Τελείωσε η φετινή Μαθηματική χρονιά και καλά να είναι να γίνεται ένας απολογισμός και η απαιτούμενη κριτική προκειμένου να διορθωθούν κάποια λάθη και να γίνουμε καλύτεροι .
Ο κ.Αλεξανδρής (τέως πρόεδρος της ΕΜΕ) σε κάποια ομιλία του είπε ότι εμείς ασχολούμαστε «ερασιτεχνικά» με τους Μαθηματικούς διαγωνισμούς σε αντίθεση με κάποιες άλλες χώρες .
Έτσι είναι .Όχι απλά «ερασιτεχνικά» αλλά θα έλεγα και λίγο "κοροιδευτικά"

Και εξηγούμαι .

1) Στα παιδιά που συμμετέχουν στους διαγωνισμούς (επί τη ευκαιρία θέλω να συγχαρώ όλα τα παιδιά που αποτέλεσαν και τις δύο εθνικές μας ομάδες γιατί χωρίς καμία βοήθεια κατάφεραν αυτά που κατάφεραν) δεν δίνεται από την ΕΜΕ καθόλου βιβλιογραφία .
Απορίας άξιο , γιατί το κάνουν .
Δεν ξέρουν τίτλους βιβλίων ?
Δεν θέλουν να δώσουν ?
Τα δίνουν σε λίγους ?

2) Δεν γίνονται μαθήματα .Αυτά τα λίγα που γίνονται είναι ασήμαντα (για να πουν ότι κάτι κάνουν) .
Σε άλλες χώρες υπάρχουν trainers και τα μαθήματα γίνονται όλη τη χρονιά δι αλληλογραφίας (!!!) .
Η πρόταση που έκανε ο nickolas (με επιστολή που έστειλε στον «άφαντο» πρόεδρο της ΕΜΕ ούτε στην τελετή του Αρχιμήδη ήταν παρών !!) να αναλάβει την προετοιμασία ο Vornicu  δεν έγινε δεκτή (φαντάζομαι ότι ούτε μπήκαν στον κόπο να τον απαντήσουν) .
Τα παιδιά πάνε στους διεθνείς διαγωνισμούς έχοντας τις ευχές των γονιών τους και της εκάστοτε επιτροπής διαγωνισμών (με ισόβια μέλη τους κ.κ.Τυρλή και Κοντογιάννη -  Η απόλυτη δυαρχία της ΕΜΕ ).
Πάμε τώρα στον Αρχιμήδη .
Τα θέματα που έβαλαν φέτος ήταν απαράδεκτα . Το 2ο και το 3ο ήταν καλά μεν γνωστά δε (έγινε τότε μεγάλη συζήτηση) .
Τότε όμως δεν ειπώθηκε ότι και το 1ο θέμα ήταν γνωστότατο !
Όπως με πληροφόρησε φοιτητής του Μαθηματικού Ιωαννίνων , το είχε βάλει ο κ. Μπόλης στην εξεταστική του Ιούνιου του 2005 και του Σεπτεμβρίου του 2005 !!
Για το 4ο δεν το συζητάμε , ήταν απαράδεκτο και ασαφές .
Στη συνέχεια είχαμε την παραίτηση του αγαπητού κ. Μπόλη από πρόεδρος της επιτροπής διαγωνισμών , την οποία η επίσημη ΕΜΕ φρόντιζαν να μη γίνει γνωστή .
Γιατί άραγε ??
Ο κ. Μπόλης καλά θα ήταν να μας έλεγε τους λόγους .Πιστεύω ότι ήταν πολύ σοβαροί !!!!
Στον Αρχιμήδη επίσης υπάρχει ένα κώλυμα της ΕΜΕ με τη διάρκεια του. Πολλοί μαθητές και γονείς ζητούν ο διαγωνισμός να έχει 4 – 4.30 ώρες διάρκεια .Άγνωστο γιατί οι κ.κ της ΕΜΕ δεν το κάνουν δεκτό .Επίσης υπάρχουν καταγγελίες από γονείς ότι η επιτήρηση στον Αρχιμήδη δεν είναι καλή . Οι μαθητές άνετα μπορούν να συνεννοηθούν και να μιλήσουν .Θα πει κανείς ότι αυτό δεν τον συμφέρει .Σωστό .
Αλλά αν τους συμφέρει ? Για σκεφτείτε το . Κάποιοι  γονείς είχαν κάνει επώνυμες καταγγελίες στο παρελθόν .
Στην Μεσογειάδα το 3β ερώτημα ήταν λάθος !!! Κανείς δεν ανέλαβε την ευθύνη .Τι έγινε , λάθος στη μετάφραση ή ήταν από την αρχή λάθος .Καλά δεν τα λύνουν τα θέματα προτού τα βάλουν ?
Επίσης θέτω υπόψιν σας για προβληματισμό και το παρακάτω post του κ. Dulac (ο οποίος φαίνεται ότι ξέρει πολλά) .
http://www.mathlinks.ro/Forum/viewtopic.php?p=503458&search_author=DuLac&show_results=posts&sort_dir=desc&return_chars=200#503458

Αυτά καταθέτω για αρχή και στη συνέχεια του διαλόγου (ήρεμου και δημιουργικού) που θα προκύψει θα προσθέσω και κάποιες άλλες παρατηρήσεις μου .

Την κριτική πρέπει να την επιζητούμε . Μας βοηθάει να γίνουμε καλύτεροι .

Φίλε Στάρτ αφού θέλεις κριτική θα έχεις κριτική η οποία θα είναι ίσως κι πολύ σκληρή σε κάποια σημεία ανατρεπτική κι χιουμοριστική  σε κάποια άλλα αλλά παντού εποικοδομητική Αλλά θα ήθελα να σου θυμίσω τι είχε γίνει κατά την βαλκανιάδα όπου εσύ είχες κάνει καλή αλλά κι υπερβολικά επιθετική κριτική κι έδωσες έδαφος σε έναν ανύπαρκτο που δεν είχε το θάρρος να δώσει το όνομά του να κατηγορήσει τους πάντες εγώ του αντιστάθηκα όσο γινόταν καθότι μου απηύθυνε βαριές εκφράσεις αλλά το τελικό αποτέλεσμα ήταν να αναστατωθεί να αποδιοργανωθεί κι να ζημιωθεί το ελληνικό φόρουμ που ο μόνος σκοπός του είναι να χτίζει κι όχι να καταστρέφει θέλω να πιστεύω ότι το έκανες κατά λάθος κι από αυτήν την πίστη μου σου εφιστώ την προσοχή έτσι ώστε να μην επαναληφθεί διότι σε αυτήν την περίπτωση θα πιστέψω ότι το κάνεις εσκεμμένα δεν θέλω να με παρεξηγήσεις όμως θα παλέψω με νύχια κι με δόντια να μην ξανασυμβεί αυτό το συμβάν γιατί εγώ εκτιμώ πολύ αυτά τα παιδιά που αγαπούν το φόρουμ το αγαπώ κι εγώ κι σε μία μαθηματικά αδύναμη σήμερα-λόγω πολιτείας- Ελλάδα το να υπάρχουν παιδιά που αγαπούν μια τέτοια  ιδέα είναι αξιοθαύμαστα κι πρέπει να προστατεύονται
Φυσικά κι εγώ είμαι στην απόλυτη διάθεση όλων για οποιαδήποτε είδους κριτική όσος σκληρή κι αν είναι διότι εγώ έχω συνηθίσει στην κριτική μέχρι κι τα όρια της συκοφαντίας σε βάρος μου ακόμα κι από καθηγητή Μαθηματικών του σχολείου μου η κριτική σας μπορεί να γίνει μέσω όποιου μέσου επιθυμείτε αλλά κι μέσω προσωπικού μυνήματος θα ήθελα όμως την κριτική εναντίον μου να την δούνε όμως κι όλοι για αυτό εγώ θα την δημοσιεύσω αν μου δοθεί σε προσωπικό μήνυμα καθαρός ουρανός αστραπές δεν φοβάται ΠΟΤΕ
Στο διατάυτα όμως αφού θες κριτική ορίστε
1.	ερασιτεχνικά δεν θα διαφωνήσω επαυξάνω σε αυτό όμως το κοροϊδευτικά είναι λίγο υπερβολικό δεν είναι κοροϊδεύουν κανένα απλώς επιδεικτικά αδιαφορούν πολιτεία κράτος κι άλλοι φορείς μεταξύ αυτών κι κάποιοι αλλά όχι όλοι από την ΕΜΕ αν θες ονόματα δεν έχεις παρά να με προκαλέσεις(με την καλή την έννοια πάντα)κι εγώ θα σου αραδιάσω ονόματα με επεξηγήσεις άρα στο κοροϊδευτικά εγώ θα έβαζα {με σχετική αδιαφορία}
2.	Στις ερωτήσεις για την βιβλιογραφία
3.	Προφανώς κι ξέρουν αλλά θεωρούν δεδομένο ότι χρησιμοποιούμε την βιβλιοθήκη της ΕΜΕ (εγώ 4 χρόνια που πηγαίνω συχνά ούτε την χρησιμοποίησα ούτε είδα ποτέ κάποιον να την χρησιμοποιεί)
4.	Δεν τους αφορά θεωρούν υποχρέωση σου όλα να τα ξέρεις όσον αφορά την βιβλιογραφία  κι όσον αφορά αυτήν κάθε αυτήν την γνώση αυτό εξηγείται από 2 παράγοντες ο πρώτος αφορά την ελληνικό γένος που δεν αλλάζει να βοηθήσει τον συμπολίτη ή τον συνάνθρωπο να ανέβει αλλά τον προτιμά χαμηλά για να μπορεί να διακριθεί ο ίδιος κι δεύτερον λόγω του «ερασιτεχνικά» αφού δεν πληρώνονται ως γνήσιοι Έλληνες πάντα δεν τους ενδιαφέρει να βοηθήσουν σκέφτονται αν θες να διακριθείς βγάλτα πέρα μόνος σου σε ένα 13-17 χρονών παιδί αν είναι ποτέ δυνατόν
5.	Πιθανώς ναι συμβαίνει σε αρκετές περιπτώσεις αν έχει κάποιος φίλο καθηγητή στην Ελλάδα φίλοι μου ισχύει το καθεστώς του «μέσου» αν έχει φίλο γνωστό συγγενή κάπου οπουδήποτε θα εξυπηρετηθείς ειδεμή τρέχα γύρευε αυτό ίσως είναι το μοναδικό πράγμα που δεν πρόκειται να αλλάξει ποτέ αφού ο Έλληνας δύσκολα αλλάζει την ιδιοσυγκρασία του μακάρι η δική μας  κι οι επόμενες γενιές να το αλλάξουν άρδην αυτό
6.	Για τα μαθήματα
7.	Έχεις δίκιο ότι είναι ουσιαστικά ανύπαρκτη η διδασκαλία από την εταιρεία αλλά από την στιγμή που δεν υπάρχει επαγγελματισμός κι όλα είναι ερασιτεχνικά να είμαστε ευχαριστημένοι που υπάρχουν έστω κι αυτά τα μαθήματα ότι ο πρόεδρος της ΕΜΕ είναι σαν να μην υπάρχει το ξέρω κι εγώ διότι τόσες φορές που πήγα δεν τον είδα ποτέ σε αντίθεση με τον ΚΥΡΙΟ με Κ κεφαλαίο Αλεξανδρή ο οποίος τακτικά  επισκεπτόταν τα γραφεία όσον       αφορά το θέμα του φίλτατου Valentin Vornicu παρότι είναι εξαίρετος κι στο φόρουμ αλλά κι στους αγώνες στους οποίους έφτιαξε το όνομά του πιστεύω ότι δεν έπρεπε να επιλεγεί διότι θα υπήρχε πρόβλημα επικοινωνίας  εκτός κι αν υπήρχε ένας πολύ καλός μεταφραστής παρόλα αυτά κρίνω κι εγώ επιτακτική την ανάγκη ενός πολύ καλού προπονητή είτε από την Ελλάδα είτε από το εξωτερικό κι την άμεση κι εντατική δουλειά του
8.	Όσον αφορά τους Κύριους Τυρλή κι Κοντογιάννη
9.	Οι περισσότεροι από το φόρουμ γνωρίζουν ότι έχω μια κόντρα κι δεν το κρύβω όπως βλέπεις ούτε κι εγώ με τον Κύριο Κοντογιάννη κι ως εκ τούτου δεν θα εκφέρω καμία γνώμη καθότι θα θεωρηθεί υποκειμενική επειδή δεν μπορώ ούτε κι να μην το σχολιάσω καθόλου θα πω ότι πάει πάρα πολύ η ιστορία με αυτόν τον κύριο ο οποίος το μόνο που ξέρει είναι να βρίζει τα παιδιά (καταγγελίες συμμαθητών κι φίλων μου κι η δική μου) κι το μόνο άλλο που ξέρει είναι να είναι ισόβια μέλος της επιτροπής διαγωνισμών χώρις να συμβάλλει κι πάρα πολύ εκτός από ένα μάθημα κι 2-3 σημειώσεις για συνδυαστική γεωμετρία κι γεωμετρία δεν έχει κάνει κι τίποτα άλλο άσχετα με τα βιβλία τα οποία ας σημειωθεί ότι τα έκανε πριν 25 χρόνια ! άρα δεν μπορεί κάποιος να με κατηγορήσει για υποκειμενισμό Όμως για τον Κύριο Τυρλή επειδή τον ξέρω κι ξέρω τι έχει κάνει για την εταιρεία πόσο αγαπάει τα παιδιά πόσο δένεται με αυτά σε ΚΑΘΕ αποστολή που συμμετέχει πόσες ώρες την ημέρα τρέχει για την εταιρεία που δεν του δίνει κέρδος ενώ θα έπρεπε θα ήθελα να μην τον σχολιάσει κάποιος κακά κι  αν θέλει να το κάνει για κάθε κακό σχόλιο ας έχει μια πολύ τρανταχτή δικαιολογία κι όχι μπουρδολογία όμως θα συμφωνήσω στον χαρακτηρισμό η απόλυτη δυαρχία μόνο στο χρονικό καθαρά κομμάτι της φράσης κι όχι απόλυτα μόνο κατά το ήμισυ στο καταγγελτικό περιεχόμενό της
10.	Για τον Αρχιμήδη
11.	Αναλυτικά έχω όλες τις πηγές των θεμάτων και αφού τις γνωστοποιήσω θα τοποθετηθώ Α πρόβλημα Εξεταστική Ιωαννίων Θεόδωρου Μπόλη Ιούνιος 2005-Σεπτέμβριος 2005 Β πρόβλημα Baltic Way 2003 problem 2 Γ πρόβλημα Baltic Way 2003 problem 3 Δ πρόβλημα Θεόδωρου Μπόλη προτεινόμενο θέμα για το μαθηματικό της Αθήνας  κι ασαφές όπως πολύ σωστά λες φίλε Σταρτ Αυτό είναι κατεξοχήν ντροπή αφετέρου λογικό από μια μαθηματικά αδιάφορη όπως προείπαμε χώρα κι μόνο σε εμάς που αγαπάμε τα Μαθηματικά κι την χώρα μας μάς πονά κι μας πληγώνει ελπίζω κάποια στιγμή να το συνειδητοποιήσουν αυτοί που διοικούν την ΕΜΕ αν κι το βλέπω  δύσκολο
12.	Για την παραίτηση του κυρίου Μπόλη
13.	Το ότι ο κύριος Μπόλης παραιτήθηκε μυστικά (για προσωπικούς λόγους όπως ανακοινώθηκε) δείχνει σκοτάδι κι προξενεί πονηρές σκέψεις κι τι θα άλλαζε όμως αν το μαθαίναμε? Εγώ πιστεύω ότι κι αν το μάθουμε ποτέ κι αν όχι (που δεν θα το μάθουμε ποτέ στην Ελλάδα ζούμε τώρα) η κατάσταση θα παραμείνει η ίδια άσχημη εικόνα άρα καλό θα είναι να κοιτάμε μπροστά στο μέλλον κι να μην ασχολούμαστε με το παρελθόν Φυσικά όμως κι θέλω κι θεωρώ υποχρέωση των υπευθύνων κι του Κ Μπόλη να ενημερώσει τους λόγους της αποχώρησής του
14.	Για την διάρκεια του Αρχιμήδη κι την επιτήρηση
15.	Προσωπικά θεωρώ ότι οι 3 ώρες είναι ενδεικτικές  κι ότι δεν θα αλλάξουν κι πολλά αν αυξηθεί με την διάρκεια του διότι ας μην λέμε ό,τι θέλουμε αν δεν μπορείς να λύσεις τα θέματα σε 3 ώρες δεν θα μπορέσεις εύκολα τουλάχιστον να το κάνεις σε 1-1.5 ώρα έξτρα χρόνο πιθανώς έτσι σκεπτόμενοι οι της ΕΜΕ  δεν το έκαναν δεκτό το αίτημα Η επιτήρηση επίσης θα συμφωνήσω ότι είναι άθλια κι θα πρέπει να προσεχθεί αυτό πολύ αν θέλουμε να κρατήσουμε σε υψηλό επίπεδο το κύρος της ΕΘΝΙΚΗΣ ΜΑΘΗΜΑΤΙΚΗΣ ΟΛΥΜΠΙΑΔΑΣ εγώ συγκεκριμένα στον Ευκλείδη ο διπλανός μου με είχε τρελάνει φωνάζοντας τον μπροστινό του (Κώστα Κώστα-Κώστα) σκέτη αηδία αυτό το πράγμα το ότι δεν τον συμφέρει όμως φίλε Στάρτ καταδεικνύει ένα άλλο στοιχείο της αιώνιας ελληνικής ιδιοσυγκρασίας (το ό, τι με συμφέρει κι στα τσακίδια όλα τα άλλα) αν δεν σε συμφέρει κύριε (γενικός χαρακτηρισμός κι κλήση δεν απευθύνεται σε συγκεκριμένο πρόσωπο) ΝΑ ΜΗΝ ΣΥΜΜΕΤΑΣΧΕΙΣ ΣΤΗΝ ΕΘΝΙΚΗ ΜΑΘΗΜΑΤΙΚΗ ΟΛΥΜΠΙΑΔΑ τώρα επειδή γίνονται υπαινιγμοί κι για συμφέροντα από την ΕΜΕ θα πω ότι από τις καταγγελίες μπορεί να συμπεράνει κανείς ότι υπάρχουν αλλά θα πρέπει να μην είναι απόλυτος αν δεν έχει απτά στοιχεία στα χέρια του γιατί δεν είναι κι σωστό θίγει έμμεσα κι τον εαυτό διότι τον θέτει στα μάτια των άλλων πιθανό ευνοούμενο της κατάστασης
16.	Για την Μεσογειάδα κι το λάθος θέμα
17.	Αυτό το πρόβλημα το αντιμετώπισαν όλες οι μεσογειακές χώρες που έδωσαν αυτόν τον διαγωνισμό κι τα θέματα τα έχουν προφανώς λυμένα άρα πιο πιθανή εκδοχή είναι το από την αρχή λάθος κι δεν πρέπει να κατηγορούμε σε τελική ανάλυση το θέμα δεν έδωσε ανισότητα μεταξύ των διαγωνιζόμενων καθότι όλοι πήραν τους βαθμούς της 3β
18.	Τέλος για τον DULAC κι του σχολίου του
19.	Αρχικά θα σου κάνω λίγο κωμική κριτική Γιατί φίλε μου Στάρτ σε πειράζει που ξέρει ο DULAC τόσα πολλά θα τον φιμώσεις ώστε να μην μιλήσει ? ή θα τον σκοτώσεις επειδή «μίλησε»? Τι το τόσο παράνομο άτιμο ή πονηρό βρίσκεις σε αυτό ? πάντως για το σχόλιο του DULAC θα το σχολιάσω ότι το α ψάχνει ο Κ Ζώτος για υπαρχηγό δεν μου λέει τίποτα τουλάχιστο τίποτα πονηρό το σχόλιο για τον κύριο Μπόλη κι τον Κύριο Φελλούρη είναι πληροφορίες που τις ξέρουν κι οι πέτρες κι όποιον κι αν ρωτήσεις τις ξέρει άρα δεν συνιστά ούτε κι αυτό κάποιον πονηρό προβληματισμό
Δεν νομίζω ότι σε άφησα παραπονεμένο φίλε μου Στάρτ κλείνοντας θα ήθελα να σου ευχηθώ καλή συνέχεια κι καλή μελέτη δουλεία  κι τα ίδια θα ήθελα να ευχηθώ σε όλους μας ενόψει της νέας μαθηματικής χρονιάς η οποία ελπίζω να είναι με επιτυχίες
Με συναγωνιστικούς χαιρετισμούς

ΟΧΙ ΠΑΛΙ ΡΕ ΠΑΙΔΙΑ  :mad:  :mad:  :cursing:

[quote="ambros"]
16.	Για την Μεσογειάδα κι το λάθος θέμα
17.	Αυτό το πρόβλημα το αντιμετώπισαν όλες οι μεσογειακές χώρες που έδωσαν αυτόν τον διαγωνισμό κι τα θέματα τα έχουν προφανώς λυμένα άρα πιο πιθανή εκδοχή είναι το από την αρχή λάθος κι δεν πρέπει να κατηγορούμε σε τελική ανάλυση το θέμα δεν έδωσε ανισότητα μεταξύ των διαγωνιζόμενων καθότι όλοι πήραν τους βαθμούς της 3β
[/quote]

Σαφώς η διατύπωση σου δεν αφήνει αμφιβολία αλλά [b] όλοι [/b] πήραν τους βαθμούς? Επειδή συμμετείχα (όχι βέβαια πως είχα αξιώσεις)...

Αγαπητέ φίλε μου έτσι είναι ΟΛΟΙ ΑΝΕΞΑΙΡΕΤΑ τους πήραν τους πόντους κι μην λές οτι δεν έχεις αξιώσεις γιατί ακόμα κι έτσι να είναι ΚΑΝΕΙΣ ΜΑ ΚΑΝΕΙΣ ΔΕΝ ΕΧΕΙ ΔΙΚΑΙΩΜΑ ΝΑ ΣΟΥ ΣΤΕΡΕΙ ΜΟΝΑΔΕΣ ΑΠΟ ΤΟΝ ΚΟΠΟ ΣΟΥ
Καλη Συνέχεια με επιτυχίες
χαιρετώ

Φίλε start,

  το ξέρεις ότι γενικά κινούμαι στο ίδιο μήκος κύματος προβληματισμού με σένα.Δυστυχώς ,αυτή την ιστοσελίδα δεν την παρακολουθούν πολλοί , από αυτούς που θα μπορούσαν να αλλάξουν κάτι.Από την άλλη μεριά οι μαθητές μας [b]επηρρεάζονται αρνητικά απο σχόλια που αφορούν την εταιρεία [/b], διότι δεν μπορούν να καταλάβουν τη λογική(παραλογισμό) των διοικούντων.Και εγώ τελευταία άκουσα συζητήσεις για την ΕΜΕ και τους διαγωνισμούς ,  που με τρόμαξαν. Δε θα σχολιάσω όμως τίποτα.Καλλίτερα να μείνουμε λίγο ρομαντικοί και τροβαδούροι των μαθηματικών και να μη φθαρούμε από τη γενικότερη διαπλοκή ή διαφθορά της κοινωνίας μας.[b]Συμφωνώ μαζί σου [/b]αλλά ευθυγραμμίζομαι με το Σιλουανό.Ας βρούμε άλλο χώρο να τα λέμε, το έχω και εγώ ανάγκη.Στο forum ας προσποιηθούμε ότι δεν ξέρουμε τίποτα και ας απολαύσουμε τα προβλήματα και τη φιλία.
    Και [color=red]σε άλλο βήμα ας είμαστε πύρινοι[/color] εναντίον αυτών που δεν κάνουν τα απαραίτητα για την παιδεία , την μαθηματική εταιρεία και αυτά τα παιδιά που αξίζουν και μπορούν πολλά.
Να είσαι σίγουρος ότι υποστηρίζω τις θέσεις σου και δε διαφωνώ μαζί σου.Στείλε μου ένα PM να τα πούμε. Θα χαρώ.
  Μπάμπης


pigeonhole principle
number theory unsolved
number theory

Prove that among $81$ natural numbers whose prime divisors are in the set $\{2, 3, 5\}$ there exist four numbers whose product is the fourth power of an integer.

It's very similar to an old IMO problem.

Each integer $ 2^a\cdot 3^b\cdot 5^c$ corresponds with a triple $ (a,b,c)$. There are 8 possible parity combinations, so there are two triples with the same parity combinations. So the product of the integers must be a perfect square, call this $ a_1^2$. From the other 79 integers, there exist two whose product is a perfect square, call this $ a_2^2$. Continuing this, we will get more than 8 perfect squares $ a_1^2,a_2^2,\ldots,a_9^2$. With the same method, there exist two of them whose product is a perfect fourth power. Q.E.D.

Each of these numbers can be represented in form $ 2^{\alpha _i} \cdot 3^{\beta _i} \cdot 5^{\gamma _i}$. Now we will study all $ \alpha _i, \beta _i, \gamma _i$ moduli $ 2$. So each of them can give one of the $ 2$ different rests, there can be $ 2 \cdot 2 \cdot 2 \equal{} 8$ different "oddities" of these powers. According to pigeonhole principe, there will be such two numbers that have the same their power "oddities", because $ \frac {81}{8} > 1$. Now we multiply them and denote the obtained number as $ a_1$ and remove it. And we can continue to obtain such $ a_i$ until the number of the rest numbers will become $ 7$. By that time we will have obtained $ \frac {81 \minus{} 9}{2} \equal{} 36$ such pairs.
Now we observe numbers $ \sqrt {a_i}$ - each of them is a natural number because $ a_i$ is a perfect square (. And again we observe their power "oddities" moduli $ 2$. Using the pigeonhole principe again, we obtain that there are two numbers $ \sqrt {a_m}$ and $ \sqrt {a_n}$ such that their "oddities" moduli $ 2$ will be equal, because $ \frac {36}{8} > 1$. Now we multiply $ a_m$ and $ a_n$ and we get the fourth power, because their are both perfect squares and their "oddities" moduli $ 2$ are the same, $ q.e.d.$

to end with, there are $37$ possible squares with two elements, $15$ possible fourth powers with $4$ elements and $4$ different eighth powers with $8$ elements if we count exactly (at least).

IMO 1985 P4
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366595&sid=5cb6ec4ca43e6f9bd2d6c2d3eb2b87d8#p366595

The bound I feel is very weak as 25 (in place of 81) works


linear algebra
matrix
geometry
geometric transformation
calculus
integration
vector

I want to draw up a list of conditions for solution of simultaneous linear equations
$Ax=b$ where $A\in{\mathbb R}^{m\times n},~x\in{\mathbb R}^{n},~b\in{\mathbb R}^{m}$
in all cases $m<n,~m=n,~m>n$ The case $m=n$ is well knwon.

If $m<n$, for example $A_{1,1}x_{1}+A_{1,2}x_{2}=b_{1}$, do we have a necessary and sufficient condition for the solution $x\in{\mathbb R}^{n}$ to be unique?

Does the statement "Solution exists if and only if ${\rm rank }[A]={\rm rank}[A,b]$" valid for $m<n$ and $m>n$ as well?

Think of the columns of $A$ is vectors. A solution exists precisely when $b$ is in the linear span of the columns of $A$, which in turn is equivalent to ${\rm rank }[A]={\rm rank}[A,b]$ (because rank = dimension of the column space).
The uniqueness of solution means that $b$ can be represented as a linear combination of columns in only one way. This requires an additional assumption that the columns of $A$ are linearly independent, i.e., ${\rm rank }[A]=n$.

Of course the case $m=n$ is simple and well known I guess for $m<n$, the necessary and sufficient condition is $A\sim A'=[A'_{1},0_{m,n-m}]$ where $A'_{1}\in{\mathbb R}^{m\times m}$ is non-singular

Given Ax = b, where A is an m x n matrix, a solution exists if and only if the row rank of A is equal to the row rank of the augmented matrix.

I discuss this here: http://www.mathlinks.ro/Forum/viewtopic.php?t=149836

Given that, your unique solution exists almost never. A unique solution exists if and only if each column of the rref(A) contains a pivot (and before we even check for this, the above condition has to hold, of course; you can't have a unique solution if there aren't even any solutions.) Note that this can only happen also if m>=n.

As a sidenote, you'll notice that if k columns of rref (A) do not contain pivots then the solution space is an affine translation with k free variables.

I am surprised no one answered this question citing a famous theorem which concerns itself with exactly this question:  the Fredholm Alternative theorem .

I just checked Wikipedia and it looks like they develop it for compact integral operators which is all valid and sufficiently abstract, but there is a version of the theorem for finite dimensional vector spaces and it goes like this:

The equation $Ax=b$ has a solution if and only if the inner product $<b,v> =0$ for every vector v satisfying $A^{*}v=0$.

In words, there is a solution iff the right-hand side (b) is orthogonal to the null space of the adjoint of the operator (A).

For uniqueness, we have the solution to $Ax=b$ is unique iff the solution of $Ax=0$ is $x=0$.

I hope this is helpful.


Find the number of the solutions of the system or prove, that it has infinite number of solutions.

x = y(mod z)
x - z = y

Just take $ x\equal{}2z$ and $ y\equal{}z$, then $ (x,y,z)\equal{}(2t,t,t)$ is solution for any integer t so system has infinitely many solutions.

I think should be find out all solution of this equation.
Because $ x\equiv y (\mod z)$ then
$ x\equal{}y\plus{}kz$ where $ k\in Z$
So
$ (k\minus{}1)z\plus{}y\equal{}y$ so $ k\equal{}1$
Solution of this equation are $ (x,y,z)\equal{}(a\plus{}b,b,a)$

The second equation implies the first, so if you want all solutions just take all points on the plane defined by the second equation.


You have a machine which, when you give it two real numbers a and b (with $b\neq 0$), gives you out the number $\displaystyle 1-\frac{a}{b}$. Prove that using this machine, you can add, subtract, multiply and divide any two real numbers.

  Darij

This is a very nice problem. I don't know why I like it so much :).

Let $\displaystyle (a,b)=1-\frac ab$. Then $(a,1)=1-a$, so $((a,b),1)=\frac ab$, so we have the division.

We also have $((a,b),a)=1-(\frac 1a+\frac 1b)$, so $(((a,b),a),1)=\frac 1a+\frac 1b$, and from the first paragraph we find that we can obtain $\frac 1a,\frac 1b$, so $(((\frac 1a,\frac 1b),\frac 1a),1)=a+b$.

We can find $1-(\frac 1a+\frac 1b)$, so we can also find $\frac 1-\frac 1a-1=-\frac 1a$, so we can also obtain $-a$, which gives $b-a$ after addition with $b$.

Multiplication is, of course, obtained as the quotient of $a$ and $\frac 1b$.

[quote="grobber"]We also have $((a,b),a)=1-(\frac 1a+\frac 1b)$,[/quote]

Hmm, at this step I get $((a,b),a)=1-(\frac 1a-\frac 1b)$, so I think you have a little mistake in your calculation. But actually the solution can be corrected easily.

Here is my solution:

I will also denote $\displaystyle \left(a,b\right) = 1-\frac{a}{b}$. Then, I can construct $0 = \left(a,a\right)$ (provided that $a \neq 0$; else we can take b instead of a, and if both a and b are 0, then we can actually add, subtract and multiply without the help of our machine  :D ). Then, $1 = \left(0,a\right)$ (or $1 = \left(0,b\right)$). Hence, we can force our machine to give out the numbers 0 and 1. Therefore, e can divide, since $\displaystyle \frac{a}{b} = \left(\left(a,b\right),1\right)$. Hence, since we have the number 1, we can find for every x the number $\displaystyle \frac{1}{x}$. Therefore, we can also multiply: $\displaystyle a \cdot b = \frac{a}{\left(\frac{1}{b}\right)}$. Also, we can subtract: $a-b = \left(b,a\right) \cdot a$. Since we have the number 0, we thus can also find for every x the number $-x = 0-x$. Therefore, we can also add: $a+b = a-\left(-b\right)$. And thus the problem is solved.

  Darij


I have a problem of my own. I would like all of you discuss it and give your result. Thank you very much!

Let ABC be a triangle. On the side AB, take out m points; on the side BC, take out n points; on the side CA, take out k points. These points are distinct and coincisive with the vertex of the triangle ABC. How many triangles of which vertex are those arbitrary points including A, B and C are there?

[hide]${m+n+k+3\choose 3}-{m+2\choose 3}-{n+2\choose 3}-{k+2\choose 3}$[/hide]

Dear Bojan Basic,
Thank you very much for your answer. Your result is the same as mine.


$\lim \frac{tan x}{x}$
when $x-----\to \ infinty$

I don't think there exists a limit, since the function is unbounded even when x is very large.


If we let $\vec{a}\neq-\vec{b}$ it holds ?
If not could anyone find a form of this inequality to hold .(maybe reserved).

It doesn't seem it should hold. You can always choose $\vec a,\vec b$ and/or $\vec v,\vec w$ so that the $\measuredangle(\vec a,\vec b)$ and/or $\measuredangle(\vec v,\vec w)$ are arbitrarily close to $\pi$. Then it's obvious that $|\vec a+\vec b|$ and/or $|\vec v+\vec w|$ are arbitrarily close to $|a-b|,|v-w|$ respectively and $\sqrt{|a-b|\,|v-w|}\geqslant\sqrt{av}+\sqrt{bw}$ doesn't hold for positive $a,b,v,w$, which is easily shown.

Thank you .Reserved signed  holds ? Or could we find a similar one to hold ?


trigonometry
geometry
perpendicular bisector
similar triangles

If the altitude AE and the median BF intersect at the point H of the triangle ABC, such that the perpendicular bisector of BC intersects BF and BC at the points O and D, then prove that OD = AH/2.

Let FX be the perpendicular from F to BC.

We have FX= H/2=h (where AE=H )
BX= a-b/2 cosC...a,b,c and A,B,C carry the usual meaning of sides and angles

In similar triangles BOD and BFX we have OD/FX=BD/BX....OD=ah/(2a-bcosC)

In similar triangles BHE and BFX we have HE/FX=BE/BX......HE=2chcosB/(2a-bcosC)...
AH= 2h-HE=2ah/(2a-bcosC)....(we have a=b cosC+c cosB).

Hence the result.

Here is my proof with no trigonometry (picture included!):

Consider the midpoint G of AB. Draw the auxiliary line DG. Since DG connects the midpoints of sides AB and BC, ABC and GBD are similar triangles. Now drop a perpendicular from G such that GP is parallel to OD and AE, where P is the midpoint of BH because of similar triangles. GBP and ABH are similar so GP = AH/2. All we need to prove now is that GP = OD. Notice that <PGD = <GDO and BF bisects GD, so by congruency of triangles, GP = OD = AH/2.

Say X is the point on OA and FB (both medians) then we know that 2 OX = OA (median theorem)

so <AXD = <OXH and <ODX = <AHX (OD//AE)
thus we can conclude that AH/OD = AX/OX = 2

I guess this is a little shorter ;)

sorry AD is the median

so AX = 2 XD

so <AXH = <OXD

and <XOD = <AHX (OD//AE)

thus we can conclude that AH/OD = AX/DX = 2

that will do better :)


Let 1<a[size=75]1[/size]<a[size=75]2[/size]<...<a[size=75]51[/size]<142.Prove that among the $50$ consecutive differences, (a[size=75]i[/size]-a[size=75][i]i-1[/i][/size]); [i]i[/i]= $2,3,...,51,$some value must occur atleast 12 times.

Note that $a_{1}\geq 2$ and $a_{51}\leq 141$ so that $a_{51}-a_{1}\leq 139.$
In another hand, if no difference appears more than $11$ times, we have
$a_{51}-a_{1}= (a_{2}-a_{1})+(a_{3}-a_{2})+\cdots+(a_{51}-a_{50})$
$\geq 1 \times 11+2 \times 11+3 \times 11+4 \times 11+5 \times 6 = 140$, a contradiction.

Pierre.


This monday I'm having a math test which will be based on statistic and there are some questions concerning the suggested subject which I'm having trouble understanding. Believe me when I say I've tried but I still can't figure them out... So if anyone would be a dear and explain how you could solve these two problems I would be forever grateful.

1. A couple of hand crafters are standing in a row. Four of them are carrying a folding rule in their overall and six of them are carrying a hammer in their belt. The chief then picks randomly five of them. Each of the handcrafters that were picked, goes back to the line. What are the odds that the chief choses exactly two handcrafters who carries a folding rule?

2. A company did a survey in which they found out that 48% of the workers had a positive attitude towards doping tests.
The company then selects 200 employees, randomly to decide how many are interested in having doping control tests at work. Calculate the probability that at least half are positively inclined to adopting a doping.

Thanks in advance to anyone who can easen my burd!

I hate to bump and I'm not even sure if its allowed but I would really appreciate if someone could take on these two questions....

Once again, sorry

[quote="Swiifty"]This monday I'm having a math test which will be based on statistic and there are some questions concerning the suggested subject which I'm having trouble understanding. Believe me when I say I've tried but I still can't figure them out... So if anyone would be a dear and explain how you could solve these two problems I would be forever grateful.

1. A couple of hand crafters are standing in a row. Four of them are carrying a folding rule in their overall and six of them are carrying a hammer in their belt. The chief then picks randomly five of them. Each of the handcrafters that were picked, goes back to the line. What are the odds that the chief choses exactly two handcrafters who carries a folding rule?
[/quote]I'm assuming there are ten people to choose from. There are $\binom{10}{5}= 252$ ways to choose five and $\binom{4}{2}\binom{6}{3}= 120$ ways to choose exactly two with a folding rule and three with a hammer. So the probability in question is $\frac{120}{252}= \frac{10}{21}$. This is an example of the [url=http://mathworld.wolfram.com/HypergeometricDistribution.html]hypergeometric distribution.[/url]
[quote]
2. A company did a survey in which they found out that 48% of the workers had a positive attitude towards doping tests.
The company then selects 200 employees, randomly to decide how many are interested in having doping control tests at work. Calculate the probability that at least half are positively inclined to adopting a doping.[/quote]Use the [b]Central Limit Theorem (CLT).[/b] Let $X$ be the number of employees in the sample with a positive attitude, then $X \sim B(200,.48)$ (binomial distribution) and therefore by CLT approximately $X \sim N(96, \sqrt{.48 \times .52 \times 200}) = N(96,7.065)$. Then
\[Prob(X \ge 100) = Prob(X > 99.5) = Prob(\frac{X-96}{7.065}> \frac{3.5}{7.065}) \approx Prob(Z > 0.50) = ... \]
(Look it up in a table, $Z$ is standard normal).

Hope this helps.

Hey Guys, thank you for helping a poor fellow.

I have another problem, which buggs me enormously. It goes like this:

10 persons work at a company.  6 of them use a computer. The boss wants to pick out 4 of them, randomly, and offer them a health care examination.

What is the probability that at most half of the persons chosen work with a computer.


i thought we weren't suppose to talk about this until after the 25th....
i feel stupid i think i got a 21 22 or 23 but my mistakes could've been avoided. what's the arml?

it doesnt matter if you say what you got or how you did, just as long as you dont discuss the problems.

max: im not sure anyone can tell you what they're looking for, since different places have different ways of picking their teams.  The best thing would just be to move on and just do better on the next contests that you do.  They wouldnt overlook you just because you did not do that well on one exam, so long as you show that you can do well on others.

[quote="cats..."]The best thing would just be to move on and just do better on the next contests that you do.  They wouldnt overlook you just because you did not do that well on one exam, so long as you show that you can do well on others.[/quote]

That makes sense. For programs that emphasize writing proofs, a test like the AMC 8 isn't going to tell the whole story. For programs of multiple-choice problem-solving, the level of the AMC 8 is different from the level of other programs, so it isn't always a perfect predictor of performance on tests of secondary-level math.

Just move on and try something else is generally a good philosophy of life. Don't give up too soon--find another way to prove yourself.

I don't know specifically about ARML contests, but I hope to have my son participating in them in a few years.

I don't think any ARML team uses the AMC 8 to pick their team members . . . but where do you live?   What team are you competing for?

For the person who asked about ARML -- it's the American Regions Math League, a nationwide math contest for (mostly) high school students.  http://www.arml.com.  If you do a search of this site, you can find much more information about it.

Probably fairly competitive, that.  You should check to see if they even look at AMC 8 scores for middle schoolers trying to make the team.  Anyhow, don't be discouraged if you can't beat out all the high schoolers from the second-largest state in the country . . .


Hello, I have discovered that I have a lack of inginuity in Number Theory.  Would you please recommend some books that will improve my problem solving skills involving Number Theory?  Thanks in advance.

Masoud Zargar

[quote="blahblahblah"]This is hardly an open question.

Anyways, try Andrews' Number Theory (published by Dover).  I'm not a huge fan of it, but it has a good selection of problems AND solutions.  The kicker is that it's like 15 bucks.[/quote]
Oh, okay.  Thanks.  I'll try ans see if Chapters has it. :)

Masoud Zargar

Well,

  a) D.Burton , Elementary Number theory. This is a text book

   b) Adler and Coury. Many solved problems. One of the best.(expensive !)

   [u]Babis[/u]

A friend of mine recently referred me to a great number theory book that has a really good mix of substance and focused problem solving at an accessible level. Elementary Number Theory with Applications by Thomas Koshy. According to my friend, all math "majors" at Duke's TIP program take a course based on this book.

There is book called the "Elementary Number Theory and Its Applications(Fourth Edition)". It was written by Kenneth H.Rosen. How do you guys think about this book?

I have "Classical introduction to Modern Number Theory" by Rosen which is an expanded and modified version of what you wrote (if I remember correctly preface  :D ) and I like it very much. Every chapter has some exercises plus I love bibliography given at the end of every chapter and the way authors direct the reader to other texts.

I haven't read too much of it yet, but I like "The Higher Arithmetic" by Davenport so far.

Has anyone read a number theory book by Sierpinski?
If so, was the book good?

I've taught a class using Rosen, and for teaching the class, I would choose Rosen again. We expect our students to take number theory before they take abstract algebra (groups-rings-fields), and this course is required for all of those going into high school teaching. For that audience, It may be a fairly difficult book for that audience (and we only try to do a little over half of the book in a semester), but it does work as a textbook.

One of the books stergiu mentioned was Burton. That's been used around here as well, but Rosen has more life to it, more applications, and better exercises.

Wait, when you guys say that the book has excersizes, are you differentiating between problems and excersizes like Zeitz?

Well, this is number theory so there is no big difference between an exercise and a problem :D

[quote="Arne"]I like Sierpinski very much, in fact :)[/quote]

If you have a Ti-84, 84+, 84 se, or 84+ se, you can make a program that creates the Sierpinski triangle. If you don't have the code, I'll give it to you.

This is not about the Sierpinski triangle :p but rather about the book Sierpinski wrote. He was a great number theorist.


Support
function
probability
geometry
linear algebra
matrix
articles

Someone in the thread about Intelligent Design (ID) was complaining that ID was not supported by evidence. I'd like to create this thread to discuss possible evidence for both theories.

Please give me some evidence for evolution that's conclusive enough to pass the tests that people are using on ID.

I think the one most commonly cited example in biology textbooks for easily observable evolution (not a long-term fossil record sort of thing, which is what we're primarily interested in, but over a time period that we can actually observe) is that one about butterflies or moths or something in Britain (they may have been some other species, it doesn't matter). The general story is that they originally used to be white because they hid on the light-colored trees; as the Industrial Revolution progressed, the trees became black with smoke and soot, and accordingly, the animals that were born had progressively darker tints until they were pretty much back. Then, environmental regulations helped the trees become light in color again, and indeed, the species returned to its previous state.

We deal with evolution every year when we make up flu vaccines.

The flu virus mutates very quickly.  There are many strains as environmental conditions (including immune system conditions and vaccinations) force dramatic shifts in evolutionary stability of various genetic patterns.

In fact, this happens in many viruses.

Another piece of evidence is the precision of the model of genetic evolution when we overlay DNA patterns.  About a decade ago I worked at the Institute for Biomedical Computing where genomes from various species were compared for deviations using a program called BLAST, which is ultimately a brute-force base pair matching program.  The differences between the DNA of various species matches up beautifully with how far apart we would expect those species to be evolutionarily.  In other words, if we assume that no matter what type of genetic selection occurs, that DNA mutates (mutations/insertions/deletions) at a particular rate based on its complexity, then, for instance, humans and chimps are expected to be genetically similar to a certain degree.  And according to BLAST...we are indeed similar to that degree!

Those who refuse to incorporate genetic evolution into their worldview are either not aware of the tremendous evidence or are intentionally turning a blind eye to science.  It's not just a matter of evidence.  We observe on a small scale something that when extrapolated to a large scale equals evolution.  We don't deny relativity because we can't observe it from moment to moment -- so why deny evolution?

[quote="ToBe"]Yeah, but that stuff only proves microevolution and adaptation. It does not prove that is how life formed.[/quote]

I'm not just talking about microevolution when I point at that point accepted mutation (PAM) time models of evolution match up perfectly with mapped genomes and it has absolutely zero to do with adaptation.

And genetic evolution is not a theory of how life formed, but rather, how it proceeds.

Q: Are humans intelligent designers?

I'd like to refer to the Creationism/Intelligent Design thread.  The only evidence (correct me if I'm wrong) presented for intelligent design was the supposed complexity of life forms today.  Life can be corresponded with computer hardware/software today.  Computers today combined with their wonderous software are so complex, they could not have been developed without devine intervention.  Does this seem reasonable?

 Another common complaint from the Creationism/Intelligent Design thread says that Intelligent Design is not a religious theory, if it can be considered a theory at all.  Look around you.  Do you see atheist ID supporters?  Now can you claim ID to be a non-religious "theory"?

I'd also like to point out, if some people dont know it, that Darwin himself was a believer in a creator, and several times in Origin of Species he says "nature does not make transitions" - nature only makes jumps. (and as a believer, the proof is that at the end he mentions something about all being possible from a breath of a creator.)

I'd also like to ask a question: why is this even an argument? If one looks closely at Genesis, it mentiones NOWHERE anything about special individual creation for each animal. The only creature that it says in specific is man. For general evolution, aside from man (i'll get back to y'all later on man) there is no real dispute, no reason for it. Everyone just 'assumes' that all creatures according to Genesis were created specially. That is not true. ~ in truth, there doesnt even have to be a dispute about religion vs. science, because when both percieved from an unbiased view, and from the view of someone that actually has researched the material on both sides, they are actually interdependent~ - I'd reccomend the book 'The Science of God' by Gerald L. Schroeder.

- yes i know this is on creationism, not quite intelligent design, but all the same.. *i realized this afterwards. maybe i should revive old threads.*

[quote="MCrawford"]I'm not just talking about microevolution when I point at that point accepted mutation (PAM) time models of evolution match up perfectly with mapped genomes and it has absolutely zero to do with adaptation.[/quote]

I'm sorry, but I didn't quite follow that sentence. Could you please rephrase it? Thanks.

[quote="mathnerd314"]Someone in the thread about Intelligent Design (ID) was complaining that ID was not supported by evidence. [/quote]
Who said that, exactly (quote the original text, please)?   ID is consistent with anything and its opposite, so there is no possibility of evidence for or against it.

[quote] I'd like to create this thread to discuss possible evidence for both theories.  [/quote]
There are very few legitimate reasons to discuss both together.  It appears more as a rhetorical tactic than shedding light on either theory.   ID doesn't rise to the level of what is usually called a "theory", by the way; to do that it would have to make at least one meaningful statement about the natural world.

[quote] Please give me some evidence for evolution that's conclusive enough to pass the tests that people are using on ID.[/quote]

One test that ID fails and evolution passes is "does it say anything meaningful whatsoever".
As it happens, the assertions of evolution are generally borne out by observation, and this is why it ends up being taught in the classroom; but all that is subsequent to the first hurdle, which ID doesn't pass, of actually saying something.  That the quality of fit between theory and observation for evolution can actually be debated is a huge, knockout-level point in its favor as compared to ID.  Even literal bible-thumping creationism does better than meaningless ID in that respect, although it ends up being discarded in light of observation.

Anyway, the thread for evolution is the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23486

Has anyone heard of the Burgess Shales? Some smithsonian guy went on a mission to canada and found fossils that proved that gradual evolution did not exist, as he found dif creatures jumping from small creatures to fully formed, fully capable ones. He didnt believe in the stuff, so he hid it in his lab for 80 years till it was discovered, made news like 10 years ago.

er, yes. I think currently accepted theory does in fact reveal that Darwin's idea of very, very gradual mutations was in fact not exactly correct and that changes are slightly more abrupt (there's a term for this, but I can't remember it ... took AP Bio 2 years ago now...)

What's the [b]evidence[/b] for punctuated equilibrium, please? As I understand it, there's more evidence for Darwinian evolution than punctuated equilibrium.

[quote="mathnerd314"]What's the evidence for punctuated equilibrium, please? As I understand it, there's more evidence for Darwinian evolution than punctuated equilibrium.[/quote]
These are not theories on the same level; they are consistent with each other, and seek to explain different facets of the history of life on earth.

Darwinian evolution: Natural selection, forcing adaptation to local conditions, is the primary cause of change in species over time. Also, history is very long, and even this inefficient method produces great change over such timescales.
The power of selection as an agent of change is undeniable- see what we have done to our domestic animals and crops. This human selection is positive rather than negative, and therefore works much faster

Punctuated equilibrium: Most change, especially speciation, happens relatively rapidly in small isolated populations. Successful and widespread species like those we see in the fossil record are extremely unlikely to change significantly except in abundance, even over millions of years.
The historical evidence here supports the second sentence- the first is an inference from the second and from statistics. This theory does not really deal in methods- it accepts natural selection.

You might have been looking for gradualism as a foil, and that is very hard to support with historical evidence, because the theory denies the "equilibrium" part of punctuated equilibrium. How would you explain the great profusion of species unique to geologically young small islands?

Some modern notes, which have come to light with DNA analysis. In closely related species, the genes which matter are much more similar than the sections of DNA that don't code for anything. This suggests that natural selection normally acts as a barrier to change by striking down variations that don't work. On the other hand, random change is predictable, moving at an easily calculated rate which depends inversely on population size- we can even estimate how long two modern species have been distinct this way.

As for the Burgess Shale- find "Wonderful Life" by Stephen J. Gould. Many of the details are now obsolete, but it's a great book. A central theme, which the title brings up, is the unpredictability of life's history, at every level. Even if the processes are deterministic, their interactions are so complex that nothing is inevitable from a broad picture.

Hey guys- in case you guys are still interested -

http://www.pennlive.com/news/patriotnews/index.ssf?/base/news/1130667685324780.xml&coll=1

[quote]The Discovery Institute, one of the nation's leading organized proponents of intelligent design, said it warned the Dover Area School District not to institute a policy on the concept because of the risk it would be found "somehow unconstitutional."[/quote]

http://aclupa.blogspot.com/
Closing statements of the K v DASD trial.
[quote]The defendants' attorney Patrick Gillen noted that the day was the fortieth day of the trial, and that night was the fortieth night. He asked if the judge had done that on purpose.

The judge said with a smile that it was "not by design." The entire courtroom broke out in laughter and then applause.

And with that, the biggest trial on evolution in two decades came to a close.[/quote]

[quote="ToBe"]I noticed that too. As I said, I didnt think that all of those questions were even valid (some were just...easy to answer too).[/quote]

Then tell us which ones you think are valid, so we don't waste our time answering all the questions on the site?

zeus_three is absolutely right.  We talked about this in the first day of science this year, and why we do not teach ID in [b] public [/b] school.  ID is pseudoscience, whether true or not.  For something to be considered science, it must be empirically demonstratable.  This is why I do not believe theories of multiple universes can be currently considered science either.  There are no tests we can perform to prove ID correct.

stop questioning questions that don't need answers

[quote="zeus_three"]
Religious teachers [i]do not teach quantum theory [/i]and [u]special relativity [/u]in [b]Religious studies lessons[/b], so [i]why are they even contemplating teaching ID [/i]in [b]science lessons[/b]?!  [b]The absurdity confounds me, and the logic is non-existent.[/b]
[/quote]

 :D My religion teacher taught us a bit about special relativity, because of Einstein and his connection with the Big Bang.  Its actually in the curriculum. edit:  its in the provincial curriculum that is.

In other words, don't say things you obviously don't know anything about.  If you're not in a religion class, how do you know?  Or if you actually are, then you sure aren't learning anything from it.

[quote="Sunny"] My religion teacher taught us a bit about special relativity .... Its actually in the curriculum ....  In other words, don't say things you obviously don't know anything about.  [/quote]

It is possible that somewhere in the world in a class labelled Religion, people are learning how to solve physics problems.  Nevertheless, the other poster's statement is clear enough, and clearly correct, that "special relativity and quantum mechanics are not taught in religion classes".   If your religion class has problem sets on relativistic velocity addition or inflationary cosmology, as opposed to the historical/philosophical/sociological matters supposedly related to modern physics, that would be unusual, but it would not justify treating his self-evident observation as some sort of absurd hallucination.

There are different selection criteria for the curricula in science and non-science classes, and (see posting 219 above) ID doesn't qualify for the former.

[quote="ToBe"]Um, why is [something both irreducible and complex] the expected result [of evolution]? Things with no goal other than survival do not have to strive for optimization, only what is required.[/quote]

That's right, which is why (as I said several times before) evolution ruthlessly optimizes systems that are critical to survival and reproduction, and does not necessarily optimize other systems.  The optimized systems will tend to be "irreducible", and most evolved systems of any kind will also be "complex", so as pointed out many times but answered zero times in this thread, "both irreducible and complex" is a signature of evolved systems.   It is brain-dead for ID to use a signature of evolution as attempted proof that evolution did not occur.

I think in general the tension in this debate arrises from a matter on which compromise must occur, but few are willing to accept.

What i am referring to is the existence of probabalistic systems. It is preferrable that science would always be able to say with certainty, this happened in so many years, or that is impossible. But when probabilities are dealt with, certainty in these issues are impossible. However, this does not mean that probabilities are never of any use! Although i am no expert, i think that in physics the uncertainty principle makes schrodinger's wave equation the best possible location of an election, but does that make quantum mechanics useless?

To return to the earlier references in this thread dealing with the likelihood of royal straight flushes... It is [i]always[/i] possible that a royal straight flush is dealt fairly, and hence it is always possible that 20000 royal straight flushes, are dealt fairly, and again that 2^20000 royal straight flushes are dealt fairly. But does this mean that if 2^20000 royal straight flushes occurred, it might not be necessary to question the validity of the occurrence?

My point is this. Probabilities cannot be viewed as 100% distinct from scientific truths. I think any reasonable person can agree, that if the evolution from the earliest form of life to the human being were supposed to occur in a period of 100 years, that evolution would be impossible.

I think that it would be perfectly scientific to analyze the probability of the occurrence of the evolutionary process, (for i think we can agree that natural selection is not of itself enough to produce the changes that exist, but that mutations subject to probability must play a part), and construct an expected value for length of the process thereby. Should this deviate significantly from the known length of time during which the process occurred, evolution would be impossible.

I have heard several ID theorests argue this point, specifically William A. Dembski, who contrary to the generalizations of ID theorests as stupid, possesses doctorates in both mathematics and philosophy, in his book [i]The Design Inference: Eliminating Chance through Small Probabilities[/i]. Those who continue to argue the ID is completely unfounded and worthless,(not that everyone here is) would do well to take a look at this book and cease discussing what they do not know.

[quote="lotrfan"]... if 2^20000 royal straight flushes occurred, it might not be necessary to question the validity of the occurrence?  [/quote]The question is how to credibly arrive at those astronomically low probabilities, not  what to do given such a very small number.   ID has to do the first before it can start the drumbeats about the second.   So far the probability models in ID literature discredit ID, not evolution.

[quote] Should this [expected length of time] deviate significantly from the known length of time during which the process occurred, evolution would be impossible. [/quote] All you have to do is to find a reasonably inclusive probability model where that can be demonstrated, then sit back and collect the Nobel Prizes.

[quote] I have heard several ID theorests argue this point, specifically William A. Dembski, who contrary to the generalizations of ID theorests as stupid, possesses doctorates in both mathematics and philosophy, [/quote]
"By their fruits shall ye know them".

Dembski's public displays of inanity are quoted and linked in this thread, so we don't need any academic pedigree to calibrate his competence.   But it would be interesting to find out where his degrees are from and what his alma mater thinks of him.

[quote]The question is how to credibly arrive at those astronomically low probabilities, not what to do given such a very small number.[/quote]

i agree completely. The task of determining the expected length of time of the evolutionary process is monolithic, but it is still an unanswered question, a question which can either prove evolution beyond all shadow of doubt, or leave it in need of deep-seated reform.

Such a significant question should be of interest to scientists of whatever religious persuasion or of whatever camp, and i see no grounds for the consideration of an entity that attempts to explore this question as unscientific! Agreed, ID attempts to do more than just answer this question, and the manner in which it goes about this qeustion can discredit its scienticity(lol, a new word). My objection is more to the general consideration of all objections to evolution as unscientific. As fleeting_guest and i both agree, the expected length of the evolutionary process is as yet an unanswered question. Why is there no room for any other theories to sit alongside evolution in our textbooks? Even if no specific theory is given, people should know that there are still defensible grounds for rejecting evolution. If no one knows the expected length of the process, a critical consideration to the establishment of its feasability, no one can know that it is true.

[quote="lotrfan"]a significant question should be of interest to scientists of whatever religious persuasion or of whatever camp, [/quote]Mathematical models of evolution are a large area of research.  The ordinary, non-ID approach is to just study the models and draw whatever conclusions are warranted.   The ID approach is to assume certain conclusions and look for models that might support those.   Since the ID starting point is almost certainly wrong, it is no surprise that the resulting models are irrelevant, nonsensical or un-quantifiable.

[quote]grounds for the consideration of an entity that attempts to explore this question as unscientific! [/quote]ID acquires its abysmal reputation by ignoring 100 years of data in favor of half-baked wishful thinking, meaningless assertions and pseudoscientific posturing.  Instead of fighting its battles with data and relevant mathematical models in the scientific literature, it fights using advertising, lawyers, politics, and in media under the sole control of ID supporters (ID journals and web sites, ID biology textbooks, etc).  This, not the particular questions it raises, are what leads to the contempt for ID.   The questions are problematic only insofar as they are too vague to be the basis of any research program.

[quote] My objection is more to the general consideration of all objections to evolution as unscientific.[/quote]ID falsely equates criticism of its aggressive spam pseudo-objections, with dogmatic refusal of all objections.  I doubt that there are any examples of "consideration of all objections to evolution as unscientific".

[quote]As fleeting_guest and i both agree, the expected length of the evolutionary process is as yet an unanswered question. Why is there no room for any other theories to sit alongside evolution in our textbooks? [/quote]Please speak for yourself.  I don't agree with your assertions in the least, and talk of "expected length" is not an argument against evolution.    The reason there is no room for other theories is that there are, as of 2005, no viable non-evolutionary ideas to be taught, whereas evolution stands on a mountain of supporting evidence.

So... by [b]ignoring[/b] data, you can come to the conclusion of ID?

[quote="dasherm"][quote="zeb"]I have something to say about an argument I saw a while back, which said something like, it is highly unlikely that you will flip ten coins in a row and they will all come up heads. Is that any more likely than flipping ten coins in a row and them coming up heads, tails, heads, heads, tails, heads, tails, tails, tails, heads?

Holy crap, I flipped ten coins and they came up tails, heads, heads, tails, heads, tails, heads, tails, tails, tails! That was a 1/1024 chance! The coin must be rigged![/quote]

I don't know what this has to do with intelligent design, but people usually look at THHTHTHTTT, and only see 4H and 6T.  In other words, people don't usually pay attention to the order of the flips, just the number of heads and tails.  Thus, 10H has a probability of 1/1024 and 5H/5T has a probability of 63/256, much greater.[/quote]
It has to do the following:
most ID-supporters here claimed that the chances (in evolution) where to small to be occured all together.
And in evolution, the ordering of the mutations does play a big role, they cant occure in any order. So also THHTHTHTTT has just a chance of 1 to 1024.

But some people here also claim that we [u]must[/u] have developed this only way to get what we are now. But if other changes would have occured, we would just be different, and not, as some claim, not exist (possibly it would have needed some 100 millions years more or less for animals with 'human'-type intelligence to occure, but noone would notice).


inequalities
search
email
inequalities unsolved

1) Let a,b,c be positive real numbers such that $ abc\equal{}1$. Prove that:
$ \frac{1}{(a\plus{}b)b}\plus{}\frac{1}{(b\plus{}c)c}\plus{}\frac{1}{(c\plus{}a)a}\geq\frac{3}{2}$
2) Prove that following ineq: $ (3a\plus{}2b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\leq\frac{45}{2}$
where $ a,b,c \in [1;2]$
3) Let x,y,z be non-nagetive real numbers. Find the min value of
$ F\equal{}(xy\plus{}yz\plus{}zx)(\frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{(x\minus{}y)^2})$

2. It's problem in Mathematics and Youth Magazine  :mad:
See here http://www.mathlinks.ro/viewtopic.php?t=197349

[quote="mehdi cherif"]1) $ a \equal{} 2$  ,  $ b \equal{} 2$  ,  $ c \equal{} \frac {1}{4}$   :wink:[/quote]
If $ a \equal{} b \equal{} 2,c \equal{} \frac {1}{4}$
$ \rightarrow \frac {1}{(a \plus{} b)b} \plus{} \frac {1}{(b \plus{} c)c} \plus{} \frac {1}{(c \plus{} a)a}\equal{} 2\frac {1}{8}$

[quote="mr.danh"]
3) Let x,y,z be non-nagetive real numbers. Find the min value of
$ F \equal{} (xy \plus{} yz \plus{} zx)(\frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(x \minus{} y)^2})$[/quote]
I think  $ F \equal{} (xy \plus{} yz \plus{} zx)(\frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(y \minus{} z)^2} \plus{} \frac {1}{(y \minus{} z)^2})$
It is an inequality form VMO 2008,the min value is $ 4$
http://www.mathlinks.ro/viewtopic.php?search_id=1330177774&t=185643

I don't understand your idea about Pr 1. The ineq is true for every x,y,z satisfying xyz=1.

[quote="mr.danh"]1) Let a,b,c be positive real numbers such that $ abc \equal{} 1$. Prove that:
$ \frac {1}{(a \plus{} b)b} \plus{} \frac {1}{(b \plus{} c)c} \plus{} \frac {1}{(c \plus{} a)a}\geq\frac {3}{2}$
2) Prove that following ineq: $ (3a \plus{} 2b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\leq\frac {45}{2}$
where $ a,b,c \in [1;2]$
3) Let x,y,z be non-nagetive real numbers. Find the min value of
$ F \equal{} (xy \plus{} yz \plus{} zx)(\frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(x \minus{} y)^2})$[/quote]
The first is in my ebook at toanthpt.net. Here is my solution for it :)
[b]Problem.[/b](VasC) Let $ a,b,c$ be positive real numbers. Prove that
\[ \sum \frac {1}{a(a \plus{} b)} \ge \frac {3}{2\sqrt [3]{a^2b^2c^2}}
\]
[i]Solution[/i]

Can you please post the exact address of your email book at at toanthpt.net, can_hang2007.
Thank you very much.

[quote="Sunjee"]plz share it us. :roll:[/quote]
Here is it (but in Vietnamese) and maybe there is some mistakes in typing, please forgive me if you found them...

Thank you very much, can_hang2007 for this BEAUTIFUL GIFT at all people very keen on inequalities :)


f is defined from $ R to R$ .find all $ F$ such that the folllowing holds,
$ f(x^2\minus{}y^2)\equal{}(x\minus{}y)( f(x) \plus{} f(y) )$

* $ y \leftarrow 0 \Rightarrow f(x^2) \equal{} x [ f(x) \plus{} f(0) ] \forall x$
* $ x \leftarrow 0 \Rightarrow f(\minus{}y^2) \equal{} (\minus{}y) [ f(0) \plus{} f(y) ] \forall y \Rightarrow f(\minus{}x^2) \equal{} (\minus{}x) [ f(0) \plus{} f(x) ]$ (change of variable).
Then
\[ f(x^2) \plus{} f(\minus{}x^2) \equal{} 0 \forall x\]
\[ \Rightarrow f(x) \plus{} f(\minus{}x) \equal{} 0 \forall x\]

* $ y \leftarrow (\minus{}y) \Rightarrow f(x^2\minus{}y^2) \equal{} (x\plus{}y) [ f(x) \plus{} f(\minus{}y) ] \equal{} (x\plus{}y) [ f(x) \minus{} f(y) ]$
From this & the hypothesis, $ 2y f(x) \equal{} 2x f(y) \forall x,y$

Then $ f(x) \equal{} \frac{2x f(1)}{2 \times 1} \equal{} x f(1) \forall x$.

* In conclusion, $ f(x) \equiv ax$.

No matter which $ a$ you choose, $ f(1) \equal{} a$.


Here is the $COWBOYS$ tale
In the land of $cowboys$ , at the doom day, $n$ $cowboys$ were dueling. we know that the distances between the $cowboys$ were all different from each other. we know that by the alarm of the city clock , $cowboys$ started shotting eachothers. each $cowboy$ shot the nearest one to himself. prove that:
1. at least one $cowboy$ survived to rule the kingdome.
2. dedicate that at least how many $cowboys$  lost thier lives?

[quote="lomos_lupin"]Here is the $COWBOYS$ tale
In the land of $cowboys$ , at the doom day, $n$ $cowboys$ were dueling. we know that the distances between the $cowboys$ were all different from each other. we know that by the alarm of the city clock , $cowboys$ started shotting eachothers. each $cowboy$ shot the nearest one to himself. prove that:
1. at least one $cowboy$ survived to rule the kingdome.[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=28690 for this problem.

[quote="lomos_lupin"]2. dedicate that at least how many $cowboys$  lost thier lives?[/quote]

Do you think there is a closed form for this number in terms of n ? I don't know, but already the case n = 10 is considerably difficult. See http://www.kalva.demon.co.uk/short/soln/sh00g7.html for this case.

  Darij

as i have posted it in $Geometry Unsolved Problems$  i didnt know the solution but i  asked one of our trainig courses proffs (the problem was suggested by iran)
as far as i found out , the problem would be enourmesly hard for greater $n$s. i am sorry for posting it here
it seems it needs stronger tools


function
integration
search
real analysis
real analysis unsolved

If $ f$ is a nonnegative, continuous, concave function on the closed interval $ [0,1]$ such that $ f(0)=1$, then \[  \int_0^1 xf(x)dx \leq \frac 23 \left[ %Error. "diaplaymath" is a bad command.
\int_0^1 f(x)dx \right]^2.\]

[i]Z. Daroczy[/i]

I think it is easier to prove the stronger version

[hide="hint"]$ \frac32 \int_{0}^1 xf(x)dx\le \int_0^1f(x)dx \minus{}\frac14$[/hide]

Does it help to think of it as:

$ \bar{x} \le \frac{2}{3} \int_0^1 f(x) \,dx$

where $ \bar{x}$ is the center of mass for $ x$?

Here is the proof for the stronger version.
http://www.mathlinks.ro/viewtopic.php?search_id=476848550&t=202028


geometry
real analysis
advanced fields
advanced fields theorems

1. Let there be 'some' points in a plane such that distance between any two them is < 1. What is the minimum area/shape which can inscribe (cover) all these points?

2. You grab an area by putting infinite points on a plane. Only condition - any two points you put should be <1 apart. What is the maximum area you can grab?

Any pointers? I think I know the answer but am struggling to justify it. Also if we are able to show (1) above (2) follows directly from that.

Thanks

2. This is called the [color=blue]isodiametric problem[/color]. It has been shown by Bieberbach that the open disk of diameter $ 1$ has largest area.

1. This  is called the [color=red]Lebesgue universal cover problem[/color] and is currently unsolved. There are too many facts to say about it - if interested, Google it.


Prove that if $a+b+c|a^2+b^2+c^2$,then there existe infinite many natural numbers $n$,such that $a+b+c|a^n+b^n+c^n$.

This problem is (if I remember well), from a Romanian contest, and has already been posted/solved.

[quote="Hawk Tiger"]let $u_n=a^n+b^n+c^n$ then $u_{n+2}=(a+b+c)u_{n+1}-(ab+bc+ca)u_{n}+abcu_{n-1}$[/quote]
And what then?


Let $ ABC$ is a triangle, let $ M, N, P$ are midpoints of $ AB, AC, BC$ respectively. Let $ (ABN), (ACM), (AMN)$ are circumcircles of $ \triangle{ABN}, \triangle{ACM}, \triangle{AMN}$ respectively. Let $ BN$ meet $ CM$ at point $ G$, let $ (ABN)$ meet $ (ACM)$ at point $ D$, let $ AD$ meet $ (AMN)$ at point $ E$. Prove that $ GE // DP$.

[color=darkblue][u][b]Note[/b][/u]: Hi, all my friend! I don't like use Argentina TST 2009 to prove this problem.[/color]

I have read problem at your link. If we use one of all solution those for prove $ \frac {AP}{AQ} \equal{} \frac {3}{2}$ then $ \Longrightarrow GE // DP$, I think it's not well and I don't like this way.


How many numbers are in the list $ \minus{}33, \minus{}28, \minus{}23, \ldots, 52, 57?$

Okay, so difference between each term is $ 5$, so the number of terms is $ \frac{57 \minus{} (\minus{}33)}{5} \plus{} 1$, or $ \boxed{19}$.

I used ernie’s strategy. It’s quick and easy and you can use it for any question like this one.

[quote=pirsquared]I like pianolover's solution...but i just subtracted 2 from all the numbers so its: -35, -30, -25,...50, 55...and divided by 5. There are 8 numbers from -35 to 0 plus the 11 positive ones = 19. Its less efficient than most but thats how my brain works :D[/quote]

Alternatively I [i]added[/i] $3$ like the Alcumus Solution.  :-D

[quote=RTB][quote=technoblade]Add 34 to each number and then get
1, 6, 11, . . . 91.
Then subtract 1 from each number
0, 5, 10, . . . 90
Then divide everything by 5 to get
0, 1, 2, . . . 18.
Then add 1 to every number to get
1, 2, 3, . . . 19
So the answer is 19[/quote]
OG username, nice. o7 Techno.

I have clinical depression. I answered $18$, then $17$.[/quote]

This was my first problem on Alcumus. I answered $18$, then $16$ :oops:.


algebra
polynomial
superior algebra
superior algebra unsolved

Explain why $ \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2)$ is not a Galois extension. Find the smallest extension field $ E/\mathbb{Q}(^3\sqrt2)$ so that $ E/\mathbb{Q}$ is Galois. Determine the isomorphism class of $ \text{Gal}(E/\mathbb{Q})$.

I know how to prove that $ \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2)$ is not a Galois extension. But I don't know how to do the next two parts of this question. Thanks in advance.

What's the splitting field of $ x^3 \minus{} 2$?

One important property of Galois extensions is that every irreducible polynomial which has a root in such an extension also splits in that extension. What's a good example of an irreducible polynomial with a root in this extension?


LaTeX
calculus
derivative
real analysis
real analysis unsolved

One more problem :
Let g(x)=f(x+2)-2f(x+1)+f(x). Suppose that f(x) is twice differentiable on R. Apply the mean value theorem twice to prove that, for any given a, there is a point b in (a,a+2) such that g(a)=f''(b).

for convenience, let a=-1, i.e. we consider f,g in (-1,1)

let h(x)=(f(x)-f(0))/x for non-zero x, and h(0)=f'(0)

recall Taylor's:
f(a+y) = f(a) + yf'(a) + y^2f''(a+ky)/2 , 0<k<1

a=x, y=-x yields:
f(0) = f(x) - xf'(x) +x^2f''(x-kx)/2 , 0<k<1

which means
(xf'(x) - (f(x)-f(0)))/x^2 = f''(mx)/2 , 0<m<1

but LHS=h'(x)

so g(a)=g(-1)=f(1)+f(-1)-2f(0)=h(1)-h(-1)=2h'(c) (-1<c<1)
                                                             =f''(b)   some b in (-1,1)  QED

I apologize for these clumsy texts as I am quite new to latex

Thanks for the nice solution but,
is there any other proof without using taylor's theorem, just by applying the mean value theorem twice, as the question suggests?

Taylor's theorem is proven by repeatedly applying the mean value theorem, so the two approaches are more or less identical.  You end up essentially reproving a special case of Taylor's theorem.

[quote="georgech"]is there any other proof without using taylor's theorem, just by applying the mean value theorem twice, as the question suggests?[/quote]
Consider $ h(x) \equal{} f(x \plus{} 1) \minus{} f(x)$. Your expression is $ h(a \plus{} 1) \minus{} h(a)$. Apply the MVT to $ h$. What do you see?

I see that the derivative of h at a is equal to the expression h(a+1)-h(a). But does this expression has anything in common with the fact i want to prove?

And what exactly is the derivative of $ h$ in terms of $ f$? ;)

ok, but, how will the second derivative will appear?

Once you have $ h(a\plus{}1)\minus{}h(a)\equal{}h'(c)\equal{}f'(c\plus{}1)\minus{}f'(c)$, you just apply the MVT to $ f'$ on $ [c,c\plus{}1]$.


function
combinatorics proposed
combinatorics

Prove the following identities:
a)
\[ \sum_{k \equal{} 0}^n( \minus{} 1)^k2^{2n \minus{} 2k}\binom{2n \plus{} 1 \minus{} k}{k} \equal{} n \plus{} 1;
\]
b)
\[ \sum_{k \equal{} 0}^{\left[\frac {p(n)}{2}\right]}( \minus{} 1)^k2^{p(n) \minus{} 2k}\binom{p(n) \minus{} k}{k} \equal{} p(n) \plus{} 1,
\]
where $ p(x)\in \mathbb{Z}[x]$ and $ p(x) > 1$ for all $ x\in \mathbb{N}.$

This doesn't feel like the "right" proof for a)  (mainly on aesthetic grounds...more on this after the argument), but here goes anyways:

We will actually prove the following:
$ \sum_{k\equal{}0}^n (\minus{}1)^k 2^{2n\plus{}1\minus{}2k} \binom{2n\plus{}1\minus{}k}{k}\equal{}2(n\plus{}1)$

We view the left hand side as follows:  How many ways are there to tile a $ 2n\plus{}1 \times 1$ strip by red $ 1 \times 1$ squares, blue $ 1 \times 1$ squares, and uncolored $ 2 \times 1$ dominoes?  To do so, we first need to pick some number $ k$ of dominoes (at least 0, at most $ n$).  There will be $ 2n\plus{}1\minus{}2k$ $ 1 \times 1$ squares, so $ \binom{2n\plus{}1\minus{}k}{k}$ ways to place the dominoes relative to the squares.  We then color the squares in one of the $ 2^{2n\plus{}1\minus{}2k}$ ways.

In other words, the left side would exactly count these arrangements, except for that $ (\minus{}1)^k$ term.  So now what we aim to do is to "pair off" the arrangements with an even number of dominoes with those with an odd number.   We do so as follows:  Start from the left of the row of blocks and move right until we encounter either

a) a domino with a $ 1 \times 1$ square immediately to its left.  We replace it by two identical squares of the opposite color of that $ 1 \times 1$ square.  Or

b) a pair of identically colored squares with a $ 1 \times 1$ square of the opposite color immediately to their left.  We replace them by a domino.

This operation is its own inverse, and changes the parity of the number of dominoes, so the sum cancels everywhere the operation is defined (as it cancels on each pair $ (x, f(x))$).  However, there are some configurations where it is not defined.

What are those configurations?  No domino can appear to the right of a colored square (else we could apply a) ), so they consist of some number of dominoes, then colored squares thereafter.  To avoid being able to apply b), the colored squares must consist of some number of identically colored squares, followed by squares alternating in color.

In other words, we can define our remaining configurations by picking some number $ k$ of dominoes to start with, picking the color of the next square (2 choices), and choosing how long the next stretch of that color is ($ 2n\plus{}1\minus{}2k$ choices).  There are therefore $ 2(2n\plus{}1\minus{}2k)$ choices with $ k$ dominoes remaining, so the whole sum is $ 2( (2n\plus{}1)\minus{}(2n\minus{}1)\plus{}(2n\minus{}3)\minus{}...\plus{}1)\equal{}2(n\plus{}1)$.

----------------------------
Now, at the beginning I said I thought this was the wrong argument.  There are two reasons for this.

1. It doesn't generalize to part $ b$.  But I don't think any pure combinatorial proof would do that anyways, so I don't mind that too much.

2. I dislike that I still had more counting and cancellation even after defining my involution.  The "right" involution would have left exactly $ 2(n\plus{}1)$ elements unpaired, all of which would have an even number of dominoes (0, perhaps?)

Therefore my challenge is

[b]Define a better pairing on the configurations I defined in my proof, in the sense of objection 2.[/b]

[quote="kevinatcausa"]

1. It doesn't generalize to part $ b$.  But I don't think any pure combinatorial proof would do that anyways, so I don't mind that too much.

[/quote]
Are you sure ?  :lol:

Not at all actually.  I guess I was misled somewhat by the notation $ p(n)$...there's no real reason to think of it as a function of $ n$ at all, at least for my proof method.  Here's a reworked proof, with what I consider to be the "right" bijection:

As before, we view the left hand side as counting colorings of a strip of length $ p(n)$ with red and blue $ 1 \times 1$ blocks and colorless $ 2 \times 1$ dominoes, with signs determined by the number of dominoes.  Given a configuration $ A$, we define $ f(A)$ as follows.

Read from left to right until the first time we encounter either a domino or a red block immediately to the left of a blue block.  We replace the domino by Red/Blue, and Red/Blue by a domino.  As $ f(f(A)) \equal{} A$, the sum over all pairs $ (A,f(A))$ is 0.

What's left is all the colorings containing no dominoes, and no red blocks to the left of blue blocks.  In other words, there is some number $ j$ of blue blocks ($ 0 \leq j \leq p(n)$), followed by red blocks.  There are $ p(n) \plus{} 1$ choices for $ j$, so the sum is $ p(n) \plus{} 1$.


are H20 and H402 the same, explain.

yeah , i know this is kind of elementary , but what do you expect from a 5th grader who can't even make it past AIME?

H4O2 is probably referring to two hydrogen bonded molecules of water.


calculus
integration
trigonometry
function
calculus computations

Evaluate the following integrals using Laplace Transform:
1. $ \int_{0}^{\infty} e^{\minus{}tx^2} dx$.
2. $ \int_{0}^{\infty} \cos tx^{2} dx$ . Hence evaluate, $ \int_{0}^{\infty} \cos x^{2} dx$

Pl help in these problems. Thanks  :)

please can anyone explain me  what is laplace transformation?

Let f(t) be a function defined for t > 0. Then, the Laplace transform of f(t), denoted $ \mathcal{L} [f(t)]$, is, by definition, given by

$ \mathcal{L} [f(t)] \equal{} \int_0^{ \plus{} \infty} f(t) e^{ \minus{} st}\,dt \equal{} F(s)$,

where F(s) is a way of saying that the transform is a function of the variable s. Of course, for the Laplace transform of f(t) to exist the integral in the definition must converge. Sufficient, but not necessary, conditions of convergence are piecewise continuity and exponential order. One the more important applications of the Laplace transform is in solving differential equations and systems of differential equations. Look [url=http://en.wikipedia.org/wiki/Laplace_transform]here[/url].

hello
1. $ \int^{\infty}_{0}e^{\minus{}tx^2} dx$
putting $ tx^2\equal{}y , dx\equal{}\frac{dy}{2\sqrt{ty}}$
=$ \int^{\infty}_{0}e^{\minus{}y}\frac{dy}{2\sqrt{ty}}$
=$ \frac{1}{2\sqrt{t}}\int^{\infty}_{0}e^{\minus{}y}y^{\minus{}\frac{1}{2}} dy$
As u know
$ \int^{\infty}_{0}e^{\minus{}x}x^{n\minus{}1}dx \equal{}$Gamma (n)
=$ \frac{1}{2\sqrt{t}}$Gamma($ \frac{1}{2}$)
=$ \frac{\sqrt{\pi}}{2\sqrt{t}}$
Thank u


[color=darkblue]Bueno........ :? me defiendo en algo!!...pero más allá todos estarán(y los que no deberian hacerlo) estudiando idiomas basicos para la difusion de articulos(como el ingles), pero buena iniciativa de consultar como va la gente de ingles :) [/color]

[quote="Pascual2005"]Bueno, queremos ver como esta el nivel de ingles en la comunidad latina para difusion de articulos:

Algun Comentario?[/quote]

Yo estoy en mi 12º año de Inglés; no soy un experto ni mucho menos, pero lo básico lo sé, y puedo tanto escuchar, como leer, escribir y hablar.

Sobre el tema de si saber un idioma o no es importante (además de la lengua principal, obviamente) creo que si, que es una herramienta tanto para enriquecerce culturalmente como para trabajar.

[color=blue]Desde principios del año pasado estudio inglés. Puedo afirmar que aprendí basante, pues entiendo bastante bien cuando alguien me habla en dicho idioma y, a pesar de que no me resulta nada sencillo, puedo también responder. Lo he podido comprobar en la I.M.O. de este año.
Por supuesto, siempre es más fácil entender el lenguaje escrito.

A pesar de ello, tengo ciertas complicaciones con el inglés matemático. (De hecho, recién pude averiaguar lo que significa [i]odd[/i]…  :blush:)[/color]

Igual me cuesta un poco. Pero con el tiempo, mejorará.

Saludos


Natalie and Asya can finish a job in 1.5 hours. Megan and Asya can finish a job in 2 hours. Natalie and Megan can finish a job in one hour. How long would it take them to finish if all three worked on the job?

My answer
[hide]$ \frac {12} {13}$[/hide]

Their answer
[hide]$ \frac {13} {12}$[/hide]

The reason I disagree
[hide]If one pair can finish it in one hour, why would the addition of a worker increase the amount of time before the job was done?[/hide]

Here is why $ \frac{13}{12}$ is right:

Let $ a, m, n$ be Asya, Megan, and Natalie respectfully.  We then have:
\[ (\frac1n+\frac1a)\frac32 = 1
\]
\[ (\frac1m + \frac1a )2 = 1
\]
\[ ( \frac1n + \frac1m )1 = 1
\]

Divide each equation by the number that is multiplying the parenthesis.  You get:
\[ \frac1n + \frac1a = \frac23
\]
\[ \frac1m + \frac1a = \frac12
\]
\[ \frac1m + \frac1n = 1
\]

Add all those and divide by $ 2$.  You get the final answer of $ \boxed{\frac{13}{12}}$

I see why you think it should be less, but really, I don't know.

the reason that it would be greater than an hour is because the pesron who is NOT involved in that one hour duo has less productivity, so the other two have to work for less time with someone who has less, but still SOME productivity, and still does some work. when you think about it, it makes sense-the duo who could do it in an hour isn't getting enough work OR time to do it.

Actually 12/13 is the right answer.  The work that 007math posted wasn't finished.  When you add the 3 equations together and divide by 2, you are finding 1/a + 1/m + 1/n = 13/12, but that is the amount of the job that the three working together complete in one hour.  If they finish 13/12 of the work in 1 hour, they finished 1 job in 12/13 hour.  One lesson here is, don't ignore common sense!   :)


Terry bicycled due north for 7 miles and due east for 24 miles. He then returned straight back to his starting point. How many miles did he bicycle on the trip?

Drawing a diagram gives us a right triangle with legs 7 and 24, because distance is defined as the length of the perpendicular line from one place to the next. You could manually do the Pythagorean Theorem to find that the answer is  25, or just recognize it as a triple.

When it says how many miles does he bicycle on his trip, is that referring to the whole entire bicycling?  If it is that, then we would have $ 7\plus{}24\plus{}25\equal{}\boxed{56}$ instead...

The answer is not 25, as they asked for the total number of miles he went. The answer, as jxl28 said, is the perimeter of the triangle, or $ 56$.


What do you mean by probability? There are infinitely many such k that the equation does have real roots, and infinitely many such that the equation does not have real roots

The problem is on The Arbelos, volume 5, chapter 3.

[quote="lo zio"]If 3*x^2 + 9*k*x + 3*x +1 = 0, where k is a real number such that k < 2, what is the probability that the equation has real roots ?[/quote]Taking the discriminant, we get $\sqrt{b^{2}-4ac}=\sqrt{(9k+3)^{2}-12}$. The equation does not have real solutions in the interval $(-\frac19 \sqrt{12}-\frac13, \frac19 \sqrt{12}-\frac13)$. Anything less than the lower bound suffices, making the probability 1. (If k had a lower bound the probability would be less than 1.)

There is a mistake in the post because the probability actually is not 1. There are infinitely many k such that the equation does not have real roots ( for example k = 0).

Since the interval $]\infty , 2[$ the desired $k$ comes from has infinite length, you have to define what probability even means (how you choose reals by random).
In a natural and intuitive way (which can also be done exact by e.g. limits), the probability of numbers that have non-real roots is indeed $0$.
This can happen, independent from the infinity of that set.

Or is the true condition something like $|k| <2$ ¿

I was going to ask as well if it was intended to be absolute value. The probability given from what my post says is the infinitely bounded interval length divided by itself plus the finitely bounded interval length, thus it is 1. This is due to the continuous nature of the variable.

[quote="Hokkage"]This is due to the continuous nature of the variable.[/quote]
To be even more precise, the cardinality of any interval $(a,b)$, with $a<b$, is the same as the cardinality of $\mathbb R$ itself. It is this property that leads to the probability being equal to 1, in this particular problem.

The probability equal to 1 is by comparing segments of finite and infinite lengths. Cardinalities don't really help... it's the length of the segment that determines it.

I just edited my post here. I think I might have been wrong.

[b]Hokkage[/b], you are right!


1/(a+2) - 2/(b-3) = -3/10

i know that we need to get a common denominator and i've tried to do the rest of the problem several ways...can someone please show me how to solve the problem...please and thanks

[hide]$\frac1{a+2}-\frac2{b-3}=-\frac3{10}$

$\frac{b-3-2(a+2)}{(a+2)(b-3)}=-\frac3{10}$

$\frac{b-2a-7}{ab+2b-3a-6}=-\frac3{10}$

$10b-20a-70=-3ab-6b+9a+18$

$3ab+16b-29a=88$

Then... I don't know. Is this supposed to be diophantine? Obviously you can't solve for two variables with one equation.[/hide]

[hide]

Simon's Favorite Factoring...

[/hide]

[hide]Just find all numbers so that $\frac{1}{x}+\frac3{10}=\frac{2}{y}$
$\frac{3x+10}{10x}=\frac{2}{y}$
$y=\frac{20x}{3x+10}$
$3x+10|20x, |21x+70 \implies 3x+10|x+70|3x+210\implies 3x+10|200$
which yields a finite amount of solutions $x$ which can be plugged in and checked for $a,b$ pairs

This is a cumbersome method, it is easier to just look at the problem and solve it but that is the method to use in general :) [/hide]

[hide="partial solution"]

We have the equation

$\frac{1}{a+2}-\frac{2}{b-3}=-\frac{3}{10}$

We get a common denominator:

$\frac{b-2a-7}{ab-3a+2b-6}=-\frac{3}{10}$

We cross multiply:

$3ab-9a+6b-18=20a-10b+70$

$3ab=29a-16b+88$

$3ab-29a+16b-88=0$

$ab-\frac{29}{3}a+\frac{16}{3}-\frac{88}{3}=0$

$ab-\frac{29}{3}a+\frac{16}{3}b-\frac{464}{9}=-32$

$\left(a+\frac{16}{3}\right)\left(b-\frac{29}{3}\right)=-32$

So there are multiple solutions. 12 to be precise, because there are two choices for which one is negative, and there are 3 pairs, and 2 ways to arrange those.[/hide]

Who wants to look for the solutions(there are more than one)?


Hi all;
 I saw an equation which I do not know how did we had it.
inputs:3 points,fist one is the center of the circle(a),second is in the circle(b)
and third is out of the circle(c). Suppose there is a line segment between  a and c and b is on the line.
I found this distance equation for which I do not know how we had it:
distance=(((xb-xc)*(xb-xc)+(yb-yc)*(yb-yc))*((xb-xc)*(xb-xc)+(yb-yc)*(yb-yc)))
xb:means x poistion for b
xc:means x poistion for c
yb:means y poistion for b
yc:means x poistion for c

Thanks

hi ;
 I do not know but I think I saw it somwhere in the math field,there must be someone who know it.Hopefully you will know the answer.

 Solversolver


2+y/2=3y/4-3

Again, I am sorry ahead of time for my inability to represent fractions better on this site.
I am having trouble with the problem above. I can not find the formula in my book for how to work with a variable and a constant in the numerator spot. Any help is greatly appreciated.

The way to write fractions, is by typing
[code]
$\dfrac{a}{b}$
[/code]

that comes out $ \dfrac{a}{b}$,

As to your question, these steps should help
First thing to do is isolate the variable,

\begin{eqnarray*}
2+\dfrac{y}{2}&=&\dfrac{3y}{4}-3\\
2+3&=&\dfrac{3y}{4}-\dfrac{y}{2}\\
5&=&\dfrac{3y}{4}-\dfrac{y}{2}
\end{eqnarray*}
Now, you will want to have the variable in an easy way to work with by eliminating denominators, try to finish it yourself, if you cannot do it, here are the last steps,

[hide="Solution"]
\begin{eqnarray*}
5&=&\dfrac{3y}{4}-\dfrac{y}{2}\\
5\cdot 4&=&\dfrac{3y}{4}\cdot 4-\dfrac{y}{2}\cdot 4\\
5\cdot 4 &=&3y-2y\\
\boxed{20}&=&y
\end{eqnarray*}
[/hide]


Putnam
number theory solved
number theory

Let x,y be the postive real numbers satisfying
     + x,y in not Q
     + 1/x +1/y=1
Prove that for all integer numbers n then
  exist the positive integer i such that [ix]=n
  or exist the positive integer j such that [jx]=n

Actually, I think it works both ways:

Let x,y be 2 positve reals. Then the disjoint sets {[nx] | n in N*} and {[ny] | n in N*} completely cover N iff x,y are irrational and 1/x+1/y=1. I remember having a proof, but it was kind of long and not very interesting (just using the properties of the integer part).

yes it is well-known as Beatty's sequences.

Pierre.

Here's one on the same topic:

Show that there are no 3 numbers x,y,z s.t. N is the disjoint union of the sets X={[nx] | n in N*} and Y and Z defined similarly for y and z respectively.

I think this is the correct form (not sure)..


When does it hold:

${2n \choose 10} \leq 2^{10}{n\choose 10}$ ?

Only when both sides are equal to 0, e.g. $n\leq 4$.


Not much useful, but it gotta think a bit more.

That's using the hint...

But it's gonna become a mess...


[b]1. The idea of a confidence interval is a crucial notion for working with polls and other kinds of sampling.

a. In your own words, explain what a confidence interval is.

b. Give specific examples of how the concept of a confidence interval might be used and how it is related to the concepts of confidence level and margin of error.[/b]

This is what I got so far but it is not in my own words because I couldn't understand it... please help!

1a) Confidence interval gives an estimated range of values which is likely to include an unknown population parameter, the estimated range being calculated from a given set of sample data.

b) Confidence level is the probability value (1-$\alpha$) associated with a confidence interval. It is often expressed as a percentage.

Margin of error: amount of random sampling error in survey's result.

Please help me connect these ideas.

A confidence interval is most certainly how confident you are in something.  It is technically though the percent you'd expect to have a (mean, mean proportion, etc) value within if you sample continuously.  It would be safe to say if your confidence is $95 \%$ that in $95 \%$ of random samples the true mean is within the given range.  If you're doing a mean though.

An example of using a confidence interval would be polling Americans for the presidential candidacy.  Polls report margin of errors to their proportion for those supporting a candidate, for example.  And your margin of error's definition is straightforward with what you put.  $(1-\alpha)$ is the probability of whatever you're testing occurring in random samples (aka your confidence level).


Let $ a,b,c,d$ be positive real numbers and let

$ Q_{n}(a,b,c,d)\equal{}\frac{a(a\plus{}b)(a\plus{}2b).....(a\plus{}(n\minus{}1)b)}{c(c\plus{}d)(c\plus{}2d)....(c\plus{}(n\minus{}1)d)}$

Evaluate

$ L\equal{}\lim_{n\to \infty}Q_{n}$

[quote="pankajsinha"]Let $ a,b,c,d$ be positive real numbers and let

$ Q_{n}(a,b,c,d) \equal{} \frac {a(a \plus{} b)(a \plus{} 2b).....(a \plus{} (n \minus{} 1)b)}{c(c \plus{} d)(c \plus{} 2d)....(c \plus{} (n \minus{} 1)d)}$

Evaluate

$ L \equal{} \lim_{n\to \infty}Q_{n}$[/quote]

\[ Q_{n}(a,b,c,d) \equal{} \frac {a(a \plus{} b)(a \plus{} 2b)\cdots (a \plus{} (n \minus{} 1)b)}{c(c \plus{} d)(c \plus{} 2d)\cdots (c \plus{} (n \minus{} 1)d)} \]
\[ \equal{} n^{\scriptstyle{\frac{a}{b}}\minus{}\scriptstyle{\frac{c}{d}}}\cdot\left(\scriptstyle{\frac{b}{d}}\right)^n\cdot \frac{\scriptstyle{\frac{a}{b}}\left(\scriptstyle{\frac{a}{b}} \plus{} 1\right)\left(\scriptstyle{\frac{a}{b}} \plus{} 2\right)\cdots\left(\scriptstyle{\frac{a}{b}} \plus{} n \minus{} 1\right) }{n!n^{\scriptstyle{\frac{a}{b}}\minus{}1}}\cdot\frac{n!n^{\scriptstyle{\frac{c}{d}}\minus{}1}}{\scriptstyle{\frac{c}{d}}\left(\scriptstyle{\frac{c}{d}} \plus{} 1\right)\left(\scriptstyle{\frac{c}{d}} \plus{} 2\right)\cdots\left(\scriptstyle{\frac{c}{d}} \plus{} n \minus{} 1\right)}\]
\[ \rightarrow \frac{\Gamma\left(\scriptstyle{\frac{c}{d}}\right)}{\Gamma\left(\scriptstyle{\frac{a}{b}}\right)}\cdot \lim_{n\rightarrow\infty}\left(\scriptstyle{\frac{b}{d}}\right)^n\cdot n^{\scriptstyle{\frac{a}{b}}\minus{}\scriptstyle{\frac{c}{d}}}\mbox{ as }n\rightarrow\infty\]


Let $ (M)_{t \geq 0}$ be a continuous martingale such that for every $ \omega \in \Omega$ and $ T > 0$, the variation of $ M_t(w)$ over $ [0;T]$ is finite. (i.e. $ \sup_{t_1<t_2< \ldots < t_n < T} \sum |M_{t_{i}}(w)\minus{} M_{t_{i\minus{}1}}(w)| < \infty$) Prove that $ M\equal{}0$.

Rq: I do not find this result intuitive at all? Can someone explain it ?

The intuitive explanation is easy but it doesn't prove anything. When you add "orthogonal" functions, their values at a fixed point tend to add much slower than their absolute values.

If your martingales were any decent (say, had 4th moments or something like that), then you would be able to formalize this idea using the square functions $ S(\omega)\equal{}\sqrt{\sum_i (M_{t_i}\minus{}M_{t_{i\minus{}1}})^2}$ to say that first, $ ES^4\le CEM_T^4$, second $ S\to 0$ a.s. if you take smaller and smaller partitions, and third that therefore $ EM_T^2\minus{}EM_0^2\equal{}\lim ES^2\equal{}0$, which would be the end of the story. I suspect that for general $ L^1$-martingales you can do something similar but I have not been able to figure it out in 10 minutes :(

thanks  :)   your proof seems to work the same for the $ L^1$ case. One can define the stopping time $ T_n( \omega_0 ) \equal{} \{ \sup s \geq 0 : Var(M_t(\omega_0), [0;s]) \leq n\}$. Then the stopped martingale $ M^{T_n}_t(w) \equal{} M_{\min( T_n(\omega), t)}(\omega)$ is bounded in $ \L^{\infty}$ (by $ n$), so you have showed that $ E((M^{T_n}_t)^2)\equal{}0$ for every $ t \geq 0$ and $ n \geq 0$. And letting $ n$ tend towards $ \plus{} \infty$ shows (because $ \lim M^{T_n}_t(\omega) \equal{} M_t(\Omega)$) that $ E(M_t^2) \equal{} 0$, as one wanted.

Rq: is there a more "probabilistic" way to "see" this result. I still do not have a lot of intuition with these continuous martingales  :maybe: when you think about it, what is your favorite example (except a brownian motion  :lol:  )


group theory
abstract algebra
superior algebra
superior algebra unsolved

$G$ a group whose all subgroups are cyclic. Is $G$ cyclic. And if $G$ is abelian ?

No. Take the dihedral group $D_{3}$.
If $G$ is abelian then yes, since every $p$-Sylow will be normal and $G$ will be the direct internal product of its $p$-Sylows, which are all cyclic, whence $G$ is cyclic.


The least common multiple of positive integers $a,b,c,d$ is equal to $a+b+c+d$. Prove that $abcd$ is divisible by at least one of $3$ or $5$.

He seems to want it moved ( http://www.mathlinks.ro/Forum/viewtopic.php?t=116004 ), please delete thie topic.

We can supose that $a\leq b\leq c\leq d$ and $a<d$.
$d|a+b+c+d$ implies or $a+b+c=2d$  or $a+b+c=d$.
In the first case, $3|3d=a+b+c+d|abcd$.
In the other case:
 $c|a+b+c+d=2a+2b+2c$ implies or I.) $2a+2b=4c$ or II.) $2a+2b=3c$ or III.) $2a+2b=2c$ or IV.) $2a+2b=c$.
I.) $3|6c=a+b+c+d|abcd$
II.) $5|5c=a+b+c+d|abcd$
IV.) $3|3c=a+b+c+d|abcd$
III.) $b|a+b+c+d=4a+4b$ implies or 1.) $4a=4b$ or 2.) $4a=3b$ or 3.) $4a=2b$ or 4.) $4a=b$,
1.) $mcm(a,b,c,d)=mcm(a,a,2a,4a)=4a<8a=a+b+c+d$
2.) $3|a|abcd$
3.) $3|3a=c|abcd$
4.) $5|5a=c|abcd$


calculus
integration
trigonometry
complex analysis
calculus computations

Hi, can someone please aid me with this problem.  Thank you.

Question:  If $ \alpha$ and $ p$ are real and such that $ 0<|p|<1$ and $ 0<|\alpha|<\pi$, prove that $ \displaystyle\int_0^\infty\frac{x^{\minus{}p}}{x^2\plus{}2x\cos\alpha\plus{}1}dx\equal{}\left(\frac{\pi}{\sin p\pi}\right)\left(\frac{\sin p\alpha}{\sin\alpha}\right)$.

Is complex analysis method allowed? Because it can be attacked by typicial method in elementary complex analysis.

But if it is not allowed, well... I'm still thinking... OTL

Hi sos440, I know complex analysis "reasonably" well.   :maybe:  If you have a solution, may you please post it.

For the convenience, I'll let $ q \equal{} \minus{}p$, $ \beta \equal{} \pi \minus{} \alpha$ and consider $ I \equal{} \int_{0}^{\infty} \frac{x^{q}}{x^2 \minus{} 2x \cos\beta \plus{} 1} \, dx$.

By change of variable $ x \equal{} t^2$, we have $ I \equal{} \int_{0}^{\infty} \frac{2t^{1\plus{}2q}}{(t^2 \minus{} e^{i\alpha})(t^2 \minus{} e^{\minus{}i\alpha})} \, dt$.

Let $ f(t)$ be the integrand of this integral and consider the contour $ C$ which travels the boundary of upper half-circle with radius $ R$ counter-clockwise. Then $ C$ consists of an arc $ \Gamma$ and diameter $ L$. Note that for $ R > 1$, the bounded region determined by $ C$ contains two zeros of $ f$, namely $ z_1 \equal{} e^{i\alpha / 2}$ and $ z_2 \equal{} \minus{}e^{\minus{}i\alpha / 2}$. Thus by residue theorem,

$ \int_{C} f(z) \, dz \equal{} 2\pi i \left[ \underset{z\equal{}z_1}{\text{Res}} f(z) \plus{} \underset{z\equal{}z_2}{\text{Res}} f(z)\right] \equal{} 2 \pi i  e^{\pi q i} \frac{\sin q (\beta \minus{} \pi)}{\sin \beta}$.

Also, since $ \textstyle \int_{\Gamma} f(z) \, dz$ vanishes as $ R \to \infty$, we have

$ \int_{C} f(z) \, dz \equal{} \int_{\minus{}\infty}^{0} f(x) \, dx \plus{} \int_{0}^{\infty} f(x) \, dx \equal{} (1 \minus{} e^{2q \pi i}) \int_{0}^{\infty} f(x) \, dx.$

Combining these two equations gives the desired equality.


In a triangle $ABC$ the median $AM$ is equall to $BC$ . On $AC$  we take point $D$ with $2AD = CD$ . If $N$ is midpoint of $BD$ , prove that
 $CN = CD$.

 [u]Babis[/u]

[quote="stergiu"]Well , I see that you have difficulties. The problem is proposed in Gazeta matematica 2005 - Romany. I have made a solution. I'll send it after some days. Try it again! It is not difficult but not  easy!

Babis[/quote]
Thank for sharing a problem. I think there is a rule here that people would not solve a recent problem from a journal. It has to be at least one year old. It's not fair that someone come here, gets the solution, and then go send the solution to journals to publish an unjust work and credit to someone who did not do the problem. It's just common sense.

From an other point of vew I agree with you.But I think that in this forum nobody wants to present a solution as own , if it not belong to him. Any way , SORRY!
  I do not send the solution yet.

 Babis

[i]Lets clear this topic with the solution. It took me 15 minutes to find  the solution , so I share it with you. [/i]- [b]No objections please ![/b]
 We have

    $MN = \frac{1}2 DC = AD$ , so  $ADMN$ is a parallelogram. If  $AM$ meets $ND$   at  $P$  , then  $P$  is the mid point of  $ND$   and  $AM$  .
 Since $PM = \frac{1}2 AM = \frac{BC}2$ , we take that $CP \perp BP$.
 But then , in triangle $\triangle CDN$, $CP$   is both altitude and median , so $CD = CN$  .

[u]Babis[/u]

a beautiful solution for a nice problem. Thanks for the problem.


Anyway, I take a Combinatorics elective at my school, and our instructor posed this questiona nd the class didnt finish in time before the class came to an end. He told me to post it here and challenge you all.

We have the set {1,2,3....,17,18,19,20}

we want to know how many subsets exist such that the sum of the numbers in the subset is divisible by five.

for example, {1,4} is such a subset, but {1,5,7} is not (random examples).

have fun.

i also apologize if this came out of a book, for my instructor never told me of a book to put a referance to. If anyone knows, they can put the book.

[hide]There are 4 subsets with one element.
There are 8 two element subsets containing 1 or 2 but since {2,3} is the same as {3,2}, there's one less for 3-7.  But for 8-12, there's two less since {8,2} and {8,7} have already been done.  For 13-17, there's three less since {13,2} {13,7} {13,12} has already been done.  For 18-20, there is none since all of its possiblities have been done in previous numbers.  2*4+5*3+5*2+5*1+3*0=8+15+10+5=38.
Notice this has a symmetry just like Pascal's triangle.  For example there's 1 20 element subset, since 1+2+...+20 is a multiple of five.  The only way to have a nineteen element subset that works is to subtract a single element set that works from the 20 element set, giving 4 possibilities.  So in other words, we only have to find the values for 1-10.  We sum the values for 1-9 element sets and multiply that by 2 since that will be the same for 11-19 element sets.  Then, we add the value for a ten element set and 1 (20 element set) and we're done.  I think the formula for subsets in x-element set for this problem is (20 choose x)/5 where the result is always rounded up to the nearest integer.  Now, we have 2(4+38+228+969+3101+7752+15504+25194+33592)+36952+1=209717 subsets.  Am I right?  If I am, can anyone tell me why my formula works, if it did work.[/hide]
I know, I had to use a calculator to do this, but I didn't want to calculate 4+38+228+969+4*19*3*17*16/5+... larger numbers that would take me even more time to solve.

[quote]I think the formula for subsets in x-element set for this problem is (20 choose x)/5 where the result is always rounded up to the nearest integer.[/quote]
I've never heard that theorem before...

Also, we do not multiply the number of n-element subsets from 1-10, the converse of a 10 element subset is another 10-element subset.  We need to multiply subsets from 0-9 by 2 and then add the number of subsets of size 10.


real analysis
topology
real analysis unsolved

$E \subset \mathbb{R}$ a measurable subset with null lebesgue measure. Does there always exist an homeomorphism $h: \mathbb{R}\to \mathbb{R}$ such that $E$ does not intersect $h(E)$ ?

Not if $\mathbb R\setminus E$ is of first category.

What if $E$ is assumed to be of first category instead of measure zero?  :maybe: