these roots? Express your answer as a common fraction in lowest terms.
A: The sum of the roots of a quadratic is $\frac{-b}{a}$, and the product is $\frac{c}{a}$. We seek to form the sum of the squares of the roots. Let the roots be a and b. We have $(a+b)^2-2ab=a^2+b^2$. We know the sum and the product:
$(\frac{-12}{8})^2-2(\frac{-14}{8})=\frac{23}{4}$.
[/hide]

[hide]One of the first things I'd do with this problem is cut all the coefficients in half to get $4x^2 + 6x - 7$.  If you need to factor it, it's easier to see, and if you want to churn it though the quadratic formula or Vieta, you'll get smaller numbers.

The other answers are right in the approach to the rest of the problem...the identity $a^2 + b^2 = (a+b)^2 - 2ab$ almost always helps when you're asked these kinds of questions.  Vieta will give you $a+b = -\frac{6}{4} = -\frac{3}{2}$ and $ab = \frac{-7}{4}$, and you just simplify:

$(a+b)^2 - 2ab$

$\left(-\frac{3}{2}\right)^2 - 2\cdot-\frac{7}{4}$

$\frac{23}{14}$[/hide]


calculus
integration
function
real analysis
topology
real analysis unsolved

Are all bounded open set  Jordan measurable?

I do not know whether the result stating that $f$ is Riemann-integrable if and only if it is continuous a.e still holds for the Riemann-Integral in dimension $n$, but if such is the case I think your statement is true (since the function $1_{O}$ (where $O$ is your bounded open set) is continuous except at the points belonging to the closure of at least one open ball $B_{i}$, where the countable union of the $B_{i}$ equals $O$. But the Lebesgue measure of those closures is $0$.

I didn't check all the stuff carefully.

A set is Jordan measurable iff its boundary is Jordan measurable and its Jordan measure is zero. This is not necessarily true for bounded open sets; not even in $\mathbb R$, actually.

Think about the complement of a so-called "fat Cantor set". You can construct a compact subset of $[0,1]$ just like you construct the usual Cantor middle-third set, except that at each step you take out increasingly smaller fractions of the set. You'll be left with a nowhere dense compact subset of $[0,1]$ with positive measure. The complement of this set is a bounded open set whose boundary (the fat Cantor described above) has positive measure.

Is a set Jordan measurable when  it is Lebesgue measurable?
and is its Jordan measure equal to its lebesgue measure?


I see that the solutions are posted, but I noted a small error on 'Solution 1' of Problem 1.
[quote]
102564 · 4 = 102564, which just happens to be the result of a digit slide on 102564.[/quote] is what is written, but actually it should be 102564 · 4 = [b]410256[/b]

Thanks for pointing that out - I fixed it.


geometry
geometric transformation
rotation

(sorry for my english)
a bar hung to an end can oscillate in a vertical plane. a bullet hits it to the other end sticking into it. the bar's lenght is 50 cm, its mass 1.5kg. the bullet has mass 0.050 kg and speed 50 m/s. the moment of inertia of the bar, with regard to its rotation axis, is I=(ML^2)/3.

i) determine the angular speed of the sistem just after the crash

ii) calculate how much the centre of mass of the sistem rises

This problem is very similar to the "balistic pendulum" problem. Anyway, for i) use conservation of angular momentum, and for ii) use, for example, conservation of mechanical energy.


http://www.artofproblemsolving.com/Forum/viewtopic.php?p=810637#810637

Two distinct positive real numbers A and B are chosen at random. B>A. What is the expected value of B/A?

Three distinct positive real numbers A, B, and C are chosen at random. C>B>A. What is the expected value of (A+B)/C? (A*B)/C?

What if the numbers must be integers?

Are there generalizations for when there are n numbers chosen?

I'm not sure how many of these questions are even possible to solve without being trivial...

As Joel told you, you need to specify a distribution over the reals or the integers. There is no uniform distribution.


Let ABC be a triangle. Angles A, B and C are acute. On sides BC, CA, AB we have points $A_1, B_2, C_3$. $AA_1$ ia a diameter of circle $T_1$, $BB_1$ - $T_2$, $CC_1$ - $T_3$. Prove that common chords of each two circles goes through the orthocentre of triangle ABC.

I can prove that this all 3 chords have common point, but I don't know how to prove that it is orthocentre..

Just focus on the first circle for now.

Let $D$ be the second point of intersection between this circle and $BC$. Since $AA_1$ is a diameter, we get $AD\perp BC$, so $AD$ is the altitude from $A$. This means that the power of $H$ wrt this circle is $HA\cdot HD$. Similarly, if $E,F$ are the feet of the altitudes from $B,C$ respectively, then the powers of $H$ wrt the other two circles are $HB\cdot HE$ and $HC\cdot HF$. On the other hand, we know that these three values are equal, so $H$ is precisely the radical center of the three circles, which is what we wanted to prove.


[hide]
Just did it digit by digit, until it repeated.

$ 0.3_{10}= 0.\overline{2046}_{7}$
[/hide]


Find all $ k\in\mathbb{Z}$ such that $ k^{3k\minus{}1}\plus{}k\plus{}1$ is a prime.

$ k^2\plus{}k\plus{}1 \mid k^{3k\minus{}1}\plus{}k\plus{}1 \implies k \equal{}1$

[quote="bboypa"]$ k^2 \plus{} k \plus{} 1 \mid k^{3k \minus{} 1} \plus{} k \plus{} 1 \implies k \equal{} 1$[/quote]

nice !


function
trigonometry
algebra unsolved
algebra

Determine the monoton functions $ f\colon \mathbb{R}\to\mathbb{R}$ which satisfy $ (f(\sin x))^{2}\minus{}3f(x)\plus{}2\equal{}0$ $ \forall x\in R$

_________________________________________
Azerbaijan Land of the Fire  :ninja:

[quote="Pirkuliyev Rovsen"]Determine the monoton functions $ f\colon \mathbb{R}\to\mathbb{R}$ which satisfy $ (f(\sin x))^{2} \minus{} 3f(x) \plus{} 2 \equal{} 0$ $ \forall x\in R$ [/quote]

Using $ x$ and $ x \plus{} 2\pi$ in the original equation, we get $ f(x \plus{} 2\pi) \equal{} f(x)$ and so $ f(x) \equal{} c$, since this is the only monotonic periodic function. Plugging this in the original equation, we get the only two solutions :

$ \boxed{f(x) \equal{} 1}$ $ \forall x$

$ \boxed{f(x) \equal{} 2}$ $ \forall x$


real analysis
inequalities
real analysis unsolved

Prove that $ L^{p}(\Bbb R) * L^{p}(\Bbb R)  \subseteq L^{r}(\Bbb R)$ where $ 1 \plus{} 1/r \equal{} 1/q \plus{} 1/p$ and $ *$ is the convolution operator. For $ r \equal{}\infty$ it is a classic problem, but for the rest... Any help is appreciated. Thanks

Young's Inequality for Convolution. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101943]Link.[/url]


For a function f : R-> R we denote
$ f_0(x) \equal{} x$
$ f_n(x) \equal{} f(f_{n\minus{}1}(x))$
Find all function f:R->R so that there exist a natural number n such that
$ f_n(x) \equal{} x$

[quote="HTA"]For a function f : R-> R we denote
$ f_0(x) \equal{} x$
$ f_n(x) \equal{} f(f_{n \minus{} 1}(x))$
Find all function f:R->R so that there exist a natural number n such that
$ f_n(x) \equal{} x$[/quote]

It's a crazy question, according to me, because there are many methods to specify such functions. The simpliest, IMHO :

Let $ \{A_z\}$ any partition of $ \mathbb R$ ($ A_z\cap A_t\equal{}\emptyset$ and $ \bigcup A_z\equal{}\mathbb R$) such that $ \exists M\in\mathbb N$ such that $ 0<|A_z|<M$

Let $ \preceq$ any total order on $ \mathbb R$ and $ \prec\_Inf$ the function giving the littlest element from a finite set of elements.

Define $ f(x)$ such that :
Let $ A(x)$ be the unique element $ A_t$ of the partition $ \{A_z\}$ such that $ x\in A_t$
Let $ B(x)\equal{}\{y\in A(x)$ such that $ y\neq x$ and $ y\succeq x\}$
If $ B(x)\equal{}\emptyset$, then $ f(x)\equal{}\prec\_Inf(A(x))$
If $ B(x)\neq\emptyset$, $ f(x)\equal{}\prec\_Inf(B(x))$

We have $ f_{M!}(x)\equal{}x$ $ \forall x$

It's easy to show that this is a general solution.


pigeonhole principle
vector
algorithm
algebra
polynomial
abstract algebra
geometry

Hi.

First consider the following trivial problem:

- given any x_1, ... , x_{n+1} from F_2^n (F_2 = integers mod 2), not necessarily distinct, we can find a non-empty subset of them which sums up to 0.

This can be easily solved by either pigeonhole principle or linear algebra over F_2. What about the following?

- given any x_1, ... , x_{2n+1} from F_3^n (F_3 = integers mod 3), not necessarily distinct, we can find a non-empty subset of them which sums up to 0.

Is it true? Thanks - it doesn't look too hard, but I can't seem to solve it.

[quote="darij grinberg"]Beautiful!!!!!!!!! More generally:

[color=blue][b]Theorem 1.[/b] If $ p$ is a prime, $ n$ is an integer, and $ x_1$, $ x_2$, ..., $ x_{\left(p \minus{} 1\right)n \plus{} 1}$ are $ \left(p \minus{} 1\right)n \plus{} 1$ elements of the vector space $ \mathbb{F}_p^n$, then there exists a non-empty subset $ T\subseteq\left\{1,2,...,\left(p \minus{} 1\right)n \plus{} 1\right\}$ such that $ \sum_{t\in T}x_t \equal{} 0$.[/color]

I have just put a proof of this on my website: See problem 2 in the note [url=http://www.cip.ifi.lmu.de/~grinberg/StPeters2003.pdf]St. Petersburg 2003: An alternating sum of zero-sum subset numbers[/url].

  darij[/quote]

Wow, thank you so much!  :lol:

This problem has been bugging me for quite a while, and I didn't realise it's so involved.

But now another thing bugs me: why does p have to be prime? If we modify the problem to (p-1)n+1 elements from (Z/pZ)^n, where p is not necessarily prime, does it not hold anymore?

Or, another interesting spin-off would be: is there an efficient algorithm to find such a subset?

[ Note : efficient = having runtime which is polynomial in n. ]

Clearly, when p=2 one can just use Gaussian elimination. What about higher p?

It turns out this question solves a very interesting problem I came across some time ago. Will post it and my thoughts later. :)

It's just another form of a known theorem.

Denote our $ N>(p\minus{}1)m$ vectors $ x_i\equal{}(a_1^i,a_2^i,\dots,a_m^i)$, $ i$ varies from 1 to $ N$. Then we need to find $ \epsilon_i\in\{0,1\}$ such that $ m$ linear forms $ l_k(\epsilon_1,\epsilon_2,\dots,\epsilon_N): \equal{}\sum_{i\equal{}1}^N a_i^k\epsilon_i\equal{}0$ vanish for $ k\equal{}1$, $ 2$, $ \dots$, $ m$. Also, not all $ \epsilon$'s must vanish.

It is a standard application of Combinatorial Nullstellensatz, maybe it is called Chevalley-Waring theorem.

Indeed, consider the polynomial \[ F(\epsilon_1,\epsilon_2,\dots,\epsilon_N)\equal{}(1\minus{}\epsilon_1)(1\minus{}\epsilon_2)\dots(1\minus{}\epsilon_N)\minus{}\prod_{k\equal{}1}^m (1\minus{}l_k^{p\minus{}1})\]

It's coefficient in $ \epsilon_1\epsilon_2\dots \epsilon_m$ does not vanish and this term is of maximal degree, hence there exists by CN some $ \epsilon_i\in\{0,1\}$ such that $ F$ does not vanish. Since all $ \epsilon$'s equal to 0 do not satisfy, some $ \epsilon$ equals to 1, hence first summand vanish, hence the second does not, hence all $ l_k$ vanish, as we desire.

Thanks, this is nice! No, the Combinatorial Nullstellensatz is not the same as Chevalley-Warning; actually, I have tried Chevalley-Warning, but it did not work. I should have tried the Combinatorial Nullstellensatz instead.

A little typo correction:

[quote="Fedor Petrov"]It's coefficient in $ \epsilon_1\epsilon_2\dots \epsilon_m$[/quote]

This should be "Its coefficient before $ \epsilon_1\epsilon_2\dots\epsilon_N$".

I am thinking about the case of non-prime $ p$. Lemma 3 in my note was formulated for rings and not just for fields $ \mathbb{F}_p$, but I am missing something like Lemma 0 for $ \mathbb{Z}\slash n\mathbb{Z}$ - a polynomial that is $ 1\mod n$ if $ n\mid x$ and $ 0\mod n$ otherwise. Well, such a polynomial (with integer coefficients) doesn't exist for any non-prime $ n$, but maybe something similar can be found.

I'll keep you informed.

  darij

Of course, it is a typo, thank you.

I do not remember the settings of all these theorems, since they follow from CN.  :)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43766]The Chevalley-Warning theorem[/url] in its strong form ("the number of solutions is divisible by $ p$") does not really follow from the CN. Also, there are [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160415]some stronger versions of the CN[/url] floating around, and if I am not mistaken, [url=http://en.wikipedia.org/wiki/Chevalley-Warning_theorem]some stronger versions of CW[/url] as well. It's not harmful to know more of these.

Anyway, did anyone try some combinatorial approaches? These could be useful for the case of non-prime $ p$.

@Lzw75: are you trying to find some paths in graphs? I'm just wondering...

  darij

Progress!

[color=blue][b]Problem 2 Extension #1.[/b] Let $ s$ be a prime power. Let $ m$ be an integer, and let $ n > \left( s \minus{} 1\right) m$ be an integer. Let $ a_{1}$, $ a_{2}$, ..., $ a_{n}$ be $ n$ elements of the $ \mathbb{Z}\slash s\mathbb{Z}$-module $ \left(\mathbb{Z}\slash s\mathbb{Z}\right)^{m}$. Prove that there exists a non-empty subset $ T\subseteq\left\{ 1,2,...,n\right\}$ such that $ \sum\limits_{t\in T}a_{t} \equal{} 0$.[/color]

What is happening here?

Actually I have tried to generalize my approach. We can consider $ a_{1}$, $ a_{2}$, ..., $ a_{n}$ as elements of $ \mathbb{Z}^m$ rather than as elements of $ \left(\mathbb{Z}\slash s\mathbb{Z}\right)^{m}$, and then we want to show that there exists a non-empty subset $ T\subseteq\left\{ 1,2,...,n\right\}$ such that $ \sum\limits_{t\in T}a_{t}$ is a vector with all components divisible by $ s$.

As an analogue of the equation (6) in my note, we have the identity
$ \sum_{T\subseteq\left\{1,2,...,n\right\}} \left( \minus{} 1\right)^{\left|T\right|} \prod_{j \equal{} 1}^m P\left(\sum_{t\in T}a_{t,j}\right) \equal{} 0$,
for any polynomial $ P\in\mathbb{Q}\left[X\right]$ of degree $ \leq s \minus{} 1$. (This follows from Lemma 3 since $ \prod_{j \equal{} 1}^m P\left(\sum_{t\in T}a_{t,j}X_t\right)$ is a polynomial of $ X_1$, $ X_2$, ..., $ X_n$ of degree $ \leq m\left(s \minus{} 1\right) < n$. Don't forget that $ a_{t,j}$ are integers here (rather than residues as in the note), and the identity above is an identity about integers.)

The reason why I had chosen the polynomial $ P\left(X\right) \equal{} 1 \minus{} X^{s \minus{} 1}$ in my note is its following property: $ P\left(x\right)$ is divisible by $ s$ if $ x$ is not divisible by $ s$, and $ \equiv 1\mod s$ if $ x$ is divisible by $ s$. (This is basically Lemma 0 of the note.) If we assume that there is no non-empty set $ T\subseteq\left\{ 1,2,...,n\right\}$ such that $ \sum\limits_{t\in T}a_{t}$ is a vector with all components divisible by $ s$, then the sum
$ \sum_{T\subseteq\left\{1,2,...,n\right\}} \left( \minus{} 1\right)^{\left|T\right|} \prod_{j \equal{} 1}^m P\left(\sum_{t\in T}a_{t,j}\right)$
would thus consist of a lot of terms divisible by $ s$ and exactly one term not divisible by $ s$ (namely, the term corresponding to $ T \equal{} \emptyset$), and hence it would be not divisible by $ s$, contradicting to it being $ \equal{} 0$.

How can this be generalized to $ s$ being a prime power, say $ s \equal{} p^{\alpha}$ for $ p$ prime? All we need is a polynomial $ P\in\mathbb{Z}\left[X\right]$ of degree $ \leq s \minus{} 1$ such that there exists an integer $ \beta$ such that

$ p^{\beta \plus{} 1}\mid P\left(x\right)$ for integers $ x$ such that $ p^{\alpha}\nmid x$;
$ p^{\beta}\parallel P\left(x\right)$ for integers $ x$ such that $ p^{\alpha}\mid x$.
(Here, $ p^{\beta}\parallel u$ is just an abbreviation for $ p^{\beta}\mid u$ and $ p^{\beta \plus{} 1}\nmid u$; in words: $ p^{\beta}$ is the highest power of $ p$ that divides $ u$.)

If we have such a polynomial $ P$, then we can argue as follows: Assume that there is no non-empty set $ T\subseteq\left\{ 1,2,...,n\right\}$ such that $ \sum\limits_{t\in T}a_{t}$ is a vector with all components divisible by $ s$. Then, the sum
$ \sum_{T\subseteq\left\{1,2,...,n\right\}} \left( \minus{} 1\right)^{\left|T\right|} \prod_{j \equal{} 1}^m P\left(\sum_{t\in T}a_{t,j}\right)$
consists of a lot of terms divisible by $ p^{m\beta \plus{} 1}$ (namely, all terms corresponding to non-empty subsets $ T\subseteq\left\{ 1,2,...,n\right\}$) and exactly one term not divisible by $ p^{m\beta \plus{} 1}$ (namely, the term corresponding to $ T \equal{} \emptyset$; the highest power of $ p$ that divides it is $ p^{m\beta}$), so that this sum cannot be $ 0$, contradicting

$ \sum_{T\subseteq\left\{1,2,...,n\right\}} \left( \minus{} 1\right)^{\left|T\right|} \prod_{j \equal{} 1}^m P\left(\sum_{t\in T}a_{t,j}\right) \equal{} 0$.

A polynomial $ P$ satisfying our needs is the following one:

$ P\left(X\right) \equal{} \binom{X \minus{} 1}{p^{\alpha} \minus{} 1}$

(remember, $ p^{\alpha} \equal{} s$). I leave it to the reader to show that this polynomial does what we want from it (hint: $ \beta\equal{}0$; use that $ \binom{X \minus{} 1}{p^{\alpha} \minus{} 1} \equal{} \frac {p^{\alpha}}{X}\cdot\binom{X}{p^{\alpha}}$).

Now the question is whether we can find similar polynomials $ P$ for generic (non-prime-power) $ s$, or whether we should try finding counterexamples instead.

  darij

Thanks for the two beautiful proofs. I'm particularly impressed with the use of the Combinatorial Nullstellensatz.

[quote="darij grinberg"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43766]
@Lzw75: are you trying to find some paths in graphs? I'm just wondering...

  darij[/quote]

The problem I had in mind was as follows (source forgotten: it's somewhere on the Net) :

- Find a positive perfect cube $ x > 1$, such that $ \sigma(x)$, the sum of the positive divisors of $ x$, is also a perfect cube.

The approach I had in mind is as follows:

(i) Prepare two lists of small primes $ S$ and $ T$, where $ S \equal{} \{2,3,5,... \}$ has $ 2n \plus{} 1$ elements and $ T \equal{} \{2,3,5,.... \}$ has $ n$ elements. They can be any sets of primes, although it's advantageous to pick smaller ones.
(ii) For each prime $ p \in S$, find an appropriate power $ p^{3r}$, such that $ \sigma(p^{3r})$ can be written as a product of primes in $ T$.
(iii) Each $ \sigma(p^{3r})$ can then be written as a product of a cube, and prime powers $ q_i^{a_i}$, where each $ q_i \in T$ and each $ a_i \equal{} 0, 1, 2$. Thus we obtain $ 2n \plus{} 1$ vectors of the form $ (a_1, \dots, a_n) \in F_3$. An algorithmic solution to the problem I posed would help us find a solution to this problem.

Note: I might be using a sledgehammer to kill a fly here, but this method was inspired from the method of quadratic sieve used in prime factoring.

Unfortunately, I have no experience with prime factoring algorithms, but I wouldn't compare prime factoring with a fly...

Anyway I am posting here again in order to say that what I proved above (the case of $ s$ being a prime power) was already shown in
http://www.mathlinks.ro/viewtopic.php?t=6015 :

[quote="vess"]Our problem is an easy corollary. We will show, more generally (the problem is a special case with $ k \equal{} p \minus{} 1$) that if we are given  more than $ k(p \minus{} 1)$ vectors in $ \mathbb{F}_p^k$, then some of them have sum $ 0$. Let $ s \equal{} 1 \plus{} k(p \minus{} 1)$, and let $ v_i \equal{} (a_{i1},\ldots,a_{ik})$, $ 1 \leq i \leq s$, be the given vectors. Consider the $ k$ polynomials on $ s$ variables
\[ f_j(x_1,\ldots,x_s) : \equal{} \sum_i a_{ij} x_i^{p \minus{} 1}.
\]
These polynomials fulfil the hypothesis of the Chevalley-Warning Theorem, for the sum of their degrees is $ k(p \minus{} 1)$, which is less than the number $ s \equal{} 1 \plus{} k(p \minus{} 1)$ of variables. Note that the system $ f_1 \equal{} \cdots \equal{} f_k \equal{} 0$ has the trivial solution $ x_1 \equal{} \cdots \equal{} x_s \equal{} 0$. By the Chevalley-Warning theorem, it has another one, say $ (\alpha_1,\ldots,\alpha_s)$. Now, set $ I : \equal{} \{i \, | \, \alpha_i \neq 0\}$. This set is non-empty and, by Fermat's Little Theorem, we have
\[ \sum_{i \in I} v_i \equal{} (0,\ldots,0),
\]
as desired. The proof is complete.[/quote]

I don't see yet what this all implies for $ s$ not being a prime power, but the topic is one single treasury.

[b]EDIT:[/b]

[b]1.[/b] Olson's theorem (see the topic linked above) is a generalization of the case of $ s$ being a prime power. I will check whether my proof generalizes to a proof of Olson's theorem.

[b]2.[/b] Another theorem by Olson states that if $ a$ and $ b$ are positive integers satisfying $ a\mid b$, then from $ a \plus{} b \minus{} 1$ vectors in $ \left(\mathbb{Z}\slash \left(a\mathbb{Z}\right)\right)\times\left(\mathbb{Z}\slash\left(b\mathbb{Z}\right)\right)$, we can always choose a non-empty subset with zero sum, at least [url=http://mathworld.wolfram.com/DavenportConstant.html]according to this link[/url]. [url=http://garden.irmacs.sfu.ca/?q=op/davenports_constant]Another link[/url] states the same but forgetting the (important!) condition $ a\mid b$.

[b]3.[/b] The case of $ s$ being not a prime power seems to be [url=http://garden.irmacs.sfu.ca/?q=op/davenports_constant]an open problem[/url].

  darij

I was delighted to discover this website, and to see such nice discoveries taking place here.

One small correction though: in Grinberg's previous post, specifically Edit #2, the following theorem of Olson is referred to:  Every sequence of $ a\plus{}b\minus{}1$ elements from the group $ {\mathbb Z}_a \times {\mathbb Z}_b$ has a nontrivial subsequence which sums to zero (here $ {\mathbb Z}_a \equal{} {\mathbb Z}/a{\mathbb Z}$).  I mention this theorem [url=http://garden.irmacs.sfu.ca/?q=op/davenports_constant]here[/url].  It is suggested that the condition $ a|b$ is needed for this theorem, but it is not.. indeed it's an easy consequence of the basic structure theorem for abelian groups that for every pair of positive integers $ a,b$, there exist $ m,n$ so that $ m|n$ and $ {\mathbb Z}_a \times {\mathbb Z}_b \cong {\mathbb Z}_m \times {\mathbb Z}_n$.

Cheers,
Matt

Hello Matt,

[quote="mdevos"]One small correction though: in Grinberg's previous post, specifically Edit #2, the following theorem of Olson is referred to:  Every sequence of $ a \plus{} b \minus{} 1$ elements from the group $ {\mathbb Z}_a \times {\mathbb Z}_b$ has a nontrivial subsequence which sums to zero (here $ {\mathbb Z}_a \equal{} {\mathbb Z}/a{\mathbb Z}$).  I mention this theorem [url=http://garden.irmacs.sfu.ca/?q=op/davenports_constant]here[/url].  It is suggested that the condition $ a|b$ is needed for this theorem, but it is not.. indeed it's an easy consequence of the basic structure theorem for abelian groups that for every pair of positive integers $ a,b$, there exist $ m,n$ so that $ m|n$ and $ {\mathbb Z}_a \times {\mathbb Z}_b \cong {\mathbb Z}_m \times {\mathbb Z}_n$. [/quote]

I think there was a slight misunderstanding here. Of course, every group of the form $ {\mathbb Z}_a \times {\mathbb Z}_b$ can be rewritten as $ {\mathbb Z}_m \times {\mathbb Z}_n$ with $ m\mid n$, but your formulation can be understood as if we don't need to rewrite it in this form. But this is wrong: $ m \plus{} n \minus{} 1$ is, in general, distinct from $ a \plus{} b \minus{} 1$. For instance, the group $ \mathbb{Z}_2\times\mathbb{Z}_3$ has Davenport constant $ 6$, although $ 2 \plus{} 3 \minus{} 1 \equal{} 4\neq 6$. That's what I was referring to. I hope it becomes clearer now (it's past midnight here and my skills at explaining are not too great).

  darij

I was writing without thinking.. you are completely correct.


Let $C$ be a circle touching the edges of an angle $\angle XOY$, and let $C_{1}$ be a circle touching the same edges and passing through the center $C$. Let $A$ be the second endpoint of the diameter of $C_{1}$ passing through the center of $C$, and let $B$ be the intersection of this diameter with $C$. Prove that the circle centered at $A$ passing through $B$ touches the edges of $\angle XOY$

Let $\mathcal C$ and $\mathcal C_{1}$ touches OX at point D and E respectively, and construct $AF \perp OX$. Since $C_{1}$ is the midpoint of AC, and $CD \parallel BE \parallel AF$, so we have $2BE=AF+CD$. Let the radius of $\mathcal C$ be $r$ and the radius of $\mathcal C_{1}$ be $R$, then AF can be expressed as $2R-r$. But since $AB=AC-CB=2R-r$, so B lies on the circle centred at A with radius AF. Because $AF \perp OX$, hence this circle is tangent to OX, and this problem is solved.


Hi!

Let $ x_1,x_2,\cdots,x_n$ be positive real numbers. Prove that

\[ \sum_{k\equal{}1}^n x_k^3\geq \sum_{k\equal{}1}^n x_k x_{k\plus{}1}^2\]

where $ x_{n\plus{}1}\equal{}x_1$.

[quote="yuyang"]Hi!

Let $ x_1,x_2,\cdots,x_n$ be positive real numbers. Prove that
\[ \sum_{k \equal{} 1}^n x_k^3\geq \sum_{k \equal{} 1}^n x_k x_{k \plus{} 1}^2
\]
where $ x_{n \plus{} 1} \equal{} x_1$.[/quote]
Using Holder inequality:
$ (x_1^3\plus{}x_2^3\plus{}...\plus{}x_n^3)(x_1^3\plus{}x_2^3\plus{}...\plus{}x_n^3)(x_2^3\plus{}x_3^3\plus{}...\plus{}x_n^3\plus{}x_1^3) \ge (x_1^2x_2\plus{}x_2^2x_3\plus{}...\plus{}x_n^2x_1)^3$

Possible solution is by summation inequalities of form: $ \frac {a^{3} \plus{} b^{3} \plus{} b^{3}}{3}\geq ab^{2}$ which is true by $ AM \geq GM$

Or just use the Rearrangement inequality ($ x^2\le y^2\iff x\le y$):)


ok number one

how many boxes are in an array of 3x3 rectangles_____________
                                                                  |___|___|____|
                                                                  |___|___|____|
                                                                  |___|___|____|

and number 2
                                                                /---
if a, b, and c are positive integers such that a\/b             9
                                                            --------= ---/-----
                                                                c         2\/ 3

what is the value of a plus b plus c? givin that a and c have no common prime factors, and b has no perfect square factor other than 1?

i seriously need help!!!!! please reply quickly thanks

Eh.. It would be really helpful if you use capital letters when starting a sentence.
Also, try to learn how to use LaTeX.. It's really hard to make understandings out of your questions..

Anyways, I somehow understood your #1, but not #2...
Pretend this picture is what you have.. even though it's really a square..
[asy]size(150); pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A=(-3,3); B=(-1,3); C=(1,3); D=(3,3); draw(A--D); E=(-3,1); F=(-1,1); G=(1,1); H=(3,1); draw(E--H); I=(-3,-1); J=(-1,-1); K=(1,-1); L=(3,-1); draw(I--L); M=(-3,-3); N=(-1,-3); O=(1,-3); P=(3,-3); draw(M--P); draw(A--M); draw(B--N); draw(C--O); draw(D--P);[/asy]

I think this was posted already...
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170089[/url]

Anyways, the formula for m by n grid is $ \binom{n\plus{}1}{2}\cdot\binom{m\plus{}1}{2}$

So use that to get the answer, $ \boxed{36}$

MathMaster:  Tell where the problems are from, so we can understand what the problem is.


calculus
integration
limit
real analysis
real analysis unsolved

Let $ \{a_n\},n\equal{}0,1,\cdots$ be real sequence such that
(1) $ \sum_{n\equal{}0}^{\plus{}\infty}a_nx^n$ converges for every $ x \in [0,\plus{}\infty)$
(2)$ \sum_{n\equal{}0}^{\plus{}\infty}n!a_n$ converges.

Prove,the improper  integeral $ \int_0^{\plus{}\infty} e^{\minus{}x}\sum_{n\equal{}0}^{\plus{}\infty}a_nx^ndx$ exist,
and, $ \int_0^{\plus{}\infty} e^{\minus{}x}\sum_{n\equal{}0}^{\plus{}\infty}a_nx^ndx\equal{}\sum_{n\equal{}0}^{\plus{}\infty}n!a_n$

I can only prove this problem with the stronger condition that $ \sum_{n\equal{}0}^{\plus{}\infty}n!a_n$ is absolutely converges.Can somebody help me to prove or disprove this theorem.Thanks in advance.

The first thing I would check is whether the sequence $ \frac 1{n!}\int_0^T x^ne^{\minus{}x}\,dx$ is monotone for every $ T>0$. If it is, you are done by Abel-Dirichlet. If it isn't, you should check if its variation is uniformly bounded in $ T$. If "yes", you are done again. If "no", the theorem is false. To check monotonicity, I would just integrate by parts once. By the way, (2) clearly implies (1), so whoever formulated the problem did a rather poor job.

Replacing $ a_n$ by $ \frac {a_n}{n!}$, you may obtain the eqivalent problem:
Let $ \{ a_n \}_{n = 0}^{\infty}$ be a sequence of real numbers such that $ \sum_{n = 0}^{\infty} a_n$ converges to $ S$, then $ \int_{0}^{\infty} e^{ - x} \left( \sum_{n = 0}^{\infty} \frac {a_n}{n!}x^n \right) \, dx$ exists and is equal to $ S$.

The answer is 'yes'.

[hide="Proof"]Consider $ f(R) = \int_{0}^{R} e^{ - x} \left( \sum_{n = 0}^{\infty} \frac {a_n}{n!}x^n \right) \, dx$, where $ R$ is a positive real number.

Since $ \sum_{n = 0}^{\infty} \frac {a_n}{n!}x^n e^{ - x}$ converges uniformly on $ [0, R]$, the order of summation and integration in $ f(R)$ can be exchanged, giving $ f(R) = \sum_{n = 0}^{\infty} \int_{0}^{R} \frac {a_n}{n!}x^n e^{ - x} \, dx$.

Then one can show that $ S - f(R) = \sum_{n = 0}^{\infty} \int_{R}^{\infty} \frac {a_n}{n!}x^n e^{ - x} \, dx = e^{ - R} \sum_{n = 0}^{\infty} \left( \sum_{k = 0}^{n} \frac {R^k}{k!} \right) a_n$ by applying integration by parts successively.

Let $ s_n = \sum_{k = 0}^{n} a_k$ be partial sum of $ a_n$, with $ s_{ - 1} = 0$. Using summation by parts, $ S - f(R) = e^{ - R} \sum_{n = 0}^{\infty} (S - s_{n - 1}) \frac {R^n}{n!}$.

Since $ s_n \to S$ as $ n \to \infty$, for $ \epsilon > 0$, we can choose $ N$ such that $ |S - s_{n - 1}| < \frac {\epsilon}{2}$ whenever $ n \geq N$. Then

$ \begin{eqnarray*}
|S - f(R)|
& \leq & e^{ - R} \sum_{n = 0}^{\infty} |S - s_{n - 1}| \frac {R^n}{n!} \\
& \leq & e^{ - R} \left( \sum_{n = 0}^{N} |S - s_{n - 1}| \frac {R^n}{n!} \right) + e^{ - R} \sum_{n = N+1}^{\infty} |S - s_{n - 1}| \frac {R^n}{n!} \\
& \leq & \underset{0 \leq n \leq N}{\max} \left\{\frac {R^n}{n!} e^{ - R} \right\} \left( \sum_{n = 0}^{N} |S - s_{n - 1}| \right) + e^{ - R} \sum_{n = N + 1}^{\infty} \frac{\epsilon}{2} \frac {R^n}{n!} \\
& \leq & \underset{0 \leq n \leq N}{\max} \left\{\frac {R^n}{n!} e^{ - R} \right\} \left( \sum_{n = 0}^{N} |S - s_{n - 1}| \right) + \frac {\epsilon}{2} \quad \quad \cdots (\ast) \end{eqnarray*}.$

Choosing $ K$ sufficiently large, we have $ \underset{0 \leq n \leq N}{\max} \left\{\frac {R^n}{n!} e^{ - R} \right\} \left( \sum_{n = 0}^{N} |S - s_{n - 1}| \right) < \frac {\epsilon}{2}$ for $ R > K$.

Together with $ (\ast)$, we have $ |S - f(R)| < \epsilon$ for sufficiently large $ R$.

Therefore, $ \int_{0}^{\infty} e^{ - x} \left( \sum_{n = 0}^{\infty} \frac {a_n}{n!}x^n \right) \, dx = \lim_{R\to\infty} f(R)$ exists and is equal to $ S$ as desired.[/hide]

Thank you,fedja and sos440.
I agree with you that (2) implies (1),your argument works well for the sequence is  monotone,Thank you.

And thank you for your detail proof sos440.

thanks for your kindness.
i asked zhaobin the problem  in china's bbs about math.
partly for my poor english,zhaobin help me asked the problem there.


Έφτιαξα σήμερα αυτό το ωραίο αλλά εύκολο πρόβλημα .

Αν $ a,b,c>0$ τέτοιους ώστε $ abc\equal{}1$ να αποδειχθεί ότι

$ \sum a\plus{}\sum\frac{32a}{a\plus{}1}\geq 51$

[quote="silouan"]Έφτιαξα σήμερα αυτό το ωραίο αλλά εύκολο πρόβλημα .

Αν $ a,b,c > 0$ τέτοιους ώστε $ abc \equal{} 1$ να αποδειχθεί ότι

$ \sum a \plus{} \sum\frac {32a}{a \plus{} 1}\geq 51$[/quote]

to provlhma anagetai sto eksis:

$ a,b,c>0$ me $ abc\equal{}1$ n.d.o.

$ \sum {\frac{a}{a\plus{}1}}\geq \frac{3}{2}$ to opoio isxyei kai boreis na to apodeikseis kai me pollaplasiastes lagrange nomizo kai kana diko sou kolpo...

den stekomai polu se auto gt to provlhma borei na lyhtei kai theorontas mia synartisi tis morfis

$ f(x)\equal{}x\plus{}\frac{32x}{x\plus{}1}\plus{}\lambda ln x$. H paragogos einai thetiki, ara i syanrtisi gnisios auksousa kai $ f(x)\geq f(1)\equal{}17$ prosthetoume kata meli kai qed.

So did I. $ \lambda \equal{}\minus{}9\Longrightarrow f'(x)\equal{}\frac{(x\minus{}1)(x\minus{}3)^2}{x(x\plus{}1)^2}$, yielding $ f(x)\geq f(1)\equal{}17$.

yes kunny... i think that this is the standard method for proving inequalities of this forms with the condition $ abc\equal{}1$.

Εγώ πιστεύω ο Σιλουανός έβαλε το πρόβλημα με σκοπό να το ανάγουμε στο στυλ αυτού:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207043[/url]
(Sil έχω δίκιο?)

Τέλος πάντων προσωπικά αυτό προσπάθησα να κάνω και βγήκε μια συναφής μορφή. Συγκερκιμένα θεωρώντας $ a \equal{} \frac {x}{y}$ και τα λοιπά έχουμε:

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \plus{} 32(\frac {x}{x \plus{} y} \plus{} \frac {y}{y \plus{} z} \plus{} \frac {z}{z \plus{} x})\geq 51\iff$

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \minus{} 3\geq 32(\frac {y}{x \plus{} y} \plus{} \frac {z}{y \plus{} z} \plus{} \frac {x}{z \plus{} x} \minus{} \frac {3}{2})$

Ισως βοηθάει επίσης ότι
$ RHS \equal{} 16(\frac {y \minus{} x}{x \plus{} y} \plus{} \frac {z \minus{} y}{y \plus{} z} \plus{} \frac {x \minus{} z}{z \plus{} x}) \equal{} 16\frac {(x \minus{} y)(y \minus{} z)(z \minus{} x)}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}$

Από εδώ δεν μπορώ να προχωρήσω... :huh:

Nick κάποιες παρατηρήσεις πάνω στο ποστ σου :

[quote="Nick Rapanos"]  to provlhma anagetai sto eksis:

$ a,b,c > 0$ me $ abc \equal{} 1$ n.d.o.

$ \sum {\frac {a}{a \plus{} 1}}\geq \frac {3}{2}$ to opoio isxyei [/quote]

Αυτό είναι λάθος . Για $ a,b,c>0$ με $ abc\equal{}1$ ισχύει $ \sum\frac{a}{a\plus{}1}\leq\frac{3}{2}$

(Μπορείτε να το αποδείξετε ?  :wink: )

Επίσης εκεί που λες ότι η f' είναι θετική δεν ισχύει αυτό .Αλλά πιστεύω ότι αυτό ήταν ένα typo και νομίζω ότι έχεις κάνει αυτό που κάνει ο kunny οπότε οκ.
@kunny : nice kunny ! :)

silouan tin patisa... :blush:  eida oti einai akrotato me tous pollaplasiastes kai to parathsa. gia to allo apla den egrapsa poio einai to katallhlo λ (stin periptosi mas to 9).

[quote="NickNafplio"]Παιδια εγω τωρα πριν λιγες μερες αρχισα να διαβαζω το βιβλιο του κ. Στεργιου για να εξασκηθω στις ανισοτητες. Στη συγκεκριμενη προσπαθησα αρχισα να αντιστρεψω αφου κανω καποιους μετασχηματισμους και να παω με AM - HM και ενω αρχικα φαινεται οτι κατι παει να βγει τελικα δεν βγαινει, το ιδιο και με Andresscu που προσπαθησα... Επειδη ειμαι λιγο αρχαριος ομως εκει με το λ τι ακριβος κανετε; Με προσθεση κατα μελη των σχεσεων το αθροισμα λογαριθμων γινεται λογαριθμος γινομενου δηλαδη ln1 = 0 και χανεται, αλλα δεν μπορω να καταλαβω γιατι στη λυση παιρνετε μονο την περιπτωση οπου x>1 αφου δεν ειναι υποχρεωτικα x>1, μπορει να ειναι πχ a=1, b=2, c=1/2...[/quote]

Α οκ, νομιζω οτι καταλαβα, για λ=-9 η f παρουσιαζει ελαχιστο στο 1

[quote="silouan"]Nick κάποιες παρατηρήσεις πάνω στο ποστ σου :

[quote="Nick Rapanos"]  to provlhma anagetai sto eksis:

$ a,b,c > 0$ me $ abc \equal{} 1$ n.d.o.

$ \sum {\frac {a}{a \plus{} 1}}\geq \frac {3}{2}$ to opoio isxyei [/quote]

Αυτό είναι λάθος . Για $ a,b,c > 0$ με $ abc \equal{} 1$ ισχύει $ \sum\frac {a}{a \plus{} 1}\leq\frac {3}{2}$

(Μπορείτε να το αποδείξετε ?  :wink: )
[/quote]
Είσαι σίγουρος για αυτό Silouan???
Παρε $ a\equal{}\frac{1}{6},b\equal{}3,c\equal{}2$  :wink:


find all integers a, b , c  such that:

 $c | ab -1$

$a | bc - 1$

$b | ac - 1$

This is already on the forum : http://www.mathlinks.ro/Forum/viewtopic.php?t=792 :)


Decode  the following quote by the famous humorist and science fiction novelist Douglas Adams.

BEMJM  XL  H  BEMTJP  ZEXWE  LBHBML  BEHB  XO  MQMJ  HIPDTGP  GXLWTQMJL  MAHWBUP   ZEHB  BEM  KIXQMJLM  XL  OTJ  HIG  ZEP  XB  XL  EMJM,  XB  ZXUU  XILBHIBUP  GXLHFFMHJ  HIG  DM  JMFUHWMG  DP  LTRMBEXIV  MQMI  RTJM  DXSHJJM  HIG  XIMAFUXWHDUM. BEMJM  XL  HITBEMJ  BEMTJP  ZEXWE  LBHBML  BEHB  BEXL  EHL  HUJMHGP  EHFFMIMG.

[hide="hint 1"]This is a simple substitution cipher.[/hide]

[hide="hint 2"]E=H[/hide]

THERE IS A THEORY WHICH STATES THAT IF EVER ANYBODY DISCOVERS EXACTLY WHAT THE UNIVERSE IS FOR AND WHY IT IS HERE, IT WILL INSTANTLY DISAPPEAR AND BE REPLACED BY SOMETHING EVEN MORE BIZARRE AND INEXPLICABLE. THERE IS ANOTHER THEORY WHICH STATES THAT THIS HAS ALREADY HAPPENED.


Given $ a, b, c > 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that:
\[ \frac{8}{3\sqrt{3}}(ab\plus{}bc\plus{}ca)\sqrt{ab\plus{}bc\plus{}ca} \ge (a\plus{}bc)(b\plus{}ca)(c\plus{}ab)\]
 :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {8}{3\sqrt {3}}(ab \plus{} bc \plus{} ca)\sqrt {ab \plus{} bc \plus{} ca} \ge (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)\]
:)[/quote]
observe $ a\plus{}bc\equal{}(a\plus{}b)(a\plus{}c)$. Putting $ x\equal{}\frac1a$ etc, we obtain after squaring
$ xyz\left(\frac{x\plus{}y\plus{}z}{3}\right)^3\ge\prod\left(\frac{x\plus{}y}{2}\right)^2$ which is Popoviciu :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {8}{3\sqrt {3}}(ab \plus{} bc \plus{} ca)\sqrt {ab \plus{} bc \plus{} ca} \ge (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)\]
:)[/quote]
For the proof enough to check only two cases:
1) $ b \equal{} c$ ;
2)$ c \equal{} 0.$ :wink:
[quote="spanferkel"]
observe $ a \plus{} bc \equal{} (a \plus{} b)(a \plus{} c)$.[/quote]
It's true for $ a\plus{}b\plus{}c\equal{}1.$ :wink:

[quote="spanferkel"][quote="nguoivn"]Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {8}{3\sqrt {3}}(ab \plus{} bc \plus{} ca)\sqrt {ab \plus{} bc \plus{} ca} \ge (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)\]
:)[/quote]
observe $ a \plus{} bc \equal{} (a \plus{} b)(a \plus{} c)$. Putting $ x \equal{} \frac1a$ etc, we obtain after squaring
$ xyz\left(\frac {x \plus{} y \plus{} z}{3}\right)^3\ge\prod\left(\frac {x \plus{} y}{2}\right)^2$ which is Popoviciu :)[/quote]
We only have: $ xyz\left(\frac {x \plus{} y \plus{} z}{3}\right)^3\le\prod\left(\frac {x \plus{} y}{2}\right)^2$  :wink:

[quote="nguoivn"]Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {8}{3\sqrt {3}}(ab \plus{} bc \plus{} ca)\sqrt {ab \plus{} bc \plus{} ca} \ge (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)\]
:)[/quote]
Using the well-known inequality $ xyz \le \frac{(xy\plus{}yz\plus{}zx)^2}{3(x\plus{}y\plus{}z)}$ with noting that
$ \sum (a\plus{}bc)(b\plus{}ca)\equal{}4(ab\plus{}bc\plus{}ca)$, we get
$ (a\plus{}bc)(b\plus{}ca)(c\plus{}ab) \le \frac{16(ab\plus{}bc\plus{}ca)^2}{3(3\plus{}ab\plus{}bc\plus{}ca)},$
so it is enough to prove that
$ \frac {8}{3\sqrt {3}}(ab \plus{} bc \plus{} ca)\sqrt {ab \plus{} bc \plus{} ca} \ge  \frac{16(ab\plus{}bc\plus{}ca)^2}{3(3\plus{}ab\plus{}bc\plus{}ca)},$
or
$ ab\plus{}bc\plus{}ca \plus{}3 \ge 2\sqrt{3(ab\plus{}bc\plus{}ca)}.$
But this is clearly true according to the AM-GM. The proof is completed.

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {8}{3\sqrt {3}}(ab \plus{} bc \plus{} ca)\sqrt {ab \plus{} bc \plus{} ca} \ge (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)\]
:)[/quote]
Using the well-known inequality $ xyz \le \frac {(xy \plus{} yz \plus{} zx)^2}{3(x \plus{} y \plus{} z)}$ with noting that
$ \sum (a \plus{} bc)(b \plus{} ca) \equal{} 4(ab \plus{} bc \plus{} ca)$, we get
$ (a \plus{} bc)(b \plus{} ca)(c \plus{} ab) \le \frac {16(ab \plus{} bc \plus{} ca)^2}{3(3 \plus{} ab \plus{} bc \plus{} ca)},$
so it is enough to prove that
$ \frac {8}{3\sqrt {3}}(ab \plus{} bc \plus{} ca)\sqrt {ab \plus{} bc \plus{} ca} \ge \frac {16(ab \plus{} bc \plus{} ca)^2}{3(3 \plus{} ab \plus{} bc \plus{} ca)},$
or
$ ab \plus{} bc \plus{} ca \plus{} 3 \ge 2\sqrt {3(ab \plus{} bc \plus{} ca)}.$
But this is clearly true according to the AM-GM. The proof is completed.[/quote]
Indeed, you really understand my thinking  :lol:
PS: [hide]Cu' nhu a doc. dc suy nghi~ cua e y', po tay a luon :|[/hide]


Let $n$ be the dot product of two vectors $A$ and $B$ (any number of dimensions). Find the minimum value for $|A|+|B|$.

[hide="hint"][url=http://wangsblog.com/jeffrey/]See blog post[/url][/hide]

[hide="hint again"]
$A \cdot B = |A||B|\cos{\theta} \le |A||B|$

wow, intermediate problems die so fast if no one posts in them...[/hide]

[quote="pkerichang"]$n \le |A||B|$
$\sqrt{|A||B|} \le \frac{|A| + |B|}{2}$
$n \le |A||B| \le \frac{(|A| + |B|)^2}{4}$
$|A| + |B| \ge \sqrt{4n}$

so the answer is $\sqrt{4n}$???[/quote]

I believe so. Pretty much just AM-GM and a little knowledge of dot products.

bonus, whats the minimum value of $|A|^2 + |B|^2$??


Has anyone received anything from NMSQT about whether they will be a sem-finalist or commended scholar yet?

i didn't yet, I think I will hear this week or pretty soon, seeing as the date for the finalist application is oct. 12.  The cutoffs were posted here:

http://www.collegeconfidential.com/discus/index.html

by the way, were you at RPI yesterday?  i thought i saw you, but wasn't sure.

Yeah, I was-- my parents forced me to go.

Anyway, I missed the cutoff by 1 point..............

i know, just about anywhere else you would have made it.

i saw like 4 people from our ARML team at rpi yesterday.


AMC
AIME
geometry
geometric transformation

I will start to translate the following sequence of words into my maternal language, Romanian (sorry american members - nothing to say here :D)

1- Hello!
2- My name is ...
3- I love you
4- MathLinks
5- This is a hard problem
6- Bye.

ROMANIAN:

1- Salut!
2- Ma numesc ...
3- Te iubesc
4- MatematicaLeaga
5- Aceasta e o problema grea
6- La revedere.

Please continue :D

Your word for "hello" in Romanian is very close to an English derogatory word you would call a wanton girl in the states (definitely not in front of her face unless you want to invoke the wrath of nails, legs, and punches that you can't get her back for (can't hit a girl remember)).

The word in the USA that I am referring to is "Slut".  I would be careful about saying salut in the states unless the pronounciation is very different and thus I don't understand it correctly.

lol

it quite different ... you can't take one for another ... :D

what's the pronunciation of a Romanian word?

Happily, I know some Spanish, so here it goes...

1- ¡Hola!; Buenos dias; or Bienvenidos.
2- Me llamo _____ or Mi nombre es _____
3- Te amo.
4- ??? er, MathLinks?
5- Este es un problema difcil.
6- Adis; Hasta luego; or Chao

Come to think of it, "salut" also works.  But no, it is not pronounced at all like the word "slut", at least not in Spanish.  The 'a' is very noticiable (it's pronounced like 'ah'), as is the 'u', which is more of a 'oooooh' sound.  In other words, in Spanish, 'salut' is pronounced "Sah-loot", which would be hard to mistake for another word.

[quote="Valentin Vornicu"]
1- Hello!
2- My name is ...
3- I love you
4- MathLinks
5- This is a hard problem
6- Bye.
[/quote]
well, Chinese is known by just a few people here.... (less than five?)
1. 你好 (ni3 hao3)
2. 我的名字是____ (wo3 de ming2 zi4 shi4)
3. 我爱你(wo3 ai4 ni3)
4. (i don't know how to translate it properly)
5. 这是一个难题(zhe4 shi4 yi1 ge4 nan2 ti2)
6. 再见(zai4 jian4)

Actually nearly everyone knows 'hello' and 'bye' and two words have been widely used in day-to-day life. More and more Chinese understand English. You can speak English in some big cities, others might understand you.

French :

1- Salut ! / Bonjour
2- Je m'appelle ....
3- Je vous aime / Je t'aime
4- Les maths rassemblent les peuples ?
5- C'est un problème difficile
6- Salut ! / Au revoir

belenos
Moderator ????

I think there are a few bugs still  :)

liyi:

how about 數學天地？
It's really hard to translate Mathlinks to chinese...  :(

but it's not the original meaning of mathlinks.
Its meaning is from "math links everyone"...

yes in fact MathLinks is supposed to be MathConnnects or MathGetsTogether rather than MathLinks :D

My turn,to give the german translation  :) :

[quote]1- Hello!
2- My name is ...
3- I love you
4- MathLinks
5- This is a hard problem
6- Bye.
[/quote]

1 - Hallo!
2 - Mein Name ist...
3 - Ich liebe dich
4 - Mathematik verbindet
5 - Diese Aufgabe ist schwer
7 - Auf Wiedersehen/Tschss

[quote="PiDeltaPhi"]My turn,to give the german translation  :) :
1 - Hallo!
[/quote]
"Hallo" is still used in Britain, but it is a dialect there! :D

Swedish:

1 - Hej!
2 - Mitt namn r ...
3 - Jag lskar dig
4 - Mattefrbindelser
5 - Detta problemet r svårt
6 - Hej då

Polish:

1 - Czesc!
2 - Ja sie nazywam
3 - Kocham ciebie
4 - Matematyczne kozenie
5 - Ten problem jest trudny
6 - Dowidzenia

I know now,from where the americans have stolen their "Hey"  :)

[quote="white_horse_king88"]2) zhe shi yi [b]dao4[/b] nan ti - wrong descriptive adjective.[/quote]
Lol, living in China and can't speak Chinese properly?
I'm just kidding of course...
The thing I notice is that a lot of native Chinese speakers don't really care about grammar that much.  And they don't follow grammar strictly while they speak or write.  (Myself included)
Hmm...  Is it because there are too many Chinese dialects?

[quote="anipoh"]The thing I notice is that a lot of native Chinese speakers don't really care about grammar that much. . . . Is it because there are too many Chinese dialects?[/quote]

The Sinitic languages ("Chinese dialects") differ from one another at least as much as English differs from German or French differs from Romanian. They are cognate languages, to be sure, but they are DISTINCT languages and have distinct grammatical rules. Only the standard national language is taught in school, so a lot of southern Chinese people (native speakers, or the children of native speakers of other Sinitic languages) can be vague about Chinese grammar in the same way that second-language speakers of English can be vague about English grammar. (Examples of bizarre English from the keyboards of United States schoolchildren abound on the AoPS forums--I usually pass them by without comment because I assume the writers of those fractured English sentences come from non-English-speaking families.)

In general Modern Standard Chinese (PUTONGHUA), "GE4" is always the correct DEFAULT measure word for just about every abstract noun but "SHI4QING2," which almost demands to be measured by "JIAN4." There is DEFINITELY dialectal variation in which measure word is preferred with other nouns. I speak a fair amount of Taiwanese (Taiwan dialect of Minnan Chinese) and more limited Cantonese and Hakka, and all the time I'm noticing measure words that don't correspond from one Sinitic language to another.

[quote="Erinda"]Interesting fact: most students hate learning French, I don't know why.[/quote]

That's because learning a language in class is horribly painful. I only enjoyed one of the 9 years of French I had, because my teacher was awesome (an alcoholic but awesome, and really funny) He got me to take in gr10, but I got this idiotic teacher who had a mental breakdown 2 years ago and changes marks like crazy (last year almost all her kids had 90's, this year the highest was 86 and that kid had a 95 last year, and went through French immersion so she sure knows the language). I ended up dropping the class and taking Art. I do hope to live in Paris when I am older and learn the language in an effective way.

now the translation in Croatian (my first language) (ijekavica dialect if anyone for some strange reason cares)

1) Dobar dan(day), Dobro jutro (morning) and Dobar vecer (night)
2) Moje ime je....
3) Volim te
4) hmmm.....
5) Ovo je teski problem
6) Do videnja

No. I think it more relates to conventions. For example, it just wouldn't make sense to say "LOOK! A pride of geese. Or LOOK! A gaggle of lions. You get my drift? Yeah. I don't think I can translate the descriptive adjectives that well, because they are quite unique, but it is conventional to say dao4 for problems. ge4 is informal reference to everything, colloquially, but technically it isn't correct(in some cases).

[quote="tokenadult"]There is at least one Singaporean AoPSer too who knows Chinese well. [/quote]
Hann Yew (rippledance) no longer frequents these forums, unfortunately.

Where I come from (northern China), the correct usage of measure words is seen as paramount. Ge4 is, in that region, definitely not an universal measure word; for example,

yi zhang zhi (a piece of paper),
yi ben shu (a book),
yi fu hua (a picture),

et cetera. Hmm, it's interesting that all of my examples involve paper in one form or another...

I can't vouch for other regions though...

[quote="Osiris"]Where I come from (northern China), the correct usage of measure words is seen as paramount. Ge4 is, in that region, definitely not an universal measure word; for example,

yi zhang zhi (a piece of paper),
yi ben shu (a book),
yi fu hua (a picture).[/quote]

I checked a dictionary about the word mentioned earlier in the thread. You will recall that I said that GE4 is the general Modern Standard Chinese default measure word for abstract nouns. (You have given examples of phrases with concrete nouns, all spoken the same way I would speak them, mostly on the example of Taiwanese people who speak Mandarin as a second language.) A specific claim has been made in this thread about a particular word, used by a native speaker of Chinese back before the AoPS-MathLinks merger.

I looked up the word "NAN2TI2" in the dictionary Han-Ying Cidian (A Chinese-English Dictionary) by Beijing Waiguoyu Xueyuan Yingyu Xi "Han-Ying Cidian" Bianxie Zu (Beijing: Shangwu Yinshuguan, 1980). Of course that dictionary was compiled in north China, and indeed one of the members of the editorial committee was a visting professor at my university while I was studying Chinese, so I am sure that he is a native speaker of northern Mandarin. Under the entry "NAN2TI2," the dictionary includes the example sentence, "ZAI ZHEYANG GANZAO DE DIGQU ZHONG SHUIDAO KE SHI YIGE NANTI" "How to grow rice in such a dry area is a difficult problem." If the dictionary compilers thought the measure word "GE4" was good enough for a sentence illustrating the usage of the term "NAN2TI2," I'm sure not going to criticize any Chinese person who writes in the same manner.

Hmmm, this whole thread is getting me interested in getting new software for typing Chinese on this computer.

Oh. Write it off to not reading the thread carefully enough - I wasn't aware you were talking about abstract nouns. ;) In that case, yeah, pretty much everything can be prefixed with Ge4.

The "yi dao ti" versus "yi ge ti" thing: "ti" in the former phrase is a concrete entity (math problem, et cetera), where as the "ti" in the latter phrase is used more abstractly. I think that's the distinction.

I think the explanation given by Osiris in the post above is correct.

Actually we don't care "quantifier" much in our dailylife here in China. But we need to pay attention to it during some exam... Instead, we use many informal "quantifier", even wrong ones, to make the language vivid.

For example, we say "5 tou2 jia1 huo3" instead of "5 ge4 jia1 hou3"(five guys) ... the correct use of "tou2" is "5 tou2 zhu1" (five pigs)

this topic started with a simple but important words in all languages :) let's continue in that way. if there is a lauguage you know but has not yet been mentioned here please add the proper translation :) (maybe the mods can help me clean out this topic, and make it all [b]on-topic[/b] and ready to be transported in the round table :D).

Chinese

1-nee houw
2-woa da ming ds shu
3-woa ei nee
4-?
5-gu shu ee ga nan won tee
6-zie gien

for no. 4, actually MathLinks should be translated as "Math Connects" (links people together :)) not as a collection of WWW math-related links :)

I think "Salut!" is more like the English word "salute"


Consider that $x_{0}$ and  $x_{1}$ are selected in ${\mathbb R}$ such that the sequence $\left(x_{n}\right)_{n=0}^{\infty}$ generated by
$\; x_{n+1}=\frac{x_{n-1}x_{n}-1}{x_{n-1}+x_{n}}\; \; ,\; \; n\in \{1,2,...\}$ is well defined . Let $\left( F_{n}\right)_{n=0}^{\infty}$ with $x_{n}=\cot{F_{n}}\; ,\; F_{n}\in (0,\pi)\; ,\; n\in \{0,1,2,...\}.$ [b]
Find the recurrence relation(s)   satisfied by the terms of $\left( F_{n}\right)_{n=0}^{\infty}$ . [/b]

[hide="hint"]
$\cot{(a+b)}= \frac{1-\cot{a}\cot{b}}{\cot{a}+\cot{b}}$

The sequence isn't called $F_{n}$ for no reason...

EDIT: Yeah, my trig is off. [/hide]

[quote="flip2004"][quote="paladin8"]...$\cot{(a+b)}= \frac{ab-1}{a+b}$
The sequence isn't called $F_{n}$ for no reason...[/quote]
Hmmm ! ???? Are you sure ? Please verify your proposed  equality ![/quote]

Heh, oops. Missed a couple of cots there...

[quote="flip2004"][quote="paladin8"].. for no reason...Hmmm !  Missed a couple of $\cot$ there...[/quote]
Please not forget that $\cot{X}=\cot{Y}$ implies.... ?[/quote]
He meant that the actual formula had a couple more cots ($\frac{1-\cot a \cot b}{\cot a+\cot b}$)

[quote="flip2004"][quote="bpms"]He meant that the actual formula had a couple more cots ($\frac{1-\cot a \cot b}{\cot a+\cot b}$)[/quote]
Please right,  .... the complete formula!
[b]However, where is the solution of  proposed question ? [/b][/quote]
I thought that we all know the solution.
[hide]$\left(F_{n}\right)_{n=0}^{\infty}$ is the generalised Fibonacci sequence: $F_{n+1}=F_{n}+F_{n-1}$. [/hide]

[quote="Rzeszut"]I thought that we all know the solution.
$\left(F_{n}\right)_{n=0}^{\infty}$ is the generalised Fibonacci sequence: $F_{n+1}=F_{n}+F_{n-1}$. [/quote]
[b][color=green] Sorry ! [/color][/b] The answer is $\begin{array}{|c|}\hline \\ F_{n+1}\equiv F_{n}+F_{n-1}\; \; \; \hbox{(mod}\; \pi) \\ \\ \hline \end{array}\; ,$  that is [b]  Fibonacci sequence "modulo $\pi$ "[/b]


I remember seeing a theorem long ago that says like, given a rectangle ABCD and a point P in the interior, Pa+ PC = PB + PD, or something like that? Also, this was supposedly generalizable to a four-sided pyramid. Is there such a theorem, or am I completely confused?

It's the British Flag Theorem: $PA^{2}+PC^{2}= PB^{2}+PD^{2}$. Can be proved by the Pythagorean Theorem and is generalizable to points $P$ not in the plane of $ABCD$.

For the proof, drop the perependicular from $P$ and it should be obvious

make a perpendicular and then do right triangle stuff

Chuck Norris can divide by 0. thats funny  :rotfl:  :rotfl:  :rotfl:


trigonometry
geometry
angle bisector
trig identities
Law of Sines
geometry proposed

Since in general the angle bisector from A doesn't  intersect EF, it semms to me a little bit strange the way to pose the question.

In any case I have a proof of the fact that if P lies on the bisector of <A then M,E and N are collinear.

I don't have yet a direct proof of the other part.
I suppose, is possible to derive it by absurd.

[quote="ThAzN1"]In acute triangle $ABC$, $AB \neq AC$.  Let $D$ be the midpoint of side $BC$ and let $E$ be a point on segment $AD$.  Let $F$ be the foot of the perpendicular from $E$ to side $BC$, and let $P$ be a point on segment $EF$.  Let $M$ and $N$ be the feet of the perpendiculars from $P$ to sides $AB$ and $AC$, respectively.  Prove that $M,E,N$ are collinear if and only if $\angle BAP = \angle CAP$.[/quote]

huh

$\cot \measuredangle EMP =$ $\frac{\sin \measuredangle AME}{\sin \measuredangle EMP} = \frac{\frac{AE}{ME} \sin \measuredangle MAE}{\frac{PE}{ME} \sin \measuredangle MPE}$ $= \frac{\sin \measuredangle MAE}{\sin \measuredangle MPE} \frac{AE}{PE} = \frac{\sin \measuredangle BAD}{\sin \measuredangle ABD} \frac{AE}{PE} =$ $\frac{BD}{AD} \frac{AE}{PE} = \frac{CD}{AD} \frac{AE}{PE} =$ $\frac{\sin \measuredangle CAD}{\sin \measuredangle ACD} \frac{AE}{PE} =$ $\frac{\sin \measuredangle NAE}{\sin \measuredangle NPE} \frac{AE}{PE} = \frac{\frac{AE}{NE} \sin \measuredangle NAE}{\frac{PE}{NE} \sin \measuredangle NPE} = \frac{\sin \measuredangle ANE}{\sin \measuredangle ENP}$ $= \cot \measuredangle ENP$; as $0 < \measuredangle AME, \measuredangle ANE < \pi \Rightarrow \measuredangle EMP = \measuredangle ENP$. If the three points are collinear, we have that $\measuredangle MAP = \measuredangle MNP = \measuredangle ENP = \measuredangle EMP = \measuredangle NMP = \measuredangle NAP$, while if we have the angle bisector condition, $\measuredangle EMN = \measuredangle AMN - \measuredangle AME = \measuredangle ANM - \measuredangle ANE = \measuredangle NME$. If these angles are not $0$, we again use the law of sines to find (I emphasize the law of sines rather than just noting isosceles triangle to make it clear we have a singularity if the angles are $0$) $ME = NE$, so as $AM = AN, \measuredangle AME = \measuredangle ANE$, $\measuredangle MAE = \measuredangle NAE$ and $E$ lies on both the angle bisector and the median, implying against the given condition that the triangle is isosceles. Hence the angles given above must both be $0$, implying collinearity.

Vry good idea Thazn1!!

I try to prove that <EMP = <ENP whitout trig.

Let Q be the intersection between c(PEM) whit AB and R be the intersection between c(PEN) whit AB. Is easy to see that Q, E and R are collinear and EQ = ER and from this it easily follows the thesis.

I think you meant good idea MysticTerminator!  :P

Anyways, I think I might have something that works (without trig?)

Let $\ell$ be the line through $E$, parallel to $BC$, and suppose it meets $AB$ and $AC$ at $X$ and $Y$ respectively.  Note that $X,M,E,P$ and $Y,N,P,E$ are concyclic.

Now, say $M$ and $N$ are on different sides of $\ell$ and without loss of generality, say $M$ and $A$ are on the same side of $\ell$.  Thus, $M,E,N$ are collinear if and only if $\angle MEX = \angle NEY$.  Note that $\angle MEX = \angle MPX$ and $\angle NEY = \angle YPN$.  Furthermore, we have $PX = PY$ so $M,E,N$ are collinear if and only if $\triangle PMX \cong \triangle PNY$ and this is equivalent to $PM = PN$, i.e. that $P$ lies on the bisector of $\angle BAC$, as desired.

Ok, well to make it rigorous, if $M$ and $N$ are on the same side of $\ell$, the two points clearly are not collinear with $E$ and $\triangle PMX$ and $\triangle PNY$ are not congruent (one side is congruent, the angles are different, which isn't too hard to show) so $P$ does not lie on the bisector of $\angle BAC$...


geometry
calculus
integration
perimeter
inequalities
number theory
geometry unsolved

Wanted is the triangle of smallest area with integral side lengths which are pairwise coprime and these sides shall have at least one factor > 1 with the perimeter in common.

Looks like Number Theory + Inequality :?
It is a third problem in a row posted by Orl more than 1 year ago. And no one of these problems by Hans-Diether Wolf got any reply.


The integers 1,2,3… [i]n[/i], are equally spaced out around a circle.  The numbers 17 and 54 are located at opposite ends of the same diameter.  What is [i]n[/i]?

64

Between 17 and 54(only one side) there are 36 terms. On the other side of 17 and 54 there should also be 36 terms. 16-1 is only 16 terms, so you need 20 more coming from the 54. There should be 72 terms in total. Remember that you should only count the integers between 17 and 54 not inclusive.

Your answer isn't correct, but you brought up an idea that I hadn't thought of.  There was just slight miscalculation in your solution.  There are the same number of people on either side of numbers 17 and 54.  There are 54-17-1=36 people on either side.  36-16 (the number of people on the other side of 17 down to 1)=20.  So, the answer is 54+20=[b]74[/b].

Here's how I did it.
There is a 54-17=37 difference between the opposites. So 1 is opposite 38. For number 37, if the opposite was less, it would be 0, but it can't be 0. So the opposite is 37 greater or 74. We know that it is the highest number because for 38 and higher numbers, the opposite is -37 not +37.

[hide]$\frac{1}{2}n$ and $n$ are endpoints of a diameter of the circle. From the given data, the difference between the two numbers should be 37.
This means that $n-\frac{1}{2}n=37$, and $n=74$. [/hide]