Show that every positive intiger $ K$ can be expressed in the form

$ K \equal{} a_n(n)! \plus{} a_{n \minus{} 1}(n \minus{} 1)! \plus{} a_{n \minus{} 2}(n \minus{} 2)!... \plus{} a_11!$

Given that and $ n$ is an intiger, and $ a_i$ are intigers with $ a_n > 0$ and $ 0\le a_i \le i$

[hide="Hint"]The [url=http://en.wikipedia.org/wiki/Greedy_algorithm]Greedy algorithm[/url] gives a valid solution for all $ K \in \mathbb{N}$[/hide]

Thanks for the hint. :)  I can see how it applies to the problem, but how would you go about writing up a proof using it? I attempted to solve it using stong induction...

Idea:

do an induction on an interval of numbers (ex. let S_n be the interval of integers whose representation has n has at most n digits)

How to use TZF's hint and write it up:
[hide]Division algorithm to the rescue! For any $ n,d$ there exists an integer $ q$ and $ r$ with $ 0 \leq r < d$ such that $ n \equal{} dq \plus{} r$. Now do strong induction on $ k$ for the intervals $ \{k!,k!\plus{}1,k!\plus{}2, \ldots (k\plus{}1)! \minus{} 1\}$. Doing the divison algorithm on any number with $ d \equal{} k!$ must give a $ q$ for which $ 1 \leq q \leq n$; this is your $ a_k$. Apply the inductive hypothesis on the remainder term.[/hide]

Yep.  See also the counting argument and last problem [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237669]here[/url].

[quote="MellowMelon"]
Division algorithm to the rescue! [/quote]

Ah, that makes things a lot easier :) thanks. So here is my solution

[hide]
Consider the case $ K\equal{}1 \Rightarrow K\equal{}1(1)!$ which is true

Now assume it is true for all intigers $ K< n!$. We will show that it is true for all intigers $ n!<K<(n\plus{}1)!$

If we divide $ K$ by $ n!$, then by the division algorithm $ K\equal{}a_nn!\plus{}r$

We know $ a_n<n\plus{}1$ because $ K<(n\plus{}1)!\equal{}(n\plus{}1)n!$

Also since $ r<n!$, we know that $ r$ can be reprsented in the form $ r\equal{}a_{n\minus{}1}(n\minus{}1)!\plus{}...a_1(1)!$. Therefore, by the inductive hypothesis, every number $ K$ can also be represented in the form
$ K\equal{}a_nn!\plus{}r \equal{} a_nn! \plus{} a_{n\minus{}1}(n\minus{}1)!...\plus{}a_11!$.

Now all that is left to show is that if $ K$ holds for every intiger less than $ (n\plus{}1)!$ then it also holds for $ (n\plus{}1)!$ Which is true because $ n(n)! \plus{} (n)! \equal{} (n\plus{}1)n! \equal{}(n\plus{}1)!$ and;

$ \left(n(n)!\plus{}(n\minus{}1)(n\minus{}1)!...\plus{}2(2)!\plus{}1(1)!\right) \plus{} 1 \equal{} (n\plus{}1)!$

And this completes our proof  :) [/hide]


$ x$ is 3 dollar plants, $ y$ is 5 dollar plants.

$ x\plus{}y\equal{}15$
$ 3x\plus{}5y\equal{}55$
from the first, $ x\equal{}15\minus{}y$
Put that into the second, $ 3(15\minus{}y)\plus{}5y\equal{}55$
$ 45\minus{}3y\plus{}5y\equal{}55$
$ 2y\equal{}10$
$ y\equal{}\boxed{5}$

[quote="GameBot"]Hal purchased 15 plants at a nursery, some at $ \$$3 and the rest at $ \$$5. The total cost was $ \$$55. How many $ \$$5 plants did he buy?[/quote]

If this was a countdown, I'd note that he must purchase a multiple of 5 of 3 dollar plants, so s is clearly not 0 or 15, so we only need to test 5,10.

Suppose that all the plants were $ \$ 3$.  Then, the total cost would be $ \$ 45$.

For every $ \$ 5$ plant that we switch a $ \$ 3$ plant with, the total will go up by $ \$ 2$.

We need to do this $ 5$ times, so there are $ \boxed{5}$ $ \$ 5$ plants.


Yea, I'm having problems with MSN's web messaneger - [url]http://webmessenger.msn.com/default.aspx?R=1[/url]

Whenever I try to log in, it loads, but then says sign in again.  And I don't know how to stop this.

SOLUTION:
don't use msn im.
use yim or aim.

Um, I can't talk to my friends that use MSN im though.

ok...
DIFFERENT SOLUTION:
Threaten your friends to use yim or aim.

i think you should shut down the program and try again(unless you already did)

Are your settings set to reject cookies or something?

Anyway...

http://www.meebo.com

meebo...ccy uses that(i think)
I don't like meebo....It looks...um..unprofessional(well not as professional as yim).


With both a Southern California forum and a Bay area forum, both flying the California state flag ... the problem is that I can't think of a flag or symbol that could be used to distiguish us. Certainly not any flag or seal for the City of Los Angeles or Los Angeles County, because too many of us live in other cities and counties. So I guess the state flag will have to do.

[quote="JBL"]I just wanted to point out [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29116]this[/url] post to any administrator who has a free minute -- by a resounding, unanimous, 5-0 vote, users of the NYC forum would prefer that we have the New York City flag (rather than the New York State flag) to represent our community.[/quote]I don't have that flag in the right dimensions. If you would be so kind as to provide it (you can easily get the dimensions of the other flags, just by looking in the left of this post :P) I will be kind in return to change it ;) :)

[quote="JBL"]Just a little reminder ....[/quote]:P ok it's done.


trigonometry
inequalities
inequalities unsolved

$\cos{x}+\cos{y}+\sin{x}\sin{y}\leq 2$
$\Leftrightarrow \sin{x}\sin{y}\leq 1+1-\cos{x}-\cos{y}$
$\Leftrightarrow \sin{x}\sin{y}+\cos{x}\cos{y}\leq 1+1-\cos{x}-\cos{y}+\cos{x}\cos{y}$
$\Leftrightarrow \cos(x-y)\leq 1+(1-\cos{x})(1-\cos{y})$
 :gleam:

[quote="andy_tok"]For reals x,y
Prove $\cos{x}+\cos{y}+\sin{x}\sin{y}\leq 2$[/quote]

I am not sure how nice this solution is but consider,
$x,y\in [0,\pi/2]$

Then we have,

$\cos x+\cos y+\sqrt{1-\cos^{2}x}\sqrt{1-\cos^{2}y}\leq 2$

Call, $a=\cos x$ and $b=\cos y$ subject to conditions $0\leq a,b\leq 1$

So,
$a+b+\sqrt{1-a^{2}}\sqrt{1-b^{2}}\leq 2$
Iff,
$\sqrt{1-a^{2}}\sqrt{1-b^{2}}\leq 2-(a+b)$
Iff (square),
$(1-a^{2})(1-b^{2})\leq 4-4(a+b)+(a+b)^{2}$
Iff,
$1-a^{2}-b^{2}+a^{2}b^{2}\leq 4-4a-4b+a^{2}+2ab+b^{2}$
Iff,
$a^{2}b^{2}+4a+4b\leq 2a^{2}+2ab+2b^{2}$
Im just not sure how to do it from here  :blush:

[quote="arqady"] $2\left((1-a)^{2}+(1-b)^{2}\right)\geq(2-a-b)^{2}\geq(1-ab)^{2}.$[/quote]
I see you used AM-GM on first inequality. What about the last one? What did you use?

[quote="sylow_theory"]  What about the last one? What did you use?[/quote]
$(1-a)(1-b)\geq0.$ :wink:

Because $0\le a,b\le 1$ then $2-a-b\ge 1-ab\ge 0\Leftrightarrow (1-a)(1-b)\ge 0$, it's trivial

[quote="Mathsniper"][quote="Extremal"][quote="Mathsniper"]Here is my own problem. Try to prove that:
For any reals x, y, z
$\sin{x}+\sin{y}+\sin{z}+\sin{x}\sin{y}+\sin{y}\sin{z}+\sin{z}\sin{x}\leq \frac{15}{4}$[/quote]
it's wrong.
put here sinx=1,siny=1,sinz=1.
and also in left equation max  value is 6. it's simpl :)[/quote]
Sorry, there's a typo. I have proved it which is true.[/quote]

Clearly the inequality is wrong. Therefore if you have a proof for it, i surely think there is a mistake in this. A couterexample is: $x=y=z=\frac{\pi}{2}$

Okay, there is a complicated thing in here. When you have edited your problem above then it's clearly true.

[quote="andy_tok"]To hien: Mathsniper has edited his post
[quote="Mathsniper"]Here is my own problem. Try to prove that:
For any reals x, y, z
$\cos{x}+\cos{y}+\cos{z}+\sin{x}\sin{y}+\sin{y}\sin{z}+\sin{z}\sin{x}\leq \frac{15}{4}$[/quote]
Proof:
USE AM-GM,$\sin{x}\sin{y}\leq \frac{\sin^{2}{x}+\sin^{2}{y}}{2}$;
Let $a=\cos{x},b=\cos{y},c=\cos{z}$
So we only need to prove$3+\sum{a}-\sum{a^{2}}\leq \frac{15}{4}$
$\iff \sum{(a-\frac{1}{2})^{2}}\geq 0$[/quote]
Right, the RHS-LHS can be equivalent to
$\sum(\cos{x}-\frac{1}{2})^{2}+\frac{1}{2}\sum(\sin{x}-\sin{y})^{2}\geq 0$


Είχα μια διαφωνία με έναν φίλο για την απάντηση μιας άσκησης συνδιαστικής και επειδή τελικά δν μπορούσε ο ένας να πείσει τν άλλο αρκετά ζητώ λίγη βοήθεια παρότι μπορεί να είναι εύκολη η άσκηση...Η άσκηση έχει ως εξής:
"Με πόσους διαφορετικούς τρόπους μπορούν να καθήσουν 5 άντρες και 5 γυναίκες γύρω από ένα κυκλικό τραπέζι, έτσι ώστε να μην κάθονται δύο άτομα του ίδιου φύλου το ένα δίπλα στο άλλο"

Ευχαριστώ προκαταβολικά...

loipon to trapezi profanos exei deka theseis....ektos an katsoun oi gunaikes pano stous antres.

tora i apantisi einai sti geniki periptosi gia $ n$ gynaikes kai $ n$ antres, $ \frac{(n!)^2}{2n}$ opote stin periptosi sou milame gia 1440 diaforetikous tropous. pisteuo oti den kano lathos...an kano lathos smebody will point it out. en perilipsi i logiki einai i eksis...xoris vlavi ti genikotitas boreis na topothetiseis enan antra se kapoia thesi sto trapezi. meta apo auto, gia oles tis alles theseis gnorizoume an kathetai antras i gunaika (oi gya den lamvanontai upopsin apo ta restrictions tou provlhmatos). opote tous antres boreis na tous topothetiseis me $ 5!$ diaforetikous tropous stis kala orismenes theseis tous (not quite true alla perimene mexri to teleftaio vhma tis apodeiksis). to idio isxyei kai gia tis gynaikes. tora kathe topothetisi etsi opos tis pirame emfanizetai akrivos 10 fores sti syllogi mas. ara to apotelesma einai $ (5!)^2/10$.
pes mou an pianeis ti logiki, giati i arxiki idea itan na grapso ena oraio post kai na eksigo polu analutika to kathe vhma oste na  voithiso kai tous neoterous na ksexorisoun tis kuklikes metatheseis kai genika na "niosoun" ti akrivos mas deinei o kathe tupos sunduastikis kai pos vgainei klp....alla dystyxos den prolavaino.

**sti syndyastiki panta uparxei o kindunos lathous, an den eisai arketa prosektikos.opote an exo kanei mlkia...fonaxte! kata ti gnomi mou o demetres den afairese apo ton tropo tou kapoies periptoseis pou exoume idia topothetisi, eno o basted pire ton tupo ton kuklikon metatheseon gi antres kai gynaikes ksexorista kai ekane vasiki arxi aparithmisis...alla den boreis na to kaneis auto giati xaneis periptoseis....(why?-opos tha elege kai to dunatotero isos melos tou forum....pou fusika exei stamatisei na grafei edo kai kairo)

[quote="Durandal"][quote="Demetres"]Εγώ βρήκα 28800. Πες τι βρήκες και με τι μέθοδο να δούμε αν συμφωνούμε.[/quote]

Παίζει typo ή να μη διαίρεσες με το 10 του κυκλικού τραπεζιού? Εγώ βγάζω $ 2\frac {5! 5!}{10} \equal{} 2880$ :huh:

(Πολύ βιαστικό το παραπάνω, sorry προκαταβολικά αν λέω χαζομάρες)

Cheerio,

Durandal 1707[/quote]

tora to eida auto...hmmm...mou aresei pou zitisame prokatavolika suggnomi kai oi duo....auta pou sas elega paidia....i ego i o durandal kanoume kapou lathos. se tetoies priptoseis tonizo oti PANTA MA PANTA eseis pou pigainete se diagonismous kante ena paradeigma m mikra noumera na deite an douleuei o tupos...an kanete paradeigma gia $ n\equal{}2$ tha deite oti auto pou leo ego einai lathos kai auto pou leei o durandal einai sosto!!!
giati...? loipon o logos einai oti i kathe topothetisi sti sullogi mou den emfanizetai $ 2n$ fores epeidi exo stabarei tis arties theseis gia tis gunaikes kai tis perittes gia tous antres etsi opos exo anaptuksei to sullogismo mou sto proigoumeno post. ara kathe topothetisi uparxei sti sullogi mou akrivos $ n$ fores sunepos o tupos einai $ (n!)^2/n$.

episis eseis pou asxoleiste me diagonismous...otan kanete to paradeigma me mikra noumera na prosexete na to kanete sosta.....oxi sa merikous merikous...pou sas lene ti na kanete kai oi idioi to kanoun xoris upomoni (kai sto telos lene nai more..etsi tha nai!)

[quote="basted"]Πόσες θέσεις έχει το τραπέζι? Αν έχει 10 τότε $ (4!)^2 \equal{} 576$ αφού πρέπει να διατάξουμε τους 5 άντρες ΚΑΙ τις 5 γυναίκες και το τραπέζι είναι κυκλικό.[/quote]
Οι κυκλικές διατάξεις ισχύουν μονο για  το ένα φύλο αφού όταν θα έχουμε τοποθετήσει το ένα φύλο έχουμε ήδη επιλέξει το starting point. ’ρα η απάντηση μου επί 5...

Άς δοκιμάσω κι εγώ, αφού βγάζω άλλο αποτέλεσμα...

Άν τους θέσουμε στήν τύχη είναι 10!
Έχουμε 5 tuples μέ άνδρες και 5 με γυναίκες. Μέσα σέ κάθε tuple έχουμε 5! διατάξεις...Άρα, τελικά :
10! / (5! * 5!) = 252
αν δεν απατώμαι...

Άντε να δούμε πόσες διαφορετικές απαντήσεις θα δοθούν.  :lol:

Εγώ δεν διαίρεσα με το 10. Ερμήνευσα το πρόβλημα ότι περιστρέφοντας το τραπέζι, αλλάζει η τοποθέτηση. Αν δεν αλλάζει, τότε συμφωνώ με την απάντηση του Durandal και την (δεύτερη) απάντηση του Νίκου.

Λοιπόν η άσκηση υπάρχει στο "Counting and Configurations" της CMS και βγάζει απάντηση (όσο υπέθεσα και εγώ) $ 2(5!)^2$....Εγω προσωπικά άνοιξα το arrangement σε ευθεία και ελλάτωσα την περίπτωση για 2 και για 3 άντρες/γυναίκες και μετά για 5 και σκέφτηκα τους συνδιασμούς $ (5!*5!)$ και μετά επί 2 επειδή είναι κύκλος...Βέβαια θα έπρεπε να διαπυστωθεί και αν υπάρχουν κάποιοι συνδιασμοί που να ξαναπροκύπτουν, αλλά...Το πρώτο αποτέλεσμα που βγάζει ο Νίκος ήταν το άλλο που προέκυπτε....Λογικά αυτό είναι (του Νίκου δλδ $ \frac {n!}{2n}$) καθώς προφανώς οι συνδιασμοί είναι λιγότεροι σε κύκλο από ότι σε ευθεία....Ποιος θα φανταζόταν τι πρόβλημα θα μπορουσε να δημιουργήσει ένα απλό arrangement εεε......

Ρε σεις σκεφτειτε απλα με λογικη και χωρις τυπους.
Στην πρωτη θεση μπορουν να κατσουν και οι 10
Στην δευτερη ολοι αντιθετου φυλου απο αυτον που καθεται στην πρωτη αρα 5
Στην τριτη ολοι οι εναπομεινοντες αντιθετου φυλου απο τον δευτερο αρα 4.
Στην τεταρτη ολοι οι εναπομεινοντες αντιθετου φυλου με τον τριτο αρα 4.
....................
Σκεφτειτε ετσι και θα καταληξετε στο
1 θεση 10 ατομα
2 θεση 5 ατομα
3 θεση 4 ατομα
4 θεση 4 ατομα
5 θεση 3 ατομα
6 θεση 3 ατομα
7 θεση 2 ατομα
8 θεση 2 ατομα
9 θεση 1 ατομο
10 θεση 1 ατομο

Αρα 10*5*4*4*3*3*2*2*1*1=28800 τροποι διαφορετικοι. Επειδη ομως το τεπεζι ειναι κυκλικο /10
Αρα τελικα 2880

Filosofimene πολύ ωραίος (δείχνεις έναν ωραίο χαρακτήρα της συνδιαστικής μακριά από τύπους που προβληματίζουν) αλλά εγώ προσωπικά προβληματίζομαι από τη διαρεση δια του 10...μάλλον εγώ κάτι δν πιάνω....θα το σκεφτώ κι άλλο...

[quote="Durandal"][quote="filosofimenos"]Ρε σεις σκεφτειτε απλα με λογικη και χωρις τυπους.
Στην πρωτη θεση μπορουν να κατσουν και οι 10
Στην δευτερη ολοι αντιθετου φυλου απο αυτον που καθεται στην πρωτη αρα 5
Στην τριτη ολοι οι εναπομεινοντες αντιθετου φυλου απο τον δευτερο αρα 4.
Στην τεταρτη ολοι οι εναπομεινοντες αντιθετου φυλου με τον τριτο αρα 4.
....................
[/quote]

Κι εγώ έτσι το έβγαλα αρχικά, απλώς μετά το μάζεψα γιατί μου άρεσε μία πιο conditional διατύπωση :)

2 επιλογές για την πρώτη θέση (άντρας - γυναίκα)
Μόλις διαλέξουμε το φύλο, έχουμε: 5, 5, 4, 4...
Διαιρούμε με το 10 για το τραπέζι... eh voila :)

Πάντως τρομερό: σχεδόν κάθε post είχε και μία διαφορετική απάντηση :rotfl:

Cheerio,

Durandal 1707[/quote]

ontos...

loipon....exei dikio o filosofimenos..sti sundiastiki prepei na skeftesai apla gia na min kaneis lahos!kai otan pairneis kapoion tupo, prepei na ksereis akrivos ti dinei autos o tupos kai na eisai 100% sigouros oti exeis idia katastai sto provlhma kai boreis na to efarmoseis. alla thelo na po oti protimo ti logiki pou periegrapsa sta proigoumena posts me tin ennoia oti einai pio geniki-oxi mono epeidi vgazei eukola genikeusi sto sugekrimeno provlhma, alla ki epeidi perigrafei mia antimetopisi gia ta pio duskola provlhmata.

Wallace δεν ξερω αν καταλαβες γιατι διαιρουμε δια 10 αλλα για καλο και για κακο ας δωσω ενα hint.
Σκεψου μια συγκεκριμενη δεκαδα και μετατοπισε την σαν συνολο(χωρις να αλλαξεις τη σειρα με την οποια καθονται οι ανθρωποι ) πανω σε διαφορες θεσεις πανω στο τραπεζι.
Για οτι περετερω εδω ειμαστε.

Παιδιά thanks σε όλους...ήταν ένα λεπτό σημείο για μένα και δν το έπιανα αρχικά....filosofimene με βοήθησε πολύ η προσέγγιση σου....Τελικά όλα θέλουν πρώτα απλές σκέψεις και ύστερα πολύπλοκες :oops: .....Ευχαριστώ και πάλι...!

Ουπς, ενα είναι σίγουρο, ότι η δικιά μου είναι η πιό λάθος...Όμως, αυτό με τήν διαίρεση δεν είναι δυνατόν να ισχύει...:

Οι γυναίκες κάθονται στις θέσεις 2,4,6,8,10 κατά 5! τρόπους. Αυτοί συνδυάζονται με άλλους 5! τρόπους των ανδρών...Άρα συνολικά κατά (5! * 5!)...Όμως οι γυναίκες μπορούν να κάτσουν καί στίς θέσεις 1,3,5,7,9...Άρα συνολικά :

2*5!*5!=28800 τρόπους.

Αυτό με την διαιρεση πού κολλάει?

Διαβασε τα οσα γραψαμε και θα καταλαβεις...

Αν ισχύει αυτό που ισχυρίζεσαι, τότε γιά 2 γυναίκες και 2 άνδρες έχουμε 2 μόνο δυνατότητες , αντί για 8 που ειναι το σωστό...

wallace...kai oi upoloipoi...make sure oti katalavaineis gt sti diki mou posegisi den prepei na diaireseis me 10 alla me 5. to eksigo se kapoio apo ta parapano posts..an den einai katharo pes mou, na to kano pio clear. episis to teleftaio post tou filosofimenou na to diavasete oloi giati etsi tha katalavete ti paizei me tis kuklikes metatheseis. ki ego otan skeftomai kuklikes metatheseis i paromoia provlhmata san auto, ta antimetopizo etsi opos anaferei o filosofimenos. to kano restate (fil de se perazei upotheto)..tous vazoume se mia thesi kai meta tous metatopizoume mia thesi ton kathena anticlockwise lets say. ara xreiazetai na to kanoume auto n fores gia na epistrepsoume ekei pou ksekinisame kai oloi autoi oi tropoi einai isodunamoi!

[quote]Ενας απ τα πολυ ''ισχυρους'' λογους που κανουν οι μαθητες καταληψεις είναι γιατι οι τυρόπιτες ειναι στρογγυλές και δεν ξερουν απο να αρχίσουν το φαγητό[/quote]  :rotfl:  Τελικά όλα είναι μαθηματικά....!

Ευχαριστώ πολύ Νίκο και κύριε $ XP$ .....Βοηθήσατε πάρα πολύ να καταλάβω....!

[quote="xp"]Ξενιτεμενε Billy σου απαντηθηκε η απορια σου απο καποιον απ τα παιδια αλλά ας σου την πω και εγω χαριτολογώντας
Ενας απ τα πολυ ''ισχυρους'' λογους που κανουν οι μαθητες καταληψεις είναι γιατι οι τυρόπιτες ειναι στρογγυλές και δεν ξερουν απο να αρχίσουν το φαγητο.............
Σκεψου δηλαδή οτι μια κυκλικη δεκαδα ενω ειναι η ιδια εχει δεκα διαφορετικες αρχες (δες τυροπιτα) και προσμετραται ως μια.......
Ολα θα αλλάζαν αν οι θεσεις ηταν αριθμημενες...[/quote]
Thanks!!! afto pistevw pws einai to zumi tis askhshs!!! :wink:


function
calculus
derivative
integration
calculus computations

Let $ a > 0$, $ f: [0,a] \rightarrow R$ be a derivable function, with a continuous derivative, such that $ f ' (0) > 0$, $ f(a) \equal{} f(0) \plus{} \frac {a^2}{2}$. Show that there exist a convergent sequence $ (x_n)_{n\geq 0}$ such that $ f ' (x_n) \equal{} (n \plus{} 1)x_n$. Then find the limit $ lim x_n$

Let's denote $ g(x)\equal{}f'(x)\minus{}x$ . We can write $ \int_0^a g(x)dx\equal{}\int_0^a (f'(x)\minus{}(n\plus{}1)x)dx \equal{} f(a)\minus{}f(0) \minus{} (n\plus{}1)\frac {a^2}2 \equal{} \minus{}n\frac{a^2}2 \leq 0$ so there is $ u_n\in ]0,a[$ such that $ g(u_n) \leq 0$
But $ g(0) \equal{} f'(0) > 0$, so (because of intermediate value property of derivative functions) there is $ x_n\in ]0,u_n]$ such that $ g(x_n)\equal{}0, i.e.\  f'(x_n)\equal{}(n\plus{}1)x_n$
From $ x_n\equal{} \frac {f'(x_n)}{n\plus{}1}$, with $ f'(x)$ bounded, you obtain that $ x_n$ is convergent : $ x_n\rightarrow 0$
 :cool:


function
advanced fields
advanced fields unsolved

Let $ f: S^1 \to S^1$ be continuous and let $ x_0 \in S^1$ be a fixed point of $ f$. Suppose that the induced map $ f^*: \pi_1\left(S^1, x_0\right) \to pi_1\left(S^1, x_0\right)$ is given by $ [g] \mapsto k [g]$ where $ k \geq 3$ is an integer.

(a) Show that $ f$ has at least $ k \minus{} 1$ fixed points.

(b) Give an example of such a function having exactly $ k \minus{} 1$ fixed points.

(c) Give an example of such a function having at least $ k$ fixed points.

(without using the result that maps $ S^n \to S^n$ having the same degree are homotopic)

(this has something to do with universal covering spaces I believe...)

Yes you're right, you can solve it using the fact $ {\mathbb R}$ is the universal covering space for $ {\mathbb S}^1$. If $ f: {\mathbb S^1}\to {\mathbb S^1}$ has degree $ k$, then its unique lifting $ \tilde{f}: [0,1]\to {\mathbb R}$ with $ e^{2\pi i \tilde{f}} \equal{} f$ and $ \tilde{f}(0) \equal{} 0$ satisfies $ \tilde{f}(1) \equal{} k$ (thats one possible definition of the isomorphism $ \pi_1({\mathbb S^1})\cong {\mathbb Z}$. Now the identity on $ {\mathbb S}^1$ lifts to the identity, and $ t\in [0,1]$ is a fixed point for $ f$ iff $ |\tilde{f}(t) \minus{} t|$ is an integer. So geometrically, the fact that $ f$ has at least $ k \minus{} 1$ fixed points corresponds to the fact that the graph of a map $ \tilde{f}: [0,1]\to {\mathbb R}, \tilde{f}(0) \equal{} 0, \tilde{f}(1) \equal{} k$ has to intersect the translates $ \{(t,t \plus{} n) | t\in [0,1)\}\subset {\mathbb R}, n\in {\mathbb Z}$ at least $ k \minus{} 1$ times.


abstract algebra
superior algebra
superior algebra unsolved

X is a group with operation * sastifying
f:X-------->X
  x|--------->$x^n$ is a surjective homomorphism .
Prove that X is a Abel group

Do you mean that the group is Abelian (commutative)?

[quote="tµtµ"]

We also have $(xy)^{-n} = x^{-n}y^{-n}$

Is-it correct ?[/quote]

Well that line bothers me, could you be more specific?


The title sounds like "us" at the IMO... :rotfl:
To make this not like a spam, the page is not working :P

[quote="xxreddevilzxx"][url]http://www.maa.org/news/072806imoresults[/url]

Congratulations to the US IMO team![/quote]

[url]http://www.maa.org/news/072806imoresults.html[/url]


Prove that $\frac{n(n+1)(2n+1)}{6}$ is always an integer (don't use the fact that its equal to the sum of the squares of the first $n$ positive integers).

[hide]well for it to be an integer then one of x ,  x+1, or 2x+1 must be divisible by 2 and one must be divisible by 3.

x = 0(mod 2) then x is a multiple of 2
x = 1(mod 2) then x+1 is a multiple of 2

x = 0(mod 3) then x is a multiple of 3
x = 1(mod 3) then 2x+1 is a multiple of 3
x = 2(mod 3) then x+1 is a multiple of 3

and since every integer must fall in to one category for each congruency class x(x+1)(2x+1) must be divisible by 6[/hide]

[hide]This is equivalent to showing that $n(n+1)(2n+1) \equiv 0 \pmod6$. To show this, just plug in the cases if n is 0,1,2,3,4,5 mod 6 (since we know in congruences that we can replace any n with one of these 6 numbers). These are all easily verified.

This is what nateharman did just a bit more simple.[/hide]


[hide]
A train, x meters long, traveling at a constant speed, takes 20 seconhds from the time it first enters a tunnel 300 meters long until it completely emerges from the tunnel. One of the stationary lights in the tunnel was directly above the train for 10 seconds. Find x.

So the front of the train travels (300 + x) meters in 20 seconds.  Also x/10 is the velocity by the 2nd fact. So

(300+x)/20 = x/10

x=300 meters.

[/hide]

[hide]
If $x$ is the length of the Train, and $s$ is the speed of the Train.

$300+x=20s$
$x=10s$
$300=10s$
$30=s$
$x=10s=10\cdot 30=300$[/hide]

Yes it was. Linear Equations section, Chapter 3.  :P


calculus
integration
limit
trigonometry
function
calculus computations

This problem is very easy, use the fact that
\[ \mathop {\lim }\limits_{n \to \infty } \frac{{\int_0^n {f\left( {\sin x} \right)dx} }}
{n} \equal{} \frac{1}
{{2\pi }}\int_0^{2\pi } {f\left( {\sin x} \right)dx}
\]
to deduce the desires result.

Yes! But few students will be able to use in exam.

[quote="kunny"]Find $ \lim_{a\rightarrow \plus{} \infty} \frac {\int_0^a \sin ^ 4 x\ dx}{a}$.[/quote][color=darkred]
[b]Lemma I[/b] (prove easily) Let $ f: \mathcal R\rightarrow \mathcal R$ be a continue and periodical function

(with the period $ T > 0$). Then for any $ a\in\mathcal R$ and $ n\in \mathcal N$ we have :

$ 1\blacktriangleright$ $ \int_a^{a \plus{} nT}f(x)\ \mathrm {dx} \equal{} n\cdot \int_0^Tf(x \plus{} a)\ \mathrm {dx} \equal{} n\cdot\int_a^{a \plus{} T}f(x)\ \mathrm {dx}\ .$

$ 2\blacktriangleright$ $ \lim_{x\to\infty }\frac 1x\cdot \int_a^xf(t)\ \mathrm {dt} \equal{} \frac 1T\int_0^Tf(t)\ \mathrm {dt}\ .$[/color]

[hide="Indication."]
[color=darkblue]For any $ x\in\mathrm R$ exist uniquely $ n\in \mathrm Z$ and $ \rho\in [0,1)$ so that

$ \frac xT \equal{} n \plus{} \rho$, i.e. $ x \equal{} nT \plus{} n\rho$ Therefore,

$ \int_a^xf(t)\ \mathrm {dt} \equal{} \int_a^0f(t)\ \mathrm {dt} \plus{} \int_0^{nT}f(t)\ \mathrm {dt} \plus{} \int_0^{n\rho}f(t)\ \mathrm {dt}$ $ \implies$

$ \int_a^xf(t)\ \mathrm {dt} \equal{} A \plus{} n\cdot\int_0^Tf(t)\ \mathrm {dt} \plus{} B$, where $ A$ is constant and $ B$ is bounded a.s.o.[/color][/hide]

[quote="kunny"]Find $ \lim_{a\rightarrow \plus{} \infty} \frac {\int_0^a \sin ^ 4 x\ dx}{a}$.[/quote]
$ \lim_{a\rightarrow \plus{} \infty} \frac {\int_0^a \sin ^ 4 x\ dx}{a}$
$ \equal{}\lim_{a\rightarrow \plus{} \infty} \frac {\int_0^a \left(\frac{3}{8}\minus{}\frac{1}{2}\cos\,2x\plus{}\frac{1}{8}\cos\,4x\right) dx}{a}$

$ \equal{}\lim_{a\rightarrow \plus{} \infty} \frac {\frac{3}{8}\;\frac{\partial}{\partial a}\int_0^a \;1\;dx}{\frac{\partial}{\partial a}(a)}\minus{}\frac{1}{2}\lim_{a\rightarrow \plus{} \infty}\frac{\sin\,2a}{2a}\plus{}\frac{1}{8}\lim_{a\rightarrow \plus{} \infty}\frac{\sin\,4a}{4a}$
(l'hospital's rule applies only to the first limit because the integrand is not bounded)

$ \equal{}\frac{3}{8}\minus{}0\plus{}0$ (since $ |\,\sin\,\phi\,|\leq 1$ for all $ \phi$)


[quote="Enya"]I will enter the National University of Singapore this summer.

I'm in School of Computing. :D

I didn't learn programming before.

I wanna learn it by myself now in case that I cannot understand the lectures in university.

Do you have some good advice? :(

Thanx. :)[/quote]

I am not sure I understand what you are asking for. Do you wish to learn a programming language, or theory?

In the case of the latter, Algorithms by Sedgewick is quite good, as is The Art of Computer Programming by Knuth.

It sounds like you have NO programming experience, therefore I reccomend getting a good background in a language versus theory (theory is algorithms and data structures and things that are general for all computer science).  Find out what language your school will be working in and either get a good book or find a good online tutorial for the language.  Once you are very comfortable with translating your ideas into code, you should start learning some theoretical things (which is way more fun).  I don't reccommend Knuth's books as they seem outdated and hard to follow (he uses his weird made up language to teach theoretical stuff), but if you're comfortable with mathematical notation, you can't go wrong with Intro to Algorithms.  It covers basic to intermediate material in very good depth.


MATHCOUNTS
geometry
3D geometry
octahedron

What is the volume of a regular octahedron whose vertices are the centers of the faces of a cube whose edge has length 6

[hide]side length i think is 3 :rt2:

so the volume is...

(3 :rt2: )^3 * :rt2: /12  * 2 = 18?[/hide]

[hide]The volume of this octahedron using cube side length s is s :^3: /6.  With the side 6, the octahedron is 6 :^3: /6=6 :^2: =36

Ergo, the volume of the octahedron is 36 units :^3: .[/hide]


[quote="shobber"]For every natural number $n$, $y=(n^2+n)x^2-(2n+1)x+1$ intersect the x-axis at $A_n$ and $B_n$. Let $|A_nB_n|$ be the length of $A_nB_n$, find $\sum_{i=1}^{2006} |A_iB_i|$.[/quote]

For $y=ax^2+bx+c$ with 2 real roots, $\Delta\mathrm{roots}=\frac{\sqrt{b^2-4ac}}{a}$

$|A_nB_n|=\frac{\sqrt{(2n+1)^2-4(n^2+n)1}}{n^2+n}$, $n>0$.

$|A_nB_n|=\frac1{n(n+1)}$

$\sum_{i=1}^{2006} |A_iB_i|=\sum_{i=1}^{2006}\frac1{i(i+1)}=1-\frac1{2006}$


Prove that $ \frac {\sum_{r\equal{}p\plus{}1}^{m}C_{p\plus{}1}^r C_q^{m\minus{}r}}{(m\minus{}p\minus{}q)\sum_{r\equal{}p}^{m}C_{p}^r C_q^{m\minus{}r}}\equal{}\frac {p\plus{}1}{p\plus{}q\plus{}2}$

[quote="makar"]Prove that $ \frac {\sum_{r \equal{} p \plus{} 1}^{m}C_{p \plus{} 1}^r C_q^{m \minus{} r}}{(m \minus{} p \minus{} q)\sum_{r \equal{} p}^{m}C_{p}^r C_q^{m \minus{} r}} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$[/quote]

What's the meaning of $ C_{p\plus{}1}^r$ where $ r>p\plus{}1$ ????

[quote="pco"][quote="makar"]Prove that $ \frac {\sum_{r \equal{} p \plus{} 1}^{m}C_{p \plus{} 1}^r C_q^{m \minus{} r}}{(m \minus{} p \minus{} q)\sum_{r \equal{} p}^{m}C_{p}^r C_q^{m \minus{} r}} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$[/quote]

What's the meaning of $ C_{p \plus{} 1}^r$ where $ r > p \plus{} 1$ ????[/quote]

you may be familiar with this notation

$ \frac {\sum_{r \equal{} p \plus{} 1}^{m}\binom {r}{p \plus{} 1} \binom {m \minus{} r}{q}}{(m \minus{} p \minus{} q)\sum_{r \equal{} p}^{m}\binom {r}{p}\binom {m \minus{} r}{q}} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$

Where $ \binom {n}{r} \equal{} \frac {n!}{r!\ (n\minus{}r)!}$

[quote="makar"][quote="pco"][quote="makar"]Prove that $ \frac {\sum_{r \equal{} p \plus{} 1}^{m}C_{p \plus{} 1}^r C_q^{m \minus{} r}}{(m \minus{} p \minus{} q)\sum_{r \equal{} p}^{m}C_{p}^r C_q^{m \minus{} r}} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$[/quote]

What's the meaning of $ C_{p \plus{} 1}^r$ where $ r > p \plus{} 1$ ????[/quote]

you may be familiar with this notation

$ \frac {\sum_{r \equal{} p \plus{} 1}^{m}\binom {r}{p \plus{} 1} \binom {m \minus{} r}{q}}{(m \minus{} p \minus{} q)\sum_{r \equal{} p}^{m}\binom {r}{p}\binom {m \minus{} r}{q}} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$

Where $ \binom {n}{r} \equal{} \frac {n!}{r!\ (n \minus{} r)!}$[/quote]

What's the meaning of  $ \binom {m \minus{} r}{q}$ when $ m\minus{}r\equal{}0$ ?

Cher pco, je vous invite regarder [url]http://www.mathlinks.ro/viewtopic.php?t=302128[/url]
C'est evident qu'on parle d'une legere generalisation pour favoriser l'ecriture de certains equations  :)

I see, I see .... brand new standards and conventions ...
I noticed too that OP used as a convention that $ C_n^p\equal{}\binom pn$ while in my (ancient) knowledge, $ C_n^p\equal{}\binom np$

So we have to show $ \frac {\sum_{r \equal{} p \plus{} 1}^{m}\binom {r}{p \plus{} 1} \binom {m \minus{} r}{q}}{(m \minus{} p \minus{} q)\sum_{r \equal{} p}^{m}\binom {r}{p}\binom {m \minus{} r}{q}} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$ with extended notation.

Let $ m\equal{}p\plus{}1$ and the equation becomes $ \frac {\binom {p\plus{}1}{p \plus{} 1} \binom {0}{q}}{(1 \minus{} q)(\binom {p}{p}\binom {1}{q}\plus{}\binom {p\plus{}1}{p}\binom {0}{q})}$ $ \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$

$ \iff$ $ \frac {\frac 1{q!(\minus{}q)!}}{(1 \minus{} q)(\frac{1}{q!(1\minus{}q)!}\plus{}\frac{p\plus{}1}{q!(\minus{}q)!})} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$

$ \iff$ $ \frac {1}{(1 \minus{} q)(\frac{(\minus{}q)!}{(1\minus{}q)!}\plus{}p\plus{}1)} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$

$ \iff$ $ \frac {1}{(1 \minus{} q)(p\minus{}q\plus{}1)} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$

And this seems quite false to me (but maybe there are some unknown conventions which make this true ...).

It doesn't hold with the standard notation $ C_n^p\equal{}\binom np$ either. Take $ m\equal{}p\equal{}q\equal{}2$.

  darij


[hide="Solution"] The left sie of the equation is similar to the binary expansion of (x+1)^5, except that the x term has a coefficient of -5 instead of 5. We fix this by adding 10x to both sides to that the equation is now

x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 = 10 + 10x

which becomes

(x+1)^5 = 10(x+1)

Therefore,

[u][b](x+1)^4 = 10 or 0[/b][/u][/hide]

oh,I think it would have been much easier for you if you had just quoted mine  :D

jli left out part of the statement of the problem; it was stipulated that $x\ne -1.$


The [url=http://courses.ncssm.edu/goebel/statecon/state.htm]North Carolina State High School Math Contest[/url], with the support of the $\text{EMC}^{2}$ Corporation, is offering scholarships for participants in the NC State Math Contest.  [url=http://courses.ncssm.edu/goebel/statecon/Inform/AOPSLetter.htm]Click here for more details about how to apply[/url].

umm i know it's kinda late...but i just found out about this like yesterday but is there a due date on this, or is it still open?

Yes, there is a due date on this. Once you do not use the certificate after this date is passed, you are not able to use it.
I believe that first place state will earn you $150.00 (I live in NC and received it for state Algebra I--I'm in 6th grade)


Who's doing it?  The A problems were posted fairly soon after the A test, but I can't say the same for the B problems so far.  If nobody else is, I could finish up the ones [b]worthawholebean[/b] hasn't posted yet.

Alright, I'm running out of excuses to procrastinate my history homework, so I'll post the rest of the problems, I guess.

Thanks for completing them guys. I had to stop and write my English paper last night.


Imagine you're playing a game that involves two scoring events - say, there are two hoops, one higher up than the other, that you can throw a basketball into. For the lower hoop, you score n points, and for the higher hoop, you score m points.

Deduce a formula for the highest unattainable score.

(Example: if n=3 and m=4, the highest unattainable score will be 5, and then you can score any number higher than that: 3+3=6, 3+4=7, 4+4=8, etc.)

[quote="FMako"][quote="chess64"][hide]
$mn-m-n$

It's called the Chicken McNugget Formula! :)
http://mathworld.wolfram.com/McNuggetNumber.html
[/hide][/quote]

that is ridiculous.

Assuming $m > n$ and $m$ and $n$ are relatively prime the greatest value that $am+bn$ cannot be is $mn-m-n = (m-1)(n-1)$
[hide]http://mathforum.org/library/drmath/view/65669.html[/hide][/quote]

Why is it ridiculous?? :?

Anyway, $(m-1)(n-1)=mn-m-n+1 \ne mn-m-n$.

oh whoops good call, I was referring to the chicken mcnugget formula as ridiculous.

Chess64: I think the Chicken McNugget Formula only works for relatively prime integers. :huh:
thebrokenlight: Correct me if I am wrong, but I don't think there is any general formula for two arbitrary integers without knowing if they are relatively prime. For example, say n=2, m=4; all odds will be unattainable.

Right, I forgot to mention that they must be relatively prime.

Wow, awesome! Just one question... how is it derived? And how can you prove that it works?  :?:


geometry
circumcircle
incenter
geometry unsolved

Assume that $ O$,$ I$ and $ I_a$ are the circumcenter,incenter and the excenter corresponding to the edge $ BC$ of the triangle $ ABC$,respectively.Let $ II_a$ intersect the segment $ BC$ and $ (O)$ at $ A_1$ and $ M$ respectively.($ M \in II_a$).Let $ N$ be the midpoint of the arc $ MCA$ and $ S,T$ be the intersections of lines $ NI,NI_a$ with $ (O)$,respectively.
Prove that $ S,T,A_1$ are collinear.

use menelaus for the line $ TA_1S$ and the triangle $ NII_{a}$:
$ T,A_1,S$ r collinear iff: $ IS.NT.I_{a}A_1 \equal{} SN.TI_{a}.A_1I$
there r SOME computations to do :D

Use the concept of radical center.
http://www.mathlinks.ro/viewtopic.php?t=291967


Determine x in the case:

r and s are parallel

[img]http://img517.imageshack.us/img517/9948/melhor.jpg[/img]

$ \sum\text{Red}\equal{}\sum\text{Green}$.
So, $ 50^\circ\plus{}(2x^\circ\minus{}60^\circ) \equal{} x^\circ\plus{}(180^\circ\minus{}100^\circ) \iff x^\circ\equal{}\boxed{90^\circ}$.

[quote="Kouichi Nakagawa"]$ \sum\text{Red} \equal{} \sum\text{Green}$.
So, $ 50^\circ \plus{} (2x^\circ \minus{} 60^\circ) \equal{} x^\circ \plus{} (180^\circ \minus{} 100^\circ) \iff x^\circ \equal{} \boxed{90^\circ}$.[/quote]

Kouichi Nakagawa,

if you look at the diagram in the original problem, you'll see that the
variable "x" *stands for* the degrees already, so that your equation
should have been instead:

$ 50^\circ\plus{} (2x \minus{} 60^\circ) \equal{} x \plus{} (180^\circ \minus{} 100^\circ) \iff x \equal{} \boxed{90^\circ}$.

Oh, I did typo.
I seem to have just done copy and paste of ^\circ without thinking about anything.

Thank you for indication.


There are five chairs on stage, arranged in a straight line parallel to the front of the stage. Big Kahuna, Major Domo, Em See, Lou Tenant and Captain Marv El will occupy the five chairs, but Major Domo cannot sit next to Big Kahuna. How many seating arrangements are possible.

Using complementary counting, we have $ 5!\minus{}4(2!3!)\equal{}\boxed{72}$ arrangements.

I believe this is in the All Time Greatest Math book from MC?  perhaps?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189&start=380[/url]

Look in this topic. Maybe agree to trade/exchange tests from other AoPSers via email if you wanna save money. :wink:

You can also google "minnesota mathcounts" and it'll link you to the Minnesota Mathcounts site. In the Coaches Corner, there's an archive of past state tests.


I think this is classical.
A person has the sum of n dollars(or euro, because it's bigger).Every day he buys one of the following:an icecrem -1S, a sandwich-2S or  a cake -2S till he gets without money.In ho many ways can he spend the sum?

For n = 3, if he buys an ice cream then a cake, is it considered as the same as if he would buy a cake then an ice cream?

Pierre.

yes, probably

it's the same kind of questions as "in how many ways can you pay 13 euros and 76 cents?" or anything like it

but this time we need a general solution... :?

let's see how this does:

as there is only one way of spending odds, 2k and 2k+1 will have the same amount of options

for 2k thus:
0 icecreams: we get ccc...cs....sss with k+1 different options
2 icecream: k options
... and continue till 1 (only icecreams)

so you get ((k+1)²+k+1)/2 options

which means: (k+2)(k+1)/2 options

I do not understand what an option is... Moroever, If you consider n = 2k, i don't understand how you manage to buy only one ice cream....

If T(n) denotes the number of ways to spend n euros with coins of 1 euros, 2 euros from France and 2 euros from Spain, then clearly :
T(1)= 1  (only 1 = 1)
T(2) = 3 (1 + 1 = 2f = 2s)
Moreover, we have T(n+2) = 3T(n) since you have to pay n and spend the last two euros in one of the three ways above.

Thus, we easily deduce that T(n) = 3 <sup>[n/2]</sup> , where [.] denotes the integer part.

Pierre.

what's wrong with mine? I know what's wrong with yours: say n = 6 then the possibilities are:
ccc
ccs
css
sss
iicc
iics
iiss
iiiic
iiiis
iiiiii

and that's still 10 as my formula says :)

[quote="pbornsztein"]I do not understand what an option is... [/quote]

:? then look it up in your dictionary

option = possibility = way it can happen, or here: possible division of things bought with the given sum

[quote="pbornsztein"]Moroever, If you consider n = 2k, i don't understand how you manage to buy only one ice cream....[/quote]
no, that should be 2 icecreams indeed

corrected it, sorry.

Ok, I understand what you are meaning. I did consider the order of the different spendings and not you.
So, I think that whatever the problem was, we have a solution. :D

Pierre.

eum, aren't you supposed to get a [b]higher[/b] result than me?

I get 10 without order and you get 9 with :?

you should have counted 33 options if I'm correct (counted very fast so could be wrong)

though the general rule for that way is hard to write (a lot of permutations with repetition)

You are right again. I've counted only the ways to pay with two consecutive ice cream each time....
Thus, it is nothing but stupidities....

Pierre.

ya well we all miss from time to time

ps: check out the new smileys  :D  :cool:  ;)  :blush:  :maybe:  :P


MATHCOUNTS
geometry
email
AwesomeMath
summer program
MathPath
HMMT

[u]Top 10 Teams:[/u]

10. [hide]California[/hide]

9. [hide]North Carolina[/hide]

8. [hide]Pennsylvania[/hide]

7. [hide]Maryland[/hide]

6. [hide]Massachusetts[/hide]

5. [hide]Illinois[/hide]

4. [hide]Texas[/hide]

3. [hide]Indiana[/hide]

2. [hide]Washington[/hide]

1. [hide]Virginia[/hide]

Top 12 Individuals

12. [hide]Arjun Puranik, IL[/hide]

11. [hide]George Yu, MO[/hide]

10. [hide]Andrew Ardito, NY[/hide]

9. [hide]Rolland Wu, CA[/hide]

8. [hide]George Silvis, MA[/hide]

7. [hide]Brian Hamrick, VA[/hide]

6. [hide]Sam Keller, WA[/hide]

5. [hide]Daesun Yim, NJ[/hide]

4. [hide]Nathan Banjamin, IN[/hide]

3. [hide]Kevin Chen, TX[/hide]

2. [hide]Daniel Li, VA[/hide]

1. [hide]Neal Wu, LA[/hide]

Count Down Summary

Round 1. [hide]George Silvis defeats Rolland Wu[/hide]

Round 2. [hide]Daesun Yim defeats Arjun Puranik.  Daesun amuses the crowd when he's asked if he's ready and responds "Do we have a choice?"[/hide]

Round 3. [hide]Sam Keller defeats George Yu[/hide]

Round 4. [hide]Andrew Ardito defeats Brian Hamrick[/hide]

Summary of first 4 rounds:
[hide]Most of the higher seeds (lower numbers) won, but Andrew Ardito moved up intimidatingly.  He was the fastest of the first few rounds by far.[/hide]

Round 5. [hide]Neal Wu defeats George Silvis[/hide]

Round 6. [hide]Daesun Yim defeats Nathan Benjamin.  The phrase "Like, 11" goes down in MATHCOUNTS history.[/hide]

Round 7. [hide]Kevin Chen defeats Sam Keller, coming back from a two question deficit.  Kevin says geometry is his least favorite area of math, but he won on a geometry problem.[/hide]

Round 8. [hide]Andrew Ardito defeats Daniel Li in a very close match.[/hide]

Round 9. [hide]A lot of missing from anticipated "twitching".  Neither student recognized an exterior angle of a nonagon due to time pressue.  Two total questions were missed by each student, but in the end, Daesun took down the defending national champion.[/hide]

Round 10. [hide]Andrew scores another intimidating win over Kevin, who was the top seventh grader in the nation.[/hide]

Round 11. [hide]Andrew Ardito gets out to a 2-1 lead, partially due to a twitched miss on the first problem by Daesun.  Daesun comes roaring back to win the national championship.[/hide]

[quote]Daesun amuses the crowd when he's asked if he's ready and responds "Do we have a choice?" [/quote]
Aw that's not original. Good job nonetheless, treething. Interestingly enough, Daesun had to go through all 3 of the top 7th graders from last year.

Wow this will be very interesting to watch on ESPN.

How come Michigan wasn't in top 10? How come no Michigander placed in top 12?

Also, twitching???

[quote="robinhe"]Wow this will be very interesting to watch on ESPN.

How come Michigan wasn't in top 10? How come no Michigander placed in top 12?[/quote]

Because Michigan is overrated  :P I'd like to personally congratulate Daniel Li and Brian Hamrick  :blush:

[quote="Zenith Xan"][quote="robinhe"]Wow this will be very interesting to watch on ESPN.

How come Michigan wasn't in top 10? How come no Michigander placed in top 12?[/quote]

Because Michigan is overrated  [/quote]

I guess that was the case, and because there are a lot of michiganers on AoPS.

Lol good job treething. I have to say, that was like the most unexpected thing ever. No offense of course :P And good job to hexadedron (Andrew) too. :yup:

What was the Masters Round about?  I have always felt that that is the most interesting part of MATHCOUNTS.  Any prediction on who won the Masters?  I remember it was pretty clear (to me, anyhow) that Sergei was going to win last year.

[quote="tz"]Hope it's michigan[/quote]

What is this referring to?

Countdown was fun.

<_<

I was expecting a lot of other people to make it - so many people I didn't expect (including me)...

I predict Michael Wu and/or Ming-Ming Tran make Countdown next year.

Well, Ming-Ming has already won a non-trivial scholarship, and she was pretty good for a 6th grader last year :D. So yes, I'll guess she'll make CD next year.

Here's something like the question.

What is 12 + 1?

Him: "Like, 11."
Judge: "Like, wrong."
Me: "Like, 121?"
Judge: "Correct."

And "Like, 11" goes down in history.

California, 10th?! I am stunned into oblivion. And I didn't even realize the states that beat us had anything resembling math cultures. At least I get a chance to question the CA captain about this on Monday.

yeah... unless he makes stupid mistakes

heh, i got an idea, let's keep this thread alive until the next national mathcounts so people get slightly confused... lol

[quote="Treething"]
heh, i got an idea, let's keep this thread alive until the next national mathcounts so people get slightly confused... lol[/quote]

sounds like spam to me :roll:

Okay.

Did you know number 3 at nationals was the highest number of people who answered the question?

number 3... was that the (6 - 3)(6 + 3) one

haha... i think i got that one wrong... maybe it was #2 though.

nationals has always been fun for me... 6th grade was fun, 7th grade was fun, 8th grade was kickass

HaHa.

I know you didn't get #3 wrong.

Number 13, Unlucky 13, was the first question I got wrong, and then I bombed the rest of the test.  ON Sprint.

[quote="mathgeniuse^ln(x)"]Yes.

Well, it was implying that.

223/228 people got it correct.[/quote]
Wasn't the lowest sprint a 2?

I did reasonable well the first 20 minutes. Answered 14 questions correctly. At that rate, I should have a 21, but instead, I decided to try and brute force 2 problems and I ended up getting one of them wrong. I also did casework instead of just multiplying by 6 on #29. On the bright side, I answered another question correctly meaning I got 16 total.

[quote="moogra"]I feel bad about those people. They shouldn't release the worst scores[/quote]

At least they don't release the names.

who knows...

michael wu from new jersey's going into 8th and i'm pretty sure he'll make countdown at nats next year

Oh yeah.

Is he the one who said that the answer to number 10 has to be a "round number"?

[quote="mathgeniuse^ln(x)"]Heh, that is true.

I hope this years test will be harder than last year. to determine who deserves it or not.[/quote]
A harder test will only determine who deserves countdown, not 1st place...


[quote="billzhao"]I just received in the mail a copy of this year's AMC results book. I'm just curious, who do they send this out to?

P.S. I noticed that the USAMO qualifier list comprised of last year's qualifiers, instead of what it supposed to list.[/quote]

Schools that register for the AMC exams all get them. If they register for both the A date and the B date, they will get 2. Some students get them, I guess if they are recognized above a certain level. A few AMC affiliates (not sure what that is). That's it. Colleges don't get them.

I also noticed the USAMO list is last years. The AMC organizers know. There were also some score omissions.  My program had a school award and some individual awards from the A date. We should have had about the same from the B date, but those were missing. Oh well. It's amazing the book is as accuate it is with all the minutiae it includes.

I think if you qualify for USAMO, you get one. I got one in 2002, and this year, when I qualified for USAMO, but not in 2003, when I didn't.

Yan Zhang


You connect the points (0, 0) and (9, 6) with a line segment.
Starting at (0, 0), you move $ \frac{1}{3}$ of the way along the segment.
What is the sum of the coordinates of the point you land
upon?

When moving from the origin, due to the triviality of the origin coordinates, a simple dividing by three will lead to the point $ \frac {1}{3}$ of the way to any point on the x-y coordinate plane. The answer is therefore $ \left(\frac {9}{3},\frac {6}{3}\right) \equal{} \boxed{(3,2)}$.

EDIT: I should learn to read, see jxl28's post below.

[quote="GameBot"]You connect the points (0, 0) and (9, 6) with a line segment.
Starting at (0, 0), you move $ \frac {1}{3}$ of the way along the segment.
What is the [b]sum of the coordinates[/b] of the point you land
upon?[/quote]
Thus, the answer is $ 3 \plus{} 2 \equal{} \boxed{5}$


Mister A has X and is flipping an ideal coin:
-he wins 1 dollar for each head
- he lost 1 dollar for each tail
The game will end if he reaches 0 or a proposed sum M.
Having known values X and M, compute the probabilities
that mister A looses all (reaches 0).

I assume that you mean that Mr. A has X dollars to begin with.

Notice that the game can take a long time to finish.

[hide="Not sure if this is a full solution"]After each flipping of the ideal coin, the probability of winning or losing a dollar is the same, so this is a fair game (in average, he won't win nor lose).
So, at the end of the game we must have the following average:
$0\cdot P(0)+M\cdot P(M)=X$
The solution is:
$P(M)=\frac{X}{M}$
$P(0)=1-\frac{X}{M}$[/hide]

Thanks LordWings, your answer is good!
 I owe you one :lol:


geometry
geometric transformation
homothety
circumcircle
geometry proposed

For M is the centre of homothety of the two circles, we have $MBC$ and $MDC$ similar,and so $BC//DE$.
(in my draw, $B,M,E$ are on the same line).
We have $\angle DCA=\angle MBC=\angle DEM=\angle DAM$,so $DAM$ and $DCA$ are similar,so $DA^{2}=DM\cdot DC=pow_{O_{1}}D$.In the same way,we have $EA^{2}=pow_{O_{1}}E$.
Now,let consider $A$ as a "circle" or radius 0.
This means that $DE$ is the radical axes of the circles $O_{1}$ and "$A$",and so,if $F$ is the intersection of $DE$ and $AC$, we have $FC=FA$.

bye bye

[quote="Huyền Vũ"]The circle $(O1)$ touches $(O2)$ externally at $M$. $A$ is a point on $(O2)$. The tangents from $A$ to $(O1)$ touch $(O1)$ at $B$ and $C$. $MB, MC$ cut $(O2)$ at $E$ and $D$. Prove that $DE$ passes through the midpoint $K$ of $AC$.[/quote]

 [color=blue]I have studied carefully the solution , but I would like to ask , if you have any different solution than frengo. I wonder if we can change the last part of his nice solution , where he considers a point as a circle. I like this idea , but I'm trying to find a ..more elementery explanation.[/color]

Thanks - Babis

Notice that $ADME$ is a harmonic quadrilateral so we can change the original problem to the following problem
 P2:Let $ADME$ be a harmonic quadrilateral. $H$ is the projection of $A$ on $DE$.Let $P,Q$ be the midpoint of $AM$ and $DE$. Prove that $HP$ and $MQ$ are parallel

[quote="Huyền Vũ"]
Let $ADME$ be a harmonic quadrilateral. $H$ is the projection of $A$ on $DE$. Let $P,Q$ be the midpoint of $AM$ and $DE$. Prove that $HP$ and $MQ$ are parallel[/quote]
Denote: $S$ be the point of intersection of two tangents at $A$ and $M$ of the circumcircle of quadrilateral $ADME$ then $D,\,E,\,S$ are collinear.
And $O$ be the center of circumcircle of quadrilateral $ADME$
We have:

$\blacktriangleright\;\angle APS=\angle AHS=90^\circ\Longrightarrow\angle PHS=\angle PAS\quad (1)$

$\blacktriangleright\;\angle OQS=\angle OAS=\angle OMS=90^\circ$$\Longrightarrow O,\,A,\,Q,\,S,\,M\;\text{lie on a circle}\Longrightarrow\angle PAS=\angle MQS\quad (2)$
From $(1)$ and $(2)$ we have: $\angle MQS=\angle PHS\Longrightarrow HP\parallel QM$


inequalities
function
inequalities unsolved

We will use Jensen's inequality for function $1/\sqrt{x}$.

We have
\[\frac{1-a}{\sqrt{a^2+bc}}+...=2\cdot\frac{(1-a)/2}{\sqrt{a^2+bc}}\]
and $\frac{1-a}{2}+\frac{1-b}{2}+\frac{1-c}{2}=1$, so
\[2\cdot\frac{(1-a)/2}{\sqrt{a^2+bc}}+...\geq 2\cdot\frac{1}{\sqrt{(a^2+bc)(1-a)/2+(b^2+ac)(1-b)/2+(c^2+ab)(1-c)/2}}\].
So we need to prove that
\[\frac{2\sqrt{2}}{\sqrt{(a^2+bc)(1-a)+(b^2+ac)(1-b)+(c^2+ab)(1-c)}}\geq\sqrt\frac{2}{a^3+b^3+c^3}\].
So we obtain inequality
\[4(a^3+b^3+c^3)\geq (a^2+bc)(1-a)+(b^2+ac)(1-b)+(c^2+ab)(1-c)=a^2+b^2+c^2+ab+bc+ca-(a^3+b^3+c^3-3abc)\].
Using inequality (*) $x^3+y^3+z^3+3xyz\geq x^2y+...$ for $a,b,c$ ($a+b+c=1$) we obtain $a^3+b^3+c^3+3abc\geq a^2-a^3+b^2-b^3+c^2-c^3$, so new inequality for consideration is following
$4(a^3+b^3+c^3)\geq a^3+b^3+c^3+ab+bc+ca$ $\Leftrightarrow$ $3(a^3+b^3+c^3)\geq ab+bc+ca=(a+b+c)(ab+bc+ca)$ -- here we only have to apply Cauchy and (*).

It is long solution, isn't it?

[quote="Maverick"]Let $a,b,c>0$ with $a+b+c=1$. Prove that:

\[\displaystyle \frac{1-a}{\sqrt{a^2+bc}}+\frac{1-b}{\sqrt{b^2+ca}}+\frac{1-c}{\sqrt{c^2+ab}} \geq \sqrt{\frac{2}{a^3+b^3+c^3}}\][/quote]

I have another solution.
Rewrite the inequality :
$\frac{b+c}{\sqrt{a^2+bc}}+\frac{c+a}{\sqrt{b^2+ca}}+\frac{a+b}{\sqrt{c^2+ab}} \geq \sqrt{\frac{2}{a^3+b^3+c^3}} $
We try to prove these two inequalities :
(1)$\frac{b}{\sqrt{a^2+bc}}+\frac{c}{\sqrt{b^2+ca}}+\frac{a}{\sqrt{c^2+ab}} \geq \sqrt{\frac{2}{a^2b+b^2c+c^2a}} $
(2)$\frac{c}{\sqrt{a^2+bc}}+\frac{a}{\sqrt{b^2+ca}}+\frac{b}{\sqrt{c^2+ab}} \geq \sqrt{\frac{2}{ab^2+bc^2+ca^2}} $

To prove (1)
$(\frac{b}{\sqrt{a^2+bc}}+\frac{c}{\sqrt{b^2+ca}}+\frac{a}{\sqrt{c^2+ab}})({b}{\sqrt{a^2+bc}}+{c}{\sqrt{b^2+ca}}+{a}{\sqrt{c^2+ab}}) \geq\ (a+b+c)^2=1 $
Use BCS's inequality we can prove
${b}{\sqrt{a^2+bc}}+{c}{\sqrt{b^2+ca}}+{a}{\sqrt{c^2+ab}} \leq\ \sqrt{2(a^2b+b^2c+c^2a)(a+b+c)} =\sqrt{2(a^2b+b^2c+c^2a)}$
Therefore
$(\frac{b}{\sqrt{a^2+bc}}+\frac{c}{\sqrt{b^2+ca}}+\frac{a}{\sqrt{c^2+ab}})\geq\ \frac{1}{\sqrt{2(a^2b+b^2c+c^2a)}}$
(2) can be proved similarly.
Use the inequality : $ a^3+b^3+c^3 \geq\ $ max{$a^2b+b^2c+c^2a,ab^2+bc^2+ca^2$}
we finish the proof.


To many to count really but some of my favorites are:
Rakkety Tam by Brian Jaques(I can't spell his last name right :blush: )
Going Postal by Terry Pratchett
and Elder the sequel to Eragon is coming out soon

Well, right now I am reading three, and the three are excelent

1) Les Liaisons Dangereuses, Chordelos de Laclos.
2) Three Degrees Above Zero, Jeremy Bernstain
3) The ****** Monologues, Eve Ensler.

Something funny about this books is that non of them is the kind of literature that I use to read, nevertheless, all of them are quite good.

(1) is interesting, is written in a somehow complicated way. If you saw Cruel Inentions (The first movie), well, the book is from the seventeen hundreds, but the movie was based on the book (you will be amazed how much!).

(2) is great (but may be a little outdated) for those who wants to study mathematics or pure sciences and don't know what they will do afterwards. It is not a novel.

(3) is kind of feminist. The truth is that I am enjoying it a lot. The ********, any one who knows about Eve Ensler should now what is the word that it is replacing. The title may sound very acward and freak out somany people, but it makes you think about the misoginistic preconscepts that were implanted in our subconcius since we were babies, and the injustices that we have been tolerating against women. May be PG-13, but it is a good book.

What good books I have ever read: Last time I counted them (several years ago) they were around 150, not counting tecnical books (YEAH YEAH, I know that some people that is five or seven years younger than me have read twice as much, but, what the heck, I enjoyed all of them).

Best,

Some of my favorite books are:

The DaVinci Code by Dan Brown
The Millennium Problems (self explanatory) by Keith Devlin
Fermat's Enigma by Simon Singh

They aren't exactly recent but they're still great books!!!

I also enjoyed Dan Brown, but my list extends beyond the DaVinci Code:

Digital Fortress by Dan Brown
The DaVinci Code by Dan Brown
Angels and Demons by Dan Brown

My maths teacher got me hooked on the Left Behind series by Tim LaHaye and Jerry B. Jenkins lately. I started a week ago and am almost done with book 6. :) If you're a Christian, I highly suggest you read the series.

[quote="MathFiend"]

My maths teacher got me hooked on the Left Behind series by Tim LaHaye and Jerry B. Jenkins lately. I started a week ago and am almost done with book 6. :) If you're a Christian, I highly suggest you read the series.[/quote]

My friend got me hooked on those too.  I find series annoying though...either I'm always waiting for the next one to come out, or trying to find the ones missing at my library.  Still good though!

[quote="jackroland"][quote="djimenez"]3) The ****** Monologues, Eve Ensler.[/quote]
If this is what I think it is then my mom and guite a lot of other people are going to be acting it in a couple of days to raise money the Magdelin project (I would explain further but there would be a lot of little stars.) It's a really big thing here.[/quote]

Probably it is what you think. The book is good, is written by a woman who wants a better world for women, a world in which ladies don't suffer what she did. I placed stars instead of the world because I thought may be someone could consider the world "dirty", even when it is just a part of the female anathomy. (Why do we have to consider dirty some words that refer to the human anatomy? :? )

Any way, what is that "Magdelin Project", there area a bunch of "Magdelen Projects" in the web. Eitherway, if it is something that can't be publically discussed here, send me a private message.

Best,

[quote="djimenez"](Why do we have to consider dirty some words that refer to the human anatomy? :? )[/quote]
I really don't know it doesn't make sense to me either.
Oh, I thought you might like to know my mom is playing the
"sad voice" in The Village monologue.

Not really new, but kind of unknown.

The Leopard - Giuseppe Tomasa di Lampedusa
Summer - Edith Wharton
Timequake - Kurt Vonnegut (his newest, but not one of his best, check out Breakfast of Champions)

I recommend reading both of Jared Diamond's books: Guns, germs, and steel; and Collapse
Together, they make for 1000 page epic. The books are all about the structure of civilization and how modern civilization came to exist.

Everyone should read them, even if you do not agree.........especially if you do not agree :D

The Sword of Truth series by Terry Goodkind.  Yes, it's fantasy, and yes, the books are slightly annoying, but you feel accomplished after having read them. (or angry.  It depends)

9 books (currently), average length of 850 pages.
I finished the entire series in 10 days.  I was proud.


Let a triangle $ ABC$ with internal bisector $ BE,CF$. Ray $ EF,FE$ cut circle $ (ABC)$ at $ M,N$. Prove that:
$ \frac{1}{BM}\plus{}\frac{1}{CN}\equal{}\frac{1}{AM}\plus{}\frac{1}{AN}\plus{}\frac{1}{BN}\plus{}\frac{1}{CM}$

Can somebody solve this problem by use Ptolemy theorem?


Hi.  Can someone please help me with 5 simple statistics questions? I would be forever greatful!

-If a single fair die is rolled, find the probability of rolling a 4 given that the number is odd.

-How many ways can a commitee of 5 be selected from a club with 10 members?

-How many 5 digit numbers can be formed using the digits 0-9, if repitition of digits is not allowed?

-Find the probability that when two dice are rolled, the score on the dice is 5 or the dice both show the same number ("doubles").

-If two cards are drawn from a deck without replacement from a deck, find the probability that the second card is red, given that the first card was a heart.

1.  Given that the number is odd, 4 is not a possible outcome of this roll, and the probability of rolling a four is zero.

2. 10C5, or 10!/(5!*5!) = 252

more info on combinatorics:
[url]http://rds.yahoo.com/S=2766679/K=combinitorics/v=2/SID=w/TID=OOP5_5/l=WS1/R=2/IPC=us/SHE=0/H=2/SIG=128gukngr/EXP=1109558511/*-http%3A//www.umkc.edu/is/mts/math116/Chapter%202-4.ppt[/url]

3.  The first digit must be 1-9 (if the first digit is zero, the number is not considered a five digit number), giving 9 choices, the second must be taken from 0-9, but cannot be the first digit, giving 9 choices for the second.  The third, eliminating the first two from the possible choices, is chosen from a set of 8, the fourth from a set of 7, and the final from 6.
The number of possibilities is given by multiplying these numbers
9*9*8*7*6 = 27216

4.  P(Event 1 or Event 2) = P(Event 1) + P(Event 2) - P(Event 1 and Event 2) or, the probability that one event or another occurs it the sum of their individual probabilities minus the probability that both occur.  The total number of possible outcomes of a roll of two fair die is 6 choices for the first * 6 choices for the second = 36 total outcomes
P(sum of two die=5) =  4/36 (four possibilities (1,4), (4,1), (2,3), (3,2) divided by the total).  P (same number on both) = 6/36 ((1,1), (2,2), (3,3), (4,4), (5,5) (6,6))
Since the two events cannot both occur in the same trial (since 5 is odd), P(both events) = 0.  4/36+6/36-0=10/36= 5/18

5.  If the first card was a heart, there are now 25 red cards in the deck (1 was removed).  Also, the total number of cards in the deck is 51, since 1 was removed.
P(drawing a red card on the second draw given that the first was a heart without replacement) = 25/51

OMG!  You rock!!  Thank you so much!!!!!!!!!!!
I know I've bothered you enough, but I have one more question... On questions 2 & 3... They seem like the same question to me... Why are the answers different?

can you please tell me if my answer is right?

Question:  Three coins are tossed 10 times to simulate the outcome of the sexes of the children in a three-child family.  If H simulates a boy and T a girl, use the results to determine the empirical possibility that the family has at least two boys.

Results:  TTH, HTT, HHT, HHH, TTH, TTH, HHT, HHH, HHT, HTH

Is the answer the empirical possibility that the family has at least 2 boys b/c the coin landed on H 60% of the time?

[quote="sleepysassy1"]Okay, nevermind.  I figured out my own answer.  It's b/c in question 3, the first digit cannot be a 0, b/c then it wouldn't be a five digit number.[/quote]o

actually, the main difference is that in number two, order doesnt matter, meaning that a committee that is Bob, Jack, Joe, Sally, and Sue is the same as Jack, Sue, Joe, Sally, and Bob
in number 3, order does matter, because 98765 is obviously a different number from 56789

so, as miraculous postmaster said, #2 is 10C5

but you are right. the first digit cannot be 0, so it is not as easy as just doing 10 permute 5 for the third question

sorry, i dont know what empirical possibility is so i cant help you out on that one  :?

sorry, i dont know what empirical possibility is so i cant help you out on that one

Suppose an experiment with a number of possible outcomes is performed repeatedly- say n times, and that a specific outcome E occurs r times.  The empirical probability that E will occur on any given trial of the experiment is given by...   oh shoot I dont even remember the formula.  But do you get what I'm saying?[/u][/list][/code][/i]

A google search says that empirical probability is actually the expected probablility; that is
empirical probability = number of favourable outcomes/number of possible outcomes.
So the empirical probability of rolling 4 on a die is 1/6, no matter how many experiments we do.
There are 4 ways to get at least 2 heads (HHT, HTH, THH, HHH) and 2 x 2 x 2 = 8 combinations in total, so the empirical probability is 4/8 = 1/2.

So if the empirical probability is 1/2, how does that answer the question "use these results to determine the empirical probability that the family has at least 2 boys"?

If the question asks you to simulate the experiment by flipping the coins or gives you the results of the flips, then sleepysassy's method is correct. Based on the simulation, there is a 50% chance that the family will have at least 2 boys becuase out of the 10 trials, 5 give two boys.

It's not 60% because we care about how many heads in one trial, then counting the number of trials in which two heads occur. We do not want to count the total number of heads because that represents the probability of getting a boy based on a simulation of 30 flips.

Saul (correctly) used thoeretical probability. However, the question specifically asks for the probability based on the simulation. It was coincidence that the answers come out the same. If (by some amazing luck) the simulation gave all heads, then the answer would be 100%.

Thanks Saul & Bictor.
I definately understand the difference now.  What I don't understand is what kind of answer the question is looking for?  "If H simulatesa boy and T a girl, use these results to determine the empirical probability that the family has at least 2 boys."
So is the answer "The family would have at least 2 boys because the outcome of the coin flips resulted in H showing up 2 or more flips in 6/10 of the sets?
Thanks again for all your help!


vector
linear algebra
matrix
abstract algebra
linear algebra unsolved

I'm having a little problem solving these 2 questions...

1.If T is a normal operator on a finite-dimensional complex inner product space such that T^9 = T^8
prove that T is self-adjoint and T^2 = T.

2.Let T be a linear operator on a vector space V such that dim(Im(T)) = n. Prove that T has at
most n + 1 distinct eigenvalues.

1)normality on a finite dimensional inner product space is exactly what you need in order to have it unitary equivalent with a diagonal matrix

in diagonal form, your other conditions quickly gives that all eigenvalues satisfy $x^9=x^8$ and thus they are 0 or 1

It follows immediately now it is self adjoint and idempotent

2) if the image has dimension n, and v is the dimension of V itself
then your kernel has dimension v-n, this means geometric multiplicity of zero is v-n, and thus algebraic dimension at least v-n
this leaves space for at most n other eigenvalues->at most n+1 eigenvalues


The first five terms of an arithmetic sequence are 25, 22, 19, 16, 13. What is the 30th term?

Common difference is $ 3$.  $ 29 \times 3\equal{}87$, so this is $ 25\minus{}87\equal{}\boxed{\minus{}62}$.


geometry
circumcircle
power of a point
radical axis
cyclic quadrilateral
geometry unsolved

I did a problem and I need to use this conclusion, but I cannot prove it:
$ ABCD$ is the inscribed quadrilateral of $ \odot O$, $ AC$ meets $ BD$ at $ E$, $ BA$ meets $ CD$ at $ P$
$ M$ is the midpoint of $ PE$, $ MS$,$ MT$ are the tangents.
Prove: $ M$ is the circumcenter of $ \triangle PST$

It suffices to show that the circle with diameter $ PE$ is orthogonal to the circle with center $ O$.

Let $ Q$ be the intersection of $ AD$ with $ BC$. Then, the polar of $ P$ is $ EQ$. This is because it is well-known (for example, proven at (URL) ) that $ PE$ goes through the intersections of the tangents from $ A,D$ and $ B,C$ to the circle with center $ O$. This implies that $ OP \perp EQ$.

Hence, the point of intersection $ T$ of $ OP, EQ$ lies on the circle with diameter $ PE$, as $ \angle PTE$ is right. But $ T$ is the inverse of $ P$, as it lies on the polar of $ P$ and lies on $ OP$. Consider the inverse of the circle with diameter $ PE$. Let it intersect the circle with center $ O$ at $ X,Y$. Then, $ P,T,X$ lie on the circle with diameter $ PE$, but they also lie on the inverse of that circle. This happens iff that circle is orthogonal to the circle with center $ O$; hence we are done.

(Note all inversions here are with respect to the circle with center $ O$)

Let M is the circumcenter of △PST, S,T are the intersections of the two circless
our mission is to prove OS is the tangent to ⊙M
1.It is well-known that OE⊥PL , denote the foot is K , K is the miquel point of  complete quadrilateral ABCD
2.according to radical axis theorem ST,BC,AD,PK are concurrent at Q
3.If we denote X,Y the tangent point of P to ⊙O, then Q,X,Y,E are on a line(you can find this conclusion in 1997CMO 1996CMO) So we can acquire that QE⊥OP , denote the intersection is R , we have R is on ⊙M(PE is the diameter of ⊙M)
Using the same method, we have PE⊥OQ,denote N is the foot
4.E is the orthocenter of △OPQ
5.from 3,we have QO*QN=QR*QE  but QR*QE=QS*QT   So  QO*QN =QS*QT  => N,O,T,S concyclic
6.∠NSO=∠NST-∠OST=∠TOV-∠OTS=∠TQO  => △SNO∽△QSO  => SO^2=NO*OQ
 But from 4 we have  NO*OQ=OE*OK   =>  SO^2= OE*OK =>  SO is the is the tangent to ⊙M

Let $ X,\ Y$ be, the intersection points of the circle $ (O),$ from the line segment $ PE$ $ ($ $ X,$ between $ P$ and $ E$ $ ).$

So, because of the points $ P,\ X,\ E,\ Y$ are in harmonic conjugation, based on the [b][size=100]Newton theorem[/size][/b], we have $ (MX)\cdot (MY) \equal{} (ME)^{2} \equal{} (MP)^{2}$ $ ,(1)$

We have also, $ (MX)\cdot (MY) \equal{} (MS)^{2} \equal{} (MT)^{2}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ ME \equal{} MP \equal{} MS \equal{} MT$ $ ,(3)$ $ \Longrightarrow$ $ \angle ESP \equal{} 90^{o} \equal{} \angle ETP$ $ ,(4)$

From $ (3),$ $ (4)$ we conclude that the point $ M$ is the circumcenter of the cyclic quadrilateral $ PSET$ and the proof is completed.

kostas Vittas.

It is easy to see that $ P,E$ are the $ 2$ conjugated points wrt $ (O)$. Therefore, if $ \{X,Y\}\equiv EP\cap (O)$ such that $ X$ is between $ P$ and $ E$, we will have $ (PEXY)\equal{}\minus{}1$, but $ M$ is the midpoint of $ PE$. Hence $ ME^2\equal{}MP^2\equal{}\overline {MX}\cdot\overline {MY}\equal{}MT^2\equal{}MS^2\Longrightarrow MP\equal{}ME\equal{}MS\equal{}MT$, which implies that $ M$ is the circumcenter of $ \triangle PST$.
Our proof is completed then.


irrational number
algebra unsolved
algebra

What's the equal of the number cos1.cos3.cos5......cos89 ?

excuse me. i am asking for exact solution.

Solution as in a representation in terms of known constants? I don't believe there is one.. May be I'm wrong...

the solution can be product of an irrational number and power of 2. This was our exam question. It was an multiple choice question. But i don't remember the choices.

Consider
\[ f(x)\equal{}\prod_{j\equal{}1}^{2n}(x\minus{}exp(\frac{\pi i (2j\minus{}1)}{2n}))\equal{}\minus{}\prod_{j\equal{}1}^{2n}(xexp(\frac{\pi i}{2n})\minus{}exp(\frac{2\pi i j}{2n}))\equal{}x^{2n}\plus{}1.\]
We get $ 2\equal{}f(\minus{}1)\equal{}\prod_{j\equal{}1}^{2n}(\minus{}1\minus{}cos\frac{\pi(2j\minus{}1)}{2n}\minus{}isin(\frac{\pi (2j\minus{}1)}{2n}))$.
Therefore $ 2\equal{}(2^n\prod_{j\equal{}1}^ncos(\frac{\pi (2j\minus{}1)}{4n}) )^2$ or
$ \prod_{j\equal{}1}^ncos\frac{\pi(2j\minus{}1)}{4n} \equal{}\sqrt 2 2^{\minus{}n}$.
Your case is n=45.


inequalities
FTW
topology
calculus
geometry
analytic geometry
trigonometry

I tried this five minutes ago and something went wrong.  Lemme try again.

You could put this in the Polls section, if you wanted.

But my favorite type of math is probably algebra. It's fun finding what [i]x[/i] is and stuff :P. Every other option on there I don't know, let alone learned.

I thought there's something like this in the Polls... I voted Number Theory.

DIE NUMBER THEORY

Where's the option for inequalities? I really don't want to check Algebra, since I hate most of that...

shouldn't this be in polls or something?

I only really learned three of them. Not sure, but a tie between combinatorics and number theory.

[quote="gauss1181"]I love Algebra. It's really easy and interesting. Last year i took in in 6th grade and it was so freakingly easy...I got 100s on all of my tests and quizzes :D[/quote]

Um... $ \text{Elementary Algebra 1}\ne \text{Algebra}$. $ \text{Elementary Algebra 1}\subset \text{Elementary Algebra}\subset \text{Algebra}$.

http://www.mathlinks.ro/viewtopic.php?t=121565

http://www.mathlinks.ro/viewtopic.php?t=134287

Algebra and Number Theory FTW!

[quote="Temperal"]DIE NUMBER THEORY

Where's the option for inequalities? [/quote]

LONG LIVE NUMBER THEORY!

I guess inequalities would fall under Algebra (i.e. an under garment  :P ).

Topology, but that's not in the poll, so I voted number theory.

[quote="miyomiyo"][quote="D3m0n Shad0w"]Every other option on there I don't know, let alone learned.[/quote]Umm, is that sarcasm?[/quote]

She's in fifth grade. Give her a chance.  :lol:

[quote="Nerd_of_the_Ages"][quote="gauss1181"]I love Algebra. It's really easy and interesting. Last year i took in in 6th grade and it was so freakingly easy...I got 100s on all of my tests and quizzes :D[/quote]

Do you mean you went to middle school to take it?
If you didn't I could say the same thing. The teacher self taught a tiny bit. But it was pathetic, mainly just simplifying linear equations.[/quote]

Oh oh oh! I took algebra 1 in 6th grade at middle school!

Now I'm in a similar position as the original poster in multivar(iable calculus) by having a 105.69%...

man, you guys are lucky
im in 7th and taking algebra 1
i mean, all elementary math is basically waste of time (after 3rd grade, that is. I admit knowing how to multiply and divide is important). Prealgebra is also useless, because it's basically a review or 6th grade (and 6th grade is really just a review of 5th grade, which is review of 4th grade). With just a little algebra background, I can imagine that it wouldn't be too hard to take Algebra 1 in 6th grade.
Algebra 1 could've been kinda interesting if the teachers explained WHY to do things instead of only WHAT to do.

I like combinatorics. At least the ones they teach at school as they can actualyl be harder later on.

It doesn't ask which is the easiest, just your favorite. So it doesn't really matter if it gets harder later on...most things do.

Since I'm unwilling to select Algebra for Inequalities, I just checked Calculus.

I would've checked Geometry, but it says "Euclidean", so I didn't.

(1) Euclidean Geometry is Geometry not bound by a coordinate plane.
(2) I don't understand how [i]anybody[/i] could like combinatorics or [blech]trig[/blech] best.

Actually, Euclidean Geometry simply refers to geometry in which one accepts the four axioms that Euclid laid so long ago, and the Fifth Postulate.

Number Theory woooooooooooo!!!!!!!!  :wow:

WTF Algebra Bar Begla FTW

I know, stupid palindrome.

Number Theory FTW

[quote="dbkarp"]You should put one: all math

that's mine.[/quote]
No.
I don't really like geometry and algebra.
Trig is all right, as are inequalities.
Number Theory and Combinatorics FTW!


calculus
integration
probability
number theory
\/closed

Is it possible to get college credit for AoPS classes?

You'd have to check with the specific school but probably not.

Agreed; it's pretty unlikely that you can use AoPS classes for collegiate credit.

Do most people get credit from their schools for AoPS?

No.  Most of our students are taking our classes for fun or for contest prep.  We have some homeschoolers taking the classes "officially", and there are a few schools that give their students credit for their work with us, but you'd have to clear that with your school if you wanted to do so.

Putting on one of my other hats: I'm the one who deals with questions of transfer credit at my institution (California State University, Long Beach). Here's what would happen:

1. We wouldn't give any actual college credit for AoPS courses. They're outside our rules, and there is no mechanism for it.

2. A person with AoPS courses would have significantly enhanced credibility when talking to me about prerequisites and what courses we might let that student take.

3. And I'd personally want to talk to any such student about some other things.

Most universities accept some "prior college-level learning" for credit if that learning can be assessed or validated in a way that is acceptable to the institution.  Credits transferred from other regionally accredited colleges and universities are governed by school policies and implemented by folks like Kent Merryfield.  When the prior learning comes from a source that lacks collegiate regional accreditation (i.e. AoPS, most high schools, and homeschools) the most commonly used forms of prior learning assessments used for lower-division (freshman/sophomore) courses are:

[url=http://www.collegeboard.com/student/testing/ap/subjects.html]Advanced Placement (AP)[/url] Exams
International Baccalaurate (IB) Exams
[url=http://www.collegeboard.com/student/testing/clep/exams.html]College-Level Entry Placement (CLEP)[/url]
Departmental placement exams at the university.

Just about anyone in high school should be able to sign up for an AP exam - there is no AP requirement that an AP course be taken prior to the exam.  However, some high schools might frown at letting one take the exam without the course so be prepared to demonstrate that you are well prepared.  In math, there are just the two Calculus exams available (AB and BC.)  These cover differential and integral calculus, but not multivariable calculus.  AP Statistics is also available and some math departments offer an equivalent course.

So, if you took the AoPS Calculus course and wanted to obtain college credits, you'd just need to take the Calculus BC or AB exam (or CLEP, if acceptable) and obtain the score required by the university (minimally acceptable scores vary from 3 to 5.)

With respect to the other common placement tests, as a general rule, you need to be in an IB course or program to take the IB exams.  Most regional schools and many others accept CLEP tests.  CLEP is open to all but typically used by older students, out of high school, with no access to AP or IB programs.  Many universities administer CLEP tests on Saturdays every month or so.  To my knowledge, there are no CLEP or IB math exams beyond differential and integral calculus.

College-level knowledge of number theory, probability, counting, etc. that might be obtained through AoPS courses can't be assessed with AP, IB, or CLEP exams.  You'd need to talk to a university math department chair or someone responsible for such assessments to see what might be appropriate.  They might be able to set up a departmental exam or independent study that could provide advanced standing (no credit but ability to use the course as a prerequisite for higher-level courses) or course credit for knowledge that can be validated in an acceptable manner.  This is not a very common situation at most regional schools so department chairs and advisors may be given substantial latitude to determine what might be in your best interest.  Top-tier schools probably encounter several students each year with such knowledge and may have more formalized mechanisms for advising and placing such students.  Advanced standing will be a much more common outcome at the top-tier schools than course credit.

[quote=BiornH]We're 7 years down the road from this original conversation/thread.  Has anything changed on this topic? Or does the the previous information still stand?[/quote]

From a few months ago in a global announcement I think that they mentioned that in California now you can get credits from AoPs courses.

Someone correct me if I'm wrong.

Not quite.  AoPS courses are now accepted by California for the A-G requirements.  These are the requirements California universities have for high school students applying for admission.  Some AoPS subject courses can be used to fulfill the math requirements.

AoPS still does not give students college credit.  Some colleges give student credit or placement based on exam results, and AoPS courses can prepare students for such exams (Calculus AP exam is probably most common).


conics
parabola
quadratics
analytic geometry
graphing lines
slope
calculus

[hide="Solution"]To find their intersection point, we substitute the $ y$, so

$ x^2 \minus{} 4x \equal{} 2x \plus{} b$

$ x^2 \minus{} 6x \minus{} b \equal{} 0$

Setting the discriminant to 0 will mean that there will be only one x value that satisfies the equation, thus one point of intersection. We have that

$ (\minus{}6)^2 \minus{} 4(1)(\minus{}b) \equal{} 0$

$ \boxed{b \equal{} \minus{}9}$[/hide]

we can write $ x^2\minus{}4x\minus{}(2x\plus{}b)\equal{}(x\minus{}3)^2$, yielding $ b\equal{}\minus{}9$.

[hide="Alternative way"]Basically, you substitute for the $ y$ and get one equation with $ x$ and $ b$. You get it into the form $ Ax^2+Bx+C=0$. The only time when there is one intersection is when there is only one $ x$, so the quadratic can be factored as a square. We complete the square and find that $ b=-9$.
\[ \begin{eqnarray*} y=2x+b &\text{   and   }& y=x^2-4x \\
2x+b&=&x^2-4x \\
x^2-6x-b&=&0 \\ \\
x^2-6x+9&=&(x-3)^2 \\ \\
b&=&-9\][/hide]

[quote="314.math"]Consider the parabola whose equation is y=(x^2)-4x and the line whose equation is y=2x+b, where [i]b[/i] is some unknown constant.  Determine the value of b such that the line and the parabola will intersect at exactly one location.

Could someone please help me with this question?  Thanks in advance![/quote]

[hide]Set the two equations equal.  $ x^2\minus{}4x\minus{}2x\minus{}b\equal{}0.$ This equals $ x^2\minus{}6x\minus{}b\equal{}0$.  Using discriminant, we know that $ b^2\minus{}4ac$ must be equal to 0.  Therefore, $ 36\plus{}4b\equal{}0.$  Solving the equation, $ b\equal{}\fbox{\minus{}9}$[/hide]

[hide="Cheating"]
If they intersect at only one point, the line must be tangent, so the slopes at the point must be the same, so the first derivatives must be equal. Then $ 2x\minus{}4\equal{}2$, so $ x\equal{}\minus{}3$. Plugging in $ \minus{}3$ to both and setting them equal (finding the constant of integration, I guess), you get $ 9\minus{}12\equal{}6\plus{}b\implies\boxed{b\equal{}\minus{}9}$

Yes, yes, I know, no calculus in the HSB section, but the calc solution is faster than the discriminant one in my opinion.
[/hide]


i really, really need your help.. cuz we have this math project about probability.. and in this project.. we need to pick a board/card game that has something to do with probability..

and we chose crazy eights..

why do i need ur help? cuz we cant find a way[simple way] to increase the probability of us winning the game.. :?

waahh!! i really need ur help.. :?

tnx :lol:

Wow, there is strategy in crazy eights?!

Strategy: fix the deck so that you deal yourself an 8.  :P

Here are a few questions for you to think about:

When it is best to play your 8's?  Is it best to not play any 8's in your hand until you absolutely have to, or are there situations in which it is a good idea to play an 8 rather than another card?

If you have many cards of the same number or suit, is it better to get rid of them so that you have a balanced hand, or is it better to hold on to them until the end?

Why didn't you just choose blackjack, a much easier to analyze game?  :P

My statistics class is analyzing "pass the pigs."  It's sort of an uneven dice game except small rubber pigs are used instead of dice.  It would certainly make for an interesting project.

I've actually played pass the pigs before.  Out of curiosity how would you go about analyzing a game with such random falls?


Let $ x\in\mathbb{R}$ and $ n\ge 2$. Calculate the determinant: \[ \det A_{n\times n}\equal{}\det\begin{pmatrix} 1&1&1&1&\cdots &1\\ 2&1&x&x&\cdots&x\\ 3&2&1&x&\cdots&x\\ 4&3&2&1&\cdots&x\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ n&n\minus{}1&n\minus{}2&n\minus{}3&\cdots&1\end{pmatrix}\]

My solution is simpler  :D

\begin{eqnarray*}\det\begin{pmatrix} 1&1&1&1&\cdots &1\\ 2&1&x&x&\cdots&x\\ 3&2&1&x&\cdots&x\\ 4&3&2&1&\cdots&x\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ n&n-1&n-2&n-3&\cdots&1\end{pmatrix}&=&\det\begin{pmatrix} 1&1&1&1&\cdots &1\\ 1&0&x-1&x-1&\cdots&x-1\\ 1&1&1-x&0&\cdots&0\\ 1&1&1&1-x&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ 1&1&1&1&\cdots&1-x\end{pmatrix}\\ &=&\det\begin{pmatrix} 1&1&1&1&\cdots &1\\ 0&-1&x-2&x-2&\cdots&x-2\\ 0&0&-x&-1&\cdots&-1\\ 0&0&0&-x&\cdots&-1\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ 0&0&0&0&\cdots&-x\end{pmatrix}\\ &=& \left(-1\right)^{n-1}x^{n-2}\end{eqnarray*}


1) How do you put one or more empty line(s) between paragraphs?

2) For some reason, every time I put in a PNG picture into my document, there's an error and it doesn't show up in the final file...

Any help would be great!  :lol:

[quote="dajoe"]1) How do you put one or more empty line(s) between paragraphs?[/quote]The simple answer is don't. Proper typesettting (unlike word processing) indents new paragraphs but does not put an empty line between them. Pick up a book and see what I mean. If you must do it then see [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=parskip]Zero paragraph indent[/url] but I'm not sure you'll like it!

[quote="dajoe"]2) For some reason, every time I put in a PNG picture into my document, there's an error and it doesn't show up in the final file...[/quote]You will need to give more information. What is the code you are using and what is the error? [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url]

[quote="SuperNerd123"]For number 1 you can use the
[code]\linebreak[/code]
command[/quote]
That can cause bad boxes, justification of short lines and, unless you force a linebreak with \linebreak[4], may not work. \linebreak is for (possibly) forcing a linebreak [i]inside[/i] a paragraph whilst keeping the justification which \\ doesn't do.


geometry
3D geometry
sphere
rotation
geometric transformation
analytic geometry
calculus

We can rotate a circle and integrate to find the volume of the resulting sphere.

Can we "rotate" the sphere and (double?) integrate to get the hypervolume of the resulting hypershere?

Rotation is not always how one computes the volume of a sphere.  Typically, especially in $ \mathbb{R}^{n}$, one sets up a multiple integral.  So for an ordinary sphere, one would evaluate $ \iiint_{x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}R^{2}}\,dV$, a computation which is made considerably simpler by spherical coordinates.  In $ \mathbb{R}^{n}$, you can still use [url=http://en.wikipedia.org/wiki/Hypersphere#Hyperspherical_coordinates]hyperspherical coordinates[/url], or you can use the [url=http://en.wikipedia.org/wiki/Hypersphere#Volume_of_the_n-ball_and_n-sphere]reduction formula[/url].


SMO 2007, Open #16

Let $O$ be the centre of the circumcircle of $\Delta ABC$, $P$ and $Q$ the midpoints of $AO$ and $BC$, respectively. Suppose $\angle CBA = 4\angle OPQ$ and $\angle ACB = 6 \angle OPQ$. Find the size of $\angle OPQ$ in degrees.

Hmm no reply yet? 60 looks like a nice answer.. divisible by both 6 and 4, and a common angle.

If it was 60 then $\angle ACB=360$ and $\angle CBA=240$, which I doubt.

let $\angle{OPQ}=\phi$
$\angle{POQ}=\angle{POC}+\angle{COQ}=\angle{AOC}+\angle{COQ}=2\angle{ABC}+\frac{1}{2}\angle{COB}=2\beta+\frac{1}{2}*2\angle{CAB}=2\beta+\alpha=180+\beta-\gamma=180-2\phi$
so $\angle{PQO}=\phi$ and $|OQ|=|OP|=R/2$ where $R$ is the circumradius. From triangle $COQ$ you get $cos \alpha =\frac{|OQ|}{|OC|}=\frac{R/2}{R}=1/2$ so $\alpha =60\to 120=\beta+\gamma=10\phi\to \angle{POQ}=12$

Whoops.. i meant $15^{O}$. But it turns out the answer is 12?


Define $ f(x) \equal{} 4x(1 \minus{} x)$. How many distinct roots does $ f^7 (x)$ have? What about $ f^n (x)$?

[b]Note:[/b] [i]$ f^2(x) \equal{} f(f(x))$, $ f^3(x) \equal{} f(f(f(x)))$, etc.[/i]

[b]Edit:[/b] [i]Please also comment on the difficulty of the problem; ex. if it were on an AIME, what number would you expect it to be at?[/i]

[quote="The Zuton Force"]Define $ f(x) \equal{} 4x(1 \minus{} x)$. How many distinct roots does $ f^7 (x)$ have? What about $ f^n (x)$?[/quote]

Hmm sorry for asking but what does $ f^n(x)$ denote?  :maybe:

$ f^1 (x) \equal{} f(x)$ and $ f^{n\plus{}1}(x) \equal{} f(f^n (x))$. So for example, $ f^3(x) \equal{} f(f(f(x)))$. In other words, it denotes a recursion.

[hide="Solution, but maybe not correct"]
f(x)=4x(x-1)
The original function has 2 distinct roots.
f^2(x)=64x^2(x-1)(x-1/2)^2
3 distinct roots. Note that 2 of them are the same.
f(f^n(x))=4(f(x))(f(x)-1)
As we can see, composing the function retains the roots of the previous function, and then adds distinct roots.
The roots of f^3(x) will be the roots of f^2(x) in addition to the new roots created. The (x-1/2) term becomes -4x^2+4x-1/2, which has 2 roots. Each of those roots can be expressed as a+b and a-b. Composing the function with f(x) again will make 2 new terms (distinct), -4x^2+4x-a-b and -4x^+4x-a+b. We can see that to each root, we must add 4x-4x^2. The polynomial created is only a perfect square when it is -4x^2+4x-1, and so we know that we double the number added to the number of roots each time. Let R(n) be the number of distinct roots of f^n(x). R(1)=2, R(2)=3, for all n>=3, R(n)=3+2(n-2).
[/hide]Not really rigorous and detailed, but I just want to see if that's correct first.
EDIT: caught a little mistake.
EDIT2: fixed?

[hide="May I also make a guess first?"]
$ 2^{n\minus{}1}\plus{}1$?
[/hide]

[quote="The Zuton Force"]Mmm a valiant effort

Nope it's not 3+2(n-2)=2n+1  :P

The problem is a bit simpler than what you did, though[/quote]
Sorry forgot to add the exponent in...I meant 3+2^(n-2). But maybe that's not right either...

[b]serialk11r:[/b] you've got the $ 2^{n \plus{} ?}$ part right, that's good, but I think you got a bit too caught up in the algebra and might've screwed something up. Look for a simpler approach. Remember, a function maps inputs to outputs. In this case, what gets mapped to what?

[b]azjps[/b], you have the right answer, go ahead and post your solution :)

[b]Sidenote:[/b] Post 1111  :)

I had had half of my solution typed up but then lost it  :( , so I decided to see if my answer was correct before re-typing.

[hide]
(*)Note that $ f(x) \equal{} 1$ when $ 4x \minus{} 4x^2 \equal{} 1 \Longrightarrow (2x \minus{} 1)^2 \equal{} 0$ so $ x \equal{} 1/2$.

We claim that $ f^{n}(x) \equal{} 0$ has $ 2^{n \minus{} 1} \plus{} 1$ roots by induction. Trivially $ f^1(x) \equal{} 0$ has $ 2 \equal{} 2^0 \plus{} 1$ roots.

Since $ f^n(x) \equal{} 4f^{n \minus{} 1}(x)(1 \minus{} f^{n \minus{} 1}(x)) \equal{} 0$, either $ f^{n \minus{} 1}(x) \equal{} 0$ or $ f^{n \minus{} 1}(x) \equal{} 1$. In the former case by the inductive hypothesis, there are $ 2^{n \minus{} 2} \plus{} 1$ roots. In the latter case,
\[ f^{n \minus{} 1}(x) \equal{} f(f^{n \minus{} 2}(x)) \equal{} 1 \stackrel{(*)}{\Longrightarrow} f^{n \minus{} 2}(x) \equal{} 1/2
\]
Suppose that $ f(x) \equal{} k$ for some real $ k$ such that $ 0 < k < 1$. Since $ 0 < k \equal{} 4x(1 \minus{} x) < 1$, then we see (by sign tests, etc) that $ f(x) \equal{} k$ has two distinct roots that also satisfy $ 0 < x_1, x_2 < 1$. Thus, from $ f^{n \minus{} 2}(x) \equal{} f(f^{n \minus{} 3}(x)) \equal{} 1/2$ we have $ f^{n \minus{} 3}(x)$ may be one of two distinct roots, and for each of those values $ f^{n \minus{} 4}(x)$ will have to assume one of two values, making $ 2^2 \equal{} 4$ values, and so forth (ok, that was a bit hand-wavy, but I'm feeling a bit lazy). Continuing, and we have $ f^1(x)$ must assume one of $ 2^{n \minus{} 2}$ different values. Thus $ f^n(x)$ has $ 2^{n \minus{} 2} \plus{} 2^{n \minus{} 2} \plus{} 1 \equal{} \boxed{2^{n \minus{} 1} \plus{} 1}$ roots.

A pretty easy way to see this is to just draw a tree starting with $ f^n(x)$ and working downwards.
Edit: Oh, and about the difficulty, I think its a nice problem, but the answer was pretty guess-able if certain things were just assumed for the roots.
[/hide]

[b]azjps[/b], if you don't mind, could you please clarify what you meant by it being guessable by making assumptions about the roots?

azjps's solution was along the line of the intended solution. The idea:

[hide="Solution outline"]
Note that $ f$ maps $ 0 \to 0$, $ 1 \to 0$, $ 1/2 \to 1$, and that $ f$ increases until $ x \equal{} 1/2$ and then decreases.

So let's track the values $ 0$, $ 1$, and $ 1/2$:

We start off with a $ y$ value of $ 0$ at $ x \equal{} 0$, a $ y$ value of $ 1$ at $ x \equal{} 1/2$, and a $ y$ value of $ 0$ at $ x \equal{} 1$, for $ f^1 (x)$

So the sequence of important $ y$ values goes $ 0, 1, 0$

Now, let's fill in a $ \frac {1}{2}$ between each of these via intermediate value theorem: $ 0, \frac {1}{2}, 1, \frac {1}{2}, 0$

Now, let's run the function $ f$ over this list: $ 0, 1, 0, 1, 0 \equiv 0, \frac {1}{2}, 1, \frac {1}{2}, 0, \frac {1}{2}, 1, \frac {1}{2}, 0$

Repeating this process, it is easy to see that between any 2 zeros, there is exactly value of 1, which will become a zero in the next iteration.

Thus, if $ f^n (x)$ has $ k$ zeros, then $ f^{n \plus{} 1}$ has $ k \plus{} (k \minus{} 1) \equal{} 2k \minus{} 1$ zeros.

$ r_{n} \equal{} 2 r_{n \minus{} 1} \minus{} 1 \iff (r_{n} \minus{} 1) \equal{} 2 (r_{n \minus{} 1} \minus{} 1)$

$ \implies r^{n} \minus{} 1 \equal{} 2^{n \minus{} 1} (r_1 \minus{} 1) \equal{} 2^{n \minus{} 1} \iff r^n \equal{} 2^{n \minus{} 1} \plus{} 1$
[/hide]


pigeonhole principle
modular arithmetic
number theory unsolved
number theory

$ 101$ positive integers are written on a line. Prove that we can write signs $ \plus{}$, signs $ \times$ and parenthesis between them, without changing the order of the numbers, in such a way that the resulting expression makes sense and the result is divisible by $ 16!$.

Let $ m$ be a positive integer. Given a sequence of $ m$ positive integers $ a_1,a_2, \ldots, a_m$, consider the partial sums $ s_1 \equal{} a_1, s_2 \equal{} a_1 \plus{} a_2, s_3 \equal{} a_1 \plus{} a_2 \plus{} a_3, \ldots, s_m \equal{} a_1 \plus{} a_2 \plus{} \ldots \plus{} a_m$. If none of these sums is $ 0$ mod $ m$, then by the pigeonhole principle two of them must have the same residue mod $ m$, say $ s_i \equiv s_j \pmod{m}$ with $ i < j$. In this case $ a_{i \plus{} 1} \plus{} a_{i \plus{} 2} \plus{} \ldots \plus{} a_j \equiv 0 \pmod{m}$. In either case we see that a certain consecutive sum $ a_{i \plus{} 1} \plus{} a_{i \plus{} 2} \plus{} \ldots, a_j$ of the $ a_i$'s is divisible by $ m$, so we can add parentheses and operators like so:

$ \underbrace{(a_1 \plus{} \ldots \plus{} a_i)}_{\mbox{whatever}} \times \underbrace{(a_{i \plus{} 1} \plus{} \ldots \plus{} a_j)}_{\mbox{divisible by m}} \times \underbrace{(a_{j \plus{} 1} \plus{} \ldots a_m)}_{\mbox{whatever}}$

and the result is divisible by $ m$.

Now, let $ m \equal{} ab$ and let $ n \equal{} a \plus{} b$. Given any sequence of $ n$ positive integers, we can insert parentheses and operators around the first $ a$ terms to make the result divisible by $ a$, and around the last $ b$ terms to make the result divisible by $ b$, and finally enclose these two within parentheses and add a multiplication operation to make the result divisible by $ m$.

Therefore if $ m \equal{} 16! \equal{} 2^{15} \times 3^6 \times 5^3 \times 7^2 \times 11 \times 13$, we can handle any set of $ 30 \plus{} 18 \plus{} 15 \plus{} 14 \plus{} 11 \plus{} 13 \equal{} 101$ integers as required.

In general, we can insert parentheses and operators to make any set of $ n \equal{} \sum p_i$ positive integers divisible by $ m \equal{} \prod p_i$, where the $ p_i$'s are not necessarily distinct primes.


Let $F$ be the set of functions $f: R^{+}\rightarrow R^{+}$ such that
$f(3x)\geq f(f(2x))+x$ for all $x\in R^{+}$

Find the greatest $a\in R$ such that $f(x)\geq ax$ for all $f\in F$ and $x\in R^{+}$

[quote="easternlatincup"]Let $F$ be the set of functions $f: R^{+}\rightarrow R^{+}$ such that
(1) $f(3x)\geq f(f(2x))+x$ for all $x\in R^{+}$

Find the greatest $a\in R$ such that $f(x)\geq ax$ for all $f\in F$ and $x\in R^{+}$[/quote]
Let a is greatest, suth that $f(x)\ge ax \ \forall x\in R_{+}$. From (1) we have $a\ge \frac{1}{3}$.
It  give $f(3x)\ge (2a^{2}+1)x$, therefore $3a\ge 2a^{2}+1$.
Therefore $a_{0}=\frac{1}{2}$ is solution. For $f_{0}(x)=a_{0}x$ we have $f(3x)=f(f(2x))+x$.

Again repost
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=43295[/url] :wink:


[quote="vaibhav2903"]ABCD is a rectangle.X is any point inside the rectangle.if XA=a,XB=b,XC=c.find XD ?[/quote]
Hello
dont dubble
see here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319540
mrmohamed


complex analysis
complex analysis theorems

Hi,

I have currently got interested in the theory of multi-index Dirichlet series, i.e. series of the form

$ F(s_1,...,s_k) \equal{} \sum_{(n_1,...,n_k)\in\mathbb{N}^k}\frac {a_{n_1,...,n_k}}{n_1^{s_1}...n_k^{s_k}}$,

so I´m now looking for some good learning sources on the subject. In the logic of things, as a more general "bonus" I´m also looking for materials/sources on the theory of multi-index series, i.e. series of the form

$ \sum_{(n_1,...,n_k)\in\mathbb{N}^k}a_{n_1,...,n_k}$

(summability, conditioned summability, regularization etc.) as well as materials/sources on the subject of multi-index power series (also in the context of generating functions).

Thanks in advance for any suggestions!

efq

For multi-index power series, look for a textbook on several complex variables, or use "several complex variables" as keywords for a search. See if you can work your way through the section on sets of convergence for such series - you'll find some features not hinted at in the theory of a single complex variable.

Thanks for the good idea! Multidimensional complex analysis was anyway something on my mind to do. And most of the books on several complex variables cover the more general topic sums in Banach spaces, which is of course relevant to my second question about the multi-index infinite sums. Unfortunately, none of those books/scripts even mention about multi-dimensional Dirichlet series :( Some guy on another forum even expressed the opinion that one must be insane to deal with such objects like multi-dimensional Dirichlet series... Anyway, I´m still looking for materials on this subject, and if I find something good, I´ll keep posted.

PS: I was able to find isolated papers on the subject, all suggesting that this is conformal field for itself in mathematics, but they were far from 'basic', I even traced the references, they didn´t make much of a difference either.

You should post this question on mathoverflow.net

Thanks for the great idea! I wasn´t aware of mathoverflow.net before. I now posted my question there too.


[url=http://www.snopes.com/college/homework/unsolvable.asp]This should be one of us.[/url]

pah. like one of us would EVER be late to class.

[quote="math92"]that was really cool.  But was the story even true?[/quote]

Yes, look at the title.

oh oh oh I did that once

except it was like this

[quote=blackboard]
(x+2)^2 = 9

UNPROVED PROBLEM

*insert four-color theorem*
[/quote]

and I got x = 1, -5 but then I realized the unproved problem was the four-color theorem :(

btw I bet x^2 = 23405897234589072345219875098^2 is an unsolved problem just b/c nobody has ever actually solved this EXACT problem! WHO WANTS TO BE THE FIRST TO SOLVE MY UNSOLVED PROBLEM??? OMG!!!

[quote="zeb"]Me! I just did! Yay, I solved an unsolved problem!!! :w00tb:[/quote]

You don't get credit for it until you publish it in a journal :D