calculus
integration
real analysis
limit

Καλημέρα

Με αφορμή τη συζήτηση
http://www.mathlinks.ro/viewtopic.php?t=256307

θα ήθελα να ρωτήσω κάτι για την ολοκληρωσιμότητα.

Μιλώντας για κατά Riemann ολοκλήρωση υπάρχουν 2 βασικοί λόγοι για να μην είναι μια συνάρτηση f '  ολοκληρώσιμη.
Ο ένας είναι να μην είναι φραγμένη οπότε στη διαμέριση υπάρχουν διαστήματα που η μικρότερη τιμή της f στο μικρό διάστημα μπορεί να γίνει οσοδήποτε μεγάλο. Τέτοιο παράδειγμα έδωσε ο r_boris
http://www.mathlinks.ro/viewtopic.php?p=1401121#1401121

Ο άλλος λόγος για μη ολοκληρωσιμότητα κατά Riemann είναι το από κάτω πολυγωνικό εμβαδό να έχει διαφορετικό όριο από το πάνω πολυγωνικό εμβαδό. Παράδειγμα μη ολοκληρώσιμης συνάρτησης για αυτό το λόγο είναι αν έχει τιμή 0 στους άρρητους και 1 στους ρητούς.

Μέχρι εδώ νομίζω ότι συμφωνούμε. Πάμε τώρα στο ερώτημα μου:

Δανείζομαι από τον R_boris τη συνάρτηση του
$ f(x) \equal{} x\sqrt {x}cos(\frac {1}{x}) , x > 0$
Με $ f'(x) \equal{} \frac {3}{2}\sqrt {x}cos(\frac {1}{x}) \plus{} \frac {1}{\sqrt {x}}sin(\frac {1}{x}) ,x > 0 . f'(0) \equal{} 0$

Έστω ότι θέλω να βρω το εμβαδό μεταξύ της f’, της ευθείας $ x \equal{} \frac {1}{\pi}$ και του άξονα $ y$.

Κάνω κάτι απλό: Υπολογίζω εύκολα το εμβαδό με παράγουσα στο $ [a, \frac {1}{\pi}]$ οπότε καταλήγω σε μια συνάρτηση του α. Μετά παίρνω το όριο όταν το α τείνει στο 0. Το εμβαδό είναι πεπερασμένο και βγάζω την τιμή του.

Τι είναι αυτό που έκανα; Αν δεν είναι κατά Riemann ολοκληρώσιμη η f’ του r_boris, υπάρχει κάποια άλλη έννοια ολοκληρωσιμότητας υπό την οποία η f’ είναι ολοκληρώσιμη;

Δηλαδή έχω 2 συναρτήσεις f και f’ με τη δεύτερη να είναι παράγωγο της πρώτης και την πρώτη παράγουσα της δεύτερης. Επίσης μπορώ να υπολογίσω το εμβαδό κάτω από την f και ξέρω ότι ο βασικός λόγος για τον οποίο ορίστηκε το ολοκλήρωμα Riemann είναι το εμβαδό (o Spivac έτσι το ορίζει).
Πως θεμελιώνεται αυστηρά αυτό που έκανα με το όριο και γιατί δεν συμπεριλαμβάνεται στην ολοκληρωσιμότητα κατά Riemann;

Από όσο ξέρω, υπολόγισα ένα γενικευμένο ολοκλήρωμα, αλλά αν καταλαβαίνω καλά, η συνάρτηση f’ είναι μη ολοκληρώσιμη. Δε θα έπρεπε η ολοκληρωσιμότητα (είτε κατά Riemann είτε κατά κάποια άλλη έννοια) να είναι έτσι ορισμένη ώστε να θεωρούνται ολοκληρώσιμες και τέτοιες συναρτήσεις (μη φραγμένες αλλά με εμβαδό που συγκλίνει); Έτσι θα λέγαμε ολοκληρώσιμες όσες τελικά βρίσκεται το εμβαδό και μη ολοκληρώσιμες θα ήταν όσες δεν μπορούμε με κανένα τρόπο να βρούμε το εμβαδό.

Μπορεί κάποιος να βάλει μια τάξη στις σκέψεις μου;

Γιώργο, απαντώ στα γρήγορα. Αργότερα ίσως το συζητήσουμε περισσότερο. Η προσέγγισή σου είναι γνωστή ως Improper Riemann Integral. Κοίταξε [url=http://en.wikipedia.org/wiki/Improper_integral]εδώ[/url]

Θα απαντήσω και 'γω σύντομα λόγω πίεσης χρόνου. Στον ορισμό της Riemann-ολοκλήρωσης θέλουμε τα πεδία ολοκλήρωσης να είναι φραγμένα (συν κάτι άλλα regularities), όπως και οι συναρτήσεις μας. Από δω και πέρα, όπως λες και 'συ, ο ορισμός επεκτείνεται και σε μη φραγμένα πεδία ορισμού/μη φραγμένες συναρτήσεις. Αυτός είναι και ο λόγος που δεν έκλεισα τη συζήτηση στο άλλο topic, γιατί εύκολα βρίσκεις μη-ολοκληρώσιμες παραγώγους επειδή δε φράσσονται, αλλά πιο δύσκολα κατασκευάζεις φραγμένες μη-ολοκληρώσιμες παραγώγους :)

Γιώργο

Δυσκολα θα καταλάβαινα τι είναι εμβαδόν κατω απο την 2η γραφική παρασταση που σου στέλνω στο συνημμένο

[quote="Demetres"][quote="Durandal"][quote="r_boris"]Δυσκολα θα καταλάβαινα τι είναι εμβαδόν κατω απο την 2η γραφική παρασταση που σου στέλνω στο συνημμένο[/quote]

Rudy, δεν μπορώ να το διαβάσω, ίσως όμως να έχω μόνο εγώ το πρόβλημα... Ποιά συνάρτηση είχες κάνει plot?

Cheerio,

Durandal 1707[/quote]

Ούτε εγώ μπορώ, αλλά μου φαίνεται πως η συνάρτηση είναι η ημ(1/χ) ή κάποια παρόμοια.[/quote]      $ f(x) \equal{} x\sqrt {x}cos(\frac {1}{x}) , x > 0$και την παράγωγό της

Θα γράψω τη γνώμη μου (χωρίς βέβαια να σημαίνει πως τη θεωρώ σωστή). Απλά για κουβέντα και καλύτερη κατανόησή μου.

Ο λόγος που δυσκολευόμαστε να αναληφθούμε το εμβαδό κάτω από την δεύτερη συνάρτηση είναι (πιστεύω) η κλίμακα. Αν μπορούσαμε να μεγεθύνουμε  κατάλληλα θα το βλέπαμε κανονικά. Βέβαια τα tools που έχουμε έχουν περιορισμούς στις δυνατότητες. Δηλαδή δεν περιμένω από το εργαλείο να μπορεί να μου δείξει με διαύγεια οσοδήποτε μικρές περιοχές. Αλλά μπορούμε να το φανταστούμε ότι έχουμε στη διάθεσή μας οσοδήποτε μεγάλη μεγέθυνση (σαν να γίναμε μικροί εμείς ξαφνικά).

Το μόνο πρόβλημα είναι στο 0. Αλλά εκεί είναι μόνο ένα σημείο (ουσιαστικά βουλώνουμε μια τρύπα) και δεν κρύβει εμβαδό από κάτω του. Απλά το αγνοούμε όπως κάνουμε και σε μια συνάρτηση με ασυνέχεια (απλό πήδημα). Νομίζω ότι αν βρούμε το εμβαδό στο [α,1/π] με α>0 και πάρουμε το όριο όταν το α πάει στο 0 είμαστε ΟΚ.

Κατά τη γνώμη μου μπορούμε να φανταστούμε και πιο δύσκολα εμβαδά. Πχ έχουμε τη μια συνάρτηση στο [0, 1] που είναι 1 στους άρρητους και 0 στους ρητούς. Έχει νόημα να μιλάμε για εμβαδό από κάτω της τη στιγμή που έχουμε να κάνουμε με τόσο ισχυρή ασυνέχεια;

Πιστεύω πως και εδώ μπορούμε να μιλήσουμε για εμβαδό και να διαισθανθούμε τι μπορεί να σημαίνει κάτι τέτοιο. Να τι εννοώ:

Ξέρουμε ότι το εμβαδό του τετραγώνου [0,1] Χ [0,1] είναι 1. Άρα το εμβαδό κάτω από τους άρρητούς συνολικά  συν το εμβαδό κάτω από τους ρητούς να είναι ίσο με 1 (αφού κάθε πραγματικός αριθμός είναι ρητός ή άρρητος). Οι άρρητοί είναι μη αριθμήσιμοι και άρα απείρως περισσότεροι από τους αριθμήσιμους ρητούς, οπότε «κερδίζουν». Σε ένα πίνακα συχνοτήτων θα είχαν απείρως μεγαλύτερη τιμή. Επειδή οι άρρητοι είναι μη αριθμήσιμοι είναι κατά κάποιο τρόπο «ισοδύναμοι» με συνεχές διάστημα και άρα έχουν εμβαδό από κάτω τους παρά την ασυνέχεια.

Έτσι το εμβαδό κάτω από τους άρρητους πρέπει να είναι 1 και το εμβαδό κάτω από τους ρητούς πρέπει να είναι 0. Προσοχή ότι μιλάω τελείως διαισθητικά χωρίς καμία αναφορά σε κάτι αυστηρό.

Θέλω να καταλήξω στο εξής:

Αυτό που δε μου κάθεται όμορφα (και ο λόγος που άνοιξα το θέμα) είναι ότι δε μου πολύαρέσει να λέμε αν μια συνάρτηση είναι ολοκληρώσιμη κατά Riemann και να αδιαφορούμε για το improper ολοκλήρωμα.

Αισθάνομαι ότι η ολοκληρωσιμότητα πρέπει να ορίζεται με τέτοιο τρόπο έτσι ώστε να θεωρούνται ολοκληρώσιμες εκείνες οι συναρτήσεις που έχουν πεπερασμένο εμβαδό από κάτω τους υπό την γενικότερη δυνατή έννοια. Να συμπεριλαμβάνονται ασυνέχειες, να συμπεριλαμβάνονται μη φραγμένες συναρτήσεις που όμως το εμβαδό να συγκλίνει, να συμπεριλαμβάνονται και άπειρα διαστήματα. Να συμπεριλαμβάνονται δηλαδή τα πάντα που έχουν πεπερασμένο εμβαδό από κάτω τους. Και τότε μόνο ένα ερώτημα θα υπάρχει: «Είναι η f ολοκληρώσιμη ή όχι» χωρίς να μιλάμε για κατά Riemann ή improperly ή οτιδήποτε άλλο. Αυτός νομίζω ότι πρέπει να είναι ο τελικός στόχος του ορισμού της ολοκληρωσιμότητας.

Ελπίζω να βγάζουν νόημα αυτά που λέω.

Γιώργο, βεβαίως και βγάζουν νόημα όλα αυτά που λες. Το προτέρημα του ολοκληρώματος Riemann είναι η απλότητά του.

Και αυτό που λες για την κλίμακα έχει νόημα. Ας κάνουμε για παράδειγμα την αντικατάσταση y=1/x. Τότε θα ψάχνουμε για ένα ολοκλήρωμα από 1 ως άπειρο. Το ολοκλήρωμα Riemann ορίζεται μόνο σε πεπερασμένα διαστήματα, αλλά σίγουρα θα θέλαμε να υπολογίζουμε π.χ $ \int\limits_1^{\infty} \frac{1}{x^2} \, dx$. Και για να τα υπολογίσουμε κάνουμε κάτι που ξέρουμε καλά. Παίρνουμε το όριο ακριβώς όπως εξηγείς.  (Υπάρχει ένα πρόβλημα στο οποίο θα επανέλθω αργότερα.)

Πάμε τώρα στο εμβαδό κάτω από τους άρρητους. Και εδώ είσαι σωστός μόνο που άργησες 100 χρόνια   :P   Αυτή ακριβώς είναι η νοοτροπία του ολοκληρώματος Lebesque. Θα εξηγήσω μερικά πράγματα γι'αυτο το ολοκλήρωμα παρακάτω. Το μειονέκτημα είναι η δυσκολία ορισμού του. (Δεν γνωρίζω πανεπιστήμιο που να το διδάσκει από το πρώτο έτος.) Τα πλεονεκτήματα είναι ότι συνοδεύεται από αρκετά ισχυρά θεωρήματα που βοηθούν στον υπολογισμό του και ότι οριζεται για πολύ περισσότερες συναρτήσεις από ότι το ολοκλήρωμα Riemann

Η βασική ιδέα αυτού του ολοκληρώματος είναι η ακόλουθη: Θα ορίσουμε πρώτα για κάθε υποσύνολο Α του R ένα μέτρο μ(Α) που ουσιαστικά θα αντιστοιχεί στο $ \int 1_A$, όπου 1_Α είναι η συνάρτηση που είναι 1 στο Α και 0 έξω από το Α. (Λέω κάπιο ψέμα εδώ αλλά ας το αγνοήσουμε προς στιγμης.)

Τι ιδιότητες θέλουμε να ικανοποιεί αυτό το μέτρο;

1) μ(Α) >= 0 για κάθε Α
2) μ([α,β]) = β-α
3) μ({χ}) = 0
4) Αν Α και Β δεν έχουν τομή τότε μ(ΑuB) = μ(Α) + μ(Β)
5) Αν Α_1,Α_2,... δεν τέμνονται ανά δύο, μ(Α_1 u A_2 u ...) = μ(Α_1) + μ(Α_2) + ...

Ο συνδυασμός του 3 και του 5 μας λέει ότι θέλουμε μ(Q) = 0. Σε συνδυασμό με το 2 και 4 παίρνουμε μ([0,1]\Q) = 1. (Ακριβώς όπως θα ήθελες)

Έχει αποδειχθεί ότι υπάρχει μοναδικό μ που να ικανοποιεί αυτές τις συνθήκες. (Συνεχίζω να λέω κάποια ψέματα)

Πως τώρα ορίζουμε το ολοκλήρωμα μιας συνάρτησης f;

Αν είναι χαρακτηριστική συνάρτηση (δηλαδή της μορφής 1_Α) το έχουμε ήδη ορίσει.
Αν είναι "απλή συνάρτηση", δηλαδή της μορφής $ \lambda_1 1_{A_1} \plus{} \ldots \plus{} \lambda_k 1_{A_k}$ με όλα τα λ_i [b]θετικά[/b], τότε το ορίζουμε σαν $ \lambda_1 \mu(A_1) \plus{} \ldots \plus{} \lambda_k \mu(A_k)$. (Ζητάω θετικά για να αποφύγω πράγματα της μορφής $ \infty \minus{} \infty$.)
Αν είναι θετική συνάρτηση την προσεγγίζω με απλές συναρτήσεις (Γίνεται και όλες οι προσεγγίσεις δίνουν το ίδιο αποτέλεσμα.)
Τέλος για μια γενική συνάρτηση f την γράφω σαν $ f \equal{} f^\plus{} \minus{} f^\minus{}$ όπου $ f^\plus{}$ και $ f^\minus{}$ θετικές, και παίρνω την διαφορά εκτός εαν είναι της μορφής $ \infty \minus{} \infty$. Σε αυτήν την περίπτωση δεν ορίζεται.

Από ότι εγραψα μέχρι στιγμής το ολοκλήρωμα Lebesque ορίζεται για όλες τις θετικές συναρτήσεις (μπορεί να είναι άπειρο). Είχα όμως πει κάποια ψέματα... Το μέτρο μ δεν ορίζεται σε όλα τα Α. Γιατί; Έχει αποδειχθεί πως με τους κανόνες της συνολοθεωρίας που χρησιμοποιούν οι περισσότεροι (αλλά όχι όλοι) οι μαθηματικοί, [b]δεν μπορεί[/b] να οριστεί τετοιο μ για όλα τα Α! Ορίζεται όμως σε όλα τα Α που θα μπορούσε να σκεφτεί ένας μη μαθηματικός. (Αν κάποιος βρει από μόνος του ένα τέτοιο Α, τότε θα τον συμβούλευα να γίνει μαθηματικός.)

Αν μια συνάρτηση είναι ολοκληρώσιμη κατά Riemann, τότε είναι ολοκληρώσιμη κατά Lebesque (και δίνουν το ίδιο ολοκλήρωμα) με μια διαφορά: Ας κοιτάξουμε την συνάρτηση που είναι 1 στο [0,1], -1 στο [1,2], 1/2 στο [2,3], -1/2 στο [3,4], 1/3 στο [4,5] κτλ. Το improper Riemann μας δίνει 0. Ο Lebesque μας λέει: Άπειρο πάνω από τον άξονα του χ, άπειρο κάτω από τον άξονα του χ άρα δεν υπάρχει.

Ελπίζω να βοήθησα.

Δημήτρη σε ευχαριστώ πολύ... Θα τα διαβάσω άμεσα και θα προσπαθήσω να καταλάβω όλα αυτά που έγραψες.

Να ρωτήσω κάτι ακόμα: Αν καταλαβαίνω καλά ίσως θα μπορούσαμε να αντιπαραβάλουμε από τη μια του Riemann+improper Riemann και από την άλλη το Lebesque. Τελικά φαίνεται πως κανένα από τα 2 δεν είναι υπερσύνολο του άλλου. Αυτό όμως δε σημαίνει πως ακόμα δεν έχουμε καταλήξει ακόμα στην τελική μορφή; Δεν είναι στόχος να φτιάξουμε τη γενικότερη δυνατή μορφή ολοκληρωσιμότητας που θα εμπεριέχει όλα τα υπό μέρους που προκύπτουν από τις άλλες μορφές και τελικά να μιλάμε μόνο για αυτή;
Θέλω να πω ότι... τελικά εμένα με νοιάζει αν μια συνάρτηση είναι ολοκληρώσιμη ή όχι και αν  μπορώ να βγάλω το εμβαδό. Δε με νοιάζει σύμφωνα με ποιον ορισμό γίνεται αυτό. Θέλω να ξέρω απλά τι ισχύει στην πραγματικότητα.

[quote="gpapargi"]
Να ρωτήσω κάτι ακόμα: Αν καταλαβαίνω καλά ίσως θα μπορούσαμε να αντιπαραβάλουμε από τη μια του Riemann+improper Riemann και από την άλλη το Lebesque. Τελικά φαίνεται πως κανένα από τα 2 δεν είναι υπερσύνολο του άλλου.
[/quote]

Σωστά.

[quote="gpapargi"]
Αυτό όμως δε σημαίνει πως ακόμα δεν έχουμε καταλήξει ακόμα στην τελική μορφή; Δεν είναι στόχος να φτιάξουμε τη γενικότερη δυνατή μορφή ολοκληρωσιμότητας που θα εμπεριέχει όλα τα υπό μέρους που προκύπτουν από τις άλλες μορφές και τελικά να μιλάμε μόνο για αυτή;
Θέλω να πω ότι... τελικά εμένα με νοιάζει αν μια συνάρτηση είναι ολοκληρώσιμη ή όχι και αν  μπορώ να βγάλω το εμβαδό. Δε με νοιάζει σύμφωνα με ποιον ορισμό γίνεται αυτό. Θέλω να ξέρω απλά τι ισχύει στην πραγματικότητα.[/quote]

Υπάρχει το Generalized Riemann Integral ή Gauge Integral ή Denjoy-Perron integral που περιλαμβάνει Lebesque+Riemann+Improper Riemann και άλλα. Δεν γνωρίζω αρκετά γι'αυτό. Ο ορισμός του είναι περίπου όπως του Riemann, μόνο που αντι να χωρίζεις σε ίσα διαστήματα επιτρέπεις κάποια διαστήματα να είναι μικρότερα και κάποια μεγαλύτερα. Δεν είναι αρκετά διαδεδομένο και δεν γνωρίζω πολλά για αυτό. (Πριν μερικές βδομάδες έμαθα για την ύπαρξή του!). Η διαφορά του από το Lebesque από ότι καταλαβαίνω, είναι ουσιαστικά η διαφορά του Ιmproper Riemann από το Lebesque. Είναι πιο εύκολο να οριστεί αλλά από ότι καταλαβαίνω δεν δουλεύει καλά σε πιο γενικούς χώρους από τον R.

Ουσιαστικά φαίνεται ότι είναι αρκετά σημαντικό ολοκλήρωμα που μάλλον θα έπρεπε να διδάσκεται. Δεν ξέρω γιατί δεν χρησιμοποιείται όσο το ολοκλήρωμα του Lebesque. Θα ρωτήσω και αν μάθω περισσότερα θα τα γράψω.

Θα ήθελα να πω πρώτα ένα πράγμα: είναι διαφορετική "ενόχληση" το απλώς να πρέπει να πάρουμε ένα όριο σε ένα μεμονωμένο σημείο ή να μας βγεί ένα άπειρο κάπου, και είναι εντελώς διαφορετική ενόχληση το να μη δουλεύει το ολοκλήρωμα μας σε κανένα σημείο του πεδίου ορισμού μας.

Σε ό,τι αφορά improper integrals κλπ... εγώ δεν μπορώ να πώ ότι τα θεωρώ ως κάτι παραπάνω από μία απλή επέκταση της ολοκλήρωσης Riemann. Π.χ. η $ \frac {1}{\sqrt {x}}$ δεν είναι Riemann ολοκληρώσιμη (by the book) στο $ [0,1]$... αλλά εγώ μια χαρά (κακώς ίσως) Riemann ολοκληρώσιμη τη λέω, γιατί βαρίεμαι να κουβαλάω το "in the improper Riemann integral sense" μαζί μου. Δηλαδή, αυτό σαν πρόβλημα ονοματολογίας το θεωρώ παρά σαν πρόβλημα ουσίας.

Το ουσιαστικό πρόβλημα είναι σε συναρτήσεις όπως η $ 1_{\mathbb{Q}}$ του Dirichlet που λέγαμε παραπάνω... από εκείνες και μετά είναι που πραγματικά καταρρέει η ολοκλήρωση Riemann.

[quote="Demetres"][quote="gpapargi"]
Να ρωτήσω κάτι ακόμα: Αν καταλαβαίνω καλά ίσως θα μπορούσαμε να αντιπαραβάλουμε από τη μια του Riemann+improper Riemann και από την άλλη το Lebesque. Τελικά φαίνεται πως κανένα από τα 2 δεν είναι υπερσύνολο του άλλου.
[/quote]

Σωστά.
[/quote]

[hide="Εδώ λέω βλακείες..."]Δε συμφωνώ εδώ πέρα. Οποιαδήποτε συνάρτηση είναι ολοκληρώσιμη κατά Riemann είναι και ολ/μη κατά Lebesgue (το αντίστροφο δεν ισχύει όπως ήδη αναφέρθηκε με την $ 1_{\mathbb{Q}}$). Περαιτέρω, αν μία συνάρτηση έχει πεπερασμένο γενικευμένο ολοκλήρωμα κατά Riemann, τότε είναι επίσης και Lebesgue ολοκληρώσιμη (το ολ/μα Lebesgue δεν το πειράζουν ασυνέχειες, singularities κλπ). Τέλος, αν σε μία θεωρία ολοκλήρωσης, μία συνάρτηση έχει άπειρο ολοκλήρωμα, θα πρέπει να έχει άπειρο ολοκλήρωμα και σε κάθε άλλη.

Συνεπώς, θα έλεγα ότι η Riemann κάθεται εξ'ολοκλήρου μέσα στη Lebesgue.[/hide]

Αυτό που σου δίνουν πιο γενικά ολοκληρώματα όπως η προσέγγιση του Daniell (εναλλακτική διατύπωση της Lebesgue πρακτικά), το Denjoy-Perron ή το Henstock-Kurzweil (που είναι το D-P τρομερά απλοποιημένο) είναι η δυνατότητα:
i. να έχεις ακόμα πιο γενικές integrands (δηλ. να μπορείς να ορίσεις ολοκληρώματα ακόμα πιο δύσκολων συναρτήσεων, που είναι π.χ. conditionally convergent), ακόμα και στην περίπτωση που έχεις $ \int f_ \minus{} \equal{} \int f_ \plus{} \equal{} \infty$ (εκεί χαλάνε και οι δύο προηγούμενες θεωρίες και συνήθως παίρνεις την Cauchy Principal Value του ολοκληρώματος).
ii. να έχεις καλύτερη σχέση ανάμεσα στην ολοκλήρωση και στην παραγώγιση, δηλ. να ισχύει το θεμελιώδες θεώρημα του λογισμού με όσο το δυνατόν λιγότερες υποθέσεις.

Τώρα, για το γιατί δε διδάσκεται το Henstock-Kurzweil, παρότι πιο απλό στο χτίσιμο από το Lebesgue? O Dem έχει δίκιο εδώ πέρα, χρειάζεσαι πάρα πολύ τη δομή του $ \mathbb{R}^n$ για να κάνεις τη θεωρία να δουλέψει (συγκεκριμένα, χρειάζεται να έχεις "κύβους" στο χώρο σου, δηλ. να μπορείς να γράψεις τα σύνολά σου ώς "γινόμενα", πράγμα που δεν είναι τόσο απλό όσο φαίνεται). Επιπλέον, το θέμα είναι ότι το μόνο benefit που σου δίνουν τελικά αυτές οι θεωρίες είναι να μελετάς περιπτώσεις συναρτήσεων που έχουν $ \int f_ \minus{} \equal{} \int f_ \plus{} \equal{} \infty$... αυτές δεν είναι τόσο συχνές και χρήσιμες ώστε να αλλάξουμε θεωρία ολοκλήρωσης.

My 2 cents, anyway :)

Cheerio,

Durandal 1707

[quote="gpapargi"][quote="Demetres"]
Ας κοιτάξουμε την συνάρτηση που είναι 1 στο [0,1], -1 στο [1,2], 1/2 στο [2,3], -1/2 στο [3,4], 1/3 στο [4,5] κτλ. Το improper Riemann μας δίνει 0. Ο Lebesque μας λέει: Άπειρο πάνω από τον άξονα του χ, άπειρο κάτω από τον άξονα του χ άρα δεν υπάρχει.
[/quote]

Παιδιά σε αυτό το σημείο φαίνεται να λέτε κάτι διαφορετικό. Η συνάρτηση αυτή τελικά είναι ή δεν είναι Lebesque ολοκληρώσιμη;[/quote]

Eeeeeeeek, o Dem έχει [i]απόλυτο[/i] δίκιο :oops:

Δεν το είδα το αντιπαράδειγμα που έδωσε και την είχα ξεχάσει τελείως αυτήν τη συμπεριφορά. Φταίει και το ότι έχω να ασχοληθώ με )@(*&)(*&-Riemann πράγματα κάτι αιώνες...

Sorry gents,

Durandal 1707

[quote="Demetres"]
Ουσιαστικά φαίνεται ότι είναι αρκετά σημαντικό ολοκλήρωμα που μάλλον θα έπρεπε να διδάσκεται. Δεν ξέρω γιατί δεν χρησιμοποιείται όσο το ολοκλήρωμα του Lebesque. Θα ρωτήσω και αν μάθω περισσότερα θα τα γράψω.[/quote]

Ρώτησα και μου είπαν τα εξής: Υπάρχουν διάφοροι τρόποι να γενικεύσεις Lebesque + Impoper Riemann. Αλλά χάνεις π.χ. το θεώρημα $ \sum\limits_{n\equal{}1}^{\infty} \int f_n \equal{} \int \sum\limits_{n\equal{}1}^{\infty} f_n$. Στο Lebesque ισχύει αυτό, στο Ιmproper Riemann όχι. Αυτό έχει ώς συνέπεια να χάνεις αρκετά από τα σημαντικά θεωρήματα της θεωρίας Lebesque. (Monotone Convergence Theorem, Dominated Convergence Theorem κτλ.) Αυτό θεωρείται αρκετά μεγάλο τίμημα για να συμπεριληφθούν στο σύνολο των ολοκληρώσιμων συναρτήσεων ορισμένες "παθολογικές" συναρτήσεις. Μερικές φορές όμως αυτές οι γενικεύσεις είναι χρήσιμες. Παράδειγμα: Γενικεύωντας μπορεί να αποδείξουμε πιο εύκολα κάποια θεωρήματα τα οποία αυτομάτως ισχύουν και για τα ολοκληρώματα κατά Lebesque [i]δεδομένου ότι[/i] οι συναρτήσεις που θα δουλεύουμε θα είναι όντως ολοκληρώσιμες κατά Lebesque.

Αυτό το αντιπαράδειγμα δε δείχνει ότι κάτι δεν πάει καλά και στη Lebesque;
Βλέπουμε δηλαδή ότι τα επιμέρους εμβαδά γίνονται ανα 2 cancel out και δείχνει το improper Riemann να έχει δίκιο.
Τελικά συμφωνούμε στο ότι μόλις τελειώσουν οι ορισμοί η ολοκλήρωση πρέπει να είναι μόνο μια και καθολικής αποδοχής και ότι δεν έχει νόημα να μιλάμε για πολλούς ορισμούς; Αυτό είναι κάτι που δεν έχω καταλάβει αν το πιστεύω εγώ μόνο ή πράγματι έτσι πρέπει να είναι.

[quote="gpapargi"]Αυτό το αντιπαράδειγμα δε δείχνει ότι κάτι δεν πάει καλά και στη Lebesque;
Βλέπουμε δηλαδή ότι τα επιμέρους εμβαδά γίνονται ανα 2 cancel out και δείχνει το improper Riemann να έχει δίκιο.[/quote]

Not really, δε θα το έθετα έτσι.

To improper Riemann συλλαμβάνει την τιμή $ 0$ σαν το όριο μίας προσεγγιστικής διαδικασίας; δεν είναι ένα ολοκλήρωμα per se, είναι ένα [i]όριο ολοκληρωμάτων[/i]. Αυτό δε χάνεται στη Lebesgue; π.χ. αν πεις $ f$ τη συνάρτηση του Δημήτρη και ορίσεις $ f_n\equiv f\cdot 1_{[0,2n]}$ να είναι ίση με την $ f$ μέχρι το $ 2n$ και $ 0$ μετά, τότε, ακριβώς όπως και στη Riemann, παίρνεις:
$ \lim \int f_n \equal{} 0 \equal{} \textrm{R}\int_0^\infty f(x) dx$
(όπου το R αναφέρεται σε improper Riemann integration). Αν θέλεις, το όριο $ \lim\int f_n$ βάφτισέ το "improper Lebesgue integral", οπότε οι δύο θεωρίες σου συμφωνούν μια χαρά και σε αυτήν την περίπτωση.

Τι εννοώ με αυτό? Για να μη συγκρίνεις μήλα και πορτοκάλια, αν κάνεις με τη Lebesgue ό,τι ακριβώς κάνεις και με τη Riemann, παίρνεις την ίδια απάντηση, και η proper Lebesgue integration περιέχει όλη την proper Riemann integration και ένα μέρος της improper (αυτό ήθελα να πω πριν). Το ενδιαφέρον πράγμα που σε κάνει να υποπτεύεσαι ότι "κάτι δεν πάει καλά" είναι ότι στο παράδειγμα αυτό δεν εναλλάσσονται οι διαδικασίες του ορίου και της ολοκλήρωσης, δηλ. ότι [i]δεν ισχύει[/i]: $ \lim \int f_n \equal{} \int \lim f_n$.

Αυτό όμως δε σημαίνει ότι κάτι δεν πάει καλά με τη θεωρία ολοκλήρωσης: στη Lebesgue έχουμε πολύ ισχυρά θεωρήματα ακριβώς για το πότε μπορείς να εναλλάξεις όριο και ολοκλήρωση (αυτό μάλιστα είναι ένα από τα σημαντικότερα ερωτήματα της θεωρίας!). Αυτό το παράδειγμα δείχνει ότι κάτι δεν πάει καλά με τη [i]συνάρτηση[/i] που μελετάς, ότι δηλαδή η συμπεριφορά της μπορεί να γίνει παράξενη και θέλει προσοχή.

[quote]Τελικά συμφωνούμε στο ότι μόλις τελειώσουν οι ορισμοί η ολοκλήρωση πρέπει να είναι μόνο μια και καθολικής αποδοχής και ότι δεν έχει νόημα να μιλάμε για πολλούς ορισμούς; Αυτό είναι κάτι που δεν έχω καταλάβει αν το πιστεύω εγώ μόνο ή πράγματι έτσι πρέπει να είναι.[/quote]

Για μένα, μία θεωρία είναι τόσο χρήσιμη όσο τα αποτελέσματα που σου δίνει και τα φαινόμενα που μπορεί να περιγράψει. Μέχρις στιγμής, φαίνεται ότι η Lebesgue είναι μακράν η χρησιμότερη (αν και όχι η γενικότερη) θεωρία που έχουμε και, όπως λέει και ο Δημήτρης, δεν αξίζει να θυσιάσουμε τα πολύ δυνατά της θεωρήματα μόνο και μόνο για να γλυτώσουμε το να γράφουμε ένα όριο ή να συμπεριλάβουμε μερικά ακόμα παθολογικά παραδείγματα. Το τί προσφέρει πραγματικά το ΗΚ, δεν έχω ιδέα, ήταν από τα εξωτικά θέματα που πάντα έλεγα να ασχοληθώ αλλά ποτέ δεν έκατσα να μάθω δυο πράγματα...

Επίσης, αυτό δε σημαίνει ότι θα έπρεπε να ξεχάσουμε τη Riemann: ως θεωρία χτίζεται πολύ πιο εύκολα και είναι πολύ πιο κατάλληλη για εισαγωγή στο θέμα της ολοκλήρωσης. Έτσι, αν ασχολείσαι μόνο με αρκετά καλές συναρτήσεις και δε σε νοιάζουν σειρές, ιδιότητες σύγκλισης κλπ, δε χρειάζεται να πληρώσεις το τίμημα για να μάθεις τη θεωρία μέτρου που χρειάζεται η Lebesgue.

Cheerio,

Durandal 1707

Νομίζω πως κατάλαβα. Υπάρχει η γενική ιδέα της ολοκλήρωσης αλλά από κάποιο σημείο και μετά μπορείς να επιλέξεις ορισμό ανάλογα με τι ιδιότητες θέλεις να κρατήσεις/θυσιάσεις.
Για μένα που αγνοώ τις πιο προχωρημένες ιδιότητες, αυτό που θέλω να κρατήσω είναι το εμβαδό και γύρω από αυτό περιστρέφομαι συνέχεια. Δηλαδή θα διάλεγα εκείνη τη θεμελίωση που μου δίνει εμβαδό όπου είναι δυνατό να δοθεί (με οποιοδήποτε τρόπο)  γιατί θεωρώ ότι αυτό είναι το αρχικό μου κίνητρο για τη θεωρία ολοκλήρωσης.

Οι ορισμοί είναι θέμα επιλογής και αν επιλέξεις κάποιον μετά πρέπει να δεχτείς τις συνέπειες. Το ποιον από όλους θα δεχτείς τελικά είναι κάπου και φιλοσοφικό ζήτημα (τι από όλα θεωρείς πιο σημαντικό).

Θα θέσω στα υπ’ όψιν κάποια στιγμή να δω μέσα από κάποιο βιβλίο τη θεωρία μέτρου και του ολοκληρώματος Lebesque.

Σας ευχαριστώ πάρα πολύ και τους 3 για το χρόνο και τη διάθεση να μου εξηγήσετε  :)

[quote="gpapargi"]Θα θέσω στα υπ’ όψιν κάποια στιγμή να δω μέσα από κάποιο βιβλίο τη θεωρία μέτρου και του ολοκληρώματος Lebesque.[/quote]

Δεν είμαι 100% σίγουρος, αλλά νομίζω ότι και ο μικρός Rudin έχει μια εισαγωγή στη θεωρία (μου φαίνεται με inner/outer measure approach). Δεν παίρνω κι όρκο όμως, απλώς στο λέω επειδή ξέρω ότι τον έχεις...

Cheerio,

Durandal 1707

Όντως το έχω σε ένα υπερ-CD που μου έδωσε ο (γνωστός) Φάνης. Και πράγματι το έχει στο κεφάλαιο 11 (τελευταίο). Εννοείται πως έτσι όπως το βλέπω δεν καταλαβαίνω τίποτις. Δεν ξέρω αν γίνεται να πάω κατευθείαν στο κεφ 11 ή πρέπει να δω και άλλα πράγματα πιο πριν. Τέλος πάντων... θα δω.


Did anyone notice the math problem when the movie was starting?
[hide="my answer"]3[/hide]

yeah i did. I watched the movie even though I am probably too old for Santa Claus kid's stuff.  :rotfl:

i watched it too :lol:

what was the math problem (was it 1+2 or something :P)

have a happy thanksgiving
-jorian

I think it was:
If one reindeer is traveling west at miles 20 per hour and another is traveling east at 50 miles per hour when will the two reindeer be 210 miles apart?
but than again I'm not sure...but it was [i]very like[/i] this problem if not it. :lol:

happy Thaksgiving!!!
The Macy's Parade is still on!!!!1 But I've been doing math the whole time so I'm not sure what's happened...


search
function
AMC
AMC 12
AIME
AMC 10
USA(J)MO

what's the grading for 2007 AMCs??
will it be the same as before??

No, the scoring will be 6 points for each correct question, 1.5 for each blank, and no points for incorrect answers. Please use the search function at the top of the page, or even read the AMC test descriptions.

Thank you.

I asked because I saw the news somewhere on the website but I can't find it anymore
the [AMC12 factoids] hasn't been updated yet either...

only 1.5 for a blank?
grr that's not cool
I like the 2.5
that's how 3 people at our school qualified for aime
they wouldnt have if the blanks were 1.5 each...

personally i like the new system..i mean, just answering 11 questions and making sure you got them right, and leaving the rest blank SHOULD NOT get them to qualify for AIME, especially since the first 10 or so AMC questions are like SAT difficulty...

This year's 14 seems more reasonable

good point
I wasnt thinking about the amc12
In the 10, for a 120, you needed 17 right (the rest left blank) to qualify
now you need 19 right (the rest left blank) for the 120

yea so no screwing up hopefully hehe  :lol:

I really don't like how much "harder" the AMC 10 becomes under the new scoring system.  I mean getting 19/25 on AMC 10 is still pretty difficult in terms of pure accuracy.  I mean I know I could now get like 22-23 on AMC 10s, but when I was in 9th grade, I know I couldn't get a 19 questions right.  But I dunno, maybe I just wasn't that good in 9th grade  :blush: .  Anyhow, you gotta live with the system, and maybe even think about taking the AMC 12.

But the top 1% will still qualify automatically from AMC10 regardless of cutoff.  In fact, I wouldn't be surprised if, after the scoring change, the top 1% cutoff regularly falls below 120.

And FYI, when the 10 first started, top 1% was the *only* way to qualify; there was no automatic cutoff until several years later.

[quote="h_s_potter2002"]Uh, isnt the test supossed to be easier this year, because of the scoring change?

That's what I thought.[/quote]

Yea, since the point system decreases, though the score cutoff stays the same, shouldn't the problems become noticeably easier? Also, shouldn't the USAMO floor score become maybe about some 7 or 8 points lower? I guess that if these statements are true, it isn't all that bad that the scoring system is changing. It might even be better...

[quote="Ignite168"][quote="lotrgreengrapes7926"][quote="Karth"]Yea, since the point system decreases, though the score cutoff stays the same, shouldn't the problems become noticeably easier? Also, shouldn't the USAMO floor score become maybe about some 7 or 8 points lower? I guess that if these statements are true, it isn't all that bad that the scoring system is changing. It might even be better...[/quote]
Unless, the AIME is drastically different, I doubt it will lower 7 or 8 points. This implies that most USAMO qualifiers leave 7 or 8 questions blank, which is unlikely.[/quote]

No it doesn't, it just implies that the lower bracket of the top 375 usually leave a moderate number blank and probably attempted a bit more this year that they got wrong.[/quote]
But they could just as easily get the ones that they attempt right...


This problem seems easy but so far I am stuck.

Show that

\[
\lim_{(x,y)\rightarrow (0,0)} \frac{x^5y}{x^6+y^2}=0
\]

Any idea, hint or suggestion is very welcome.

make $x=rcos $ and  $y=rsen$ so $ r\to  0 $ and compute the limit...
any corrections, plz, post...


Question 2.

[hide="el solution"]Given a number of this form, it can only be represented in one way. (look at it mod 3,9,27,81 respectively)
For each possibility, we can match it to another one, namely the one with all the signs reversed. If one of these is negative, the other must be positive. The only exception is when this is zero. there are 3 choices for each $a_{i}$ and so we have

$\frac{3^{5}-1}{2}=121$ different positive numbers.[/hide]

[quote="bugmens"]1.
p,q,r are primes.$p^{3}=p^{2}+q^{2}+r^{2}$ find $(p,q,r)=?$[/quote]

well look at it mod 3 and break it up into a bunch of cases

if none of p, q, and r are 0 mod 3 (or if all of them are) then p has to be 3 and the only time that works is if $p=q=r=3$

now if p is 1 mod 3 then q and r both have to be 3 and $p^{3}= p^{2}+18$ and that just says p = 3 which doesn't really work out

and if p is 2 mod 3 then one of q or r has to be 3 so if you write it out that basically means $\sqrt{p^{3}-p^{2}-9}$ has to be an integer, which kinda doesn't work out unless p = 3

so yeah, $(3,3,3)$ is the sole solution

sorry if it doesn't make much sense


Positive integers x and  are such that 3x^2 - 8y^2 + 3(x^2)(y^2) = 2008. Find the product of x and y.

[hide="solution"]$ 3x^2 \minus{} 8y^2 \plus{} 3x^2y^2 \equal{} 2008 \\
\Rightarrow 3x^2(y^2 \plus{} 1) \minus{} 8y^2 \equal{} 2008$
clealry rhs and the second term on lhs is divisible by 8, so we need x as even and y as odd let $ x \equal{} 2k$ and $ y \equal{} 2m \plus{} 1$ we get
$ 3k^2(2m^2 \plus{} 2m \plus{} 1) \minus{} (2m \plus{} 1)^2 \equal{} 251$
clearly m=0 gives us a rhs divisible by 9 but lhs of 3
clealry m=1 is not a solution
m=2 gives no solution chk divisibility
m=3 gives k=2
m=4 no solution
m=5 none
m=6  gives none
clearly for a larger m we will get a smaller k which is not possible
so only one pair of $ (x,y) \equal{} (4,7)$

so $ xy \equal{} 28$[/hide]

[quote="pda"]Positive integers $ x$ and $ y$ are such that $ 3x^2 \minus{} 8y^2 \plus{} 3(x^2)(y^2) \equal{} 2008$. Find the product of x and y.[/quote]
[hide]$ 3x^2y^2 \plus{} 3x^2 \minus{} 8y^2 \minus{} 8 \equal{} 2000 \iff (3x^2 \minus{} 8)(y^2 \plus{} 1) \equal{} 2000$ Brute force listing of the factors of 2000 gives 1,2,4,5,8,16,10,20,25,40,50,80,100,125,200,250,400,500,1000,2000. Of these, only 2, 5, and 50 can be equal to $ y^2 \plus{} 1$. Then we solve $ 3x^2 \minus{} 8$ for 1000, 400, and 40. Of these, only $ 40$ gives a solution. So $ y^2 \plus{} 1 \equal{} 50, 3x^2 \minus{} 8 \equal{} 40 \Rightarrow (x,y) \equal{} (4, 7)\Rightarrow xy\equal{}28$.[/hide]
Someone beat me.  :wink:


pigeonhole principle
vector
combinatorics proposed
combinatorics

Give a 16-digit integer. Prove that we can find a string of consecutive numbers such that they product is a square.

This and equivalents have appeared before on the forum.  (There was one recently in either Preolympiad or Intermediate.)

Suppose otherwise: then certainly the digits 0, 1, 4 and 9 do not appear.  Then it's a standard pigeonhole argument: consider the first digit, the first two, the first three, etc., and the first 16, prime factorized.  Each of these numbers can be characterized by a vector $ (a_{2}, a_{3}, a_{5}, a_{7})$ where $ a_{p}$ is the largest power of $ p$ dividing the number, reduced modulo 4.  If there are no perfect squares among our number then the vector $ (0,0,0,0)$ does not appear.  But then two of our numbers are assigned the same vector, and we divide to get our perfect square.


A while back I noticed that both Times Square and Union Square are places in New York City with very mathy names. Are there any others?

well, TriBeCa--Triangle Below Canal--TRIANGLE!!

and the Financial District--FINANCIAL--finance requires math

and Battery Park--BATTERY--electricity requires math

and Upper East Side, Upper West Side---EAST WEST--directions and stuff require coordinates--math.
same for SoHo, and NoHo.

[quote="bubala"]EAST WEST--directions and stuff require coordinates--math.[/quote]

Then.. 5th avenue.. 214th street.. and many more will require coordinates - math  :)

We can treat the whole street/avenue grid system as one instance. By the way, did you know that New York City was the first city in the world to use the grid system?

I looked through the buildings list and there's a Random House Building! (the headquarters of the publishing company)

The Four Seasons Hotel and the Doubletree Guest Suites are not mathy enough for my taste...


vector
topology
real analysis
real analysis theorems

Hi,

I am extremely unexperienced with systems of seminorms.  Sorry  if my question was classical.

If $(V,||.||_1)$ and  $(W,||.||_2)$  are normed spaces, and u is a linear operator from V to W :

u is continuous if and only if there is a constant $C>0$ such that :

$\forall v\in V : ||u(v)||_2\leq C ||v||_1$

Now is it true if $(V,P)$ and $(W,Q)$ are both spaces with a system of seminorms, that a linear operator u from V to W is continuous if and only if :

$\forall q\in Q : \exists p\in P\exists C>0 : \forall v\in V : q(u(v))\leq C p(v)$

basically you can imitate the whole proof... but a nasty subtlety I get is when $p(v)=0$
With norms I don't get this problem.

Any thoughts?

The right fix is to move a quantifier; we need to choose the seminorm $p$ after choosing $v$.

that I don't get

you mean the seminorm in V dependent upon the vector V

that wouldn't help me anymore

my definition of a weaker set of seminorms if you have two on the same space, is that for every seminorm in the weakest, there is a seminorm in the stronger, dominating it (with some nonzero constant)

I was hoping that weaker seminorm implied that going from the space to itself with identity, starting with the strongest and ending with the weakest, was continuous topologically, and vice versa  :(

What's the problem? This change can only make it easier to be continuous. If the same choice always works, make the same choice every time.

but first of all, my statement in my first post in incorrect?

second of all : I wanna go from identity from space to space but starting with stronger system and ending with weaker system is continuous ---->  every seminorm in the weak system has a constant and a seminorm in the strong system, such that the product is bigger for every v

that is what I derive from my textbook

I really don't see, why you don't see my problem  :?

What do I overlook and what in my claims is incorrect?

Let's look at the identity from $(V,P)$ to $(V,Q)$, where $P$ is stronger than $Q$. For each $q\in Q$, there exists a norm $p_q$ and a constant $C_q>0$ such that $C_q\cdot p_q(v)\ge q(v)$.
Is the identity continuous? Choose $q\in Q$, and set $C=C_q$. For all $v\in V$, there exists $p=p_q\in P$ such that $C\cdot p(v)\ge q(v)$. That's it; this works for any $q$, and the identity is continuous.

You're not going to get a converse; two systems of seminorms can define the same topology without either dominating the other in your sense. Consider these two families on $\mathbb{R}^2$: $p_1(x,y)=|x|,p_2(x,y)=|y|$ and $q_1(x,y)=|x+y|,q_2(x,y)=|x-y|$.

Actually, there is a way to get the converse too; we just need to weaken the "dominates" condition to match the definition of continuity. The family $P$ dominates $Q$ when for all $q\in Q$, there exists $C_q>0$ such that for all $v$, there exists $p_{q,v}\in P$ such that $C_q\cdot p_{q,v}(v)\ge q(v)$.

Aah i see your response is very clear now.  What I wanted to prove is not true at all and counterexamples are not that hard to find!

So why is this definition used then so frequently?  It's not topological. :oops:

Jmerry come to think of it, I really do not understand your counterexample in $\mathbb{R}^2$.  Your two seminorms do not have the filtering property.

Also what is your definition of the topology induced by a system of seminorms?

In the mean time,  while it wasn't at the time I posted this question, this has become a hot topic in class as my professor was in doubt too as he deduced equivalence of seminorms from the topology. :?

Disclaimer: I don't have any textbook to reference here.

The topology $T$ defined by a countable family of seminorms $p_i$ is the weakest topology such that a sequence $x_i$ converges if and only if it converges in each seminorm $p_i$. This topology is metrizable, although the metric may not be a vector space norm.

A metric for this topology: $d(x,y)=\sum_i 2^{-n}\frac{p(x_i-y_i)}{1+p(x_i-y_i)}$.

What is the filtering property?

if p is a seminorm and v a vector, denote with $p(v,r)$ the 'open semiball of radius r around v' : all vectors such that p of their difference with v is less than r

we say a set $U$ is open if for any point, i can draw an open semiball around it for some seminorm, completely in U

a system of seminorms is a set of seminorms on the same space that is
filtering : for any finite set of seminorms in that set $p_1,\cdots,p_n$ there is $C>0$ and a $p$ in the total set of seminorms such that $\forall i: p_i\leq C p$
separating : if all seminorms on a particular vector are zero, the vector is zero

I think I found it :

let $P$ be a system of seminorms on V
and $Q$ on W

let $u: V\rightarrow W$ be continuous linear map

choose a $q\in Q$

now the set $\{q<s\}$ is open , and thus its inverse will be open in V, and thus there will be a open semiball around the origin of V mapped into this open set :

there is a r and $p\in P : u (\{p<r\})\subseteq \{q<s\}$

one easily finds constants now to write $q(u(v))\leq C p (v)$
the only difficulty is when $p(v)=0$ but that implies $p(k v)=0$ for all k and thus
$q(u(k v))<s$  for all k which implies $q(v)=0$

What do you think?


function
real analysis
real analysis solved

Find an example of measurable function whose inverse is
not measurable.

$f(x) = x + \Psi(x)$, where $\Psi(x)$ is Cantor function.

liyi, that example is not going to do it. We could exhibit both your example and its inverse as the pointwise limit of continuous (piecewise linear) functions, and the pointwise limit of a sequence of measurable functions is measurable.

Something to consider: Any subset of a set of measure zero is measurable, and the Cantor function maps a set of measure zero onto the whole interval. That way the inverse image of a nonmeasurable set under the Cantor function would be measurable. That's not the answer to this problem, but it might be a place to start from.

Here's an answer that uses liyi's idea. In what follows, let $\Psi(x)$ be the Cantor-Lebesgue "devil's staircase" function, which is a non-decreasing continuous function that maps the Cantor measure-zero ternary set $K$ onto [0,1].

Let $S$ be a nonmeasurable subset of [0,1] "constructed" in the usual way. That is, split [0,1] into equivalence classes of the relation that two numbers are equivalent if they differ by a rational, then (by the Axiom of Choice) choose one member of each equivalence class to be in $S$. Make sure that the one rational member of $S$ is not a dyadic rational (an integer divided by a power of 2).

Let $A = \Psi^{-1}(S)$. Since $A \subset K$, its outer measure is zero, hence it is measurable.

Now let $f(x) = x+\Psi(x)$ for $x \in ([0,1]-A)$ and $f(x) = x+\Psi(x)+10$ for $x \in A$. This function is clearly one-to-one, hence invertible. Since it agrees almost everywhere with a measurable function, it is measurable. However, its range is not a measurable set and its inverse is not measurable.

It should be possible with more applications of the Axiom of Choice to "construct" a permutation of $A$, and to use this to define another version of this example that maps [0,1] onto [0,2] in a one-to-one way that is measurable with a nonmeasurable inverse.

Any argument with this much Axiom of Choice in it should make you vaguely uneasy.


I assume you mean $-1 < \sum\limits_{cyc} (\dfrac{a-b}{a+b})^{1993} < 1$.

We can write the terms in the brackets as (ie.) $-1 + \dfrac{2a}{a+b}$ or $+1 - \dfrac{2b}{a+b}$ which shows the desired immediately.

[quote="Singular"]which shows the desired immediately.[/quote]why? ;)

Since nobody had ever posted a proof:

Without loss of generality, we can assume that $a = \max\left\{a, b, c\right\}$. Define \[ f(a, b, c) = \left(\frac{a - b}{a + b}\right)^{1993} + \left(\frac{b - c}{b + c}\right)^{1993} + \left(\frac{c - a}{c + a}\right)^{1993}.  \] It is easy to check that $f(a, b, c) = -f(a, c, b)$. Hence $-1 < f(a, b, c) < 1 \Leftrightarrow -1 < f(a, c, b) < 1$. This allows us to assume that $b \geq c$, so that $a \geq b \geq c > 0$. It is obvious that \[ \begin{aligned} 0 & \leq\ \ \left(\frac{a - b}{a + b}\right)^{1993} & < 1 \\ 0 & \leq \ \ \left(\frac{b - c}{b + c}\right)^{1993} & < 1 \\ -1 & < \ \ \left(\frac{c - a}{c + a}\right)^{1993} & \leq 0 \end{aligned}  \] and the only thing which we still need to show is \[ \left(\frac{b - c}{b + c}\right)^{1993} + \left(\frac{c - a}{c + a}\right)^{1993} \leq 0.  \] This reduces to \[ \left((b - c)(c + a)\right)^{1993} + \left((c - a)(b + c)\right)^{1993} \leq 0  \] or \[ (b - c)(c + a) \leq -(c - a)(b + c)  \] or \[ bc + ab - c^2 - ac \leq -bc - c^2 + ab + ac  \] or \[ 2c(a - b) \geq 0  \] which is obviously true. $\blacksquare$

Yes, that's how I did it too. :)

The official solution went along the line of Singular, but I like this one more.

No, it didn't. The official solution very similar to my solution...

By the way, Singular, do you actually *have* a solution? I mean, does your idea lead to a solution?


Set A_1,A_2,....,A_35 are given with the property that each of the A_1,A_2,...,A_35 has 27 elements and such that the intersection of any three of them has

exactly one element.show that there is a element belongs to all the given sets.

Found by searching for 25 and 37 on the combinatorics forum:
http://www.mathlinks.ro/Forum/viewtopic.php?t=140614

  darij


geometry
perimeter
inequalities proposed
inequalities

Let $a,b,c,d$ be the lenghts of the sides of a quadrilateral with perimeter $P$, prove that:

$\frac{abc}{d^{2}}+\frac{bcd}{a^{2}}+\frac{cda}{b^{2}}+\frac{dab}{c^{2}}\geq P$

[quote="sinajackson"]Let $a,b,c,d$ be the lenghts of the sides of a quadrilateral with perimeter $P$[/quote]
you dont need that condition. you can prove it for all positive reals $a,b,c,d$
$\frac{abc}{d^{2}}+\frac{bcd}{a^{2}}+\frac{adc}{b^{2}}\geq 3c$ by $AM-GM$
sum this cyclicaly to get the desired inequality...


Let $ a,b,c,d$ be positive real numbers such that $ abcd \ge 1$ and $ a\plus{}b\plus{}c\plus{}d\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}.$ Prove that
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\plus{}\frac{1}{d^2} \ge a^2\plus{}b^2\plus{}c^2\plus{}d^2.$

[quote="can_hang2007"]Let $ a,b,c,d$ be positive real numbers such that $ abcd \ge 1$ and $ a \plus{} b \plus{} c \plus{} d \equal{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}.$ Prove that
$ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \plus{} \frac {1}{d^2} \ge a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2.$[/quote]
Clearly,the inequality is equivalent to $ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \geq \frac {1}{ab} \plus{} \frac {1}{ac} \plus{} \frac {1}{ad} \plus{} \frac {1}{bc} \plus{} \frac {1}{bd} \plus{} \frac {1}{cd}$ which is clealy true because $ ab \geq \frac {1}{cd}$, $ ac \geq \frac {1}{bd}$,$ ...$  :maybe:

It is totally the same as my solution, rachid. Actually, the Vasc's problem (open) is asked to prove that for any positive integer $ k,$ then
$ (1 \minus{} abcd)\left( a^k \plus{} b^k \plus{} c^k \plus{} d^k \minus{} \frac {1}{a^k} \minus{} \frac {1}{b^k} \minus{} \frac {1}{c^k} \minus{} \frac {1}{d^k}\right) \ge 0,$
where $ a,b,c,d$ are positive numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}.$

The link of the problem is here: http://www.mathlinks.ro/viewtopic.php?t=162541

It is so hard, and I have no solution for it.


Does there exist a polynomial $p(x)$ with integer coefficients such that if $p(x)=x^3+ax^2+bx+c$, then $a,b,c$ are the roots of $p(x)$, where $a,b,c$ are integers, not necessarily distinct?

By Vieta: $a+b+c=-a$;$ab+bc+ac=b$;$abc=-c$

Then we have:
$c=-2a-b$
$ab=-1$

Replacing in $ab+bc+ac=b$ we get:
$b+2=\frac{2}{b^2}-b^2$, then: $b=-1$

$a=1;c=-1$

So: $p(x)=x^3+x^2-x-1$

Not complete. From $abc=-c$ it also follows $c=0$. This gives two more solutions: $p(x)=x^3$ and $p(x)=x^3+x^2-2x$


[hide]1/50, cuz there's a pattern.
The product up to the first number is 2/3
Second is 2/4
Third is 2/5
Fourth is 2/6
....
The product up to the 98th one (which is also the last one) is 2/100 which reduces to 1/50.[/hide]

GoBraves, thanks for showing how it's done. . .I'm too lazy to write out my work on the computer. . . ;)

[quote="MCrawford"]Express as a common fraction the product of:
\[ \left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\left( 1 - \frac{1}{5}\right) \cdot \ldots \cdot \left(1 - \frac{1}{100}\right). \][/quote]
[hide="solution"]
(2/3)(3/4)...(99/100)
Everything cancels out except for the 2 and the 100.
2/100
[b]1/50[/b]
[/hide]

[hide="Click to reveal hidden content...if you dare"]That's the same as 2/3 * 3/4 * 4/5 * 5/6 * ... * 99/100. If you put that all in one big fraction, you could see that the numbers 3 through 99 are all factors of both the numberator and the denominator. Factor them out. You are left only with 2/100 or 1/50.[/hide]


Find function $ f$ which is strict increasing on $ \mathbb R$ and $ f^2(x) = x^2, \forall x \in \mathbb R$.

 Babis

I think you mean $ f^2(x)\equal{}f(f(x))$.
But there are not suth function, because if $ f,g$ increasing, then $ f(g(x))$ increasing too. $ x^2$ - is not increasing in all R.

[quote="Rust"]I think you mean $ f^2(x) \equal{} f(f(x))$.
But there are not suth function, because if $ f,g$ increasing, then $ f(g(x))$ increasing too. $ x^2$ - is not increasing in all R.[/quote]

No !

 $ f^2(x)$ means the second power of $ f$ , that is $ f^2(x) \equal{} f(x) \cdot f(x)$ and the problem is easier.

[quote="TTsphn"]A relative problem from my school contest.
find all function such that
$ f(f(x)) \equal{} x^2 \minus{} 2008,\forall x\in R$
For your problem
$ f(x) \equal{} x,\forall x\in I$
$ f(x) \equal{} \minus{} x,\forall x\in R \minus{} I$[/quote]

 Mmm! But here $ f$ is increasing. You will be surprised , if you try a little more. The solution is ... half line text!
 Of course , I would like any new solution , even not short.

Oh yes.
From condtion we must have
$ f(x)\equal{}f(\minus{}x)$ or $ f(x)\equal{}\minus{}f(x)$
But because $ f(x)$ is increasing then $ f(x)\equal{}\minus{}f(x)$
Suppose exist $ a$ is diffent from 0 and $ f(a)\equal{}\minus{}a$ then
$ f(\minus{}a)\equal{}a$ .
[u]Case 1[/u] $ a>0$ then $ f(a)>f(\minus{}a)$ so $ \minus{}a>a\Rightarrow  a<0$ (tradition).
[u]Case 2[/u]$ a<0$ then $ f(\minus{}a)>f(a)$ so $ a>0$ (tradition).
It means that $ f(x)\equiv x,\forall x\in R$

$ (f(x)-x)(f(x)+x)=0$ gives $ \forall x\in\mathbb{R}: \ f(x)=x\ or\ f(x)=-x$
we take $ E=\{x\in \mathbb{R}|\ f(x)=x\},S=\{x\in \mathbb{R}|\ f(x)=-x\},E\cup S=\mathbb{R}$
if $ \exists x>y\in S$ then $ f(x)=-x<-y=f(y)$ (imppossible) so $ S=\{0\}$ gives $ E=\mathbb{R}$

[quote="aviateurpilot"][quote="stergiu"][quote="aviateurpilot"]$ (f(x) - x)(f(x) + x) = 0$ gives $ \forall x\in\mathbb{R}: \ f(x) = x\ or\ f(x) = - x$
we take $ E = \{x\in \mathbb{R}|\ f(x) = x\},S = \{x\in \mathbb{R}|\ f(x) = - x\},E\cup S = \mathbb{R}$
if $ \exists x > y\in S$ then $ f(x) = - x < - y = f(y)$ (imppossible) so $ S = \{0\}$ gives $ E = \mathbb{R}$[/quote]

 Mmm :) .
 But if $ S$ has one only non zero element , how  will you manage the problem  to prove that this is imposible?[/quote]
impossible because we now that f(0)=-0 then $ 0\in S$[/quote]

 And if only,  for example ,  $ x = -1 \in S$ we have $ f(1) = -1$ ? Why is this a contradiction ?

[quote="stergiu"]
 And if only,  for example ,  $ x \equal{} \minus{} 1 \in S$ we have $ f(1) \equal{} \minus{} 1$ ? Why is this a contradiction ?[/quote]
$ 1>0$ but $ f(1)\equal{}\minus{}1<f(0)$ (impossible) because $ f$ is strict increasing

$ f(\minus{}1)\equal{}1$ then $ f(1)\equal{}\minus{}1$ this is tradition as I have posted. :wink:


[quote="Arvind_sn"]The other day when I was playing around with some numbers, I found 2 things.

1. When converting $\frac{x}{9}$ to a decimal, it is always going to be $0.xxx...$. For example, $\frac{2}{9}$ is 0.222... and $\frac{4}{9}$ is 0.444...

2. I know that this one helps middle schoolers a lot. If you square a number that ends with five (with a ones digit of five), then there is an easier way to find it than to actually multiply out. Let us try using my method with $25^2$. We take all the numbers except 5 (in this case, 2) and add one to that number (in this case, getting three). We multiply the new number by the first number we got (2, yielding a product of 6), and we add 25 to the end of the number (625). We can also do this to any number. Try 655. We do 66 (which is actually 65+1) multiplied by 66, and we get 4290. Now, we add 25 to the end of the number, yielding 429025. If you punch that into a calculator or something it is the same number.

Pretty kyool, huh? what do you guys think?
 :coolspeak:
  :coolspeak:
  :coolspeak:[/quote]

 :yup: nice :D
Same goes with any denominator with all 9's
So $\frac{46}{99}=.4646464646...$,$\frac{460}{999}=.460460460460...$, and $\frac{46}{999}=.046046046046...$...yeah.  :D

[quote="Arvind_sn"]Yeah, here is a really cool proof for that :D .
$x=0.99999999...$
Multiplying both sides by 10,
$10x=9.99999999...$
Subtracting $x$ from both sides,
$9x=9$
Dividing each side by 9,
$x=1$
Oooooh. Cool, huh? My teacher pointed this out to us at the beginning of the year... :coolspeak:[/quote]
Or you could just say $0.99999999...=\frac{9}{9}=1$.  :P Like Ignite said. And my friend showed me the second one when I was around 7 or something. It was my secret to superior mathematics.  :ninja:

[quote="funcia"]:idea: Could anyone make a proof so that we know that #2 is correct? Unluckily, I can't.[/quote]
The number can be expressed as $(10x+5)$. We want to square that, and expanding results in $100x^2+100x+25$. With some factoring and manipulating, we can express it as $100(x(x+1)+25$. Since $x(x+1)$ is being multiplied by 100, this automatically makes the 25 go into the last two digits spots. This also makes $x(x+1)$ go into the digits places in front of the 25.
Thus $(x5)^2=x(x+1)25$

[quote="surge"]Extra fun:

Express $.0751313131313...$ as a fraction.[/quote]

[hide]Subtract $.075$ and then you get $\frac{13}{99000}$ so then the answer is $\frac{13}{99000}+\frac{3}{40}=\boxed{\frac{3719}{49500}}$.[/hide]

[quote="Ignite168"]$\frac{9}{9} = .99999999999999999...$ Therefore, $.999999... = 1$ :D[/quote]

Actually, this was argued in MathLinks time before, search for more information.

I just wanted to point out another cool thing =D

When theres 2 consecutive numbers multiplied together, it is always the square of the number between minus 1. It's pretty cool, but you really don't even notice that its really a widely known fact. Try examples and see if they're true! Try any numbers and plug them in!

If x was the middle number, then it should be (x-1)(x+1) + 1 = x^2. That really just becomes x^2 - 1 = (x-1)(x+1).

If you follow through the logic, it should also be x^4 = (x-2)(x+2) + 4 and so on. It's pretty handy when you don't have a calculator for some specific problems. heheh  :lol:

Some examples:
12 and 14: 13^2 = (13-1)(13+1) + 1 = 12 * 14 + 1 = 168 + 1 = 169
13^2 = 169

Well, actually there's a simple proof for what you're saying... a trivial one for high school, but not for middle school...

[quote="byrdgyrl"]1÷81=0.0123456789

A friend figured this out in Spanish class, sitting there dividing 1 by random numbers on his calculator...[/quote]

See that $\frac{1}{81}$ is $0,012345679$ periodic..... without the $8$

Well, if yu want that to be a periodic fraction, you just have to do $\frac{123456789}{9999999999}$, which is almost $\frac{1}{81}$...


Let $ AH,BE,CF$ denote the altitudes of triangle $ ABC$. $ M,N,P$ are the midpoints of $ AH,BE,CF$. The incircle $ (I)$ tangent to $ BC,CA,AB$ at $ A_1,B_1,C_1$ respectively. Let $ O$ be the center of the circumcircle $ (ABC)$. Show that $ MA_1,NB_1,PC_1,OI$ are concurrent.

It's from a Vietnam TST:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42412[/url]

And give your posts better titles.


calculus
integration
floor function
national olympiad

[b]Day 1[/b]

[url=http://www.mathlinks.ro/viewtopic.php?t=154072]Problem 1[/url]
Find integral solutions to the equation
$(m^{2}-n^{2})^{2}=16n+1$.

[url=http://www.mathlinks.ro/viewtopic.php?t=154073]Problem 2[/url]
Prove that the sequence
$a_{n}=\lfloor n\sqrt 2 \rfloor+\lfloor n\sqrt 3 \rfloor$
contains infintely many even and infinitely many odd numbers.

[url=http://www.mathlinks.ro/viewtopic.php?t=154415]Problem 3[/url]
Let ABC be a triangle such that |AC|>|AB|. Let X be on line AB (closer to B) such that |BX|=|AC| and let Y be on the segment AC such that |CY|=|AB|. Intersection of lines XY and bisector of BC is point P. Prove that angle BPC + angle BAC = 180.

[url=http://www.mathlinks.ro/viewtopic.php?t=153795]Problem 4[/url]
Given a finite string S of symbols X and O, we write @(S) for the number of X's in S minus the number of O's. (For example, @(XOOXOOX) = -1.) We call a string S balanced if every substring T of (consecutive symbols) S has the property -2 (less than or equal to) @(T) (less than or equal to) 2. (Thus XOOXOOX is not balanced since it contains the sub-string OOXOO whose @-value is -3.) Find, with proof, the number of balanced strings of length n.

[b]Day 2[/b]

[url=http://www.mathlinks.ro/viewtopic.php?t=154703]Problem 5[/url]
Let there be two circles. Find all points M such that there exist two points, one on each circle such that M is their midpoint.

[url=http://www.mathlinks.ro/viewtopic.php?t=84141]Problem 6[/url]
$2n$ students $(n \geq 5)$ participated at table tennis contest, which took $4$ days. In every day, every student played a match. (It is possible that the same pair meets twice or more times, in different days) Prove that it is possible that the contest ends like this:
- there is only one winner;
- there are $3$ students on the second place;
- no student lost all $4$ matches.
How many students won only a single match and how many won exactly $2$ matches? (In the above conditions)

[url=http://www.mathlinks.ro/viewtopic.php?t=154411]Problem 7[/url]
$a,b,c>0$
$a+b+c=1$
Prove:
$\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2})$

[url=http://www.mathlinks.ro/viewtopic.php?t=154412]Problem 8[/url]
Let $x, y$ be natural numbers such that
$2x^{2}-1=y^{15}$
Show that if $x>1$, then $5\mid x$.

Writing time was 4 hours per day.

Also, I'm curious to know if any of the participating countries have had their selection tests yet? If so, could someone post the problems, please?

Austria selects their competitors in the National Olympiad (I think Yimin Ge already posted the problems!)

Ranks 7 to 12 are going to MEMO for Austria, one of them is a good friend of mine :)

We have a bit of a different selection process. After the national olympiad a couple of best students get invited to the IMO team (4 this year) and other best scorers go to the so-called "small olympiad" which is a lot tougher than national olympiad. There the rest of the team for IMO is chosen and, this year, 10 people were chosen to write MEMO selection test, which ended up being written about a month after the national olympiad. After that, 6 people were chosen for the MEMO team (and I'm among them  :D ).

[quote="mornik"]
[url=http://www.mathlinks.ro/viewtopic.php?t=154415]Problem 3[/url]
Let ABC be a triangle such that |AC|>|AB|. Let X be on line AB (closer to B) such that |BX|=|AC| and let Y be on the segment AC such that |CY|=|AB|. Intersection of lines XY and bisector of BC is point P. Prove that angle BPC + angle BAC = 180.[/quote]

I think that instead of "Let X be on line AB (closer to B)" it should state "Let X be on line AB (closer to A)", sorry.

Problem 4 of the first test was asked in the Indian TST 2 as #3.

[quote="Valentin Vornicu"]Are these tests used for IMO selection as well?[/quote]
No. Some IMO team members are picked directly after the national olympiad, without the selection test and some write the "small olympiad" (which is also the pre-selection test for MEMO, meaning people are picked both for IMO team and for writing MEMO selection test). Yup, I know it's confusing.  :blush: I would post problems from the small olympiad, but I'm afraid I forgot them... Maybe someone else from Croatia will be able to post them.

Small olympiad is equal to drzavno competition Isn't it?

Ako je tako kakvi su vam onda ovo zadaci na onoj stranici [i]e.math.hr [/i]
Posalji mi ba one particije sto sam te pitao?

Nope. Small olympiad comes after national competition.

I've talked to [b]djuro[/b] about programming your partition problem (manual labor seems too difficult as the difficulty rises exponentially - it would take at least an hour if not an hour and a half for each one of these n's and I'd make a mistake by then for sure), but I don't have the slightest idea how to do that. He seemed to have some idea, so send him a PM and maybe he has something. Sorry. On e.math.hr you can see the problems from national competition and not from small olympiad ("mala olimpijada").

OK I have asked Djuro about partition and I hope that he will help me.
If you found that problems please post them in forum.   Pozdrav.


Hi. I'm currently an AP Calc student near Miami, FL, and I was interested in competing in some of these events, except I've never even heard of half the math theories that are discussed in this forum, such as combinatorics. Where could I find a guide as to what all of these theories originate from, and on what competitions they're tested in. I'd like to compete from Florida for national competitions, but for that I'd have to know more about what exactly these topics are. I'm sure I could learn and comprehend them rather fast, it's just that I've never seen anything like this before.

Art of Problem Solving isn't just a website; it's also [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php]a series of books about precisely these subjects[/url].


geometry
circumcircle
trigonometry
geometric transformation
reflection
angle bisector
geometry unsolved

In acute triangle $ ABC$, $ CD$ is the altitude. A line through the midpoint $ M$ of the side $ AB$ meets rays $ CA,CB$ at points $ K,L$ respectively such that $ CK\equal{}CL$. Point $ S$ is the circumcenter of triangle $ CLK$. Prove that $ SD\equal{}SM$.

Proof:
Suppose $ \angle ABC > \frac {\pi}{2}$,other cases are totally similar.
Let $ l_C$ be an angle bisector of $ \angle ACB$,and $ E$ is the foot of perpendicular from $ M$ to $ l_C$,obviously $ K,E,L$ are collinear.
Denote $ AB = c,AC = b,BC = a,\angle ABC = 2\beta,\angle ACB = 2\gamma,\angle BAC = 2\alpha$,$ \angle ECM = \phi$.

We will prove problem as follows:if $ S\in CE$,such that $ SM = SD$,and $ S'$ is circumcenter of $ \triangle CKL$,then $ CS = CS'$.
Since $ CSME$ and $ SNEM$ are cyclic quadrilaterals,where $ N$ is the midpoint of $ DM$,we conclude that $ \triangle DNE\sim\triangle CSM$,hence $ \frac {CS}{CM} = \frac {DN}{DE}$.Obviously $ DN = \frac {DM}{2} = \frac {DA + \frac {c}{2}}{2} = \frac {2b\cos{2\alpha} + c}{4}$ and $ \frac {DE}{EM} = \frac {\cos{(2\alpha + \gamma)}}{\sin{\phi}}$,thus $ CS = \frac {CM}{ME}\cdot\frac {\sin{\phi}\cdot(2b\cos{2\alpha} + c)}{4\cos{(2\alpha + \gamma)}}$.Since $ \angle CEM = \frac {\pi}{2}$,thus $ \sin{\phi} = \frac {EM}{CM}$.So $ \boxed{CS = \frac {2b\cos{2\alpha} + c}{4\cos{(2\alpha + \gamma)}}}$.
Using [b]Sine law[/b] for $ \triangle MLB$,we conclude $ \frac {LB}{BM} = \frac {\cos{(2\alpha + \gamma)}}{\cos{\gamma}}$,so
$ CL = CB - LB = a - \frac {c\cos{(2\alpha + \gamma)}}{2\cos{\gamma}}$.
If $ S'$ is the circumcenter of $ \triangle CKL$,then ${ CS' = \frac {CL}{2\cos\gamma}}$.Now it is enough to check that $ \frac {a}{2\cos{\gamma}} - \frac {c\cos{(2\alpha + \gamma)}}{4\cos^2{\gamma}} = CS = \frac {2b\cos{2\alpha} + c}{4\cos{(2\alpha + \gamma)}}$,what is not so hard.

[quote="Carlez Tevos"]Thanks, but what about a synthetic solution?[/quote]
I'm thinking on it.If i will find it,i'll definitely tell you.

[hide]Let $ Z$ be, the intersection point of the circumcircle $ (O),$ of the given triangle $ \bigtriangleup ABC,$ from its angle bisector of $ \angle C.$

We draw the circle $ (S),$ centered at midpoint $ S$ of the segment $ AZ,$ with radius $ SC \equal{} SZ,$ which intersects the sidelines $ CA,\ CB$ of $ \bigtriangleup ABC,$ at points $ K,\ L,$ respectively and so, we have $ CK \equal{} CL.$

It is easy to show that the points $ K,\ M,\ L$ are collinear, where $ MA \equal{} MB,$ because of $ \angle AMK \equal{} \angle AZK \equal{} \angle BZL \equal{} \angle BML,$ from the congruent right triangles $ \bigtriangleup KAZ$ and $ \bigtriangleup LBZ$ $ ($ because of they have $ ZA \equal{} ZB$ and $ ZK \equal{} ZL$ $ )$ and the cyclic quadrilaterals $ AKZM$ and $ BLMZ.$

$ \bullet$ Let $ T$ be, the orthogonal projection of $ S,$ on $ AB$ and so, we have that $ ZM\parallel ST\parallel CD.$

Because of now, $ SC \equal{} SZ,$ by [b][size=100]Thales theorem[/size][/b], we conclude that $ TM \equal{} TD$ $ ,(1)$

Hence, from $ (1)$ we conclude that $ SM \equal{} SD$ and the proof is completed.[/hide]
Kostas Vittas

[size=150][color=darkred][b][u]Indeed, a very nice triangle problem ![/u][/b][/color][/size]

[quote] [size=134][color=darkred][b][u]An equivalent enunciation I.[/u] Let $ \triangle\ ABC$ be $ A$ - isosceles with the circumcircle $ w = C(O)$ . For $ M\in (BC)$ let

 $ N\in AC$ , $ P\in AB$ so that $ M\in NP$ , $ MN = MP$ . Denote the projection $ D$ of $ A$ to $ PN$ . Prove that  $ OD = OM$ .[/b][/color][/size] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the diameter $ [AR]$ of the circle $ w$ and the midpoint $ S$ of $ [MD]$ . Apply the [b]Menelaus' theorem[/b] to $ \overline {BMC}\ /\ \triangle\ APN$ :

$ \|\begin{array}{c} \frac {CN}{CA}\cdot\frac {BA}{BP}\cdot\frac {MP}{MN} = 1 \\
 \\
(\ \begin{array}{c} CA = BA \\
\ MP = MN\end{array}\ )\)\end{array}\|$ $ \implies$ $ \|\begin{array}{c} BP = CN \\
 \\
(\begin{array}{c} RC = RB \\
\widehat {PBR}\equiv\widehat{NCR}\end{array}\ )\end{array}\|$ $ \implies$ $ \triangle\ PBR\ \equiv\ \triangle\ NCR$ $ \implies$ $ \|\begin{array}{c} RP = RN \\
\ (\ MP = MN\ )\end{array}\|$ $ \implies$

$ RM\perp \overline {PMN}$ $ \implies$ $ RM\parallel AD$ . Thus, $ \|\begin{array}{c} OA = OR \\
\ (SM = SD)\end{array}\|$ $ \implies$ $ OS\parallel MR\parallel AD$ $ \implies$ $ \|\begin{array}{c} OS\perp MD \\
\ (SM = SD)\end{array}\|$ $ \implies$ $ OD = OM$ .

[b]Remark.[/b] The quadrilaterals $ PBMR$ , $ NCMR$ are cyclically. If $ W$ is the midpoint of $ [AM]$ , then $ OW\perp PN$ . Hence can state the our problem so :[/color]

[quote] [size=134][color=darkred][b][u]An equivalent enunciation II.[/u] Let $ \triangle\ ABC$ be $ A$ - isosceles with the circumcircle $ w = C(O)$ . For $ M\in (BC)$ let

$ N\in AC$ , $ P\in AB$ so that $ M\in NP$ , $ MN = MP$ . Denote the diameter $ [AR]$ of $ w$ . Prove that  $ RM\perp PN$ .[/b][/color][/size] [/quote]

[color=darkblue][b][u]Proof I.[/u][/b] Denote $ U\in AB$ , $ V\in AC$ so that $ MU\parallel AC$ , $ MV\parallel AB$ . Observe that the midpoint $ W$ of $ AM$ belongs to $ UV$ , $ WU = WV$ and $ UV\parallel PN$ .

The points $ U$ , $ V$ [b]have the same power w.r.t. the circle [/b]$ w$ . Thus, $ OW\perp UV$ , i.e. $ OW\perp PN$ . Hence and $ RM\parallel OW$ $ \implies$ $ RM\perp PN$ .

[b][u]Proof II.[/u][/b] Apply the [b]Menelaus' theorem[/b] to $ \overline {BMC}\ /\ \triangle\ APN\ \ : \ \ \frac {CN}{CA}\cdot\frac {BA}{BP}\cdot\frac {MP}{MN} = 1\ \Longrightarrow\ CN = BP$ .

Let $ S$ the reflection of $ P$ w.r.t. $ B$ . Thus, $ SB = NC$ $ \implies$ $ RS = RN$ $ \implies$ $ RP = RN$ $ \implies$ $ RM\perp PN$ .

[b][u]Proof III (shortest !).[/u][/b] Apply the [b]Menelaus' theorem[/b] to $ \overline {BMC}\ /\ \triangle\ APN\ \ : \ \ \frac {CN}{CA}\cdot\frac {BA}{BP}\cdot\frac {MP}{MN} = 1\ \Longrightarrow\ CN = BP$ .

Since $ \|\begin{array}{c}
CR=BR\ ,\ CN=BP\\\\
CR\perp CN\ ,\ BR\perp BP\end{array}$ obtain $ \triangle\ CRN\ \equiv\ \triangle\ BRP$ .Thus, $ RN=RP$ , i.e. $ RM\perp PN$ .

[b]Remark.[/b] The upper property allows the construction of the line $ PN$ for which $ M\in PM$ , $ MN = MP$ .

otherwise than with parallels $ MV$ , $ MU$ to $ AB$ , $ AC$ respectively and afterwards the parallel through $ M$ to $ UV$ .[/color]


This problem is not so hard. Anyway,

Let $O_1$ and $O_2$ be two circles with different radius. $l_1$ and $l_2$ are the excommon tangent lines of $O_1$ and $O_2$. $l_1$ touches $O_1$ and $O_2$ at $T_1$ and $D_2$, respectively. $l_2$ touches $O_1$ and $O_2$ at $D_1$ and $T_2$, respectively. The line $D_1D_2$ intersects the circles $O_1$ and $O_2$ at $K_1$ and $K_2$, respectively. If $T_1K_1 \cap T_2K_2 = \{M\}$ , prove that $|MD_1| = |MD_2|$.

It's easy to argue that $\angle D_2K_2T_2=\pi-\angle T_1K_1D_1$ (so I won't do that; it's a simple angle chase :)), meaning that $T_1K_1,T_2K_2$ are equally inclined wrt $D_1D_2$, or, in other words, $MK_1K_2$ is isosceles. The result would then follow if we could show that $D_1K_1=D_2K_2\ (*)$.

We have $D_2K_1\cdot D_2D_1=D_2T_1^2=D_1T_2^2=D_1K_2\cdot D_1D_2\ (\#)$, so $(*)$ follows by equating the first and last term in $(\#)$ and dividing the equality by $D_1D_2$.


geometry
rectangle
absolute value
combinatorics unsolved
combinatorics

Let $m$ and $n$ be positive integers and suppose that an $m×n$ rectangle is coloured checkerboard style with unit squares such that the lower left corner is black. Draw the diagonal of the rectangle from upper left to lower right and let $f(m, n)$ denote the absolute value of the difference between the area of the black part and the white part of the triangle below the diagonal. Compute $f(m, n)$ if both $m$ and $n$ are even or both are odd. Show that for any constant $C > 0$ there are m and n such that $f(m, n) > C$.

If $m$ and $n$ are both odd or both even, then rotating the rectangle about the centroid switches the two triangles, while preserving the unit squares and their colors.  So each triangle contains half the area of black squares and half the area of white squares.  If $m$ and $n$ are both even, there are an equal number of black and white squares, so $f(m,n) = 0$.  If $m$ and $n$ are both odd, then there is one more black square than white square, so $f(m,n) = \frac{1}{2}$.

Let $m = 2n$.  Then there are a the same number of black and white squares that lie entirely under the diagonal, and the squares that pass through the rectangle can be divided into $n$ $1 \times 2$ rectangles, for which $3/4$ of the black square and $1/4$ of the white square lie under the diagonal.  So $f(m,n) = \frac{n}{2}$.


A $9\times 9$ array is filled with integers from 1 to 81. Prove that there exists $k\in\{1,2,3,\ldots, 9\}$ such that the product of the elements in the row $k$ is different from the product of the elements in the column $k$ of the array.

the 10 primes 41, 43, 47, 53, 59, 61, 67, 71, 73, 79 can't all be on the diagonal of the array. one of them is off the diagonal; call it $p$. $p$'s row and the same-numbered column can't have the same product because that column isn't divisible by $p$ while $p$'s row is.


Any Illinois AoPSers go to ICTM?  I'm pretty sure winluo went if no one else.  I went too.

If there were any, feel free to post your thoughts/scores/whatever.  Otherwise, this thread is probably just going to drown :P.

I went. I stank. Got 36 last year (tied 1st, item analysis gave 2nd), and this year the junior test whomped on me. My TI-89 kept refusing to solve three equations in three variables, and a program of mine conked out on the two person. My junior team lost 1st by 1 question...and one of my mistakes was writing down an answer wrong because I was flustered...Wah. So 2nd junior team, 3rd calculator, 5th 2 person :( , 3rd overall was school.

I went the past two years, last year as a non scorer and this year as a scorer in Algebra II Written, and Calculator. We got 2nd in the state in Algebra II, and calculator team got 8th. ( I only got a 20 on my written  :( ) but thats good for me because i have alot of gaps in my math education, i had to teach myself algebra II the summer after my freshman year to be on math team, and i didnt really teach myself, i just used the algebra III algebra II review to learn it.
As a team (I think)
Algebra I-5th
Geometry-3rd
Algebra II-2nd
PreCalc-7th
Calculator-8th
F/S Relay-1st (35)
J/S Relay-1st (35)
F/S 2 Person- 7th
J/S 2 Person-3rd
F/S 8 Person- 5th
J/S 8 Person-4th
Oral Topic-3rd (43/50)

Overall- 4th
Last Year- 3rd
2003- State Champs

Hope this pattern doesn't continue.  :P

*hears name*

Yeah, I went. We did well, I guess, though our team events were better than individuals (sophomores: 78). For frosh-soph, we got 1st in 2-person, 1st in 8-person, and 1st for one of the relay teams. I also got 3rd individually. (34, and I made two incredibly stupid mistakes: solving a linear system in two variables, and misreading 2 for 1).

Probability: How did you do? I don't think I saw you there...

OK, so my school is small (a mere 300) and not exceptional at math, so we competed in the inferior 3AA.

I took precalc (1st individual), and my frosh-soph 2 person team won.  My partner wrote down an answer wrong though, and I thought A+B+C = 360 in a triangle.  That pretty much sums up the significant awards of our school lol.  We got a bunch of 5th places and such, but I can't remember for what.

BTW Winston, I don't think you know what I look like lol.  If you do, then... *shudders* :D  well maybe from a picture somewhere

Nope, I have no idea what you look like.  :lol: Guess I'll find out at MOP, eh? And congrats, 1st place individual is impressive, as is two-person. Two person went surprisingly well for me though - in practices I always made silly mistakes, but I didn't make any at the actual contest.

Edit: And, do you know what I look like?

You're the short asian kid with the MIB shirt.  :D

Wow, I actually saw the MIB shirts, but I didn't know you were among them!

Yep. MIB, that's us. Though, now that I think of it, I'm the only *short* Asian on our team... hmm.

i dont know if you saw any of us...we had the awful napoleon dynamite math team shirts on.


probability
combinatorics unsolved
combinatorics

From a deck of 52 cards we pick 5 cards at once ( randomly).
Assuming we play a game of poker,what is the probability
we get "nothing"?

[quote="rem"]What do you mean?

So we will only get a high card?[/quote]
I mean that we get no pair,no colour,no pokers ,no full house,...,not
anything at all valuable in  game of Poker.You can call it a "high card" in a competitive game.Otherwise,it's just 5 cards  :!:

Ok. Here is the solution.
If all the cards are different, we will get no pairs, three of a kind, etc. The number of sets we can form of five different cards is $13*12*11*10*9*4^{5}$, we multply by $4^{5}$, because every card can be one of four suits.
However we have to substract the times we have a flush. The number of ways we have that is $13*12*11*10*9*4$, because we must have $5$ different cards and $4$ different suits, of which the flush can be.
Now we need to also substract the time we have a straight. There are $9*4^{5}$ sets of those straights.
However we have substracted the times when we have a straight flush twice. So we need to add the numebr of straight flush sets we can get. The number of such sets is $9*4$.
Therefore the answer is:
$13*12*11*10*9*4^{5}-13*12*11*10*9*4-9*4^{5}+9*4$


Who here made USAMO?

I'm the youngest one from Ohio, again.   :lol:

Also, post here if you are shocked or upset that you didn't make it.

I did not due to missing both #2 and #5 stupidly on AIME II. Yay.

i knew I wouldn't make it due to stupidity on aime 2

Congratulations to Travis and Tong both for making Honorable Mention! WOO!

 :winner_first:

Are they both going to MOP, or is Tong not because he is a senior?

I'm going to blue mop. Tong is a senior and cannot go because he needs to be a winner. :(

Hey Travis, are you not doing ARML this year?  You weren't at the practice today and some people said that you weren't coming.

I was originally planning to come, but I can't anymore. I was really disappointed...  :(


AMC AIME geometry inequalities USA(J)MO USAMO

Geometry and Inequalities are the most common on USAMO.
Are there any types of problems that are the most common on AIME?

I'm not sure that you're right on what you said about geometry and inequalities, but the AIME has a lot of geometry and counting.

there are sometimes optimization problem... for inequality, of course.

I also find that solid geometry often appears on the AIME, but it never appears on the USAMO.  Both have a lot of plane geometry--that was already mentioned.

[quote="BornForMath"][quote="paladin8"]NT is common on the USAMO, as you can see from this years...

Geometry is the most common on the AIME, but counting is a close second. NT shows up once in a while and there's usually some algebraic manipulation question, whethers it's with pure numbers or variables.[/quote]

Is NT number theory?[/quote]

Yes.

I think the USAMO is slightly less predictable than the IMO. Geometry and Inequalities are, in my opinion, the most common on IMO's, whereas the USAMO tends to have years where they just have a strange distribution of problem types (like in 2002, which really didn't have a real classical geometry or inequality problem at all, with #2 being an algebra problem really and #6 being a tiling problem)

they had no classic inequality this year, also. Sometimes USAMO has graph theory, as well.... for example #1 this year.

I think USAMO made an effort not to repeat a classical Ineq three years in a row, whereas IMO makes an effort to maintain a certain level of consistency. Of course, IMO is harder, and that contributes to these differing philosophies, and also Ineqs and Geo often lend themselves to experienced problem solvers, whereas part of USAMO's goal is to identify younger students who have "the ability to think" but not necessarily a lot of training. It's especially hard to just "think out" an inequality if you know nothing, but for example 2002's #6 was such a problem.


function trigonometry algebra domain limit calculus logarithms

I need help with the following.. .thanks!

[hide]the only time g is discontinuous is at x=o

theres not much to discuss really...[/hide]

Wow, you have the picture, so I have a question, too. Hope you don't mind. What does that solid circle/point represent? It's not on the graph, but it's under the hollow circle. I am really confused with that circle.

:D

Notice that in the given, we also have:

$ g(\theta) \equal{} 1/2$ for $ \theta\equal{}0$

So basically, that solid circle "graphed" that part of the function

[quote="jeez123"]:D

Notice that in the given, we also have:

$ g(\theta) \equal{} 1/2$ for $ \theta \equal{} 0$

So basically, that solid circle "graphed" that part of the function[/quote]

well, so you meant that solid circle was also part of the function? I thought the function was just that big curve?! So if anything outside of that curve, I didn't consider it was part of the function. In given, it says: g(angle)=1/2 for angle=0. Is that completely the other different function? So like in this graph, there are two functions?

You've fallen into an 18th-century trap. You [i]want[/i] "function" to mean the same thing as "curve." You want things to be smooth. But that's not what a function is. And the truth is not in the picture - that's just a hint to your mind. Don't take it literally.

Let $ f(x)\equal{}\begin{cases}(|x|)!&\text{if }x\text{ is an odd integer}\\2^{x}&\text{if }x\text{ is an even integer}\\ x\sin\left(\frac{1}x\right)&\text{if }x\text{ is irrational}\\ x^{2}\plus{}3&\text{otherwise}\end{cases}$

Any attempt to "graph" this thing will drive you nuts - but this [b]is a function[/b] whose domain is the whole real line. Just not a very "nice" one.

For your example above, $ \lim_{\theta\to0}g(\theta)\equal{}1.$ that means that by making values of $ \theta$ arbitrarily close to (but not equal to) zero, we can make $ g(\theta)$ arbitrarily close to $ 1.$ $ g(0)$ also exists and equals $ \frac{1}2.$ (That's what's indicated by the solid circle.) The value of $ g(0)$ is "unexpected" in the sense that it does not match the limit. Very well, $ g$ is not continuous at zero - that's the very definition of continuity.

[quote="Kent Merryfield"]For your example above, $ \lim_{\theta\to0}g(\theta) \equal{} 1.$ that means that by making values of $ \theta$ arbitrarily close to (but not equal to) zero, we can make $ g(\theta)$ arbitrarily close to $ 1.$ $ g(0)$ also exists and equals $ \frac{1}{2}.$ (That's what's indicated by the solid circle.) The value of $ g(0)$ is "unexpected" in the sense that it does not match the limit. Very well, $ g$ is not continuous at zero - that's the very definition of continuity.[/quote]
So you mean the solid circle is also part of the graph of the function $ g(\theta)$? But I don't know the definition of continuity, if I interpret 'continuity' literally, then does it mean like continuous. but you just said $ g$ is not continuous at zero, so it's not continuity? BTW, the hollow circle means this point doesn't exist, then the solid circle tells us when the angle is 0, g has the answer for that. The point actually exists but just not on that line?  Does limit mean the value would make the function undefined?

The definition of continuity (as phrased without dragging out the epsilons and deltas):

$ g$ is continuous at a point $ a$ provided the following three conditions are [b]all[/b] met:

(1) $ a$ is in the domain of $ g$ - that is, $ g(a)$ is defined,

(2) The limit  $ \lim_{x\to a}g(x)$ exists, and

(3) These items agree, namely $ \lim_{g\to a}g(x) \equal{} g(a).$

[i]Any[/i] failure in any of (1), (2), or (3) means the function is not continuous at $ a.$

[This could all be stated more efficiently. Efficiency doesn't serve my current pedagogical purposes.]

What about the intuition that "it's continuous if you can draw its graph without lifting your pencil from the paper"? First, it is impossible to make that statement rigorous. Second, to the extent that we can guess at its approximate meaning, it's [b]wrong,[/b] anyway!

Here's a famous example:

$ f(x) \equal{}\begin{cases}x\sin\left(\frac{1}{x}\right), & x\ne0\\ 0, & x \equal{} 0\end{cases}.$

This function [b]is[/b] continuous. Everywhere. [b]Especially[/b] at zero. And you can't draw it at all, not in a neighborhood of zero.

[quote="Kent Merryfield"]Here's a famous example:

$ f(x) \equal{}\begin{cases}x\sin\left(\frac{1}{x}\right), & x\ne0\\ 0, & x \equal{} 0\end{cases}.$

This function [b]is[/b] continuous. Everywhere. [b]Especially[/b] at zero. And you can't draw it at all, not in a neighborhood of zero.[/quote]

This example made me think of its graph. The first branch, x is not equal to 1, it can be drawn as a hollow circle, then the second branch x=1, it can be drawn as a solid circle probably right under that hollow circle. So can we say that it is continuous by just looking at these two circles on the graph?

[quote="Kent Merryfield"]The definition of continuity (as phrased without dragging out the epsilons and deltas):

$ g$ is continuous at a point $ a$ provided the following three conditions are [b]all[/b] met:

(1) $ a$ is in the domain of $ g$ - that is, $ g(a)$ is defined,

(2) The limit  $ \lim_{x\to a}g(x)$ exists, and

(3) These items agree, namely $ \lim_{g\to a}g(x) \equal{} g(a).$

[i]Any[/i] failure in any of (1), (2), or (3) means the function is not continuous at $ a.$

[/quote]

Dear Professor,

If a function $ f$ is not defined at $ \ {x_{0}}$, can we say that it is not continuous at $ \ {x_{0}}$ ? For example, saying that "$ f(x) \equal{} lnx$ is not continuous at $ \minus{}1$ " is true ? In several calculus books there are different explanations for the continuity of $ g(x) \equal{} 1/x$. On the one hand, Robert Adams' Calculus book says :
"The function $ g(x) \equal{} 1/x$ is also a continuous function. This may seem wrong to you at first glance, because its graph is broken at $ x \equal{} 0$. However, the number $ 0$ is not in the domain of $ g$, so it doesn't make sense to talk about $ g$ being discontinuous there. A function can only be continuous or discontinuous at a point of its domain (3rd Edtion, p 76)."

On the other hand most of calculus books say that $ g(x) \equal{} 1/x$ is discontinuous at $ x \equal{} 0$.

Sir, I don't really understand how to talk about continuity - discontinuity of a function at a point at which it is not defined. Here is my reason : Let's take $ \epsilon\minus{}\delta$ approach. There is no $ g(0)$, so we can't talk  about $ \epsilon\minus{}neighboorhood$ of $ g(0)$. Hence we can't talk about continuity or discontiunity of $ g$ at $ x \equal{} 0$.

If I am wrong, I would like you to correct me.

$ Saygin\ Dincer$

category: you are correct. In fact, speaking properly, there are three mutually exclusive possibilities for the behavior of $ f$ at $ a$:

(1) $ f$ is continuous at $ a,$ or
(2) $ f$ is defined at $ a$ but discontinuous, or
(3) $ a$ lies outside of the domain of $ f.$

There's even the possibility that $ f(a)$ might be defined but $ a$ is not a limit point of the domain of $ f,$ in which case $ f$ is continuous at $ a$ but the notion of a limit at that point is meaningless.

So, $ \frac{1}x$ is continuous everywhere on its domain, and $ \ln x$ is continuous everywhere on its domain. But $ \frac{1}x$ is not continuous on $ \mathbb{R}.$

On the other hand, if $ f$ is not defined at $ a$ but is defined in a punctured neighborhood of $ a,$ the casual usage "$ f$ is discontinuous at $ a$" might not quite agree with my first paragraph but does no harm - it is an important part of our understanding of the behavior of the functions.

If you'll read the whole of this and one other topic, you'll see that the point you are trying to make would be utterly lost on jeez123 and MIA6; I was trying, without success, to speak to them.

[quote="Kent Merryfield"]
So, $ \frac{1}{x}$ is continuous everywhere on its domain, and $ \ln x$ is continuous everywhere on its domain. But $ \frac{1}{x}$ is not continuous on $ \mathbb{R}.$

If you'll read the whole of this and one other topic, you'll see that the point you are trying to make would be utterly lost on jeez123 and MIA6; I was trying, without success, to speak to them.[/quote]

Dear Sir,

I have read the whole of this and one other topic. Rudin writes in Principles of Math. Analysis "If $ x$ is a point in the domain of definition of the function at which f is not continuous, we say that $ f$ is discontinuous at $ x$, or $ f$ has a discontinuity at $ x$". This is exactly why I am confused. Because of this, I think we can not talk about continuity of $ 1/x$ on $ \mathbb{R}$. What I am trying to say : We are looking for continuity on the domain of  a function.

If I am asked "Find discontinuity points of $ 1/x$", should I say $ 0$ or there is no such point?

I hope I explain the point in my mind :) Thank you, Sir.

$ Saygin\ Dincer$

There's a difference of circumstance: did we propose the domain first, and then ask about the function? Or did we propose the function first, and then let the domain follow from that?

A "propose the domain first" question would be to ask if $ f(x) \equal{}\frac{1}{x},$ is $ f\in C(\mathbb{R})?$ And the answer is no, because $ f$ is not even defined on all of $ \mathbb{R}.$ Furthermore, the problem is not a removable singularity: no possible extension of the function to being defined at zero would make it continuous on $ \mathbb{R}.$

If we propose the function first, we say, what about $ f(x) \equal{}\frac{1}{x}?$ We see that its domain is $ \mathbb{R}\setminus\{0\},$ and that it is continuous on that set.

Both of these statements are correct; the difference is one of approach or attitude.


Noah's favorite flower shop sells 5 different types of flowers. How many different ways are there to buy 5 flowers if there are no restrictions on how many of each flower may be purchased?

[hide="Hint"]
Call the different types #1, #2, #3, #4, and #5.

Let $ a$ be the number of #1 you buy, $ b$ be the number of #2, $ c$ be the number of #3, $ d$ be the number of #4 , $ e$ be the number of #5.

Since you buy 5 flowers total, $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}5$

What is the restrictions on $ a,b,c,d,e$? Given these restrictions, how many solutions to the equation are possible?[/hide]

[hide="Solution"]Using balls and urns, $ n\equal{}\binom{5\plus{}5\minus{}1}{5}\equal{}\binom{9}{5}\equal{}\boxed{126}$.[/hide]

[hide="basically the same solution"]
Count the ways to split 5 flowers into 5 different types using 4 dividers, for example:

ff-f-f--f

there are 4 dividers, 5 flowers, and so the answer is $ {9 \choose 5} \equal{} {9 \choose 4} \equal{} 126$

In other words, as i_like_pie said, this is a balls and urns problem.
[/hide]

Hmm....I remember learning this principle in school, with the whole dividers thing, but the teacher never really explained it.  Would someone provide an explanation as to why method works because I don't really understand it.

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1323456#1323456]this; case 4.[/url]


[quote="Silverfalcon"]Show if it is possible to write simplified term for the sum

$ \frac {1}{1 \cdot 2 \cdot 3 \cdots n} \plus{} \frac {1}{2 \cdot 3 \cdot 4 \cdots (n \plus{} 1)} \plus{} \cdots \plus{} \frac {1}{n \cdot (n \plus{} 1) \cdot (n \plus{} 2) \cdots (2n \minus{} 1)}$

in terms of $ n$ where $ n \geq 3$.[/quote]

$ \frac {1}{1 \cdot 2 \cdot 3 \cdots n} \plus{} \frac {1}{2 \cdot 3 \cdot 4 \cdots (n \plus{} 1)} \plus{} \cdots \plus{} \frac {1}{n \cdot (n \plus{} 1) \cdot (n \plus{} 2) \cdots (2n \minus{} 1)}\equal{}\frac {U_n\minus{}U_0}{1\minus{}n}$

where $ U_r\equal{}\frac {1}{(r\plus{}1)(r\plus{}2).....(r\plus{}n\minus{}1)}$

[hide="Hint"]Note that with partial fraction decomposition, we get that:
$ \frac{1}{k(k\plus{}1) \ldots (k\plus{}h)} \equal{} \frac{1}{h!} \sum_{j\equal{}0}^{h} \left[ (\minus{}1)^{j} \binom{h}{j} \frac {1}{k\plus{}j} \right]$

Here $ h \equal{} n\minus{}1$ and $ k$ sums from $ k\equal{}1$ to $ k\equal{}n$.

Can we make these terms telescope?
[/hide]


Three distinct numbers are chosen at random from $1, 2, 3, ..., 1000$. What is the probability that the sum of these 3 numbers is divisible by 3?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110157]here[/url] for a very similar topic.

[hide]We are looking for numbers where the sum of their remainders modulo 3 is divisible by 3.  Since all numbers are equivalent to 1, 0, or -1 modulo 3, the possible combinations or remainders are (1,1,1), (0,0,0), (-1,-1,-1), or (1,0,-1).  There are 334 numbers in the set that are equivalent to 1 mod 3, and 333 each that are equivalent to 0 or -1 mod 3.  Thus, the desired probability is $\frac{2\binom{333}{3}+\binom{334}{3}+\binom{333}{1}\binom{333}{1}\binom{334}{1}}{\binom{1000}{3}}\approx .48$[/hide]

DanK, you were good up until you estimated the sum.  In fact, it's very close to (but slightly larger than) $\frac{1}{3}$, which is (I think) not very surprising at all.

Oh, oops.  Must've mis-entered something on the calculator...

[quote="yunbo"]Yea. Consider that all numbers divisible by 3 less than 1000 are the sum of 3 numbers, except for 3 itself. So probability is 332/1000.[/quote]

No.  You appear to be answering an entirely different question.

Ya you are right JBL. I must consider my denominators too.

Actually, we aren't just looking for numbers divisible by 3 that are less than 1000.  The largest number we could get by adding 3 numbers from that set is 1000+999+998=2997...


linear algebra
matrix
abstract algebra
vector
LaTeX
absolute value
linear algebra unsolved

Let $ A\equal{}(a_{ij})_n \in M_n(C)$. Assume that  for all $ i \equal{} 1,2,...,n$
          $ |a_{ii}|> \sum |a_{ki}|$ with k=1,2,3..,n and $ k$#$ i$
Prove that $ A$ is invertible

[hide="Hint"] Show that the kernel of $ A$ is trivial. Consider, given a non-trivial vector $ x\in\ker(A)$, the index $ i_0$ with maximal absolute value of $ x_i$ and deduce a contradiction.[/hide]

[quote="Third Edition"] Show that the kernel of $ A$ is trivial. [/quote]

kernel $ A$ is trivial, it doen't mean that $ det A$ # $ 0$. Can you prove $ det A$ # $ 0$n if kernel $ A$ is trivial

What you need to show is that $ A$ is invertible: For this show that the range of $ A$ - if $ A$ is considered as endomorphism of $ \mathbb C^n$ - is a $ n$-dimensional subspace of $ \mathbb C^n$. Hence ...

This is Lévy-Desplanques Theorem. A proof making use of the concept of linear transformation has already been given. Let us give another proof. For that, let us have a matrix $ A\equal{}[a_{ij}] \in \mathbb{C} ^{n \times n}$ and define

$ \sum_{k\equal{}1, k \neq j}^n |a_{jk}| : \equal{} \rho _j$, for $ j \equal{} 1,2,...,n$.

Now, let $ \lambda$ be an eigenvalue of A: then, there is at least one Gershgorin disk to which $ \lambda$ belongs, which implies that there is at least one $ j$ for which

$ \rho_j \geq |\lambda \minus{} a_{jj}| \geq |\,\,|\lambda| \minus{} |a_{jj}|\,\,| \equal{}> \minus{} \rho_j \leq |\lambda| \minus{} |a_{jj}| \equal{}> |\lambda| \geq |a_{jj}| \minus{} \rho_j > 0$.

Therefore, $ \lambda \equal{} 0$ cannot be an eigenvalue of A, which proves the theorem. $ \Diamond$

[quote="loc1991"]kernel $ A$ is trivial, it doen't mean that $ det A$ # $ 0$. Can you prove $ det A$ # $ 0$n if kernel $ A$ is trivial[/quote]
For a square matrix - and the context here is that of a square matrix, clearly - the kernel (null space) of $ A$ being trivial [b]does[/b] imply that $ \det A\ne 0.$

It's an exercise in vast overkill, but the textbook I'm teaching from this semester (by Kenneth Lay) gives an "invertible matrix theorem" which lists all of the following as equivalent:

a. $ A$ us an invertible matrix.
b. $ A$ is row equivalent to the $ n\times n$ identity matrix.
c. $ A$ has $ n$ pivot positions.
d. The equation $ Ax = 0$ has only the trivial solution.
e. The columns of $ A$ form a linearly independent set.
f. The linear transformation $ x\mapsto Ax$ is one-to-one.
g. The equation $ Ax = b$ has at least one solution for each $ b$ in $ \mathbb{R}^n.$
h. The columns of $ A$ span $ \mathbb{R}^n.$
i. The linear transformation $ x\mapsto Ax$ maps $ \mathbb{R}^n$ onto $ \mathbb{R}^n,$
j. There is an $ n\times n$ matrix $ C$ such that $ CA = I.$
k. There is an $ n\times n$ matrix $ D$ such that $ AB = I.$
l. $ A^T$ is an invertible matrix.
m. The columns of $ A$ for a basis for $ \mathbb{R}^n.$
n. $ \text{Col}\,A = \mathbb{R}^n.$
o. $ \dim\,\text{Col}\,A = n.$
p. $ \text{rank}\,A = n.$
q. $ \text{Nul}\,A = \{0\}.$
r. $ \dim\,\text{Nul}\,A = 0.$
s. $ \det A\ne 0.$

(All of those references to $ \mathbb{R}^n$ could just as easily be $ \mathbb{F}^n$ for an arbitrary field $ \mathbb{F}.$)

$ \text{\LaTeX}$ tip: the "not equal to" sign is \neq or \ne, not #. $ \ne.$

Key word insertion: [b]strictly diagonally dominant[/b] matrix.

[quote="Kent Merryfield"][quote="loc1991"]kernel $ A$ is trivial, it doen't mean that $ det A$ # $ 0$. Can you prove $ det A$ # $ 0$n if kernel $ A$ is trivial[/quote]
(1)For a square matrix - and the context here is that of a square matrix, clearly - the kernel (null space) of $ A$ being trivial [b]does[/b] imply that $ \det A\ne 0.$

(2) all of the following as equivalent:

d. The equation $ Ax \equal{} 0$ has only the trivial solution.
s. $ \det A\ne 0.$

[/quote]

I think (1) and (2) contrast each other? Can  you prove that d and s are equvalent?
Thanks for your reply above  :)

Both of my statements say the same thing. Read them carefully. As for the proof - do you have any reason to doubt this theorem? Try showing that (d) and (s) are both equivalent to (b), row equivalence with the identity matrix.


invariant
abstract algebra
vector
Ring Theory
superior algebra
superior algebra unsolved

We say that a ring $ R$ has an invariant basis number (IBN) if for every free left $ R$-module $ M$ every two bases of $ M$ have the same cardinality. Prove that every left noetherian ring has an IBN.

Let $ R$ be a left noetherian ring, let $ M$ be a left $ R$-module (necessarily noetherian) and let $ \varphi \in End_R(M)$ be surjective. If $ \varphi$ is not injective, let $ 0 \neq x \in ker(\varphi)$ and for each $ n \in \mathbb{N}$, choose $ y_n \in M$ such that $ \varphi^n(y_n) \equal{} x$. It follows that that $ y_n \in ker(\varphi^{n\plus{}1}) \minus{} ker(\varphi^n)$. In particular, we have a strictly increasing chain of submodules of $ M$, contradicting the fact that $ M$ is noetherian. Therefore, $ \varphi$ is an isomorphism. In particular, every surjective endomorphism of $ R^n$ (for $ n \in \mathbb{N}$) is an isomorphism.

Now, let $ N$ be a left $ R$-module such $ R^n \cong R^n \oplus N$ for some natural number $ n$. Let $ \psi$ denote an isomorphism from $ R^n$ to $ R^n \oplus N$ and let $ \pi$ be the projection map from $ R^n \oplus N$ onto $ R^n$. Then $ \pi \psi \in End_R(R^n)$ is surjective and hence an isomorphism. In particular, $ \pi$ is also an isomorphism. But this implies $ N \equal{} ker(\pi) \equal{} 0$.

So, if $ m \leq n \in \mathbb{N}$ such that $ R^m \cong R^n \cong R^m \oplus R^{n\minus{}m}$, then $ R^{n\minus{}m} \equal{} 0$, i.e., $ n \minus{} m \equal{} 0$, and so $ n \equal{} m$.

[quote="-oo-"]and the infinite case?[/quote]
It's the same as with vector spaces.
Assume we have two infinite bases $ B_1,B_2$ of the free module $ M$.
Then each $ x\in B_1$ is the linear combination of some finite $ C_x\subseteq B_2$.
Then $ M$ is generated by $ \bigcup_{x\in B_1}C_x\subseteq B_2$, which gives $ B_2\equal{}\bigcup_{x\in B_1}C_x$ and $ |B_2|\le |B_1\times\mathbb{N}|\equal{}|B_1|$.
And analog for $ |B_1|\le |B_2|$.


A standard 6-sided die is rolled repeatedly. Let
Xi be the number of 6s seen in the first i rolls. Let Yi be the largest number seen in the first i rolls.
a) Explain briefly why X1,X2,X3, . . . is a Markov chain and give the transition probabilities and the transition graph.
b) Explain briefly why Y1,Y2,Y3, . . . is a Markov chain and give the transition matrix.
The die is modified so that it behaves normally on the first roll but threafter it can never produce the same number twice in succession (with the other 5 possibilities being equally likely). The random variables Xi and Yi are defined as before.
c) Is X1,X2,X3, . . . a Markov chain?

d) Is Y1,Y2,Y3 . . . a Markov chain?

Can you help me with part c and d please? Thanks


Determine the number of ways to arrange the letters of the word TARTAR

There are 6 letters in total, 2 pairs of T's, two pairs of A's, and two pairs of R's. Thus our answer is:

$ \frac{6!}{2!2!2!} \equal{} \boxed{90}$


Without use of caculator, find $x$ in equation below:

$x^3 = 357911$

[hide="My not so elegant answer"]I know that $70^3$ and $69^3$ are pretty close to that number. I try 70 and it's a little less.

Cubing 71, I get the same number, so $x=71$[/hide]

I think it could be done this way :

First, notice that x needs to be a 2 digits number (100^3 = 1000000..). Plus, it's strictly between 50^3 and 100^3. (50^3 = 125000)
Second, since the number ends with an "1", x also ends with 1.
So, it remains to test : 51, 61, 71, 81 and 91.

Eventually, you get : x = 71

;)

[hide]Dividing by 1000 we get that $(\frac{x}{10})^3=357.911$. We want to find a perfect cube near 357. $7^3=343$ is pretty close. $70^3=343000$, so x>70. However, $80^3=512000$, so x<80. Since the cubed number ends in 1, the number being cubed must also end with 1. The only numberwith units digit 1 such that $70<x<80$ is 71. [/hide]


Given triangle $ABC$ with angle bisectors $l_{a},l_{b},l_{c}$ and semiperimeter $s$ forall $x,y,z>0$ prove that
$\frac{y+z}{x}\frac1{l_{a}^{2}}+\frac{z+x}{y}\frac1{l_{b}^{2}}+\frac{x+y}{z}\frac1{l_{c}^{2}}\ge\frac{18}{s^{2}}$ :)


function
floor function
number theory
least common multiple
number theory unsolved

a,b are two positive reals which has the property that :
[na]+[nb]=[n(a+b)] for all n,where n is a positive integer.
Prove that at least one of a,b is a positive integer.
(HERE [x] means the largest integer which is lower than x)

When if $ n\equal{}10$, $ a\equal{}b\equal{}1.01$?

Here is something a little more general that was put on a Black MOP test last year:

Prove that if $ \lfloor {na} \rfloor \plus{} \lfloor {nb} \rfloor \equal{} \lfloor {nc} \rfloor$ for all $ n$, then $ a$ or $ b$ is an integer.

[hide="Solution"]
Assume that neither $ a$ nor $ b$ is an integer.

[b]Lemma:[/b] $ c \equal{} a \plus{} b$.

[i]Proof:[/i] Assume not.  Let $ a \plus{} b \minus{} c \equal{} d$ for real $ d$.  Note that for any reals, $ x$ and $ y$, we have that $ \lfloor {x} \rfloor \plus{} \lfloor {y} \rfloor$ is either $ 1 \plus{} \lfloor {x \plus{} y} \rfloor$ or $ \lfloor {x \plus{} y} \rfloor$.  Thus, $ \lfloor {na} \rfloor \plus{} \lfloor {nb} \rfloor \equal{} \lfloor {n(a \plus{} b)} \rfloor$ $ (*)$ or $ \lfloor {na} \rfloor \plus{} \lfloor {nb} \rfloor \equal{} \lfloor {n(a \plus{} b)}\rfloor \plus{} 1$ $ (**)$ for all $ n$.  However, setting $ n > \frac {5}{d}$ gives us that Right Hand Side of $ (*)$ is $ \lfloor {nc \plus{} nd} \rfloor$, which is not equal to $ \lfloor {na \plus{} nb} \rfloor$ nor $ \lfloor {na \plus{} nb} \rfloor \plus{} 1$, which is a contradiction.  Similarly, the Right Hand Side of $ (**)$ is $ \lfloor {nc \plus{} nd} \rfloor \plus{} 1$, which is also not equal to $ \lfloor {na \plus{} nb} \rfloor \plus{} 1$ nor $ \lfloor {na \plus{} nb} \rfloor$.  Both yield contradictions, so $ d \equal{} 0$, meaning that $ c \equal{} a \plus{} b$.

[b]Lemma 2:[/b] Consider two reals, $ 0 < s < t < 1$.  There exists an $ n$ so that $ \{ ns \} > \{ nt \}$.
[i]
Proof: [/i]

[b]Case 2.1: [/b] $ s$ and $ t$ are rational.

In this case, let the denominators of $ s$ and $ t$ (in reduced form) be $ s'$ and $ t'$ respectively.  Set $ n \equal{} lcm (s', t') \minus{} 1$.  This yields that $ \{ ns \} \equal{} 1 \minus{} s > 1 \minus{} t \equal{} \{ nt \}$.

[b]Case 2.2: [/b] $ s$ is rational, $ t$ is irrational.

Since $ \{ nt \}$ is dense, set $ \{ nt \} < \frac {1}{d}$, where $ d$ is the denominator of $ s$ in reduced form.  This immediately entails that $ \{ nt \} < \{ ns \}$.
[b]
Case 2.3:[/b] $ s$ is irrational.

Assume to the contrary.  Since $ \{ ns \}$ is dense, let $ 1 \minus{} \delta < \{ ns \} < 1$ for $ \delta < t \minus{} s, 1 \minus{} t$.  Hence, $ \{ (n \minus{} 1)s \} > 1 \minus{} s \minus{} \delta$.  Furthermore, $ \{ nt \} > \{ ns \}$ by assumption, so $ 1 \minus{} \delta < \{ nt \} < 1\implies \{ (n \minus{} 1)t \} < 1 \minus{} t$.  As $ \delta < t \minus{} s$, we see that $ \{ (n \minus{} 1)s \} > \{ (n \minus{} 1)t \}$, which is a contradiction, so the Lemma is proven.

Now for the problem itself.  Assume that $ \lfloor {na} \rfloor \plus{} \lfloor {n(c \minus{} a)} \rfloor \equal{} \lfloor {nc} \rfloor$ $ (***)$ for all $ n$, so $ \{ na \} \plus{} \{ n(c \minus{} a) \} \equal{} \{ nc \}$.  We may also replace $ a$ with $ \{ a \}$, $ b$ with $ \{ b \}$, and $ c$ with $ \{ c \}$ since the integer parts of $ (***)$ add up to each other.  Putting in $ n \equal{} 1$ yields that $ c > a$.  Now, by Lemma 2, we have that there is an $ n$ so that $ \{ na \} > \{ nc \} \implies \{ n(c \minus{} a) \} < 0$, which is a contradiction.  Hence, $ a$ or $ b$ is an integer.   [/hide]

[quote="rofler"]When if $ n \equal{} 10$, $ a \equal{} b \equal{} 1.01$?[/quote]

rofler, you have set a special case for $ n$. However, the quesiton demands for all $ n$.
$ a \equal{} b \equal{} 1.01$ is definitely not a counterexample, by setting $ n \equal{} 50$.


first tell me thru which format u attach diagrams...

there r 2 perp walls (l shaped) each with coeff of restitution e
a ball is projected from the ground,hits the vertical wall perpendicular to it
then it rebounds ,bounces once on ground
and then reaches its original point

Calculate e :oops:

question is not clear.......post a diagram!
save the image in JPEG format  :D

i just don't kno y i posted that ques--its easy enuf  :o

neways i'll post the diag later if u want.....getting late 4 skul :(  :P