Θα μπορούσα να ανοίξω 2 διαφορετικά threads αλλά δε νομίζω ότι χρειάζεται. Είναι 2 καλούλια ασκησάκια που ξέθαψα τις προάλλες. Ας τα έχουν να σκέφτονται όσοι δεν τα ξέρουν ήδη.

- Πόσα, το πολύ, σχήματα Τ μπορούμε να ζωγραφίσουμε στο επίπεδο;
- Πόσα, το πολύ, οκτάρια; (το οκτάρι περιέχει μόνο το boundary, όχι και το εσωτερικό)

Προφανώς τα πιο πάνω σχήματα στο επίπεδο δεν πρέπει να τέμνονται (αλλά μπορεί να είναι μεγεθυμένα, περιστραμμένα, κλπ)

Δεν μπαίνω στη διαδικασία να ορίσω ακριβώς πώς είναι τα  Τ/8, αφήστε τη φαντασία σας να καλπάσει.

Λοιπόν, χτες είχαμε ανοίξει μία συζήτηση με έναν φίλο και θυμήθηκα αυτό το πρόβλημα που έχει μείνει καιρό άλυτο (και είναι κρίμα γιατί είναι πολύ cute). :)

Προσπαθώντας να το ξαναζωντανέψω, να ξεκινήσουμε με τα εντελώς προφανή: $ \aleph_0 \leq \#\{8\} \leq \#\{T\}\leq\mathfrak{c}$ (κάθε 8 περιέχει ένα Τ μέσα του).

1. Ακριβώς μία από αυτές τις ανισότητες είναι γνήσια (αλλιώς σκοτώσαμε την υπόθεση του συνεχούς :D ). Αν τη βρούμε, τελειώσαμε.

2. Είναι η μεσαία ανισότητα γνήσια? Ή μπορούμε να θέσουμε σε 1-1 αντιστοιχία τα 8 και τα Τ?

Cheerio,

Durandal 1707

[quote="Christiano89"]Aν μπορείς εξήγησε λίγο αυτό που γράφεις...[/quote]

Δεν μπορώ να το σχεδιάσω, αλλά... κάθε 8 περιέχει ένα Τ, εκει που γίνεται η γωνία (απλώς παίρνεις τρεις κλάδους στη διασταύρωση αντί να πάρεις και τους 4). Συνεπώς, αν μπορείς να ζωγραφίσεις ένα συνεχές πλήθος από 8 στο επίπεδο, θα έχεις αυτόματα ζωγραφίσει και ένα συνεχές πλήθος από Τ στο επίπεδο. Contrapositively, αν δείξεις ότι μπορείς να ζωγραφίσεις το πολύ αριθμήσιμα Τ στο επίπεδο, θα έχεις δείξει ότι μπορείς να ζωγραφίσεις το πολύ αριθμήσιμα 8 στο επίπεδο.

Makes better sense now?

Cheerio,

Durandal 1707

Edit: Να πω και κάτι ακόμα, εξίσου τετριμμένο :P Προφανώς, δε μας νοιάζει όλο το επίπεδο, αρκεί να δούμε πόσα Τ/8 μπορούμε να χωρέσουμε σε ένα τετράγωνο/κύκλο ή άλλο πεπερασμένο χωρίο :)

[quote="filosofimenos"]Βρε παιδια, τι εγινε εδω :huuh: . Αφου το επιπεδο ειναι απεπεραστο δεν μπορουμε να ζωγραφισουμε οσα Τ και οσα οκταρια θελουμε :?:[/quote]

Ναι, άπειρα μεν... άλλα [i]αριθμήσιμα[/i] άπειρα ($ \aleph_0$, δηλ. σε 1-1 αντιστοιχία με τους φυσικούς) ή [i]συνεχώς[/i] άπειρα ($ \mathfrak{c}$, σε 1-1 αντιστοιχία με τους πραγματικούς)?

Cheerio,

Durandal 1707

Αριθμήσιμα άπειρα. Βοηθάει όπως είπε ο Παναγιώτης να σκεφτείτε τί γίνεται μέσα σε ένα τετράγωνο. Αυτά προς το παρόν...

[quote="silouan"]Για τα οκτάρια παρατηρείστε ότι σε κάθε λωβό υπάρχει ένα ζεύγος με ρητές συντεταγμένες (συνεπώς τα οκτάρια είναι το πολύ αριθμήσιμα ) . Σκεφτείτε κάτι παρόμοιο και για τα Τ  :wink:[/quote]

Edited γιατί δεν έβγαζε νόημα αυτό που είχα γράψει: γιατί είναι αυτό που λές προφανές? Σίγουρα δεν ισχύει π.χ. για τους κύκλους (ο κύκλος $ x^2 \plus{} y^2 \equal{} e^2$ δεν έχει κάποιο ρητό σημείο), οπότε γιατί είναι προφανές ότι ισχύει για τα 8?

Cheerio,

Durandal 1707

Παναγιώτη , εννοώ στο εσωτερικό κάθε λωβού (για παράδειγμα τα σημεία Α,Β του σχήματος ) (τότε παρατηρείστε ότι δύο οκτάρια δεν μπορούν να μοιράζονται τα ίδια ζεύγη ρητών σημείων  :wink:

Ααααα, τώρα μάλιστα!! Εγώ είχα λύσει μόνο το πρόβλημα με τα Τ και ούτε καν είχα κοιτάξει τα 8 μόνα τους (επειδή δε χρειάστηκε :P), γι'αυτό σ'έχασα στο πρώτο post :)

Well done, quick'n'cute! Have an icecream :icecream:

Cheerio,

Durandal 1707

Edit: Βέβαια, strictly speaking, επικαλείσαι εδώ το θεώρημα του Jordan που λέει ότι κάθε απλή κλειστή καμπύλη κόβει το επίπεδο σε δύο ξένα χωρία, ένα εσωτερικό κι ένα εξωτερικό... αλλά χαλάλι σου :D

(Προφανώς αστειεύομαι :P)

[quote="silouan"]Για τα οκτάρια παρατηρείστε ότι σε κάθε λωβό υπάρχει ένα ζεύγος με ρητές συντεταγμένες (συνεπώς τα οκτάρια είναι το πολύ αριθμήσιμα ).[/quote]

Nice. Τι γίνεται, τώρα, αν επιτρέπεται να βάζουμε ένα οκτάρι μέσα σε άλλο, αλλά μόνο στο εσωτερικό ενός λωβού; :) (δηλαδή, κάθε λωβός οκταριού μπορεί να έχει μικρότερα ολόκληρα οκτάρια μέσα του).

[quote="silouan"]Σκεφτείτε κάτι παρόμοιο και για τα Τ  :wink:[/quote]

Η λύση (τουλάχιστον η δική μου) για το Τ είναι κάπως διαφορετική, και πολύ πιο involved από τη λύση για τα οκτάρια. Πώς ακριβώς εννοείς "παρόμοιο"; Μην ξεχνάς ότι μπορεί να είναι περιστραμμένα τα σχήματα.

[quote="Durandal"][quote="cmad"][quote="silouan"]Για τα οκτάρια παρατηρείστε ότι σε κάθε λωβό υπάρχει ένα ζεύγος με ρητές συντεταγμένες (συνεπώς τα οκτάρια είναι το πολύ αριθμήσιμα ).[/quote]
Nice. Τι γίνεται, τώρα, αν επιτρέπεται να βάζουμε ένα οκτάρι μέσα σε άλλο, αλλά μόνο στο εσωτερικό ενός λωβού; :) (δηλαδή, κάθε λωβός οκταριού μπορεί να έχει μικρότερα ολόκληρα οκτάρια μέσα του).
[/quote]

Δεν αλλάζει κάτι στη λύση του Sil... το θέμα είναι ότι δεν μπορούν δύο 8 να αντιστοιχισθούν στο ίδιο [i]ζεύγος[/i] ρητών... και το $ \mathbb{Q}^2$ είναι επίσης αριθμήσιμο ;)

Cheerio,

Durandal 1707[/quote]

Πώς αποδεχόμαστε a priori ότι επιλέγουμε διαφορετικούς ρητούς για τα οκτάρια, εφόσον υπάρχει η πιθανότητα επικάλυψης;

[quote="cmad"]Πώς αποδεχόμαστε a priori ότι επιλέγουμε διαφορετικούς ρητούς για τα οκτάρια, εφόσον υπάρχει η πιθανότητα επικάλυψης;[/quote]

Η έλλειψη μαυροπίνακα δε βοηθάει :(

Επίσης, πριν όταν έλεγα "ζεύγος ρητών" εννοούσα "ζεύγος σημείων με ρητές συντ/νες". Sorry αν αυτό προκάλεσε σύγχυση...

Πάρε ένα 8, το αντιστοιχείς σε δύο ρητά σημεία, ένα σε κάθε διαφορετικό λοβό. Ok so far.

Πάρε οποιοδήποτε άλλο 8. Αφού δεν μπορούν να τέμνονται υπάρχουν οι εξής πιθανότητες.
i. το ένα 8 είναι μέσα σε ένα λοβό του άλλου 8
ii. κανένα 8 δεν είναι μέσα σε λοβό του άλλου.
(εδώ είναι που λέω ότι ο Sil επικαλείται πρακτικά το θεώρημα Jordan)

Σε κάθε περίπτωση, το ζεύγος των ρητών σημείων στο οποίο έχει αντιστοιχηθεί το 8 θα διαφέρει από το ζεύγος του άλλου 8 σε τουλάχιστον μία συντεταγμένη.

Ελπίζω να μη χάνω κάτι...

Cheerio,

Durandal 1707

PS: Η λύση για τα Τ που έχω δεν είναι πολύ πιο δύσκολη και με tweaking μου φαίνεται ότι δουλεύει για οποιοδήποτε σχήμα ομοιομορφικό με Τ. Αλλά μπορεί και να έχω κάνει πατάτα, ιδίως στα ομοιομορφικά...

Ώπα, my bad, I misread το ποστ του sil! Νόμιζα διάλεγε [b]ένα[/b] σημείο (ένα ζεύγος ρητών, που είπες και συ και μπερδεύτηκα :lol: )--το οποίο δεν είναι αρκετό από μόνο του. Good job :)

Η ιδέα της λύσης για τα Τ:

Ας είναι Α_ν το σύνολο όλων των Τ για τα οποία όλες οι ευθείες που το σχηματίζουν έχουν μήκος τουλάχιστον 1/ν. Αν όλα τα Α_ν είναι αριθμήσιμα, τότε τελειώσαμε. Μπορούμε να υποθέσουμε ότι υπάρχει ν ώστε Α_ν είναι υπεραριθμήσιμο. Χωρίζω το επίπεδο σε τετραγωνάκια μήκους 1/100ν. (Το 100 εδώ είναι κάπως αυθαίρετο.) Πρέπει να υπάρχει τέτοιο τετραγωνάκι που να περιέχει υπεραριθμήσιμα "κέντρα" των Τ που ανήκουν στο Α_ν, άτοπο (με γεωμετρία βρίσκω ότι μπορεί να περιέχει το πολύ δύο.)

[quote="Durandal"]Η δική μου ιδέα: πάρτε το ρόμβο (ή κάποιο άλλο κατάλληλο τετράπλευρο) που σχηματίζεται από τις κορυφές του Τ (ως κορυφές) και τη διασταύρωση του Τ (ως κέντρο του ρόμβου). Μετά εφαρμόστε την ίδια λογική όπως πριν για να αντιστοιχίσετε σε κάθε Τ 3 ρητά σημεία, όπως ο Sil αντιστοίχησε 2 ρητά σε κάθε 8.

Ελπίζω να μην το έχω χάσει κάπου...

Cheerio,

Durandal 1707[/quote]

Πολύ ωραία.

[quote="Durandal"]Thanks Dem, likewise :) Ωστόσο πιστεύω ότι η δική σου λύση είναι πιο κατάλληλη για την παρακάτω γενίκευση του προβλήματος:

Ας λέμε ότι ένα υποσύνολο του $ \mathbb{R}^2$ είναι T-shaped όταν είναι ομοιομορφικό με το σχήμα Τ. Μπορεί να κατασκευασθεί ένα μη αριθμήσιμο πλήθος από ξένα μεταξύ τους T-shaped υποσύνολα του $ \mathbb{R}^2$?

Εδώ χρειάζεται λίγη περισσότερη προσοχή επειδή τα σχήματα μπορούν να γίνουν πολύ άσχημα (τα 8 δεν έχουν αυτό το πρόβλημα... γιατί?). Τι λέτε, συνεχίζει να ισχύει το ίδιο αποτέλεσμα?

Cheerio,

Durandal 1707[/quote]

Πιστεύω πως ισχύει αλλά ακόμα δεν κατάφερα να το αποδείξω. Στην δική μου λύση σίγουρα δεν υπάρχει το φράγμα των δύο Τ ανά τετραγωνάκι πλέον. Δεν υπάρχει καν πεπερασμένο φράγμα. Δεν μπορώ να σχεδιάσω, αλλά αν όλα τα "Τ" είναι τρεις ευθείες που τέμνονται σε ένα σημείο με πολύ μικρές γωνίες τότε μπορώ να έχω όσα Τ θέλω σε αυτό το τετραγωνάκι. (Αλλά μάλλον αριθμήσιμα.)

[quote="Demetres"]Πιστεύω πως ισχύει αλλά ακόμα δεν κατάφερα να το αποδείξω. Στην δική μου λύση σίγουρα δεν υπάρχει το φράγμα των δύο Τ ανά τετραγωνάκι πλέον. Δεν υπάρχει καν πεπερασμένο φράγμα. Δεν μπορώ να σχεδιάσω, αλλά αν όλα τα "Τ" είναι τρεις ευθείες που τέμνονται σε ένα σημείο με πολύ μικρές γωνίες τότε μπορώ να έχω όσα Τ θέλω σε αυτό το τετραγωνάκι. (Αλλά μάλλον αριθμήσιμα.)[/quote]

Ναι... ούτε εγώ το έχω καταφέρει τελικά. Αυτό που είχα στο μυαλό μου πήγε περίπατο γιατί κοντά στη διασταύρωση μπορεί να έχουμε συμπεριφορές όπως της $ x \sin\frac{1}{x}$. Μάλλον εδώ είναι που θα γίνονται πιο involved τα πράγματα που έλεγε και ο cmad...

Grrr... δεν έχω άλλο χρόνο όμως :(

Cheerio,

Durandal 1707

[quote="Demetres"]Νομίζω το βρήκα. Τελικά η ιδέα του Σιλουανού είναι που γενικεύεται.

Για κάθε τέτοιο Τ, παίρνουμε ένα σημείο με ρητές συνταταγμένες πολύ κοντά στο σημείο τομής, ένα κύκλο με πολύ μικρή ρητή ακτίνα γύρω από αυτό το σημείο έτσι ώστε οι τρείς "ευθείες" του Τ να τέμνουν τον κύκλο σε ακριβώς ένα σημείο η κάθε μία. Επιπλέον, παίρνουμε τρία σημεία στο εσωτερικό του κύκλου με ρητές συντεταγμένες, ένα σε κάθε χωρίο που σχηματίζεται από τον κύκλο και το Τ.

Μένει να δειχθεί ότι κάθε τέτοιος κύκλος με τα τρία σημεία στο εσωτερικό του αντιστοιχεί σε ένα μοναδικό Τ. Φαίνεται εύκολο αλλά για αυστηρή απόδειξη μάλλον χρειάζεται Jordan curve ή κάτι παρόμοιο.[/quote]

Dem, αυτή ακριβώς ήταν η ιδέα της κατασκευής μου, αλλά δεν έχω πεισθεί ότι δουλεύει όταν οι κλάδοι των Τ είναι τόσο άσχημοι $ x \sin\frac{1}{x}$... namely δηλαδή ότι μπορούμε να εξασφαλίσουμε ότι οι κύκλοι δύο Τ δεν αλληλοεπικαλύπτονται και στις τρεις περιοχές.

Granted, αν έχουμε κάποιο scale factor, μπορούμε να διαλέξουμε τους κύκλους "αρκετά μικρούς" όπως λες... αλλά αν όχι, δεν είμαι σίγουρος.

Δεν το έχω σκεφτεί διεξοδικά πάντως οπότε ίσως έχεις δίκιο...

Πίσω στο γράψιμο του paper Σάββατο βράδυ, grrrrrrrr :furious:

Cheerio,

Durandal 1707

Να πω την αλήθεια όταν την έγραφα δεν είχα ελέγξει όλες τις λεπτομέρειες. Βάζω λοιπόν τώρα λίγο περισσότερες.

Για κάθε Τ ας είναι χ το κέντρο του και α,β,γ τρία οποιαδήποτε σημεία πάνω στις 3 "ευθείες". Υπάρχει δ ρητός ώστε τα α,β,γ απέχουν τουλάχιστον δ από το χ. Παίρνω σημείο με ρητές συνταταγμένες σε απόσταση το πολύ δ/3 από το χ, και κύκλο με ακτίνα δ/2 γύρω από αυτό. Το χ βρίσκεται μέσα στον κύκλο αλλά τα α,β,γ έξω από αυτόν άρα η κάθε "ευθεία" τέμνει τον κύκλο σε τουλάχιστον ένα σημείο. Αν τον τέμνει σε πολλά σημεία παίρνω το infimum. (Εδώ εννοώ αν η ευθεία είναι η $ f: [0,1] \to X$, να πάρω το infimum όλων των $ t \in [0,1]$ ώστε $ f(t)$ ανήκει στον κύκλο. Από συμπάγεια, αν $ s$ το infimum τότε $ f(s)$ ανήκει στον κύκλο άρα $ s \neq x$. Τώρα μικραίνω το Τ ώστε οι "ευθείες" να τέμνουν τον κύκλο ακριβώς μία φορά και συνεχίζω όπως προηγουμένως.

Κάθε κύκλος με τα τρία εσωτερικά ρητά σημεία (όπως τα διάλεξα προηγουμένως) αντιστοιχεί σε ακριβώς ένα Τ: Αν αντιστοιχούσε σε δύο το κέντρο του ενός θα ήταν μέσα σε ένα από τα τρια χωρία που δημιουργεί το άλλο, αλλά για να διαχωρίσει τα τρία σημεία θα έπρεπε να τεμνει το άλλο Τ.

[quote="Demetres"]Να πω την αλήθεια όταν την έγραφα δεν είχα ελέγξει όλες τις λεπτομέρειες. Βάζω λοιπόν τώρα λίγο περισσότερες.[/quote]

Yeah, I think this works... nicely done :)

[quote]Εδώ εννοώ αν η ευθεία είναι η $ f: [0,1] \to X$, να πάρω το infimum όλων των $ t \in [0,1]$ ώστε $ f(t)$ ανήκει στον κύκλο. Από συμπάγεια, αν $ s$ το infimum τότε $ f(s)$ ανήκει στον κύκλο άρα $ s \neq x$. Τώρα μικραίνω το Τ ώστε οι "ευθείες" να τέμνουν τον κύκλο ακριβώς μία φορά και συνεχίζω όπως προηγουμένως.[/quote]

Heh :) Για όσους ενδιαφέρονται, να πω ότι αυτή είναι πολύ σημαντική ιδέα... στη στοχαστική ανάλυση, αυτοί οι "χρόνοι" λέγονται hitting times και περιγράφουν για παράδειγμα πότε ένα random walk ή μία κίνηση Brown θα φτάσουν σε μία δεδομένη απόσταση από το σημείο που ξεκίνησαν. Υπάρχει μία τρομερή κατασκευή με την οποία μπορείς να λύσεις το πρόβλημα Dirichlet (λ.χ. εξίσωση Laplace με συνοριακές συνθήκες) χρησιμοποιώντας αυτούς τους χρόνους... αν βρω χρόνο θα δω μήπως μπορέσω να φτιάξω ένα πρόβλημα που να δείχνει αυτήν την τεχνική (αν και παραθέλει πολλά tools :( ).

Cheerio,

Durandal 1707


(F.Nilov) Given isosceles triangle $ ABC$ with base $ AC$ and $ \angle B \equal{} \alpha$. The arc $ AC$ constructed outside the triangle has angular measure equal to $ \beta$. Two lines passing through $ B$ divide the segment and the arc $ AC$ into three equal parts. Find the ratio $ \alpha / \beta$.


Let $ k, m$ and $ n$ be integers so that $ 1 < n \le m\minus{}1 \le k$.
Determine maximum number of elements of subset $ S$ of set $ \{ 1,2,...,k \}$ so that sum of any n different elements of $ S$:
a) is not equal to $ m$,
b) does not exceed $ m$.

Anybody solved this "attractive" problem?


calculus
derivative
real analysis
real analysis unsolved

How would I show that if f is differentiable and |f'(x)|<=M, then f is uniformly continuous?

Use the Mean Value theorem to prove that $ |f(x) \minus{} f(y)|\le M|x \minus{} y|.$

Bounded derivative $ \implies$ Lipschitz $ \implies$ uniformly continuous $ \implies$ continuous.

The converses of all of those implications are false.


Solve this system of equalities in integers:
$x(y+z+1)=y^2+z^2-5$
$y(z+x+1)=z^2+x^2-5$
$z(x+y+1)=x^2+y^2-5$

You can find problems from this year Slovak math olympiad in
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22376]national olympiad's[/url] section.

Here's my solution

[hide]Adding the 3 equations together we get
2(xy+yz+zx)+(x+y+z)=2(x^2+y^2+z^2)-15
Adding (x^2+y^2+z^2) to both sides we get
(x+y+z)^2+(x+y+z)=3(x^2+y^2+z^2-5)
or (x+y+z)(x+y+z+1)=3(x^2+y^2+z^2-5).
Next substitute u=x+y+z+1 and v=x^2+y^2+z^2-5 then the equation above becomes
1/3u(u-1)=v.  From the first equation we have x(u-x)=v-x^2 or ux=v=1/3u(u-1) which simplifies to 3x=u-1=x+y+z.  Similarly we can get 3y=x+y+z and3z=x+y+z which implies that x=y=z.  insert into one of the eqautions to get x=y=z=-5[/hide]

First mistake (or let's call it wrong argumentation) in your proof is in sentence beginning "From the first equation we have" in 7th line of your solution. Can you see that mistake? Maybe then you could fix your solution. Or can somebody else write complete solution?
There are certainly more then 1 solution (exactly 7 solutions :) ).

And don't forget: you're solving in integers!

fix to the solution in spoiler:
[hide]
he forgot the case u = 0, from which we get v = 0 or

$x^2+y^2+z^2 = 5$ from which we get the solution (1, -2, 0) and all of its permutations.
[/hide]

Yes, [b]zschool[/b], your're right.

This is solution I've made on olympiad:
Let's call that equations (1),(2) and (3).
Then:
$(1)-(2)$ is $(x-y)(x+y+z+1)=0$
$(2)-(3)$ is $(y-z)(x+y+z+1)=0$
$(3)-(1)$ is $(z-x)(x+y+z+1)=0$
There are 2 cases:

[b]first case:[/b] $x+y+z+1$ is not 0.
Then we easily see that there's only one possibility: x=y=z. Using this in (1) we will get $x=y=z=-5$

[b]second case:[/b] $x+y+z+1=0$.
Now we will make small modification in (1):
$y^2+z^2-5=x(y+z+1)=x(x+y+z+1-x)=-x^2$
So: $x^2+y^2+z^2=5$
Equations are symmetric, so we can deduce that only possible solutions are:
$(2,1,0),(2,-1,0),(-2,1,0),(-2,-1,0)$ and their permutations.
But only for $(-2,1,0)$ holds $x+y+z+1=0$
This solution and permutations are succesful.


:? I am interested when will be next contest and how I can participate in it, also I want to know whther there are some interesting online mathematical olympiads like it is in Informatics, I'll be very thanksfull if you answer me,

thanks a lot, looking for the reply...

Read about it on the [url=http://www.mathlinks.ro/Forum/contest.php]Contest[/url] page. It is all explained there.

As far as I know MathLinks Contest is the only regularly held Online Math Contest at the IMO level in the world. There are a few others, like USAMTS held by Art of Problem Solving Foundation, but their level is around pre-olympiad.

[quote="silouan"][quote="Valentin Vornicu"]As far as I know MathLinks Contest is the only regularly held Online Math Contest at the IMO level in the world..[/quote]

Surely. The most problems are more difficult than the IMO  ;)[/quote]
I am not think so :( ,becaus if that's true,then
I must take [b]GOLD[/b] Medal in Slovenia :flex:  :gathering:

[quote="silouan"]Sorry Tiks but you forget something .It is not the same to solve a problem in your house with all the time yours and solving a problem in competition where you have both stress and the time against you. ;)[/quote]
Yes,you rigth Silouan :(.

Is the mathlinks olympiad harder than IMO is???


Show that if $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt{\frac {2c}{c \plus{} a}} \ge 3\left(\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\right)^{\frac {5}{12}}.$

[quote="can_hang2007"]Show that if $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt {\frac {2c}{c \plus{} a}} \ge 3\left(\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\right)^{\frac {5}{12}}.$[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-110-v-q-b-can.html


I have a pretty solid foundation in algebra problems but i am trying to cover all the textbooks and hopefully classes, if i need them. I was wondering what are some of the more challenging AMC problems that would be covered in Intro to Algebra?

Would skipping right to Intermediate Algebra be a bad move?

I recommend using the diagnostic tests for the Algebra 1, 2, and 3 classes (you can find these in the Online Classes section of the site) to help you decide which book would be right for you.  The Algebra 1 and 2 classes together cover the Intro Algebra book, and the Algebra 3 class covers the Interm Algebra book.


integration
calculus
Pythagorean Theorem
geometry
calculus computations

I've never been shown an example with an x^2 or at points, how do i do a question like this?

Find the length of the arc formed by:
x^2 = 10y^3
from point A to point B, where A is (0,0) and B is (100,10)

Thanks so much for help =)

For $ x\geq 0$, the arc length is $ \int_0^{10} \sqrt{1\plus{}\left(\frac{dx}{dy}\right)^2}dy$

If you want to know why that theorem works, this is why:

First you integrate the pythagorean theorem
This sums the linear distance over small intervals to get the arc length.

$ \int \sqrt{(dx)^2 \plus{} (dy)^2} \equal{} \int \sqrt{1 \plus{} \left(\frac{dy}{dx}\right)^2}dx \equal{} \int \sqrt{1 \plus{} \left(\frac{dx}{dy}\right)^2}dy$

If you choose the $ dx$ integral, then your bounds must be $ x$ values

If you choose the $ dy$ integral, then you must use the $ y$ values for bounds

I find that the nicest version is the parametric form $ \int_a^b \sqrt{(x')^2+(y')^2}\,dt$. The two cases chickendude mentioned are the parametrizations $ (t,f(t))$ and $ (g(t),t)$, where $ f$ and $ g$ are inverses.

Three parametrizations of this curve, and the integrals they lead to:
$ (x,(\frac{x^2}{10})^{\frac13}))$ and $ \int_0^{100}\sqrt{1+(\frac23\cdot 10^{-\frac13}x^{-\frac13})^2}\,dx$
$ (\sqrt{10}y^{\frac32},y)$ and $ \int_0^{10}\sqrt{(\sqrt{10}\cdot\frac32y^{\frac12})^2+1}\,dy$
$ (100t^3,10t^2)$ and $ \int_0^1 \sqrt{(300t^2)^2+(20t)^2}\,dt$

Any of these is doable; this particular curve works well with arclength. The second form is probably the easiest to work with, and the first is the hardest.


inequalities
Cauchy Inequality
inequalities unsolved

[quote="pvthuan"]1. Let $x,y,z$ be nonegative real numbers such that $x^{2}+y^{2}+z^{2}=1$, prove that

$\frac{x^{2}}{1+x-y}+\frac{y^{2}}{1+y-z}+\frac{z^{2}}{1+z-x}\geq1.$
[/quote]
It is not true: $z=0,$ $x=y=\frac{1}{\sqrt2}.$

I think solution of bel.jad5 very nice, through it and by example of arqady this problem needs $x,y,z>0$(not equal)   :)

[b]gemath[/b]: Still if $z$ very small and $x, y$ close to $\frac{1}{\sqrt{2}}$ will cause the same problem.

edit: in fact, $y=z=0, x=1$ suggests the RHS should be $\frac{1}{2}$. Can it be proven? :)

ok you are right Xevarion thus there is a mistake on proof of bel.jad5 but I can't see it can you show me?  :maybe:  :maybe:

The first has two equality cases. Can anyone check with Maple if the reverse of the second one is true?

we can also use cauchy inequality , S=x²/(1-x+y)+y²/(1-y+z)+z²/(1-z+x)
S(x²(1-x+y)+y²(1-y+z)+z²(1-z+x))>=(x²+y²+z²)²
S(1-(x^3+y^3+z^3)+x²y+y²z+z²x)>=1
S>=1

equality when 1-x+y=1-y+z=1-z+x-->y-x=z-y=x-z-->x=y=z

I think there's no mstake in my solution  :!:  :(

ok ok sorry bel.jad5  :D I read wrong the first is true and your solution is very good but with your solution needs $x,y,x>0$ but the second is really false,sorry again!  :)

sorry it isn't second solution of you  :D , I said about second problem of pvthuan by example of Xevarion ok?! Your second solution is true,too! Thank you for your good solutions  :)

[quote="jef"]i think both solutions are correct, but only with the condition $1-x+y$, $1-y+z$ and $1-z+x > 0$[/quote]

it s easy to prove that 1-x+y >0 ( because x<=1  so 1-x+y>=0 and the sum is not defined when 1-x+y=0 )


trigonometry
geometry
circumcircle
geometry unsolved

[img]http://pic40.picturetrail.com/VOL298/8929090/16470363/255041532.jpg[/img]
let A, B, C are the sides of the triange with the opposite angle  a, b, c. if a=3b,
prove that
( A + B )( A - B )² = BC²
please help me prove, thanks a lot.

[quote="yrhc@eht"]let $a,b,c$ are the sides of the triange with the opposite angle $A,B,C$.
if $A=3B$, prove that
$( a+b )( a-b )^{2}= bc^{2}$ [/quote]
[hide="trigonometric solution"]
$\sin C = \sin (180-4B) = \sin (4B)$
Suppose that the circumradius of $\triangle ABC$ is $R=\frac{1}{2}$. Then

$a = \sin (3B) = \sin B(3-4\sin^{2}B)$
$b = \sin B$
$c = \sin (4B)$

$a+b = \sin B(4-4\sin^{2}B) = 4\sin B(1-\sin^{2}B) = 4\sin B\cos^{2}B$
$a-b = \sin B(2-4\sin^{2}B) = 2\sin B(1-2\sin^{2}B) = 2\sin B\cos 2B$

$b = \sin B$
$c = \sin 4B = 2\sin 2B \cos 2B = 4\sin B \cos B \cos 2B$

$(a+b)(a-b)^{2}= 16\sin^{3}B \cos^{2}B\cos^{2}(2B)$

$bc^{2}= 16 \sin^{3}B \cos^{2}B\cos^{2}(2B)$
$\hline$[/hide]
[hide="trigo-geometric"]
Let $D\in [BC]$ such that $\angle BAD = B$. Then $\triangle DBA$ is isosceles ($DB=DA$).
But $\angle CDA = 2B = \angle CAD$, so  $\triangle CDA$ is isosceles too ($CD=CA$). So we have $BD = BC-DC = a-b$

$( a+b )( a-b )^{2}= bc^{2}\iff \left(\frac{c}{a-b}\right)^{2}= \frac{a+b}{b}$

Let $M$ be the midpoint of $AB$. From the right triangle $DBM$ we get $\cos B = \frac{BM}{BD}= \frac{1}{2}\cdot\frac{c}{a-b}$

Then $\frac{c}{a-b}= 2\cos B$ and all we have to prove is that $(2\cos B)^{2}= \frac{a+b}{b}\iff$

$4 \cos^{2}B = \frac{a+b}{b}\iff$
$2+2\cos 2 B = \frac{a+b}{b}$

From the triangle $CDA$ we get $2\cos 2B = \frac{AD}{AC}= \frac{BD}{b}= \frac{a-b}{b}$

Then $2+2\cos 2 B = 2+\frac{a-b}{b}= \frac{2b+a-b}{b}= \frac{a+b}{b}$ and we are done
$\hline$[/hide]
[hide="and the geometric one"]
Let $D$ the point as defined above. We will use it later

Let $E\in AB$ such that $\angle BCE = 2B$
The point $E$ is between $A,B$ iff $2B<C = 180-4B \iff B<30$

So, assume $B<30^\circ$. Then $\angle CEA = B+2B = \angle CAE \Rightarrow \triangle CEA$ is isosceles ($CE=CA=b$)

The parallel line to $AD$ through $E$ intersects $[BC]$ at $F$. Then of course $\angle BEF = B \Rightarrow \triangle FBE \text{ isosceles }(FB=FE)$

But $\triangle ECF$ is also isosceles because $\angle EFC = 2B = \angle FCE$.
So $FE=CE$. Finally $FB=FE=CE=CA=b$

[color=olive](We'll get the same results if we assume that $B>30$. The only difference is that we take the angles $\angle CEA = 180-3B = \angle EAC$ to prove that $\triangle CEA$ is isosceles)
On the other hand, if $B=30^\circ$ then the triangle is right at $A$ and needs no further discussion[/color]

Now draw the circle $w=(C,b)$. This circle must pass though $A,E$

The power of $B$ w.r.t. $w$ is $BA\cdot BE = a^{2}-b^{2}\ \ (1)$

From Thales in the parallel lines $FE,DA$ we get $\frac{BE}{BA}=\frac{BF}{BD}\Rightarrow \frac{BE}{c}=\frac{b}{a-b}\Rightarrow$

$BE = \frac{b}{a-b}\cdot c$ and from $(1) \Rightarrow$

$\frac{b}{a-b}\cdot c^{2}= (a+b)(a-b) \Rightarrow \boxed{bc^{2}= (a+b)(a-b)^{2}}$
[/hide]


I have a question that I can't find the exact result. Here's the question:
A steel cable carries a 1400kg load which is descending with a speed of 4m/s. It slows down uniformly and comes to a stop in 3m. Compute the tension in the cable while slowing down?
PS: I got the wrong answer, which is: 1400.8/3.Can anybody shows me why i got it wrong?

Let the $ z$ axis point upwards, with the origin at the place where the block stops. According to the problem it slows down uniformly, hence acceleration is constant. Let the acceleration be $ a$.
Then by definition:
\[ a\equal{}\dfrac{dv}{dt}\equal{}\dfrac{dv}{dz}\dfrac{dz}{dt}\equal{}v\dfrac{dv}{dz}\]
Since the velocity by definition is $ v\equal{}\frac{dz}{dt}$

Integrating both sides of the DE, se get:

\[ v(z)^2\equal{}2az\plus{}C\]

Where $ C$ is the integration constant.
Let the path it takes it stop be $ s$ i.e : $ s\equal{}3$m
So the conditions are : $ v(s)\equal{}v_0 \equal{} 4m/s$. from here the constant: $ C\equal{}v_0^2\minus{}2as$
So the velocity:
\[ v(z)\equal{}\sqrt{v_0^2\minus{}2a(s\minus{}z)}\]
but we also know that $ v(z\equal{}0)\equal{}0$, since we chose the origin to be where the block stops, hence we get:
\[ v_0^2\equal{}2as\]
That is the acceleration:
\[ a\equal{}\dfrac{v_0^2}{2s}\]

Now the tension in the cable can be calculated using newton's second law:
\[ T\minus{}mg\equal{}ma\]
Now using the expression we obtained for the acceleration, the $ T$ tension:
\[ T\equal{}m\left(\dfrac{v_0^2}{2s}\plus{}g\right)\]


Let n be a positive integer.Consider the continuous function f(x):[0,n]\rightarrow R such that f(0)=f(n). Prove that there exists n pair of real numbers \[a_i,b_i\] (i=0,1,2,...,n) there \[a_i,b_i \] belongs to [0,n] such that \[b_i-a_i\] is positive integer and \[f(a_i)=f(b_i)\] \forall i=1,2,...,n

For each $k \in \{1,2, \cdots, n \}$ consider the function $g_k(x) = f(x+k)-f(x).$
clearly $g_k$ is continuous on $[0,n-k]$ and $g_k(0)+g_k(1) + \cdots + g_k(n-k) = f(n)-f(0) = 0.$
Then, either one of these numbers is $0$ or $g_k$ takes negative values and takes positive values. In the latter, using continuity and the intermediate value theorem, it follows that $g_k$ takes value $0$.
Thus, in any case, there exists $a_k \in [0,n-k]$ such that $g_k(a_k)=0$. Now choose $b_k = a_k +k$, and we are done.

Pierre.


I apologize because I don't have latex, and typing the problem out here would be hard to understand.
http://www.princeton.edu/~ahesterb/PUMaC07Problems/FinalsB
How would I do #2 on this page?

You don't need to have $ \text{\LaTeX}$ installed, you can type $ \text{\LaTeX}$ code directly into the reply text. Look at the FAQs.

Here's the problem (slightly rephrased since, in my mind, the original presentation of the problem is awful):

Call a set $ A$ of positive integers "good" if no two distinct members of $ A$ sum to a perfect square. Namely, $ A$ is good iff for every $ a,b \in A$ with $ a \neq b$, $ a + b$ is not the square of an integer.

Prove that for every $ N \in \mathbb{N}$ we can find a good set $ A \subseteq \{1,2,\ldots,N\}$ such that $ A$ has at least $ \lfloor \frac {11}{32} N \rfloor$ elements.

[hide="Solution"]
The square of any integer has one of the following residues mod $ 32$: $ S = \{0,1,4,9,16,17,25\}$. The set of residues $ B = \{1,5,9,10,13,14,17,21,25,29,30\}$ has the property that the sum of any two members of the set (distinct or not) is not in $ S$. Therefore if $ A$ is a set of integers such that every member of $ A$ has a residue mod $ 32$ which is a member of $ B$, then $ A$ will have the desired property. So we can just take $ A$ as the set of all elements in $ \{1,\ldots, N\}$ who have one of those residues mod $ 32$, and this $ A$ has exactly $ \lfloor \frac {11}{32}N \rfloor$ elements since $ |B| = 11$.

Note, by the way, that $ A$ has the property that no two elements add to a square even if you take the same element twice.
[/hide]

[hide="How to find this solution"]

When trying to control a sum of integers so that it is not a perfect square, the easiest way is to look at the residues modulo $ 4$ or some other small power of $ 2$, since the quadratic residues are fairly sparse modulo such numbers. The fraction $ \frac {11}{32}$ in the problem is a strong hint that you should look at residues mod $ 32$.

Determining the set $ S$ of quadratic residues (those residues that are possible for integer squares) is a routine calculation. Finding a set $ B$ with the property that $ B + B$ is disjoint from $ S$ can be done by trial and error; a more orderly way is to start with a set of residues that works modulo $ 4$. The only such set is $ \{1\}$. Moving on to mod $ 8$, we can duplicate the $ 1$ to $ \{1,5\}$ since those are both $ 1$ modulo $ 4$. The quadratic residues mod $ 8$ are $ \{0,1,4\}$ so $ \{1,5\}$ works fine but adding any additional element destroys the property.

Moving on to mod $ 16$ we can start with $ \{1,5,9,13\}$ which is just that $ \{1,5\}$ we had before plus the same residues with an added $ 8$. The quadratic residues mod $ 16$ are $ \{0,1,4,9\}$, so we see that adding the residue $ 14$ to our list still keeps the sums away from being quadratic residues. Now we have a ratio of $ 5/16$ which is almost what we need. Mod $ 32$ we grow the list to $ \{1,5,9,13,14,17,22,25,29,30,33\}$ and once again we seek just one additional residue we can add without allowing a sum in $ \{0,1,4,9,16,17,25\}$. It's easy to find this extra residue, $ 21$, and we're done.
[/hide]


could any one prove "the butter fly" theorem  (if i'm right!!)   whene the line is out of the circle???
 ;)

I mean this :
  If  $px=py$  then $pm=pn$

This is so easy ? no Pooyan ? :D  :D (I can't send more)

hi mr/ms John nash
thanks for your your giude!!!
I think  you had to post a p.m to me!!! :!:


Find all the value òf x,y,z which sastify that:
\[ \begin{array}{l}
 \sqrt {x \plus{} 3}  \equal{} {y^3} \minus{} 6 \\
 \sqrt {y \plus{} 2}  \equal{} {z^3} \minus{} 25 \\
 \sqrt {z \plus{} 1}  \equal{} {x^3} \plus{} 1 \\
 \end{array}\]

hello, i have got $ x\equal{}1,y\equal{}2,z\equal{}3$.
Sonnhard.

Could ther be another triplet that works?


trigonometry
geometry
angle bisector
exterior angle
trig identities
Law of Cosines

Circle $ O$ is inscribed in triangle $ ABC$.  The circle is tangent to $ BC$ at $ A'$.  Is it possible to find $ \angle BAA'$ in terms of known angles?

thanks

[quote="jli"]Is it possible to find $ \angle BAA'$ in terms of known angles?[/quote]

And what might those known angles be?

[quote]And what might those known angles be?[/quote]

We can just substitute those known angles with variables for now.

Note that an inscribed triangle's center (in this case O) is the intersection of each angle bisector of the triangle. If we set angle OBA as [i]x[/i] and angle ACO as [i]y[/i], then we can find angle BOC by 180 - x - y. Then find that angle's exterior angle, which would be 180 + x + y. Since the quadrilateral BACO is two triangles put together the interior angles' sums would equal 360 degrees. So to find angle BAC, 360-(180+x+y)-x-y=180-2x-2y. After that we just divide that by two to get [b]90-x-y[/b].

[quote="anonymous03"]Styrofoam, you found angle BAO, but not BAA'. How many angles are known?[/quote]

just $ \angle A, B, C$

So I'm thinking A/2 but I'm pretty sure I interpreted the question wrong.

yes you did.  that would be the angle bisector

Dammit.

In light of the newly interpreted question, yes, it definitely is possible. However, the only way I see to a solution involves brute-forcing trig and I'm not willing to write it all out in terms of a real solution.

[hide]Draw a perpendicular line from O to AC and name the new point M. Set OM as 1. Express AO in terms of angle A (trig). Express angle AOC in terms of angle C and angle A. Law of Cosines to find AA', and then Law of Cosines to find angle OAA'. Add to angle BAO for the answer.

There's definitely a way to avoid finding either one of AA' or OA, but I'm too lazy to find it.[/hide]


trigonometry calculus integration AMC AIME ratio inequalities

In triangle $ABC,$ angle $C$ is a right angle and the altitude from $C$ meets $\overline{AB}$ at $D.$  The lengths of the sides of $\triangle ABC$ are integers, $BD=29^3,$ and $\cos B=m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

So far, this is what I've gotten it down to...
[hide]
Let $DC=d, AD=b, BD=c, AC=a$ obviously all the triangles are similar, so we can call the corresponding angles, the angle whos cosine we want to find, $\theta$, so $\cos{\theta}=\frac{d}{a}=\frac{29^3}{c+b}=\frac{c}{29^3}$, and we also know that $\frac{c+b}{2}=d$ and $(c+b)d=29^3 \cdot a$ and of course $29^5+a^2=(c+b)^2$, but I don't know if solving for each of these variables is the best thing to do, but it's the only thing I can think of unless someone has a better solution :? [/hide]

[hide="Solution"]
To make this very easy, set $29^3 = 29^2 (m^2 - n^2)$ by Pythagorean triple generating formulae, and we arrive at the conclusion $m = 15, n = 14$, which gives us $AB = 29^2 (15^2 + 14^2)$.  Then we have $\cos B = \frac{29}{15^2 + 14^2} = \frac{29}{225 + 196} = \frac{29}{421}$, so our answer is $29 + 421 = \boxed{450}$.
[/hide]

There does not seem to be enough information in the question - I know that there are at least two other possible sets of values for $AC$ and $AB$, and my only reason for choosing this one is that this is an AIME question and I suspect the other two sets of values produce answers $> 1000$  :D

What is the Pythagorean triple generating formula?

The Pythagorean triple generating formula is $a^{2}-b^{2},2ab,a^{2}+b^{2}$.

[hide]Consider a similar triangle $\triangle A'B'C'$ with all side lengths $\frac{1}{29^{2}}$ of the given triangle.  $B'C'=29$.  The sides of the triangle can be represented by $a^{2}-b^{2},2ab,a^{2}+b^{2}$.  Let $a^{2}-b^{2}=29$.  Then $(a-b)(a+b)=29\Rightarrow a=15, b=14$.  The hypotenuse of the triangle is equal to $a^{2}+b^{2}=15^{2}+14^{2}=421$, and $\cos B=\cos B'=\frac{29}{421}\Rightarrow m+n=450$.[/hide]

By the way, how do you know there are more solutions?

There was a typo in the problem statement, specifically $BD$ NOT $BC$ equals $29^3.$  It is now fixed.  :oops:

[quote="matt276eagles"]
By the way, how do you know there are more solutions?[/quote]

Well, we can take $29^3 = (m+n)(m-n)$ and we can use $m+n = 29^3, m-n = 1$ to get us a different triangle than the one we used, and we can use $29^3 = 29(m+n)(m-n)$ as well.

[hide="Solution given BD=29^3"]
We have $\triangle BCD %Error. "similar" is a bad command.
\triangle ABC$ and $ABC$ must have integer sides.  Call the ratio between their sides $r$.  Then $AC = r (DC), BC = r (BD) = 29^3 r, AB = r (BC) = r^2 29^3$.  Then $r = \frac{k}{29}$ for some integer $k$.

So far, we know $AB = 29k^2, BC = 29^2 k$, and so $\cos B = \frac{BC}{AB} = \frac{29}{k} = \frac{1}{r}$.

Additionally, we have that $BC^2 + AC^2 = AB^2 \implies (29^2 k)^2 + AC^2 = (29 k^2)^2$, so $(29k) | AC$ in order that $AC$ remain an integer.  We write $AC = 29kn$ for some integer $n$, and divide to obtain $29^2 + n^2 = k^2 \implies 29^2 = (k+n)(k-n)$, which only has positive integer solutions for $k+n = 29^2, k-n = 1 \implies 2k = 842 \implies k = 421$.

This gives us $\cos B = \frac{29}{421}$ and our answer $29 + 421 = \boxed{450}$.  Still.   :D
[/hide]

[hide="Answer"]$\frac{29^3}{BC}=\frac{BC}{29^3+AD}\Rightarrow BC^2=29^3(AB)$
Since all sides are integers, we have that $AB$ contains at least one factor of $29$ and $BC$ at least two. However, $AB>BC>29^3$ by triangle inequality, and $AB^2-BC^2$ is a perfect square. If $BC=29^2\cdot k$, then $AB=29\cdot k^2$, and $29^2\cdot k^4-29^4\cdot k^2=(29k)^2(k^2-29^2)$ is a perfect square. Therefore, $k^2-29^2$ is a perfect square, and the smallest possible value for $k$ is therefore $k=421$ by Pythagorean Theorem. Therefore, $BC=29^2(421)$, and $\cos B=\frac{29^3}{BC}=\frac{29}{421}\Rightarrow 29+421=\boxed{450}$.[/hide]

[b]Solution:[/b] Using similar triangles and letting the altitude be $h$, we have that $BC = \sqrt{h^2 + 29^6}$. Since this is an integer, we see that the only possible value of $h$ (after doing some algebra) is $\frac{29^4 - 29^2}{2}$. Noting that $\cos B = \frac{29^3}{\sqrt{h^2 + 29^6}}$, we have our final answer of $\frac{29}{421} \Rightarrow \boxed{450}$.

[i]Comments:[/i] When you see an altitude in a right triangle, your brain should scream similar triangles. In this case, it kills the problem. Another helpful thing to know is the "difference of squares" trick for calculating possible perfect squares. In this case, we set $sqrt{h^2 + 29^6} = k$, which gave us $(k+h)(k-h) = 29^6$, and we could proceed from there.


HAPPY MAY!!!

Who's birthday is in May? Mine is.

*sobs uncontrollably*
mine isn't

happy bday whenever in may yours is

happy birthday whenever in may yours may be


Let $a,b,c>0$ such that $a^2+b^2+c^2=1$, and let
$A=\frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca},$
$B=\frac1{1-\bigl(\frac{a+b}2\bigl)^2}+ \frac1{1-\bigl(\frac{b+c}2\bigl)^2}+ \frac1{1-\bigl(\frac{c+a}2\bigl)^2},$
$C=\frac1{1-a^2}+\frac1{1-b^2}+\frac1{1-c^2}.$
It is well known that $A\le B\le\frac92$ and $C\ge\frac92$.

Prove that
(1)	$A+C\ge9$,
(2)	$7B+3C\ge45$.
(I think both are the sharpest possible) :)

I had almost similar questions.

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%5C%5Cfrac1%7B1-ab%7D&

Yeah, yours are $3A+5C\ge36$ (which is a bit weaker than (1)) and $5A+3B\le36$ which is a trivial consequence of the (highly non-trivial :) ) $B\le\frac92$.

Did you solve $A+C\geq9$, spenferkel. In my final step I used $p,q,r$ substitution and reduced it to proving that \[1+6pr\geq q+2q^{2},\] where $p=a+b+c$, $q=ab+bc+ca$, and $r=abc$.

[quote="pvthuan"]Did you solve $A+C\geq9$ ?[/quote]No, both are still unsolved :maybe: unless somebody can prove $1+6pr\geq q+2q^{2}$.

Long time ago,I wonder if my solution is valid.
Thuan,spanferkel,Have you check it? :)

Yes it seems OK to me. Nice idea to bring it into a Vornicu-Schur form. :P I wouldn't have expected Vornicu-Schur to be strong enough for that.

My second one must be much harder though. :maybe:

[quote="spanferkel"]
My second one must be much harder though. :maybe:[/quote]

The second is harder since at least Cauchy failed. Before coming to $1+6pr\geq q+2q^{2}$, we had used Cauchy in Engel form.

[quote="can_hang2007"][quote="spanferkel"]Let $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, and let
$ A \equal{} \frac1{1 \minus{} ab} \plus{} \frac1{1 \minus{} bc} \plus{} \frac1{1 \minus{} ca},$
$ B \equal{} \frac1{1 \minus{} \bigl(\frac {a \plus{} b}2\bigl)^2} \plus{} \frac1{1 \minus{} \bigl(\frac {b \plus{} c}2\bigl)^2} \plus{} \frac1{1 \minus{} \bigl(\frac {c \plus{} a}2\bigl)^2},$
$ C \equal{} \frac1{1 \minus{} a^2} \plus{} \frac1{1 \minus{} b^2} \plus{} \frac1{1 \minus{} c^2}.$
It is well known that $ A\le B\le\frac92$ and $ C\ge\frac92$.

Prove that
(1)	$ A \plus{} C\ge9$,
(2)	$ 7B \plus{} 3C\ge45$.
(I think both are the sharpest possible) :)[/quote]
(1) is easy (it can be proved by Cauchy-Schwarz), (2) is harder. My solution does not use any big calculations but it is rather long. See it here: http://can-hang2007.blogspot.com/2010/01/inequality-64-w-berndt.html[/quote]
Yes , we can prove (1) easily by Cauchy-Schwarz my brother.

[quote="can_hang2007"]
(1) is easy (it can be proved by Cauchy-Schwarz), (2) is harder. My solution does not use any big calculations but it is rather long. See it here: http://can-hang2007.blogspot.com/2010/01/inequality-64-w-berndt.html[/quote]
Thank you so much, dear Can_Hang. I wouldn't expect an elegant proof for such a hard inequality, but yours is at least creative - as usual!  :thumbup:

For (1), see here: http://can-hang2007.blogspot.com/2010/01/inequality-68-w-berndt.html


analytic geometry
geometry theorems
geometry

How to identify the infinity point by the barycentric equation???
And what is the Conway's formula ?

The line $ px \plus{} qy \plus{} rz \equal{} 0$ has infinite point $ P_{\infty}(q \minus{} r: r \minus{} p: p \minus{} q).$

The Conway formula gives the barycentric coordinates of a point $ P,$ for instance, defined by the balanced angles $\angle CBP \equal{} \phi,$ $ \angle BCP \equal{} \theta$ $\Longrightarrow$ $ P \equiv ( \minus{} a^2: S_C \plus{} S_{\theta}: S_B \plus{} S_{\phi}).$

Keeping in mind the general Conway notation $ S_{\theta} \equal{} S \cot \theta.$


Hello.

I am learning methods of numerically integrating of stiff differential systems.

Now I need example of second order stiff differential equation.

It can looks as

y''+p(x)y'+q(x)y=f(x)   .

I tried made the example by myself, But difficult functions p(x), q(x), f(x) was recieved.

Can anyone help me?

[quote="treething"]wrong forum, getting started is for problems such as "what is 35 * 29? show work", there's a calculus forum designed for calculus problems[/quote]

Getting Started is NOT for what is 29*35.

The LaTeX forum is, you would have a hard time writing your work in LaTeX.

Getting Started is for problems like:

There is a right triangle ABC where AC is the hypotenuse.  When a line is drawn from C until it hits AB, it splits AB into two portions, AD and DB, where AD=10 and DB is 6.

Meanwhile, another similar line is drawn from A to BC, where the point where the two lines intersect is labeled E, BE=8 and EC=AD.

The intersection of AE and CD is labeled F, find the area of quadrilateral  BDFE.

Could anybody actually solve that?   :rotfl:


parameterization
calculus
trigonometry
analytic geometry
geometry
quadratics
algebra

The strength $ s$ of a rectangular beam varies as the product of the width $ w$ and the square of the depth $ d$ of the beam , i.e, $ s=kwd^{2}$ for some constant k. Find the dimensions of the strongest rectangular beam which can be cut from a cylindrical log iwth diameter $ 48cm$

[url=http://img518.imageshack.us/my.php?image=17953275bj5.png][img]http://img518.imageshack.us/img518/8090/17953275bj5.th.png[/img][/url]

anyone?                                                                     .

This is a max-min problem with two variables ($ d$ and $ w$) and one constraint (another equation which $ d$ and $ w$ must satisfy). Constrained max-min problems can be solved in one of three ways:

(1) By solving the constraint equation for one variable in terms of the other, and substituting that into the expression that is to be optimized. Then you have a one-variable optimization problem, and you can proceed in a familiar way.

(2) Replacing the constraint equation by an equivalent parametric equation. This leads to a one-variable optimization problem, with the parameter as the variable.

(3) By the technique of Lagrange multipliers.

If you know Lagrange mulitpliers, I highly recommend the technique - it often provides the answers in the most elegant form. But most of the good problems (such as this one) are placed into calculus books many chapters before the Lagrange technique.

Your first step: what is the constraint equation? That is, what else must $ w$ and $ d$ satisfy?

Alternate version of the first step, aiming towards (2) above: can you express all of the pertinent points or distances in terms of some angle $ \theta,$ using the definitions of sine and cosine?

ok thanks. I think the last method (lagrange multipliers) is a university technique. I looked in some books but I find it difficult to understand the notations etc.

;) i got it using technique 1...I'll try technique 2 as well

thanks for the help Prof.

-----------
[hide] The constraint eqn $ d$ and $ w$  is from the diagram.
$ d^{2}+w^{2}=2304$ and its given $ s=kwd^{2}$

$ d^{2}=48^{2}-w^{2}$

$ s=kw(48^{2}-w^{2})$

$ \frac{ds}{dw}=k(48^{2}-w^{2})-2kw^{2}$

$ \frac{ds}{dw}=48^{2}k-3kw^{2}$

$ \frac{ds}{dw}=k(48^{2}-3w^{2})$

at turning pt. $ \frac{ds}{dw}=0$

solving for w, $ w$=$ 16\sqrt3$

so when I sub in to find whether this point is max. or min., does it not depend on whether k is pos. or neg.?? If neg. this point becomes a minimum?![/hide]

Just to exhibit the Lagrange technique:

We're trying to maximize $ kwd^{2}$ for nonnegative $ w$ and $ d$ subject to the constraint $ w^{2}+d^{2}=48^{2}.$

We obtain the following two equations:

$ kd^{2}=2\lambda w$
$ 2kwd=2\lambda d$

Multiply the first equation by $ d$ and the second equation by $ w:$

$ kd^{3}=2\lambda wd=2kw^{2}d$

Hence $ d^{3}=2w^{2}d,$ so either $ d=0$ or $ d^{2}=2w^{2}.$

$ d=0$ is an extreme case resulting in zero beam strength. (The other extreme case, $ w=0$, also results in zero beam strength.)

Hence $ d^{2}=2w^{2}.$ But that into the equation $ d^{2}+w^{2}=48^{2}$ and we get

$ d=\frac{48\sqrt{2}}{\sqrt{3}},\ w=\frac{48}{\sqrt{3}}.$

(Notice how passively we're using the diameter of the log and how easily we could change that. That tends to be a feature of the nicer Lagrange solutions: they get us directly to the [i]shape[/i] of the answer.)

Lagrange multipliers also sees very little difference between the problem as stated and its natural dual problem. In this case, here's the dual problem: we want to obtain a beam of a certain specified strength. What is the smallest diameter of log we could use?

I see why this method is preferred.

how about if constraint is difficult to find?

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

The point $ A$ lies on the positive half of the x-axis, the point $ B$ lies on the positive half of the y-axis, and the interval AB passes through the point $ P(5,3)$. Find the coordinates of $ A$ and $ B$ so that $ \bigtriangleup$ $ AOB$ has minimum area

$ \boxed{A=\frac{1}{2}xy}$

From diagram, $ AM = x-5$ , $ ML = 3$                                   [url=http://img529.imageshack.us/my.php?image=40125748gp5.jpg][img]http://img529.imageshack.us/img529/3681/40125748gp5.th.jpg[/img][/url]

Therefore, $ AL = (x-5)^{2}+3^{2}$

$ AL = x^{2}-10x+34$

$ ///ly$ $ BC = y-3$ , $ CL=5$

Therefore $ BL = (y-3)^{2}+25$

$ BL = y^{2}-6y+34$

$ x^{2}+y^{2}= AL+BL = x^{2}+y^{2}-10x-6y+68$

$ -10x-6y+68 = 0$

$ \boxed{y=\frac{-5x}{3}+\frac{34}{3}}$

$ A=(\frac{1}{2}x)(\frac{-5}{3}x+\frac{34}{3})$

$ A=\frac{-5x^{2}}{6}+\frac{17x}{3}$

$ \frac{dA}{dx}= \frac{-5x}{3}+\frac{17}{3}$

at stat. pt $ \frac{dA}{dx}$ = 0

$ \frac{-5x}{3}+\frac{17}{3}$

$ x = \frac{17}{5}$ however answer says x=10 and y = 6 ???????????

------
could you demonstrate the lagrange method for this qn??

It's pretty clear that the formula for $ A$ can't be a quadratic polynomial. One thing I always emphasize when I teach: have "plausibility checks." Don't just proceed from one formula to the next, but [i]look[/i] at your expressions, and keep looking at them, and ask the question "Does this make sense?" in as many different ways as you can imagine.

Two related questions: for what values of $ x$ does this make sense, and what do the extreme cases look like?

To answer that: $ x$ must be greater than $ 5,$ but could be very large. That is, $ x\in(5,\infty).$ And as for the extreme cases, the area must go to infinity as either $ x\to5^{+}$ or $ x\to\infty.$ (No quadratic polynomial looks like that, hence your formula fails to satisfy a plausibility check.)

Your answer of $ x=\frac{17}{5}$ is outside the domain of the function. The line that passes through $ \left(\frac{17}{5},0\right)$ and $ (5,3)$ does not look like that diagram and does not create a triangle in the first quadrant.

Let's do it right. By similar triangles, $ \frac{x-5}{3}=\frac{5}{y-3}.$

Then $ y-3=\frac{15}{x-5}$ or $ y=3+\frac{15}{x-5}=\frac{x}{x-5}.$

Now we have $ A=\frac{1}{2}\,\frac{x^{2}}{x-5}.$ Note that this does agree with the prediction that it should tend to infinity as $ x\to5^{+}$ and $ x\to\infty.$

You should be able to finish it from here.

The Lagrange multiplier method:

Minimize $ A=\frac12\,xy$ subject to $ (x-5)(y-3)=15.$

$ \frac{y}2=\lambda(y-3)$

$ \frac{x}2=\lambda(x-5)$

So $ 2\lambda=\frac{x}{x-5}=\frac{y}{y-3}.$

$ x(y-3)=y(x-5)$

$ xy-3x=xy-5y$

$ 3x=5y.$

Put $ y=\frac{3x}5$ into the constraint equation:

$ (x-5)\left(\frac{3x}5-3\right)=15$

This simplifies to $ (x-5)^{2}=25.$

Since only $ x>5$ has any geometrical meaning, we must have $ x-5=5$ or $ x=10.$ That leads us to $ x=10,\ y=6$ and an area of $ 30.$

This leads easily to the generalization: if the point is $ (a,b),$ then the for the minimum triangle area, we have $ x=2a,\ y=2b,$ and the area is $ 2ab.$

[quote="Kent Merryfield"]It's pretty clear that the formula for $ A$ can't be a quadratic polynomial. One thing I always emphasize when I teach: have "plausibility checks." Don't just proceed from one formula to the next, but [i]look[/i] at your expressions, and keep looking at them, and ask the question "Does this make sense?" in as many different ways as you can imagine.

Two related questions: for what values of $ x$ does this make sense, and what do the extreme cases look like?

To answer that: $ x$ must be greater than $ 5,$ but could be very large. That is, $ x\in(5,\infty).$ And as for the extreme cases, the area must go to infinity as either $ x\to5^{+}$ or $ x\to\infty.$ (No quadratic polynomial looks like that, hence your formula fails to satisfy a plausibility check.)

Your answer of $ x=\frac{17}{5}$ is outside the domain of the function. The line that passes through $ \left(\frac{17}{5},0\right)$ and $ (5,3)$ does not look like that diagram and does not create a triangle in the first quadrant.

[/quote]
I see why my answer was totally absurd now  :oops:  :oops:  :blush:

thanks for that

bye


[quote="kunny"]How many circles are there which touch given three distinct circles at a time that any two circles don't have intersection points?[/quote]

I don't know if I understand well what you mean Kunny  :blush:   but what about  the Apollonius problem?
http://mathworld.wolfram.com/ApolloniusProblem.html

Yes, thta's it. Thank you, Nebraska boy. :lol:


You don't have to graph it; I can do that. However, I couldn't get past the second step. I couldn't remember how to factor it.

$f(x) = x^{3}+x^{2}$

$x(x^{2}+x)$

$x(x ... ???$

Do I need to complete the square? Is there something I'm forgetting?

That doesn't seem right. I have to factor it all out to get the x-intersects.

wouldn't the x-intercepts be
[hide]
$0$
and
$-1$[/hide]

Rightly so. Alright, I accept the communicators' implied premises and explicit conclusions. I was thinking I would have to do something with the $x^{2}$, but I guess I don't have to do something with it.

sorry for not explaining it as well

i was kinda busy
with AOPS 2 :juggle:


The length of a rectangle is three times the width. The perimeter is 96 cm. What is the length of the rectangle, in centimeters?

The formula for a  rectangle's perimeter is $ P\equal{}2(l\plus{}w)$. We have $ l\equal{}3w$. So, $ P\equal{}8w$. The perimeter is $ 96$. $ w$ is $ \frac{96}{8}\equal{}12$, and $ l\equal{}3w$, so $ l\equal{}12\times3\equal{}\boxed{36}$cm.


The square of an irrational is not always rational.  For example,

pi^2 = 9.86960440... = irrational
e^2 = 7.389056099... = irrational
phi(the golden ratio)^2 = (3 + sqrt(3))/5 = irrational

They are transcendental numbers.  Yeah, just missing the "n."

The numbers don't even have to be transcendental. Take the cube root of three. It's algebraic (the opposite of transcendental). Square that, and you still get an irrational (the cube root of 9).

it only works for the square root of a rational number

[quote="Silverfalcon"]Is it possible to say that square of irrational is ALWAYS or POSSIBLY rational number?

I think in this case:

[tex]\sqrt 3[/tex]  is irrational but [tex](\sqrt 3)^2[/tex]  is rational number since it's 3.[/quote]
easiest way: cube root 3

Or something like [tex]1+\sqrt{2}[/tex].

An irrational number^2 does not always equal rational, because  (sqrt6)^2 is rational but pi^2 is irrational.

[quote="JBL"]How do you all know that [tex]\pi^2, e^2[/tex], or the other numbers you are talking about are irrational?[/quote]

I thought someone proved that. Might've been something else though.

You're right, someone has proven it.  It isn't an obvious fact at all, though.  Just a notice that people should be careful about stating things that seem clear, without having any way to back them up.

You can read about transcendiality [url=http://mathworld.wolfram.com/TranscendentalNumber.html]here[/url].

Then here's question:

Exception of e and pi, is the result rational in this case:

[tex](^b\sqrt a)}^b[/tex]

The bth root of a raised to the bth power is always a:
(a^(1/b))^b=a^(b*1/b)=a^1=a.

If a is rational, your expression is rational. If a is irrational, like sqrt(2), then your expression is irrational.

Yeah, what they said.  And a little tip on your tex you can make the nth root look better by doing this:   [tex](\sqrt[b]{a})^b[/tex]  What I did was put the index (the b) in brackets after the square root so it would look like this \sqrt[b]   Just thought I would let you know.  It is trivial but it looks better.

[quote="tetrahedr0n"]Well, what are a and b? Integers? Rationals? Reals? I mean, you need to specify for us to answer your question.[/quote]

A and B are real rational numbers.


modular arithmetic
function
complex numbers

Let $F(z)=\frac{z+i}{z-i}$ for all complex numbers $z\not= i,$ and let $z_n=F(z_{n-1})$ for all positive integers $n.$  Given that $z_0=\frac 1{137}+i$ and $z_{2002}=a+bi,$ where $a$ and $b$ are real numbers, find $a+b.$

[hide]
After a brilliant mass of algebra, we find that $F(F(F(z))) = z$. Then $z_i = z_j$ iff $i \equiv j \pmod{3}$ so

$z_{2002} = z_1 = F(z_0) = \frac{\frac{1}{137}+i+i}{\frac{1}{137}+i-i} = 1+274i \Rightarrow a+b = 1+274 = 275$.[/hide]

[hide]Seeing that we're looking for the 2002th term, we try to find some sort of periodicity in the function. Finding  yields . Then, finding  yields , so the function is periodic every 3 terms. Since , then we only need to find , which is , and thus .[/hide]


Let $ a$ and $ b$ be nonzero complex numbers and $ f(z) = az+bz^{-1}$ . Determine the image under f of the unit circle

[hide]Let's break it up into components; $ f( e^{i t}) = x(t)+y(t) i, \; t \in [ 0,2 \pi )$.

$ x(t)=\text{Re}(a e^{i t}+b/ e^{it})$ $ = \text{Re}(a e^{i t}+be^{i(-t)})$ $ = a \cos t+b \cos (-t) = (a+b) \cos t$

$ y(t) = \text{Im}(a e^{i t}+b/ e^{it})$ $ =\text{Im}(a e^{i t}+be^{i(-t)})$ $ = a \sin t+b \sin (-t) = (a-b) \sin t$

Hence, the image under $ f$ of the unit circle is defined by $ f( e^{i t}) = (a+b) \cos t+(a-b)i \sin t$, which forms an ellipse in the complex plane.[/hide]


modular arithmetic
Diophantine Equations

Prove that for each positive integer $n$ there exist odd positive integers $x_n$ and $y_n$ such that ${x_{n}}^2 +7{y_{n}}^2 =2^n$.

The exact result is false, since $ n\ge3$ is required.

Define $ x_{3}= y_{3}= 1$, and define recursively
\[ (x_{n+1},y_{n+1})=\{\begin{array}{ll}(\frac{|7x_{n}-y_{n}|}{2},\frac{x_{n}+y_{n}}{2}) &\text{ if }x_{n}\equiv y_{n}\pmod4,\\ (\frac{7x_{n}+y_{n}}{2},\frac{|x_{n}-y_{n}|}{2}) &\text{ if }x_{n}\not\equiv y_{n}\pmod 4.\end{array}\]
Then obviously $ x_{n+1}^{2}+7y_{n+1}^{2}= 2^{n+1}$, and both are odd, so we have constructed such integers.

[solution neatly rewritten from [url=http://www.mathlinks.ro/viewtopic.php?p=918168]here[/url]]

It is a problem from Bulgarian MO 1996 - Regional Round.
http://imomath.com/othercomp/Bul/BulMO396.pdf
You can find a solution on the Bulgarian MO offical site - http://www.math.bas.bg/bcmi/ too.

Let $(x, y) = x^2 + 7y^2$ and define the composition rule $(x_1, y_1) \cdot (x_2, y_2) = (x_1x_2 - 7y_1y_2, x_1y_2 + x_2y_1)$ (corresponding to the identity $(x_1^2 + 7y_1^2)(x_2^2 + 7y_2^2) = (x_1x_2 - 7y_1y_2)^2 + 7(x_1y_2 + x_2y_1)^2$). Since both $x_1x_2 - 7y_1y_2$ and $x_1y_2 + x_2y_1$ are even (given that we are composing odd solutions), we want $x_1x_2 - 7y_1y_2 \equiv_4 2$ and $x_1y_2 + x_2y_1 \equiv_4 2$. We can obtain odd solutions for $n+1$ (which [b]Peter[/b] listed above) by carrying out the composition $(x_n, y_n) \cdot (1, 1)$ if $x_n \equiv_4 y_n$ and $(x_n, y_n) \cdot (-1, 1)$ if $x_n \not\equiv_4 y_n$.


algebra
function
domain
geometry
geometric transformation
dilation
real analysis

If \[ f(t)\longrightarrow F(j\omega)\] then how is the FT of $ f(\sqrt t)$ related to $ F(j\omega)$?

There's no clean or simple relationship.

Worse: the very statement is nonsense. In order to have a Fourier transform, $ f$ must have as its domain $ (\minus{}\infty,\infty)$ (or at least almost every point of that domain). $ f(\sqrt{t})$ is not defined for $ t<0.$ We can't have that.

Even if you fixed that by talking about $ f(\sqrt{|t|}),$ you'd still run into trouble. We could find find $ f\in L^1(\mathbb{R})$ such that $ f(\sqrt{|t|})\not\in L^1,$ or $ f\in L^2$ such that $ f(\sqrt{|t|})\not\in L^2.$

There are some things the Fourier transform handles very well: translation ($ f(t\minus{}b)$), dilation ($ f(at)$), differentiation. But that doesn't mean it can handle any arbitrary manipulation.

But I have seen in a recent paper that we can find inverse Fourier transform of the function $ F(\sqrt{j\omega})$. What's your comment on that?

regards
bhupala


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group. Let $ H, N$ be its subgroups where $ N\lhd G$. If $ G\simeq H\times N$ then we will have $ G/N\simeq H$.
Does the inverse implication also hold? Namely if $ G/N\simeq H$, then does this imply $ G\simeq H\times N$?

The motivation of my question comes from the following. If $ A,B,P$ are modules. If $ B \simeq A\oplus P$ then $ B/A\simeq P$.
However, if $ B/A\simeq P$ [b]and[/b] $ P$ is projective then we would have $ B \simeq A\oplus P$ (please correct me if I'm wrong).

So I was thinking there might be a group analogy to this. Maybe it's true affter we impose some extra analogous
conditions like "projective" to $ H$?

Thanks for your help :)

For $ G \equal{} S_3$ or $ \mathbb Z /4 \mathbb Z$, it doesn't hold, for example.
The corresponding property you look for may be a splitting exact sequence with additional properties.
Or you could look at abelian groups = $ \mathbb Z$-modules (so you can apply module theory).

In fact, one can apply the exact same answer. What should the analog of projective be? It should be, as for modules, that every short exact sequence $ G\rightarrow{P}\rightarrow{1}$ splits. Since every group is a quotient of a free group, one gets (as in the case of modules) that this definition is equivalent to being a subgroup (resp. submodule) of a free group. However, one knows that every subgroup of a free group is itself free, so here being projective means being free. Thus, the analog of projective is just simply free.

I guess there are a few more words to be said. One thing is that the notion of splitting for groups is different then for modules because the category is non-abelian. What you end up with is the notion of a semi-direct product, not just a direct product. So, first you should replace your analog of direct sum with that. Once one accepts that, the most complete answer to your question comes from the theory of group cohomology, which among other things, describes the obstruction to an exact sequence splitting.

Thanks very much. I will think about it  :maybe:

Thanks Magnara, I found that exceptionally clear and instructive.


calculus
analytic geometry
graphing lines
slope
limit
calculus computations

Let $ P(3,4)$ be a point on the circle $ x^2\plus{}y^2\equal{}25$.

(a) What is the slope of the line joining $ P$ and $ O(0,0)$?
(b) Find the equation of the tangent line to teh circle at $ P$.
(c) Let $ Q(x,y)$ be another point on the circle in the first quadrant. Find the slope $ m_x$ of the line joining $ P$ and $ Q$ in terms of $ x$.
(d) Calculate $ \lim_{x\to 3} m_x$.

Part a is trivial
Part b can be done by finding the negative reciprocal of (a) for slope, and plugging in (3,4) into $ y\minus{}y_1\equal{}s(x\minus{}x_1)$.


probability
real analysis
function
limit
probability and stats

consider $ \{X_n\}_{n > 0}$ an iid sequence of Bernoulli random variables: $ P(X_n \equal{} 1) \equal{} 1 \minus{} P(X_n \equal{} 0) \equal{} p$ and define
\[ S_p \equal{} \sum_{n \equal{} 1}^{\infty} \frac {X_n}{2^n}
\]
and $ \mu_p$ the probability measure on $ [0;1]$ induced by $ S_p$. (ie: for every Borel set $ E$ define $ \mu_p(E) \equal{} P(S_p \in E)$)

Could you prove that for different value $ p \neq p'$, the measure $ \mu_{p}$ and $ \mu_{p'}$ are mutually singular? I think that I have a proof but it uses nontrivial results (Kakutani theorem on singular measures). I suspect that there should be a simpler approach.

ps: and anybody with a nice expression of $ \mu_p([0;a[)$ ? I got something very nasty  :maybe:

[edit]: in fact, no need of strong theorems. Nice problem :-)

Hi Alekk I might say something stupid but do not the partial sums converge a.s. to some real $ x_p$ only depending on p, which would mean that $ \mu_p\equal{}\delta_{x_p}$ ?

:maybe:  I don't think so. For example, if $ p\equal{}\frac{1}{2}$, then $ \mu_{p}$ is the Lebesgue measure on $ [0;1]$. I really do not see why you would think that it would converge toward a same real number, no matter what the values of the $ X_n$ are ??

I was thinkig about characteristic functions. note that :
\[ f_{S_{p}} \equal{} \prod\limits_{j\equal{}1}^{\infty}(1\minus{}p(1\minus{}e^{ipj}))\]
it's apporximately like $ e^{ipt}\plus{}o(g(t))$  but information about $ o(g(t))$ it's important in this case...  :huh:

-Extremal
I don't find the same characteristc function than you
I have
$ F_{S_n}(\lambda)\equal{}\Pi_{j\equal{}0}^{\infty}(1\minus{}p.(1\minus{}e^{i\frac{\lambda}{2^j}}))$

For which I don't have any clue to prove why for any two p,p' the measures are singulars.

-Alekk
I don't see why for p=1/2 $ S_n$ it converges to Lebesgue measure would you tell me why please

thx

[quote="TheBridgeTwo"]-Alekk
I don't see why for p=1/2 $ S_n$ it converges to Lebesgue measure would you tell me why please [/quote]

This is essentially the construction of Lebesgue measure from 'coinflipping'.  It's in virtually any introductory graduate probability text.

yes: it is exactly as if you were sampling the binary expansion of a real number, each digit being equally likely to be $ 0$ or $ 1$. Or look at the distribution of $ S_n \equal{} \sum_1^n \frac{X_k}{2^k}$: it should convince you that as $ n \to \infty$ it of course tends towards the uniform measure on $ [0;1]$.

[quote="TheBridgeTwo"]-Extremal
I don't find the same characteristc function than you
I have
$ F_{S_n}(\lambda) \equal{} \Pi_{j \equal{} 0}^{\infty}(1 \minus{} p.(1 \minus{} e^{i\frac {\lambda}{2^j}}))$

...[/quote]
Sure you are correct. I just have written not that about what I thought. I thought of the answer and consequently have written   in the product $ \prod .... 1\minus{}e^{ipj}$ where, there is no variable $ t$  :lol: .
but the assymptotical expression is coorect .
\[ f_{S_{p}} \equal{} \prod\limits_{j\equal{}1}^{\infty}(1\minus{}p(1\minus{}e^{it/2^{j}})) \equal{} e^{ipt} \plus{}o(g(t))\]

By the SLLN, we have $ \frac 1k\sum_1^k X_i\to p$. For every $ t\in[0,1]$, we can write $ t$ as $ \sum_1^{\infty}\frac {t_i}{2^i},\ t_i\in\left\{0,1\right\}$. Define $ D_p \equal{} \left\{t\in [0,1]: \ \lim\limits_{N\to\infty}\frac {\mathop{card(i\leq N: t_i \equal{} 1)}}{N} \equal{} p\right\}$. Then, $ P(X\in D_p) \equal{} 1$. But for $ p \ne p'$, $ D_p$ and $ D_{p'}$ are disjoint.
The explicit expression for $ \mu_p$ can be given by $ \mu_p([0,t)) \equal{} \sum\limits_{k\geq 1}\left[t_k (1 \minus{} p)^{t_k}\prod\limits_{i \equal{} 1}^{k \minus{} 1} p^{t_i}(1 \minus{} p)^{1 \minus{} t_i}\right]$(indeed, a little bit ugly).


I'm looking for a program that can create graphs (it may or may not be a comprehensive mathematics program). It would be nice if this program can create a format that is easily integrated into microsoft word (makes an image of the graph).

I probably do not need actualy linear "Y=" graphing (or parametric, complex etc.), but rather I just need to plot coordinate points (for economics class). It would be really nice if the program can interpolate the points creating a connected function.

Yes, I know (:D). But I need to send it to my teacher via email.

I have graph 1.4 or something, google it.

I used it for my IB projects.


During a party, a total of 78 handshakes occurred. If each person shook hands once with each of the other people, how many people were at the party?

[quote="PerpetualMotion"]The number of handshakes among n people can be expressed as n(n+1)/2. We set this equal to 78. Solving for n gives us [b][i]12[/i][/b].[/quote]
it should be (n-1)(n)/2.

Yeah, if there were 12 people, then there would be 66 handshakes because you cannot shake hands with yourself. There would actually be 13 people at the party.

$78 \rightarrow \text{12th triangular number}$

$12+1=13$

$\boxed{13}$

[quote=KevinYang07]Since each person shakes hands with each other person, every pair of people will shake hands once. So, 78 represents the number of pairs, which we can count as ${n \choose 2}$ where $n$ is the number of people at the party. So, $n(n-1) = 2 \cdot 78 = 2 \cdot 6 \cdot 13 = 12 \cdot 13$. So, $n=13$ gives us $\boxed{13}$ people at the party.[/quote]

You literally copied the exact solution alcumus gives. :D

[quote=KevinYang07]Since each person shakes hands with each other person, every pair of people will shake hands once. So, 78 represents the number of pairs, which we can count as ${n \choose 2}$ where $n$ is the number of people at the party. So, $n(n-1) = 2 \cdot 78 = 2 \cdot 6 \cdot 13 = 12 \cdot 13$. So, $n=13$ gives us $\boxed{13}$ people at the party.[/quote]

I don't think a regular kid would start off with that wording. Plus, All these steps you gave were from Alcumus itself.

[quote=Mathpluspianoequalslife]how did he even find the sol[/quote]

Do you want to know how you solve this problem or how he found this thread?

[quote=Mathpluspianoequalslife]how did he copy the solution from alcumus[/quote]

discuss button

I think he did this problem before, and he knew exactly where it was, so here's what he did:

1) He went to Resources $\rightarrow$ Alcumus
2) Went into his report, and found the answer.
3) Copied Pasted and went back to this thread.
4) Quoted Alcumus.


Solve for $ x$: $ |x \plus{} 1| \minus{} |1 \minus{} x| < 2$.

[hide="solution"] Please check if I've got it right:

$ |x \plus{} 1| < 2 \plus{} |1 \minus{} x|$, so simplifying and doing casework I get

I get $ \minus{} 4 < 0$ and $ x < \minus{} 1$. So the solution set is $ ( \minus{} \infty, \minus{} 1)$. [/hide]

How do I solve $ |x \plus{} 1| \minus{} |1 \minus{} x| \equal{} 2$ and $ |x\minus{}1|\plus{}|x|\equal{}1$?

I'm okay at solving inequalities with absolute values but I'm not great at absolute value equations.

I am guessing that from your hint: [hide="method"] If $ g(x) \equal{} |x \minus{} a| \plus{} |x \minus{} b| \equal{} h$ and $ a < b$, I can have the cases $ x < a$, $ x \equal{} a$, $ a < x < b$, $ x \equal{} b$, and $ x > b$. Then I will re-express $ g(x)$ without the modulus function for each case and the answer will then be staring into my eyes.
[/hide] / [hide="answer"] I tried this on my second problem and got $ 0 \leq x \leq 1$, which the solution manual says is correct. [/hide]

Could you please tell me more about how to solve equations with modulus functions?

You can notice that if you have $ |x|$, then by removing the absolute value, you have $ \pm x$ (because $ |\pm x|\equal{}|x|$).
So you can simply expand it out and use casework.

For example:
$ |x\minus{}1|\plus{}|x|\equal{}1$

$ \pm(x\minus{}1)\pm x\equal{}1$

And you get four equations:

$ x\minus{}1\plus{}x\equal{}1\implies 2x\minus{}1\equal{}1\implies \boxed{x\equal{}1}$

$ \minus{}x\plus{}1\plus{}x\equal{}1\implies 1\equal{}1$

$ x\minus{}1\minus{}x\equal{}1\implies \minus{}1\equal{}1$

$ \minus{}x\plus{}1\minus{}x\equal{}1\implies \minus{}2x\equal{}0\implies \boxed{x\equal{}0}$

So the solutions are $ \boxed{\boxed{x\equal{}0,\;1}}$.

Try the other one on your own.


Solve the equality in natural numbers $ nm \plus{} mk \plus{} kn \minus{} nmk \equal{} 2$

[hide]WLOG $ m\le n \le k$.
Either $ m \equal{} 1$, $ m \equal{} 2$, or $ m \ge 3$.
If $ m \equal{} 1$
$ n\cdot 1 \plus{} 1\cdot k \plus{} kn \minus{} n\cdot 1 \cdot k \equal{} n \plus{} k \equal{} 2$
and since $ n,k \ge 1$,
$ n \equal{} k \equal{} 1$. We have the solution $ (m,n,k) \equal{} (1,1,1)$.
If $ m \equal{} 2$

$ 2n \plus{} 2k \plus{} kn \minus{} n\cdot 2 \cdot k \equal{} 2$

$ 2n \plus{} 2k \minus{} nk \equal{} 2$

$ 2 \equal{} nk \minus{} 2n \minus{} 2k \plus{} 4$

$ 2 \equal{} (n \minus{} 2)(k \minus{} 2)$
Solving this equation in integers we get $ n \equal{} 3$, $ k \equal{} 4$, for the solution $ (m,n,k) \equal{} (2,3,4)$.
If $ m\ge 3$
Dividing the given equation by $ nmk$,
$ \dfrac{1}{m} \plus{} \dfrac{1}{n} \plus{} \dfrac{1}{k} \minus{} 1 \equal{} \dfrac{2}{mnk}$
so
$ \dfrac{1}{m} \plus{} \dfrac{1}{n} \plus{} \dfrac{1}{k} \equal{} 1 \plus{} \dfrac{2}{mnk} > 1$.
However, since $ 3\le m \le n \le k$,
$ \dfrac{1}{3} \ge \dfrac{1}{m} \ge \dfrac{1}{n} \ge \dfrac{1}{k}$
so
$ \dfrac{1}{m} \plus{} \dfrac{1}{n} \plus{} \dfrac{1}{k}\le \dfrac{1}{3} \plus{} \dfrac{1}{3} \plus{} \dfrac{1}{3} \equal{} 1$,
leading to a contradiction. Thus $ m$ cannot be greater than or equal to $ 3$, so $ m$ can only be $ 1$ or $ 2$.
We have the solutions $ (m,n,k) \equal{} (1,1,1)$ and $ (2,3,4)$, and relaxing the condition $ m \le n \le k$ we get five more solutions, $ (2,4,3), (3,2,4), (3,4,2),(4,2,3)$ and $ (4,3,2)$. We are done.[/hide]


I play on a private server on World of Warcraft (WoW) and it is getting kinda lonely. You guys can join too post here if u have any questions or joined already. BTW, this is free while regular WoW is $4 a month

Connecting to WoW is easy.

The private server I play on is WoWscape.
Level Max: 70
Player Cap: 1000

Drop rates
Rate.Drop.Items = 8x
Rate.Drop.Money =8x

XP rates
Rate.XP.Kill = 10x
Rate.XP.Quest = 10x
Rate.XP.Explore = 10x

The first step is to download WoW. (this must be installed first)

Then download The Burning Crusade. (It is required to play on this server)

After it is installed, patch up to 2.1.2 by opening WoW. (it will patch automatically.)

Then go to your WoW directory in C:program files

and open up the folder (realmistwhat) open with.. notepad and delete everything in the folder and paste (set realmlist 75.126.214.116) inside withouth the parentheses.

Then close and save.

Register your account at the site, open up WoW and enjoy!

If you have any questions you are free to ask (there should be no problems though, its garanteed to work)

Burning Crusade Download=https://www.worldofwarcraft.com/account/download/clients/pc/WoW-BurningCrusade-enUS-Installer-downloader.exe
WoW Download=http://www.worldofwarcraft.com/downloads/wowclient-download.html
Register here=http://www.wowscape.net/crack-register/
Setup%20v1.1.exe=http://www.wowscape.net/WoWscape[/url]

Unfortunately, your copy/paste didn't copy/paste the links for the downloads.

I seriously doudt that they have WoW for free..

oh srry for the links ill try and fix them
then doubt it. seriously, im not gonna grudg against u or anything :oops:

It's a private server, and it's illegal. I'd recommend keeping it in a forum that, you know, tolerates illicit activities.

[quote="azhang"]BTW, this thread should not be deemed file sharing or promotion of companies. It is meant for educational purposes only.

~Andrew~[/quote] mmorpgs will eat your brain

[quote="azhang"]BTW, this thread should not be deemed file sharing or promotion of companies. It is meant for educational purposes only.

~Andrew~[/quote]
WTF?

Since when is WoW educational?

I mean seriously, people waste their entire lives playing MMORPGs, and they have no educational value whatsoever.

[quote="aspiringchemlete"]So does this really work?

I don't want to download 3.16 GB and take like 3 hours for it not to work.[/quote]
I highly doubt it, but even if it does, don't do it. WoW will, as PI-Dimension said, eat your brain. Trust me (based on personal experience).

And don't use me as an example of "I can quit whenever I want." You can't. Only Chuck Norris can.

hey peeps, stop flaming me gosh. just trying to help :blush:

Okay, how about everyone just stop POSTING in this thread. happy?


[quote="xxreddevilzxx"][quote="amirhtlusa"]this is well known

if $2^k-1$ is prime and $N=2^{k-1}(2^k-1)$ then prove, $N$ is perfect.

(a whole number is perfect if it is equal to the sum of its proper divisors.)[/quote]

I dont think we should talk about this. This was on the IT for HCSSiM. Mod, please consider locking this.[/quote]

Hmm, I wonder why they use such a common problem on the IT. But the application period is over, right?

i am not sure... they take applicants until they are filled up, which is what PROMYS does, ithink, except they actually have a due date. (you are a PROMYS-er so you should know better). Last time I checked, they were filled only half.

[quote="amirhtlusa"][quote="xxreddevilzxx"][quote="amirhtlusa"]this is well known

if $2^k-1$ is prime and $N=2^{k-1}(2^k-1)$ then prove, $N$ is perfect.

(a whole number is perfect if it is equal to the sum of its proper divisors.)[/quote]

I dont think we should talk about this. This was on the IT for HCSSiM. Mod, please consider locking this.[/quote]
thats bull, what are you taking about, i have the proof
it's from the book called Euler Master of us all, by William Dunham, page 3.
i donno what HCSSIM is but please think before insulting others, cause i feel ofended... :mad:[/quote]

It's ok. It probably is on the IT for HCSSiM; the application quizzes for many summer programs take questions that might be famous (like this one or the PROMYS one - what numbers are sums of consecutive positive integers) so they inevitably get posted every once in a while. xxreddevilzxx was just pointing out the coincidence as a contributing member of AoPS - not insulting you at all. We just don't want like people either getting their answers to those tests here or the HCSSiM people coming to get us.

[quote="amirhtlusa"][quote="xxreddevilzxx"][quote="amirhtlusa"]this is well known

if $2^k-1$ is prime and $N=2^{k-1}(2^k-1)$ then prove, $N$ is perfect.

(a whole number is perfect if it is equal to the sum of its proper divisors.)[/quote]

I dont think we should talk about this. This was on the IT for HCSSiM. Mod, please consider locking this.[/quote]
thats bull, what are you taking about, i have the proof
it's from the book called Euler Master of us all, by William Dunham, page 3.
i donno what HCSSIM is but please think before insulting others, cause i feel ofended... :mad:[/quote]

i am sorry if you were being offended. I was just suggesting, however, and you shouldnt use language like that. It's not like I used bad language. A lot of locking because of the problems seen on the test presently going on happens on this forum. I hope you dont take this as offensive and serious. I was only being suggestive, again. Personally, I dont think I insulted you in any way. Just my thinking. If you have other opinion, you can PM me. Sorry again if you were offended.

I dont think he did anything wrong. Of course, he didnt know that your post had nothing to do with IT( what is that btw, an acronym for something?), but many times in the forum i have seen threads locked because of posting problems that might be in a contest at the moment. reddevil neither  made fun of you nor called you a cheater. What he said did not deserve to be blamed.

that's okay,
i'm sorry, just having a bad night. :spider:
i'm going to bed :stretcher:

This problem is also in the Art of Problem Solving.  I'm not which volume, though...

I'm guessing it's
if 2^k-1 is prime, then N=a prime*(2^[k-1]).  Let 2^k-1=x
So then N's factors, not including itself, are
1
2
(2^2)
.
.
.
2^(k-1)
and
x
2x
(2^2)x
.
.
.
(2^[k-2])x

We know that the first set of terms adds up to 2^k-1.  The second set of terms adds up to (2^[k-1]-1)x.  Sub back in for x and you get
(2^k-1)(2^[k-1]-1+1)
=(2^k-1)(2^[k-1])
=N
Therefore, N is perfect.

EDIT: Just read paladin8's post and realized that his proof was basically the same thing...oh well.


Let $p>3$ be prime. Prove that $p^2$ divide \[ \binom{p}{1} + \binom{p}{2} + ... + \binom{p}{[\frac{2p}{3}]}  \]


Find all pairs of positive integers x and y that safisfy the equation:

1/x - 1/y = 1/2005

I'm so sick of these problems that I don't know how to do them. Is there a quick way to find the pairs? I can do it by guess and check...

I think I ve got a way...
[hide="ANSWER"]
$\frac{1}{x}-\frac{1}{y}=\frac{1}{2005}$
So by this you get the equation....
$2005(x-y)=xy$
$5.401(x-y)=xy$
there are a no. of possibilities....now taking value of only $x$,
$x=5 or x=401 or x=(x-y)$ from all these you can find all the possible values of $x$
But I dont think this the right way probably by factorising....
[/hide]

[hide="cmon guys"]clearly $y>x$. solve for $y$ and get

$y=-2005+\frac{2005^2}{2005-x}$. since $y>0$ is an integer, $\frac{2005^2}{2005-x}$ must be a positive integer greater than 2005

so $2005-x$ is a divisor of $2005^2$ but we need divisors smaller than $2005$. we know $2005^2=5^2\cdot 401^2$, then

$2005-x=1,5,5^2,401$ and so $x=2004,2000,1980,1604$ and you go find the rest of the $y's$[/hide]

I tried this I couldnt find any aswell :?  :(

[quote="JesusFreak197"]Umm... sorry to break it to you, Palytoxin, but when you find the values of $y$ that should go with those values of $x$, none of them work. The closest was $x=2004$, which gave a $y$ value that then gave $-\frac{1}{2005}$ rather than positive.[/quote]
wow you got to be kidding me... please double check your work, don't just read my solution!!! have a pencil and paper handy and do the problem by reading along with my solution, don't just go looking at it and then do some error proning calculations. my solution is correct. did you plug in correctly?

$y=-2005+\frac{2005^2}{2005-x}$

english version: $y$ is equal to negative $2005$ plus the fraction with numerator $2005$ squared and denominator $2005-x$

 :stretcher:

edit: if people still dont get it, i'll post an entire solution with all the $y$ values

[quote="mathnerd314"]I'm probably going off on opposite/adjacent (tangents, hahahahaha), but...

If 1/a + 1/b = 1/c, is there a necessary relationship between a, b, and c? We discussed this in algebra last year, but we didn't get very far. And how many pairs of (a, b) are there where all three are integers, in terms of c?

Just related questions.[/quote]

I'm not sure, but I think this is harder than getting started level, but here it goes..

If a, b and c are positive integers (if they are negative I think problem is not much harder, just write it in few cases), then it's obvious that $a>c$ and $b>c$, so put $a=c+m$ and $b=c+n$ where m and n are both positive integers. Then you'll get $c^2=m*n$, and by writing c as $c=p_1^{a_1}*p_2^{a_2}*\ldots*p_k^{a_k}$ you can see that number of solutions is equal to the number of divisors of number c, and that is $(1+a_1)*(1+a_2)*\ldots*(1+a_k)$.

Assuming $x$ is still the same value as it was originally, I should be able to use the equation $\frac{x-y}{xy}=\frac{1}{2005}$, and the $y$-values resulting from that never worked in the original equation. Therefore, the $x$ values must be incorrect. So, you must have done something wrong in getting to that other equation.

polytoxin's answers are right. He just didn't explain himself fully.

In general these type of questions always turn into divisibility questions.

[quote="JesusFreak197"]$\frac{x-y}{xy}=\frac{1}{2005}$So, you must have done something wrong in getting to that other equation.[/quote]
good that you posted your solution because i can tell you right now what is wrong

$\frac1{x}-\frac1{y}=\frac{y}{xy}-\frac{x}{xy}=\frac{y-x}{xy}=\frac1{2005}$

got it?  :D


function
LaTeX
floor function
algebra
polynomial

In how many different ways can 30 cents be made from any combination of quarters, dimes, nickels or pennies?

Using quarters:

1q + 1n
1q + 5p

Using dimes:

1d + 4n
1d + 3n + 5p
1d + 2n + 10p
1d + 1n + 15p
1d + 20p
2d + 2n
2d + 1n + 5p
2d + 10p
3d

Using nickels:

1n + 25p
2n + 20p
3n + 15p
4n + 10p
5n + 5p
6n

Using pennies:

30p

As you can see there are [color=white]18 different ways[/color] to make 30 cents out of quarters, dimes, nickels, and pennies.

Is there any way other than listing the possibilities to solve this? It's extremely hard to do such a long one in 45 seconds.

Or if you're in my case, 12 seconds.  :D  And I believe there is [b]not[/b] another way.  I probably am wrong.

Thank you for reading my 400th post.  Have a pleasant day. (not really :harhar:)

If there was a shorter solution that I thought of, i wouldve posted that instead.

I wish there were some easier way, like a formula. It's a pain to find all those combinations when there's so many!

I just try to think of the numbers: 25, 10, 5, and 1 and figure it out that way.

Which is the way dragon used...

And, no, I am almost certain there is no other way.

I believe there is a formula:

$ 3 \plus{} \frac {(int(\frac {n}{5}))(int(\frac {n}{5})\minus{}1)}{2}$

By the way, I'm using int for the floor function, cause I don't know Latex for it... :blush:

What's the floor function? I don't really understand... :oops:

[quote="mewto55555"]It's greatest integer less than or equal to, so like floor of

1.2=1
0=0
-5.6=-6[/quote]

Oh, [i]that[/i]. Isn't it $ \lfloor$ and $ \rfloor$? [dollar sign]\lfloor[dollar sign] and [dollar sign]\rfloor[dollar sign]

Yeah...those. Sorry. And n is the number of cents.

Interesting formula. Do you know how to derive it, or is really complicated?

That's a cool formula. Gotta remember that!

I never used floor and ceilings, so if I wanted the floor of x, do I do $ \lfloor{x}\rfloor$? (also, correct me if my grammar using floor and ceiling is wrong.)

Am i wrong or does this formula not work for 5 cents?

It only works for if you can use all kinds of coins - like quarters, etc. 5 cents can only be made up of a nickel or pennies.

Oh...then I'm wrong. Ignore what I said.  :blush:

[quote="mewto55555"]It works from 10-49[/quote]

Then is there a formula for less than 10 cents or more than 49 cents?

Another way to do this problem:

You take
$ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6 \plus{} x^7 \plus{} x^8 \plus{} x^9 \plus{} x^{10} \plus{} x^{11} \plus{} x^{12} \plus{} x^{13} \plus{} x^{14} \plus{} x^{15} \plus{} x^{16} \plus{} x^{17} \plus{} x^{18} \plus{} x^{19} \plus{} x^{20} \plus{} x^{21} \plus{} x^{22} \plus{} x^{23} \plus{} x^{24} \plus{} x^{25} \plus{} x^{26} \plus{} x^{27} \plus{} x^{28} \plus{} x^{29} \plus{} x^{30}$ (pennies)

and then

$ 1 \plus{} x^5 \plus{} x^{10} \plus{} x^{15} \plus{} x^{20} \plus{} x^{25} \plus{} x^{30}$ (nickels)

and then

$ 1 \plus{} x^{10} \plus{} x^{20} \plus{} x^{30}$ (dimes)

and

$ 1 \plus{} x^{25}$ (quarters)

and multiply them all together, and the coefficient of x^30 is your answer. It's called a generating function.

If you wanted to do this with, say, 50 cents instead, you would do this...

$ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} ..... \plus{} x^{50}$

multiplied by

$ 1 \plus{} x^5 \plus{} x^{10} \plus{} x^{15} \plus{} x^{20} \plus{} ... \plus{} x^{50}$

multiplied by

$ 1 \plus{} x^{10} \plus{} x^{20} \plus{} x^{30} \plus{} x^{40} \plus{} x^{50}$

multiplied by

$ 1 \plus{} x^{25} \plus{} x^{50}$

multiplied by (can't forget the 50-cent coins!)

$ 1 \plus{} x^{50}$

and your answer should be the coefficient of x^50 in that product of 5 polynomials.

What if you wanted to find all possible ways to make 35 cents? No, don't worry, you will not have to multiply any more polynomials. The answer is the coefficient of x^35 in the same product that you used to find the possible ways to make 50 cents.

If you then wanted to find all possible ways to make 40 cents, you would look up the same product, and just find the coefficient of x^40.

And this will work for anything less than or equal to fifty. If you wanted, then, to look up all the ways to make change for a dollar, you would certainly have to multiply more polynomials.

$ (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} ... \plus{} x^{100})\cdot(1 \plus{} x^5 \plus{} x^{10} \plus{} x^{15} \plus{} x^{20} \plus{} x^{25} \plus{} ... \plus{} x^{100})\cdot(1 \plus{} x^{10} \plus{} x^{20} \plus{} x^{30} \plus{} ... \plus{} x^{100})\cdot(1 \plus{} x^{25} \plus{} x^{50} \plus{} x^{75} \plus{} x^{100})\cdot(1 \plus{} x^{50} \plus{} x^{100})$ and find the coefficient of x^100... good luck!

And then after you do that, you will have the answers for any amount of cents less than a dollar.

Edit: I multiplied out the polynomials (using Mathematica) and found this:
$ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} 2 x^5 \plus{} 2 x^6 \plus{} 2 x^7 \plus{} 2 x^8 \plus{} 2 x^9 \plus{} 4 x^{10} \plus{} 4 x^{11} \plus{} 4 x^{12} \plus{} 4 x^{13} \plus{} 4 x^{14} \plus{} 6 x^{15} \plus{} 6 x^{16} \plus{} 6 x^{17} \plus{} 6 x^{18} \plus{} 6 x^{19} \plus{} 9 x^{20} \plus{} 9 x^{21} \plus{} 9 x^{22} \plus{} 9 x^{23} \plus{} 9 x^{24} \plus{} 13 x^{25} \plus{} 13 x^{26} \plus{} 13 x^{27} \plus{} 13 x^{28} \plus{} 13 x^{29} \plus{} 18 x^{30} \plus{} 18 x^{31} \plus{} 18 x^{32} \plus{} 18 x^{33} \plus{} 18 x^{34} \plus{} 24 x^{35} \plus{} 24 x^{36} \plus{} 24 x^{37} \plus{} 24 x^{38} \plus{} 24 x^{39} \plus{} 31 x^{40} \plus{} 31 x^{41} \plus{} 31 x^{42} \plus{} 31 x^{43} \plus{} 31 x^{44} \plus{} 39 x^{45} \plus{} 39 x^{46} \plus{} 39 x^{47} \plus{} 39 x^{48} \plus{} 39 x^{49} \plus{} 50 x^{50} \plus{} 50 x^{51} \plus{} 50 x^{52} \plus{} 50 x^{53} \plus{} 50 x^{54} \plus{} 62 x^{55} \plus{} 62 x^{56} \plus{} 62 x^{57} \plus{} 62 x^{58} \plus{} 62 x^{59} \plus{} 77 x^{60} \plus{} 77 x^{61} \plus{} 77 x^{62} \plus{} 77 x^{63} \plus{} 77 x^{64} \plus{} 93 x^{65} \plus{} 93 x^{66} \plus{} 93 x^{67} \plus{} 93 x^{68} \plus{} 93 x^{69} \plus{} 112 x^{70} \plus{} 112 x^{71} \plus{} 112 x^{72} \plus{} 112 x^{73} \plus{} 112 x^{74} \plus{} 134 x^{75} \plus{} 134 x^{76} \plus{} 134 x^{77} \plus{} 134 x^{78} \plus{} 134 x^{79} \plus{} 159 x^{80} \plus{} 159 x^{81} \plus{} 159 x^{82} \plus{} 159 x^{83} \plus{} 159 x^{84} \plus{} 187 x^{85} \plus{} 187 x^{86} \plus{} 187 x^{87} \plus{} 187 x^{88} \plus{} 187 x^{89} \plus{} 218 x^{90} \plus{} 218 x^{91} \plus{} 218 x^{92} \plus{} 218 x^{93} \plus{} 218 x^{94} \plus{} 252 x^{95} \plus{} 252 x^{96} \plus{} 252 x^{97} \plus{} 252 x^{98} \plus{} 252 x^{99} \plus{} 292 x^{100}... (etc.)$

But ignore everything after x^100, it has no relevance.

So:
The answer to the original question (30 cents) is 18
There are 50 ways to make 50 cents
And... there are 292 ways to make change for a dollar.

And for those who wanted to know... here is the generating function for anything up to 2 dollars (including dollar coins, this time!)
[hide]$ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} 2x^5 \plus{} 2x^6 \plus{} 2x^7 \plus{} 2x^8 \plus{} 2x^9 \plus{} 4x^{10} \plus{} 4x^{11} \plus{} 4x^{12} \plus{} 4x^{13} \plus{} 4x^{14} \plus{} 6x^{15} \plus{} 6x^{16} \plus{} 6x^{17} \plus{} 6x^{18} \plus{} 6x^{19} \plus{} 9x^{20} \plus{} 9x^{21} \plus{} 9x^{22} \plus{} 9x^{23} \plus{} 9x^{24} \plus{} 13x^{25} \plus{} 13x^{26} \plus{} 13x^{27} \plus{} 13x^{28} \plus{} 13x^{29} \plus{} 18x^{30} \plus{} 18x^{31} \plus{} 18x^{32} \plus{} 18x^{33} \plus{} 18x^{34} \plus{} 24x^{35} \plus{} 24x^{36} \plus{} 24x^{37} \plus{} 24x^{38} \plus{} 24x^{39} \plus{} 31x^{40} \plus{} 31x^{41} \plus{} 31x^{42} \plus{} 31x^{43} \plus{} 31x^{44} \plus{} 39x^{45} \plus{} 39x^{46} \plus{} 39x^{47} \plus{} 39x^{48} \plus{} 39x^{49} \plus{} 50x^{50} \plus{} 50x^{51} \plus{} 50x^{52} \plus{} 50x^{53} \plus{} 50x^{54} \plus{} 62x^{55} \plus{} 62x^{56} \plus{} 62x^{57} \plus{} 62x^{58} \plus{} 62x^{59} \plus{} 77x^{60} \plus{} 77x^{61} \plus{} 77x^{62} \plus{} 77x^{63} \plus{} 77x^{64} \plus{} 93x^{65} \plus{} 93x^{66} \plus{} 93x^{67} \plus{} 93x^{68} \plus{} 93x^{69} \plus{} 112x^{70} \plus{} 112x^{71} \plus{} 112x^{72} \plus{} 112x^{73} \plus{} 112x^{74} \plus{} 134x^{75} \plus{} 134x^{76} \plus{} 134x^{77} \plus{} 134x^{78} \plus{} 134x^{79} \plus{} 159x^{80} \plus{} 159x^{81} \plus{} 159x^{82} \plus{} 159x^{83} \plus{} 159x^{84} \plus{} 187x^{85} \plus{} 187x^{86} \plus{} 187x^{87} \plus{} 187x^{88} \plus{} 187x^{89} \plus{} 218x^{90} \plus{} 218x^{91} \plus{} 218x^{92} \plus{} 218x^{93} \plus{} 218x^{94} \plus{} 252x^{95} \plus{} 252x^{96} \plus{} 252x^{97} \plus{} 252x^{98} \plus{} 252x^{99} \plus{} 293x^{100}$ $ \plus{} 293x^{101} \plus{} 293x^{102} \plus{} 293x^{103} \plus{} 293x^{104} \plus{} 337x^{105} \plus{} 337x^{106} \plus{} 337x^{107} \plus{} 337x^{108} \plus{} 337x^{109} \plus{} 388x^{110} \plus{} 388x^{111} \plus{} 388x^{112} \plus{} 388x^{113} \plus{} 388x^{114} \plus{} 442x^{115} \plus{} 442x^{116} \plus{} 442x^{117} \plus{} 442x^{118} \plus{} 442x^{119} \plus{} 503x^{120} \plus{} 503x^{121} \plus{} 503x^{122} \plus{} 503x^{123} \plus{} 503x^{124} \plus{} 571x^{125} \plus{} 571x^{126} \plus{} 571x^{127} \plus{} 571x^{128} \plus{} 571x^{129} \plus{} 646x^{130} \plus{} 646x^{131} \plus{} 646x^{132} \plus{} 646x^{133} \plus{} 646x^{134} \plus{} 728x^{135} \plus{} 728x^{136} \plus{} 728x^{137} \plus{} 728x^{138} \plus{} 728x^{139} \plus{} 817x^{140} \plus{} 817x^{141} \plus{} 817x^{142} \plus{} 817x^{143} \plus{} 817x^{144} \plus{} 913x^{145} \plus{} 913x^{146} \plus{} 913x^{147} \plus{} 913x^{148} \plus{} 913x^{149} \plus{} 1022x^{150} \plus{} 1022x^{151} \plus{} 1022x^{152} \plus{} 1022x^{153} \plus{} 1022x^{154} \plus{} 1138x^{155} \plus{} 1138x^{156} \plus{} 1138x^{157} \plus{} 1138x^{158} \plus{} 1138x^{159} \plus{} 1267x^{160} \plus{} 1267x^{161} \plus{} 1267x^{162} \plus{} 1267x^{163} \plus{} 1267x^{164} \plus{} 1403x^{165} \plus{} 1403x^{166} \plus{} 1403x^{167} \plus{} 1403x^{168} \plus{} 1403x^{169} \plus{} 1552x^{170} \plus{} 1552x^{171} \plus{} 1552x^{172} \plus{} 1552x^{173} \plus{} 1552x^{174} \plus{} 1714x^{175} \plus{} 1714x^{176} \plus{} 1714x^{177} \plus{} 1714x^{178} \plus{} 1714x^{179} \plus{} 1889x^{180} \plus{} 1889x^{181} \plus{} 1889x^{182} \plus{} 1889x^{183} \plus{} 1889x^{184} \plus{} 2077x^{185} \plus{} 2077x^{186} \plus{} 2077x^{187} \plus{} 2077x^{188} \plus{} 2077x^{189} \plus{} 2278x^{190} \plus{} 2278x^{191} \plus{} 2278x^{192} \plus{} 2278x^{193} \plus{} 2278x^{194} \plus{} 2492x^{195} \plus{} 2492x^{196} \plus{} 2492x^{197} \plus{} 2492x^{198} \plus{} 2492x^{199} \plus{} 2728x^{200}...(etc.)$
Yeah, you read that right... 2728 ways. The reason the coefficient of x^100 went up is because I included dollar coins.[/hide]

Yes, but Mathematica took less than a second to expand it.

[quote="isabella2296"]What's Mathematica?[/quote]It's a math program

I'm not sure if it's free or not, though

I believe you can either buy it or get a free trial.

I like mewto's formula better... :maybe:

...especially spending your time on making that incredibly long $ \text{\LaTeX}$ formula, even though you shouldve got rid of the etc... and continued typing it up.  :D

Erm... I just copied it as LaTeX from Mathematica.

If I continued it, it would say... "Unparseable or potentially dangerous LaTeX formula. The formula is too long." Anyway, the part after the etc. is irrelevant anyway, because I only included up to $ x^{200}$ in the polynomials I multiplied out.

And I didn't really write out the polynomials that I multiplied into Mathematica, either, I used the built in Sum[f(i),{i,i_min,i_max,i_step}] function.

P.S. I tried to find the number of ways to make $ \$$100 using pennies, nickels, dimes, quarters, half-dollars, dollars, 5 dollar, 10 dollar, 20 dollar, and 50 dollar banknotes, but I sat there waiting for about 2 minutes for my computer to complete the calculation and I decided it would take too long.

Um... this is the longest thread on a sole problem.

Yes; if I were a mod, I'd lock it. I'm not, so please let this die.


analytic geometry
geometry
3D geometry
sphere
parameterization
vector
calculus

A friend of mine relayed this challenge to me. I don't have an answer.

Consider all (rectifiable?) paths in $ \mathbb{R}^2$ from $ (0, \minus{} 1)$ to $ (0,1).$ Minimize the time required to traverse the path if speed depends only on position and is given by the law
\[ \frac {ds}{dt} \equal{} x^2 \plus{} y^2.\]
The friend's conjecture (which I cannot verify): there is no minimum time path, but the infinum over all paths of the time required is $ 2.$ To demonstrate that infinum, choose $ R > 1,$ (and we will let $ R\to\infty$), take the line segment from $ (0, \minus{} 1)$ to $ (0, \minus{} R),$ then the semicircle centered at the origin of radius $ R$ from $ (0, \minus{} R)$ to $ (0,R),$ then the line segment from $ (0,R)$ to $ (0,1).$ The time taken to traverse this path is $ 2 \plus{} \frac {\pi \minus{} 2}{R}.$

So here's the question: can you beat that? Is there a path with time $ \le 2?$

Yes. I've confirmed the change of variables; the velocity is indeed one everywhere in the new coordinate system. Thanks, Joel.

Let me repeat here what I've already sent via PM:

Very glad to be of help! My very first thought was that this was the same (under some projection) as walking on a sphere at constant speed; I realized that was wrong (for a variety of reasons), but it led to the idea of other transformations that would preserve circles centered at the origin and so to inversion. I actually decided that this would result in a constant-speed image without thinking at all about the possibility that one might end up traveling in different speeds in different directions, so I got a bit lucky :)

You know, when I learned inversion as an olympiad problem-solving tool, I was always rather skeptical that it would ever prove useful for anything. I guess I should feel chastened now :)

Actually, one additional question: it's not terribly hard to verify that the velocity of the image in both the radial and tangential directions will be exactly 1 -- is that enough to conclude that the image velocity is 1 in every direction?

[quote="JBL"]
Actually, one additional question: it's not terribly hard to verify that the velocity of the image in both the radial and tangential directions will be exactly 1 -- is that enough to conclude that the image velocity is 1 in every direction?[/quote]
Even one direction is enough because inversion is a conformal mapping.

What's the full theorem you're invoking?  Conformal maps preserve the property "speed at a point is independent of direction"?

[quote="JBL"]Apply inversion around the origin, sending $ (x, y) \mapsto \frac {(x, y)}{x^2 \plus{} y^2}$.  If I'm not mistaken, the velocity of the image point is just 1 everywhere (but it cannot pass through the origin) and your friend's conjecture follows immediately.  This also lets us answer all sorts of other questions about this model.[/quote]

I have a little trouble following this: For my sake, define $ x: \equal{}x(t),\mbox{ and }y: \equal{}y(t)$ and put $ \vec{r}\, (t): \equal{}<x(t),y(t)>$ for $ 0\le t\le \alpha,$ where the parameter $ t$ denotes time and the given boundary conditions hold at the end points so that the time taken is $ \Delta t \equal{} \alpha$. We have then

\[ \dot{s} \equal{} | \vec{r}^{\,\prime } | \equal{} \sqrt{\dot{x}^2\plus{}\dot{y}^2}\]

from the vector calculus, where dots denote time derivatives.  Also, ds/dt was given, hence

\[ \Delta t \equal{} \int_{0}^{\alpha}\, dt \equal{} \int_{0}^{\alpha} \frac{ds}{x^2\plus{}y^2} \equal{} \int_{0}^{\alpha} \frac{\sqrt{\dot{x}^2\plus{}\dot{y}^2}\, dt}{x^2\plus{}y^2}\]

I can rewrite this as (and it really seems like this should be an easy integral like this..)

\[ \Delta t \equal{}\int_{0}^{\alpha} \frac{|\vec{r}^{\, \prime}\, (t)|}{|\vec{r}\, (t)|^2}\, dt\]

Here I get stuck, (I've only done one other calculus of variations type problem: the tautochrone.)

Can I apply JBL's change of variables to the prior integral?  Say $ (u,v) \mapsto \frac{(x,y)}{x^2\plus{}y^2}$ which (if my calculus fail me not,) gives a Jacobian of $ \left| \frac{\partial (u,v)}{\partial (x,y)}\right| \equal{}  \frac{1}{x^2\plus{}y^2}$ provided $ (x,y)\neq (0,0)$ .

Here's a excessively grubby version of why it works:

$ \frac{du}{dt}\equal{}\frac{\partial u}{\partial x}\,\frac{dx}{dt}\plus{}\frac{\partial u}{\partial y}\,\frac{dy}{dt}\equal{}\frac{y^2\minus{}x^2}{(x^2\plus{}y^2)^2}\,\frac{dx}{dt}\minus{}\frac{2xy}{(x^2\plus{}y^2)^2}\,\frac{dy}{dt}$

Similarly, $ \frac{dv}{dt}\equal{}\minus{}\frac{2xy}{(x^2\plus{}y^2)^2}\,\frac{dx}{dt}\minus{}\frac{y^2\minus{}x^2}{(x^2\plus{}y^2)^2}\,\frac{dy}{dt}$

If we square these and add them together, we find that

$ \left(\frac{du}{dt}\right)^2\plus{}\left(\frac{dv}{dt}\right)^2\equal{}\frac{1}{(x^2\plus{}y^2)^2}\left(\left(\frac{dx}{dt}\right)^2\plus{}\left(\frac{dy}{dt}\right)^2\right)$

But we have $ \left(\frac{dx}{dt}\right)^2\plus{}\left(\frac{dy}{dt}\right)^2\equal{}(x^2\plus{}y^2)^2.$

We conclude that $ \left(\frac{du}{dt}\right)^2\plus{}\left(\frac{dv}{dt}\right)^2\equal{}1.$


hab ich ein paar frage dazu ?

wie schnell kannst du kopfrechnen ?

was soll man tun zu schnell rechnen ?

Naja, ich kann zwei zweistellige Zahlen in annehmbarer Zeit multiplizieren (10 - 30s) , es sei denn sie haben dieselbe linke Ziffer, dann gehts schneller. Bei der Division größerer Zahlen hab ich bereits meine Probleme.

Um gut im Kopfrechnen zu sein muss man das einfach oft üben. Sorry, aber ne bessere Methode fällt mir nicht ein. Warum willst du schnell kopfrechnen können?


I use texmaker, and every time i try to compile anything, it gives the error: could not start the command. help?

I really like texmaker tho, is the thing's corrupted the only possibility?

Almost certainly the paths to pdflatex etc is wrong. Check that all the paths are correct and change the ones that are wrong or missing. Don't forget to use quotes when the path contains a space.

I don't understand- do you mean the stuff under "Configure texmaker"?