AwesomeMath
summer program
Princeton
college
FTW

I only came to the U.S.A 2-3 years ago, and only 2 summers, both of which I had no time to think about camp. Being a freshman in high school, I have yet to have been away from home/family for more than a week. This sounds pathetic, but I have never been a summer camp/program either. What trivial or basic things should I bring? What no-so-trivial things should I bring. I wish to know this before I apply.

Also, I live in New Jersey, thus needing a flight to UTD. I guess I would need to carry or bring a lot of stuff there, how much stuff should I bring and is it really worth it to go there if I'm from New Jersey; I have been on less than 20 flights in my life - none on my own. So along those lines, what else am I to expect?

How easy is it to get from the Airport to the Campus? I remember reading about a transportation system, but how easy is it to find the transportation system?

Also, I'm very short for my age, only 4' 7" (I know some 6th graders are this tall and they go to AwesomeMath). How easy or hard would you say this whole process would be for me? I really want to go. but I'm no too confident in the non-academic parts.

Finally, what are the campus/ apartments like? There isn't like a map of it on the website (or the UTD website)

And I read they will give you a packet, but I'd like to know a little bit beforehand.

Well. Bring a towel, swim suit (if you want), clothes to last a week or so (they have laundry machines), MONEY (for drinks, food, etc.), cards (!), cell phone, and err well. depends on what you want to do.

For the math, bring pencil, paper, and self =D. Your parents can help you with packing =P.

Ok so I'll actually organize this...

Toothbrush, toothpaste, lotion (o.o), shampoo, conditioner (cough...), shaving cream, razor (ok just ignore me), deodorant, a weeks worth of clothes, money (~$ \$$100-150 depending on your spending habits), cell phone, cell phone CHARGER, towel, swimming stuff, cards (all you need... trust me), pencils, paper, and yourself.

Not sure if that's a comprehensive list but w/e.

I went last summer from Indiana... it doesn't really matter where you come from, just go if you want to.

Is no one else from New Jersey going? If that's so, you might have problems... but it shouldn't be too hard. When you get there, just go to baggage claim and there will be counselors waiting there. They'll take you to the buses.

Your height doesn't really matter =P. The people at camp are generally friendly (though there were a few exceptions that were quite annoying) and it's not that hard to make friends. Just find a few people you're comfortable with and boom =D lots of fun.

[quote] Being a freshman in high school, I have yet to have been away from home/family for more than a week.[/quote]
hey! i was exactly in your situation last year :D

if you burn easily... sun screen.
hygienic stuff is very important -.-
extremely EXTREMELY hot in texas :O

Yeah, I was one of two people from Wisconsin to attend. but it didn't matter where i was from :) I was the only one on my flight ... but there wasnt any trouble at all...
NJ should have more people attending.... i knew some people from AAST that went last year ... and other people too :P lol.
Don't worry too much about the flight [unless you're 14 or under; because then a lot of airlines have an underage flying policy where you have to pay a little extra for flight attendants to watch /take care of you-- rather dumb] but if you're 15 or older you'll be fine. When you arrive at the airport, counselors will be looking for you, then you'll be taking a shuttle/bus to UTD
oh, make sure to know like the standard airport policies... like that liquid rule, and weight limit on your luggage. [as long as you're not a really heavy packer ... or a girl that cant part with any of her summer clothes <-- my problem. hahaha.  :P you'll be fine]

yeah, height doesnt matter . im only 3 inches-ish taller than youuu. and i had the time of my lifeee :D people at camp are pretty amazing! you'll have fun! =]]

Ok, thanks,
Now that I have talked it over with my parents and stuff, I will go..
but is it possible to get the test done by Friday?
And should I try in May instead?.. since I probably won't get this one done in time
What if the recommendation letters were already sent?
I just need to contact them and I can take the test again for $ \$$20 right?

Again, thanks for the info.
Just realized I'm not 15 yet, so another bunch of money on the plane ticket too.. soo close  :mad:

Anything else I should know about campus? I know I will be getting a packet and stuff.

How many bags of stuff did you bring?
lol AAST

Thanks for the reply.

btw,
campus map :
http://www.utdallas.edu/maps/
im really bad with maps. so i cant tell where we lived last year. anyone?

the apartments are pretty nice! air conditioning works amazingly well [lol you're going to need it] normally there are four bedrooms, a living room, a kitchen, two bathrooms in a dorm. each bedroom has two beds. so most likely you'll have a room mate. there also tends to be a counselor for every dorm or every two dorms. and they get their own room.
dorms are also based on age in general. [like if your 18 your not goign to get a 13 year old roommate lol :P ]

anyways, hope this helped :D
yayyyy cool!
ummm. im not sure. im not going back this year T_T so i havent looked at the test... but you think you can finish go ahead! :)
dont worry bout rec letters . so long as in the end they get your admission test you'll be fine. i think you only need to pay 20 bucks once.
hmmm. well different airlines have different policies. check them out. northwest [ what i flew on last year has the 14 year old one] i think united is different  but im not sure. ]
there's going to be one "restaurant" on campus. called "the pub" sells food & starbucks. :D
ummm. just one oversized, huge suitcase, my tennis racket & my backpack
wait. do you go to AAST?

noproblem :)

You should be able to finish 6-8 problems of the admission test in a day or two, inclding written solutions.  :)

Is anyone going to santa cruz?  everyone here is going to TX...

Regarding AMC/AIME scores, how do they receive them? Do they know them already from like a database or something?

I think Santa Cruz is well.. I dunno.. its just even further for me to even consider it.
Haha.. 6-8 problems... I don't think so.. But I'll try.

They need one in like princeton or like new york somewhere.. or somewhere a little northwester of the country
But CA is a start.

Well, they can't have 3 AMSP's in the same summer...that takes up 9 weeks, and some of the instructors also have to teach at MOP :\

ummm. i think you just tell them you're AMC/AIME scores...

they need to have it somewhere a little cooler. i loved texas. but the weather was insane. >< california's hot too lol

Canada FTW!

Santa Cruz actually isn't sweltering in the summer, just warm.

Most people would find Santa Cruz quite comfortable, although it's too sunny for my tastes. Heck even Palo Alto is too hot for me, I want to go live in Foster City or San Francisco darn.

hmm...what if like, you get sick there? 'cause i live in DE, which is quite far from either of the sites, and (assuming i actually get in), my parents are worried that i may get sick there and not have proper care.

i actually dont know anyone who got sick last year during camp...

but im sure proper care will be provided... like if its really bad, hospitals not too far away.

This is going to be my first year @ AMSP...

But since everyone is talking about the weather, I will give you a heads-up on that, BECAUSE I LIVE 30 MINUTES FROM UTD..

IT IS SO HOT. Temperature goes up to 110 degrees sometimes, and in like 3 minutes in the sun will cook you.. literally...

And hint hint: BRING A LOT OF SUNSCREEN, especially if you enjoy outside. But I mean, Texas is just really hot during the summer, and today, the temp went up to 87 and I hear tomorrow it's going up to 91... So yea...

I just want to know about what it's like actually at the camp.. How much freedom you have and stuff like that.. 21 days IS a rather long time.. and so.. yea.. any experts? (obviously, going to UTD)

[quote="mathtyro"]
Also, I live in New Jersey, thus needing a flight to UTD. I guess I would need to carry or bring a lot of stuff there, how much stuff should I bring and is it really worth it to go there if I'm from New Jersey; I have been on less than 20 flights in my life - none on my own. So along those lines, what else am I to expect?

[/quote]

yeah im from Nj as well and i think the flight might be like 1000 dollars its self but i think its definately worth it.

[quote="sjyoo93"]
IT IS SO HOT. Temperature goes up to 110 degrees sometimes, and in like 3 minutes in the sun will cook you.. literally...

I just want to know about what it's like actually at the camp.. How much freedom you have and stuff like that.. 21 days IS a rather long time.. and so.. yea.. any experts? (obviously, going to UTD)[/quote]

agreed bout the weather.... -.-

actually, when you get there 21 days IS NOT a long time. 3 weeks is really short when you're  having fun.
anyways. i went last year. and for the most part you have quite a bit of freedom. no going off campus though. there is a curfew i think it was 11 pm on weekdays. midnight on weekends. i think. during your free time you can do whatever you want. go to the pool/ hang out / mafia/cards. etc. there's also technically boy/girl limitations. like you cant be in a guys' room if your a girl w/o adult/counselor supervision. & vice versa. but i have to say... that rule wasnt exactly enforced very well lol :P and in addition, last day is a all nighter. do whatever you want. whenever :D oh, and no dress code =]] i think i covered most of it?

[quote="iYOA"]
yeah im from Nj as well and i think the flight might be like 1000 dollars its self but i think its definately worth it.[/quote]

whoa. wait. what?! a 1000 dollar flight??? lol. no. couple hundred max. unless you're like flying ultra high first class lol.
okay so i just did a expedia search. and from newwark nj to dallas texas is 249. so yeah :// u might have to add another hundred or so for unaccompanied minor depedning on airline...

Thanks about the heads-up on the life, I mean, yea.. I like a lot of freedom..  :D

And the all nighter  :ninja: haha.. yea.. I like that too.. I guess 21 days is not that long after all.

I'm really excited about it, so yea.. I'm very yea..y today haha..

no problem.
have funnn :D im sure you will =]] hahaha

How were the room mates? It seems like I always have the bad luck!
Last camp I was roomed with this one kid who brushed his teeth after every snack and had no fun and this other kid who looked at 'naughty' stuff on his phone.

I'm presuming we won't have too many naughty kids at this camp but were there a lot of uptight kids?

from what i hear,

theres always alot of 'naughty' kids.....

Do you select who to room with, or is it random?

LOL. there are always people who are ... sort of weird? lol :P
hahahahaha.
but anyways.
your roomies are mostly based on age & who u put down in ur app for who u want to be ur roomie.
so like. if your 15 you're going to be with mostly 14,15,16 year olds.

well you'll be having like. 5-6 roomies. and im pretty sure some of them will be pretty coool ;]
i had 4 [cuz im a girl and we had less people] and two of them were like freaking amazing and one girl just really got on my nerves sooooo badddd. the other one was okay. so, most people are really cool. and you can hang out in other people's rooms too.

Hmm? You were supposed to put down who you want to room with in your application? I don't remember doing that.

sorry. not application.
the packet that you will later get.

just leave it blank. i did that.
a lot of people do.

and the directors' will put you with other people who are around your age.

Hm, I also got into COSMOS (google it) at UC Santa Cruz. This will be interesting. Basically, COSMOS is a math AND science camp. It's the month or so before AwesomeMath and overlaps the first two days. Technically I could just finish COSMOS (or skip the last two days), walk across campus, and sign into AwesomeMath. xD

Though, which one should I cut two days from? COSMOS has like a project thing that I have to present on the last day and stuff, so there are complications. Or I could try to be in two places at once for two days.

You should contact the directors of both programs and ask.. But as you know, it's really up to you.

I'm going to look at it from both sides.

If you have to present something at COSMOS, you are going to miss the orientation, which they say is pretty important, in AMSP.
But then again, you are not really LEARNING anything in the orientation, but presentation is an important part in COSMOS (guessing??)

Yeah, at COSMOS you need to present the presentation to "officially" complete the camp. So...hm.

orientation isnt all that important at amsp i didnt think.....
just talking.
missing the first 2 days of interaction is sorta sad.
but if u need to finish last two days of cosmos to complete camp go for it.

Me, lifeisatrapezoid, and ihateplants0 will help you if you come late :P


Από ένα εξωτερικό σημείο Α ενός κύκλου C1 φέρνουμε τα εφαπτόμενα τμήματα ΑΒ και ΑΓ. Έστω Δ το συμμετρικό του Α ως προς το Β. Ένας κύκλος C2 που εφάπτεται της ΑΒ στο Α, διέρχεται από το Γ και τέμνει τον C1 στο Ε. Η ΑΕ τέμνει τον C1 στο Z. Να αποδείξετε ότι το τετράπλευρο ΑΓΖΔ είναι εγγράψιμο.

Περιμένω τις λύσεις σας επειδή πιστεύω ότι υπάρχει μία σημαντική γενίκευση για το τετράπλευρο ΑΓΖΔ που δεν υπάρχει στη λύση του κύριου Στεργίου αλλά δεν είμαι σίγουρος.

Κάτι δεν παέι καλά. Αν είναι όντως εγγράψιμο όπως αποδεικνύεται στο βιβλίο τότε προφανώς οι μεσοκάθετοι συντρέχουν.

Ωραία, όμως το κέντρο του C1, έστω Ο1, ανήκει στη μεσοκάθετο των ΑΔ και ΓΖ άρα όλες πρέπει να συντρέχουν στο Ο1. Άρα αν Ι το μέσο της ΑΓ, τα Ο1,Ο2,Ι είναι συνευθειακά και η Ο1Ο2 είναι κάθετη στην ΑΓ. Όμως η Ο1Ο2 είναι κάθετη και στη ΓΕ, άρα τα Α,Γ,Ε είναι συνευθειακά (ψιλοάτοπο αφού ανήκουν στον ίδιο κύκλο). Χάνω κάτι κάπου? :?:

Η μόνη περίπτωση να μη βγει άτοπο είναι να τα ΓΖ και ΑΔ να έχουν κοινή μεσοκάθετο οπότε τότε έχουμε τραπέζιο.

Αυτό είναι που λες Θέμη 'γενίκευση'?

[quote="Protonios"]
Η μόνη περίπτωση να μη βγει άτοπο είναι να τα ΓΖ και ΑΔ να έχουν κοινή μεσοκάθετο οπότε τότε έχουμε τραπέζιο.

Αυτό είναι που λες Θέμη 'γενίκευση'?[/quote]

Βέβαια είναι τραπέζιο! (και πιστεύω αυτό εννοεί ο Θέμης!)

Προκύπτει πολύ εύκολα αφού (γωνία) ΔΑΖ=ΑΓΕ=ΑΖΓ κι άρα ΑΔ//ΓΖ :)

[quote="Ipparchos"]Ακριβώς. ΓΖΑ=ΑΓΕ (υπό χορδής και εφαπτομένης) και ΖΑΔ=ΑΓΕ (υπό χορδής και εφαπτομένης). Αρα, ΑΔ//ΓΖ, οπότε το ΑΓΖΔ είναι τραπέζιο. Οι ΓΖ και ΑΔ έχουν κοινή μεσοκάθετο ΒΙ (αφού το απόστημα ΟΙ διχοτομεί τη ΓΖ), οπότε το τετράπλευρο είναι ισοσκελές τραπέζιο (οπότε και εγγράψιμο) ανεξάρτητα από την επιλογή του σημείου Α. Η λύση του Στεργίου είναι εντελώς διαφορετική και στο σχήμα που έχει φτιάξει το τετράπλευρο κάθε άλλο πάρα τραπέζιο θυμίζει.[/quote]

Παιδιά , σας ευχαριστώ !

 Θα πάω μέσα να το δω αμέσως και να το φτιάξω καλά ! Θα είναι χαρά μου να μου υποδεικνύετε κάθε τέτοια παρατήρηση, ώστε σε μια επανέκδοση(αν γίνει ποτέ!) , να τις περάσω .
 Είναι άλλωστε αξιωματική παραδοχή ότι ένα βιβλίο βελτιώνεται μόνο με τη βοήθεια των αναγνωστών του.

 Χρόνια πολλά και καλή μελέτη !

 Μπάμπης

[quote="Ipparchos"]......................
  .................................... στο σχήμα που έχει φτιάξει το τετράπλευρο κάθε άλλο πάρα τραπέζιο θυμίζει.[/quote]

Εντάξει, είσαι λίγο υπερβολικός  :) (για αστείο το λέω ), αλλά όπως βλέπεις ξέφυγε λίγο από τη σχεδιάστρια ο κύκλος C_2 .Και σε αυτό το σχήμα μμμ ... ψιλοδείχνει ότι ΓΖ // ΑΔ, αλλά δε μαρτυράει και παραλληλία !.  Θα έπρεπε  ο κύκλος C_2 να γέρνει λίγο ακόμα προς τα δεξιά ώστε το Ε να πλησιάσει την ΑΔ. Αν είχα προσέξει την παραλληλία9θεωρητικά), πιθανόν να της έλεγα να κάνει τη ΓΖ να φαίνεται ....  τελείως παράλληλη  με την ΑΔ.

Λοιπόν , πρόσθεσα την παρατήρηση που μου είπατε  ότι ΓΖ // ΑΔ και ότι τελικά το ΑΓΖΔ είναι ισοσκελές τραπέζιο !

 Ευχαριστώ ξανά και χαίρομαι που βοηθάτε ώστε να βελτιώσουμε το βίβλίο ακόμα και σε τέτοιες λεπτομέρειες.

 Και για να λέμε και τίποτα καινούριο, ένας μεγάλος μαθηματικός είπε :

 [size=150][b][color=red]Γεωμετρία είναι η επιστήμη των ορθών συλλογισμών σε λάθος σχήματα ! [/color][/b][/size]

 Με αυτη την παραδοχή, είστε καλοί γεωμέτρες , διότι στο δικό μου ''ανακριβές '' σχήμα , εσείς με ορθούς συλλογισμούς

βρήκατε τις σωστές σχέσεις των στοιχείων !Χαίρομαι για αυτό !

Τελικά , ουδέν κακόν αμιγές καλού !

 Νάστε καλά !!!

 Μπάμπης(ελπίζω να μου συγχωρείτε τα λάθη, είτε μεγάλα είτε μικρά!)

(με αφορμή το μήνυμα αυτό σκέφτομαι να βάζω λίγες - λίγες τις σελίδες του τόμου 1 που ετοιμάζω σιγά σιγά, ώστε να

μελετάτε τις λύσεις, να διορθώνετε τα λάθη , να βρίσκετε νέες (που θα τις συμπεριλάβω στο βιβλίο)  και να έχουμε στο τέλος το καλύτερο αποτέλεσμα ! Έτσι και αλλιώς αυτά τα βιβλία γράφονται για σας και ο εκδότης τα βγάζει μόνο επειδή έχει δύο παιδιά που σπουδάζουν μαθηματικά και δεν θέλει να τον  .... κατηγορήσουν ότι δεν στηρίζει τις ολυμπιάδες. Ε! Τον παρακαλώ και γω λιγο και ελπίζω να μας βγάλει και τους άλλους τόμους(έναν τον χρόνο).

Εντάξει δεν είναι και τόσο σημαντικό το σχήμα. Απλώς όταν λύνει κάποιος μία άσκηση και η λύση του είναι διαφορετική από τη λύση του βιβλίου (πόσο μάλλον όταν βρίσκει κάτι που δε φαίνεται στο σχήμα), ο άπειρος λύτης (όπως εγώ) πανικοβάλλεται λίγο και ψάχνει σαν παλαβός να βρει αν και πού έχει κάνει λάθος. Αν αυτό συμβεί δε σε διαγωνισμό χάνει κανείς πολύτιμο χρόνο. Πάντως το συγκεκριμένο σχήμα του βιβλίου του κυρίου Στεργίου αποδεικνύει περίτρανα ότι για πάντα το ανθρώπινο χέρι με κανόνα και διαβήτη είναι το μόνο που μπορεί να αγγίξει τη γεωμετρική τελειότητα. Όσον αφορά τους επόμενους τόμους της γεωμετρίας του κύριου Στεργίου έλπίζω οι νεότεροι να τους απολαύσουν όπως τώρα εμείς χαιρόμαστε τον 3ο και να εκμεταλλευτούν την αξία τους. Δυστυχώς όταν εκδοθούν εμείς που είμαστε τώρα τρίτη λυκείου θα είμαστε εκτός "ολυμπιακής" δράσης.

Χρόνια πολλά στους Γιώργηδες! (Γιώργος Μπασδέκης από Καρδίτσα, ...)


Hey do you guys know how to make your own e-card ? Quickly, it is for 8/3 . :D


Find the minimum $ k$ that it is possible to split a square into $ n$ non-overlapping squares for every $ n\ge k$

The answer is $k=\boxed{6}$. Clearly it is impossible to partition a square into $5$ squares (since one square must be in each corner), so we will prove that we can partition any square into at least $6$ non overlapping squares.

Note that if it is possible to split a square into $k$ squares, we can also split it into $k+3$ squares by replacing any square with $4$ smaller squares.

We have a construction for $k=2m$ for $m\ge 2$. consider a $m\times m$ square formed by one $(m-1)\times(m-1)$ square in a corner and $2m-1$ unit squares filling the rest of the square. For example, $m=4$ is shown below.

[asy]
for (int i = 0; i <= 4; ++i) {
draw((0,i)--(1,i));
draw((i,0)--(i,1));
}
draw((1,0)--(4,0)--(4,4)--(0,4)--(0,1));
draw((4,1)--(1,1)--(1,4));
[/asy]

Then we can achieve all $k=2m,2m+3$ for $n\ge 2$, so all $k\ge 6$ is possible, which completes the proof.
-------------------------
basically we can get everything except for $2,3,5$


this may lead to interesting debate... who thinks fidel castro is a good ruler and who thinks he is bad?

he is good in some ways and he is bad in some ways

Well, the matter is a little more delicated. I agree that Comunism is a great idea in the paper, but unfortunatelly, it turned to be a not that good idea in practice, basically because it turned to be Totalitarism in some ways. Fidel, read about him he was an idealist when he started his revolution. Honestly I believe that him, as well as Ernesto Guevara, when they started the revolution they did, they were really interested in the people. Latinamerica sixty years ago, specially out of the cities, was not the place it is now.

I have never been in Cuba (I would love to), but I know several people who have lived there, cubans and not cubans, and they say that the people live with only the scential, and sometimes even less. Part of it may be because the economical sizing that they have, but also a goverment that is inside a lot of corruption.

On the other hand, Castro has been the head of the Cuban state for over fourty years. Eventually every person becomes allienated with his power.

Any way, we can apply the Yin-Yang philosophy, there is good an evil on all of us!

Best,

[quote="djimenez"]they say that the people live with only the scential, and sometimes even less. [/quote]

I think you mean "the essential."  Judging Castro by the economic status of his citizens is a silly exercize since we've effectively shut off any opportunities for the kind of trade that could in fact create wealth in Cuba.

castro.... what an angel. him and kruschev were pals and played a little joke on the US with some fake bombs. fidel never meant any harm cuz his heart pumps kool-aid.
definitely good

on a more serious note....djimenez, you said that the government in cuba is corrupt. what government isn't corrupt?

yes, it has sorta become a totalitarian state, but look at cuba b4 castro... i think he's not a bad guy, its just that there is no other way. He has to take full control or else who knows whatll happen.. he's improved lots of things in cuba, healthcare, education, etc.
and he's just like a cute little teddy bear lol, an oversized one

i wonder whatll happen to cuba when he dies

[quote="Dark_Shadow"]There are levels of corrupt, you know.

I think that no matter how left you are, Fidel Castro can't be considered a very good person.[/quote]

true....but becuase he's not a very good person, doesn't make him a bad person.

[quote="ln(dx/dy)"]I wish I knew more about this topic :(
I love debating.[/quote]
you could always pretend you knew what you where talking about.  :)

Communism is a paradox in itself..as djimenez it turns out to be totalitarianism becuase of the amount of power the government is given.  The corruption is only a part of the governmental system.  Hand a man a juicy steak on a silver platter, he will surely eat it if not in that moment another (unless he is a vegitarian :D ).  Power just eats away at you whether you like it or not.  The only type of person that could withstand a country being served to him is someone with an amazing amount of selfcontrol.  A person like that is hard to come by. Fidel was not one of those people. So his corruption is only natural.


calculus
integration
function
trigonometry
calculus computations

\[ \int {\frac {1}{{\cos x \plus{} \csc x}}} dx
\]

hello, you can rewrite your integral in the form $ \int\frac{\sin(x)}{\sin(x)\cos(x)\plus{}1}\,dx$, here will help the substitution $ t\equal{}\tan(\frac{x}{2})$.
Sonnhard.


tamame javab haye moadeleye zir ra, baraye adad haye av'al $p,q$ va $r$ bedast avarid  :) \[p(p+1)+q(q+1)=r(r+1)\]

hey guys use english as a common language pls.. :!:   :wink:

Find all of the answer equlation below for numbers $p$, $q$ and $r$ that $(p,q,r)=1$ :) \[p(p+1)+q(q+1)=r(r+1)\]
If you don't understand thing, That's my falt (english=no english)! :lol:

The fact that your English translation is incorrect is ipso facto of the equivalence of "av'al" and "prime." (However, my understanding of Persian mathematical terminology is inadequate for translation [Persian-->English].)

javabha ebaratand az:(chon rahalesh toolani bood hoselam nagereft ke benevisam)
(p,q,r)=(2,2,3),(5,3,6),(3,5,6)
 ke age p,q,r aval bashand (p,q,r)=(2,2,3)


Find the distance between the parallel planes given by the equations (with proof)

$ ax \plus{} by \plus{} cz \plus{} d_1 \equal{} 0$ and
$ ax \plus{} by \plus{} cz \plus{} d_2 \equal{} 0$

[hide="Hint"]$ \langle a,b,c \rangle$ is the normal vector. A vector from plane 1 to plane 2 with the shortest possible magnitude (that magnitude being the desired value) will be a scalar multiple of $ \langle a,b,c \rangle$.[/hide]

[hide="Solution"]Because $ \langle a,b,c \rangle$ as a vector is normal to each plane, there are reals $ s,t$ such that the point $ (sa,sb,sc)$ is on plane 1, and $ (ta,tb,tc)$ is on plane 2. By putting these into the equation, we get that $ s \equal{} \frac{d_1}{a^2\plus{}b^2\plus{}c^2}$ and $ t \equal{} \frac{d_2}{a^2\plus{}b^2\plus{}c^2}$. By the hint, the desired value is \[ |\langle sa,sb,sc \rangle \minus{} \langle ta,tb,tc \rangle| \equal{} |s\minus{}t| \cdot | \langle a,b,c \rangle | \equal{}\]
\[ \frac{|d_1\minus{}d_2|}{a^2\plus{}b^2\plus{}c^2} \sqrt{a^2\plus{}b^2\plus{}c^2} \equal{}\]
\[ \boxed{\frac{|d_1\minus{}d_2|}{\sqrt{a^2\plus{}b^2\plus{}c^2}}}\][/hide]


I'm taking introductory computer science next semester.  The course is designed primarily for people who've never programmed, and the main objective is to get people programming in C++, and to instill a rudimentary understanding of general computer science.  I'm not satisfied with the textbook, or the course objectives, so I'm wanting a new book.  The one used in the course is long-winded and elementary.  I want something terse, deep, and challenging, much like baby Rudin is for Analysis.  Any suggestions?

Thanks guys.

There is no book like Rudin's [i]Principles of Mathematical Analysis[/i] in computer science because computer science and math are just too fundamentally different.

I like this book

http://www.amazon.com/Primer-4th-Stanley-B-Lippman/dp/0201721481/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1239930073&sr=8-1

which is written by someone who knows C++ like the back of his hand. It is pretty terse and challenging in my opinion. Calling it "deep" would be pushing it though...I doubt you will find a good computer science book that teaches C++ and introductory programming and that could be called "deep". But there are probably some "deep" books about theoretical computer science and NP-completeness, but I would recommend just learning how to program first.

Just learning to program first sounds good.  I'll keep that book in mind, thanks.

I believe C++ is a terrible first language for a beginner. It's a terrible language for an expert;-)

My recommendation is teach your students C using the original K & R book which is a small book with lots of good exercises. Teach them how to use 'make' and handle their way around a shell console (bash on Linux/Mac or cygwin on Windows). Use gcc for the compiler. The nice thing about C is almost all modern languages are descendent from it. Also, it allows you to get close to the machine.

If the course is open for non-computer science majors, you may want to consider going with a modern object-oriented scripting interpreted language such as Ruby or Python (this generation's Pascal) or Groovy. For Ruby, "Programming Ruby" by Thomas and Hunt is a great book. O'Reilly has a good book on Python. These languages are really the future for modern development and are so much more productive than third generation languages. What you give up in pointers and memory allocation, you will gain in associative arrays, maps, and using closures.

I'm actually not teaching the course, I'm taking it :-).  I'm wanting a book that goes above and beyond the course material, but my main priority is to get an A in the course, so it needs to be relevant, and I'm afraid I'm stuck with C++.

LOL - in that case it's a wonderful language ;-)

In that case, I was going to recommend getting the descriptive book by the author of the language itself, Bjarne Stroustrup. But I see he has a new book that may be just up your alley:

http://www.research.att.com/~bs/programming.html

I think its worth a check out.

A friend of mine had Bjarne Stroustrup as a professor, used that book, and absolutely hated the thing; though he refuses to explain why.  I'd expect that him and I would have quite different taste in textbooks, but it still worries me.  Any commentary?

I pulled my Stroustrup "The C++ Programming Language" out of storage boxes. After going through it, it's definitely not a gentle read but rather quite the opposite. No where near as nice as the short and sweet K&R book or the early versions of Gosling's Java guide. I am still glad to own it as a reference but its a rather large book (granted for a large language) and it goes quickly into advanced OO concepts such as iterators, etc that would be difficult for less experienced programmers. I'm curious if his new book is any gentler.

Let me know which book you settled on.


I use an apple computer, so I downloaded TexShop. I have a couple of questions.
1. What is the bar near the top that's supposed to say LaTeX => PDF?
2. Where is the build menu bar?
Thanks in advance for any help.

Have you tried just running it? It defaults to building pdfs, and the button to do that is called "Typeset". My copy, which works fine, doesn't have anything by either of those names.

If you're using TexShop, probably you should stop reading instructions for TeXnicCenter.   :wink:

So...
Where are the instructions for TexShop? Are they on AOPS?

The program does come with documentation. Look at the readme, and then at the help menu inside it if you have any further issues.


(a) Find the smallest positive integer n such that $ \phi(x) \equal{} n$ has no solutions x, exactly two solutions, and exactly three solutions.

(b) Prove that there is no solution to the equation $ \phi(x) \equal{}14$, and that 14 is the smallest even integer with this property.

(c) Apart from n =14, what is the next smallest even integer such that $ \phi(x) \equal{} n$ has no solution?

(c) Typically, for $ p$ odd prime and $ 2p \plus{} 1$ not a prime, $ n \equal{} 2p$ cannot be a $ \phi(x)$, so $ 26 \equal{} 2\cdot 13$ is a good candidate, and in fact can be checked to be the next (the same idea applied for b), since $ 14 \equal{} 2\cdot 7$, and $ 15$ is not a prime).

That makes sense, but how do we know that there are no more n of different form which gives no solution too?

I never said that; I said "typically". Of course there are other such numbers (no formula!).


Prove that $ (2, 4)$ is the only solution over positive integers to $ x^y \equal{} y^x$, given that $ x < y$.

[hide="Think this is right..."]$ x^y = y^x$
Dividing both sides by $ x^x$
$ x^{y-x} = \left(\frac{y}{x}\rbrack et \right)^x \Rightarrow x\vert y \Rightarrow y = kx$
Plugging $ y$ back in gives:
$ x^{x(k-1)} = k^x$
Now take the $ x^{th}$ positive root of both sides to give
$ x^{k-1} = k$
We can prove by induction that there are no solutions for $ k \geq 3$, so $ k = 1$ or $ 2$.  If $ k=1$ we get $ x=y$ which is contradictory to the original problem statement $ x < y$.  Therefore the only possiblity is $ k=2$ which gives $ x=2$ and therefore $ y=4$.[/hide]

Thanks much. I was toying around with the prime factorization and got nowhere. <_<


induction
combinatorics solved
combinatorics

Two people, $A$ and $B$, play the following game with a deck of 32 cards. With $A$ starting, and thereafter the players alternating, each player takes either 1 card or a prime number of cards. Eventually all of the cards are chosen, and the person who has none to pick up is the loser. Who will win the game if they both follow optimal strategy?

We will say that a number $n$ is winning if a person which has to play while there are exactly $n$ cards has a winning strategy. Otherwise, $n$ is loosing, which means that the second player to play has a winning strategy.

Clearly $1,2,3$ are winning since we can take all the cards and win. But $4$ is loosing, because the only possibilities to play is to choose $1,2$ or $3$ of the cards, leaving a winning position for the opponent.
By induction, assume that for some $p$, the winning positions are all the positions except those of the form $4k$ where $k \leq p.$
Then, $4p$ is a loosing one. Thus, $4p+1,4p+2,4p+3$ are winning positions since we only have to take $1,2$ or $3$ cards to leave a loosing position for the opponent.
And $4p+4$ is a loosing position because it is not $1$ or a prime, so that we cannot take all the cards, and if we take $1$ or an odd prime number of cards, we leave an odd number of cards which is clearly a winning position (according to the induction hypothesis). And if we take two cards, we leave $4p+2$ cards which is a winning position as proved before.
This ends the induction step.

Thus, the first player has a winning strategy if and only if the initial number of cards is not a multiple of $4$. Since $32$ is a multiple of $4$, it follows that the second player has a winning strategy.

Pierre.


[quote="star99"]Im going to take the SAT this year.  i was wondering how much time you would spend on it, daily or weekly, especially the non math parts.  like most around here, the math is pretty easy for me, but the reading questions completely boggle me.  any tips?  i know that reading helps, but what material is helpful?  thanks.[/quote]

I take the SAT next year.
Monday: 30 min Vocabulary
Tuesday: 30 min Reading
Wednesday: 30 min Math
Thursday: 30 min grammar
Friday - Saturday: Essay

Sometimes, I mix the schedule (Mon. – Thursday) but that is essentially it. However, I have yet to see the effectiveness of my “method” :P . Just checkout a SAT book and do a little every day.


[hide="hint"]let the sum be $S.$ what is $Sx$? What can you do with this?
[hide="try"]$S-Sx$[/hide]
[/hide]

[quote="me@home"]Just finishing your solution

$1+\frac{x}{1-x}=\frac{1}{1-x}=S(\frac{1}{x}-1)\implies S=\frac{x}{(1-x)}^{2}$
The usual way to solve this is to just take the derivative of $\sum x^{k}$

which is why I think there was a mistake in the above because it should be $\frac{-1}{(1-x)^{2}}$[/quote]

Except the derivative of $\sum x^{k}$ is $\sum kx^{k-1}$. So we actually want $x$ times the derivative, which is what you got.

EDIT: Actually, you took your derivative wrong. It's positive, so then what I said above would be true.

Oops *stops doing basic calculus in head*
that's a $-x$ on the inside...
of course.

I made the same mistake in calc the other day in the monster derivative but in the end I caught it...

The question: find point of maximum and mimum curvature of $y=e^{x}$... its an easily posed problem that gets messy...

[hide="bonus"]
prove that the maximum is $\sqrt{2}/2$ (i.e. without calculator)
[/hide]

[hide="solution"]That would be

$x+2x^{2}+3x^{3}+4x^{4}+/cdots=S$

$Sx=x^{2}+2x^{3}+3x^{4}+\cdots$

$S-Sx=x+x^{2}+x^{3}+x^{4}+/cdots=\frac{x}{1-x}$

We divide by 1-x to get

$S=\frac{x}{1-2x+x^{2}}$[/hide]

So yeah, the derivative would be positive.


A mathematics contest has 20 questions. 7 points are awarded for a correct answer and 25 points are awarded for an unanswered question. No points are awarded for an incorrect answer. How many different points for the whole test can a student receive?

I'm curious; if you get more points not answering a question than answering it, then why would you answer any questions? :P

[hide="From the given information"]
Notice that the $ 0$ points allotted for not answering a question means we can basically ignore this part.

We have that $ (7,25)\equal{}1$, and since there are $ 20$ questions, there are no overlaps.  This means if we apply any form of counting, we will not overcount.

Say one question gives points.  There are two discrete point values possible for it: $ 7$ or $ 25$.
If two questions give points, there are three discrete points values: $ 14, 32, 50$.
And in general, if there are $ n$ questions that grant points, there are $ n\plus{}1$ point values that the person can receive.

The sum becomes $ \sum_{i\equal{}0}^{20}(n\plus{}1)$.  Applying summation formulas yields $ \frac{20*21}{2}\plus{}20*1\equal{}230$.[/hide]

I meant 2.5 points for blank. Is the answer different?

I don't think the first answer is right, but if it is then, yes. $ GCD(7,2.5) \equal{} 1$
The reason I do not think the previous answer is right is because of this.
The first question has 3 possible values. The mathematician can either get 0,7,or 2.5 not just 7 or 2.5. For the next question he can get 5 values.
I think the number is $ 25\plus{}24\plus{}23\plus{}...\plus{}3$, or 322. This is probably wrong though.

[hide="I think this is right"]
Total points can be linearly expressed as $ 7a\plus{}2.5b\plus{}0c$.
Since $ a\plus{}b\plus{}c\equal{}20$ we just have to find the number of different solutions and this is equal to $ \binom {22} 2$ but if $ b\equal{}14$,  $ 7a$ and $ 2.5b$ have a common divisor so we need to subtract those solutions (because we already counted them for $ b\equal{}0$). Since there are already $ 14$ numbers distributed to $ b$, we need to distribute the remaining $ 6$ among $ a$ and $ c$ and this is equal to $ \binom 7 1$ so the number of different points that can achieved is $ \binom {22} 2 \minus{} \binom 7 1\equal{}\boxed{224}$.[/hide]

[hide]So we've established that there are 231 different ways to get your score. But this overcounts by 28. If the number unanswered $ a \ge 14$, there are two values for each possible score. Let's say $ a \equal{} 14$, then the number of answered correctly and incorrectly in the corresponding score $ (b,c)$ beyond the 5 already answered correctly could be $ (6,0),(5,1)...(0,6)$*. For $ a \equal{} 15, (b,c)$ could be $ (5,0),(4,1)...(0,5)$. Thus the total number of overcounts is $ 1 \plus{} 2 \plus{} ... \plus{} 6 \equal{} 28$. $ 231 \minus{} 28 \equal{} \boxed {203}$
*Note: Let (a,b,c) be (#unanswered, #answered correctly, #answered incorrectly). Thus we have (14, 5, 0) = (0, 5+5, 0), (14, 4, 1) = (0, 4+5, 1) etc.[/hide]

[quote="mathwizarddude"][quote="crazypianist1116"]If the number unanswered $ a \ge 14$, there are two values for each possible score. [/quote]
How do we prove this?[/quote]

OK so: Let's say (a,b,c) is (#unanswered, #answered correctly, #answered incorrectly). Now we have the restrictions $ a,b,c \in \mathbb N, a\plus{}b\plus{}c \le 20, a,b,c \ge 0$. One's score is $ 2.5a \plus{}7b$. Now for any $ a \ge 14$, the corresponding scores are $ (n\minus{}14, 5\plus{}m, k) \equal{} (n, m, k)$. We can prove this thusly:
$ 2.5(n\minus{}14) \plus{} 7(5\plus{}m) \equal{} 2.5n \plus{}7m$
$ 2.5n\minus{}35 \plus{}35 \plus{}7m \equal{} 2.5n \plus{}7m$
$ 2.5n \plus{}7m \equal{} 2.5n \plus{}7m$


let ABC be a triangle.let $ l$ be a line on ABC plane.let $ A_1$ be the feet of the height of A to $ l$.define $ B_1,C_1$ in this way.draw the heights through $ A_1,B_1,C_1$ to BC,AC ,AB, respectively.prove that these 3 heights concur. :)

[quote="rts2007"]Ceva's theorem?[/quote]

how would you apply ceva in this problem?

[quote="rts2007"]Ceva's theorem?[/quote]

actually, i proved it usung  radical axis. i don´t know if ceva will be useful here...


Alternate $ a_i$ so $ a_1\leq\ a_2\leq\ . . . \leq\ a_n$ then left part of ineq will increase then ineq follow from
$ \frac{2}{a_1\plus{}a_2}\leq\ \frac{1}{a_1}$
$ \frac{2k\minus{}1}{a_1\plus{} . . . a_{2k\minus{}1}}\leq\frac{2k\minus{}1}{a_k\plus{} . . . a_{2k\minus{}1}}\leq\frac{2k\minus{}1}{ka_{k\minus{}1}}<\frac{2}{a_{k\minus{}1}}$ for $ 2\leq k\leq (n\plus{}1)/2$
$ \frac{2k}{a_1\plus{} . . . a_{2k}}\leq\frac{2k}{a_{k\plus{}1}\plus{} . . . a_{2k}}<\frac{2}{a_{k}}$ for $ 2\leq k\leq n/2$

Because $ \frac{1}{b_1}\plus{}\frac{2}{b_1\plus{}b_2}\plus{}...\plus{}\frac{n}{b_1\plus{}b_2\plus{}..\plus{}b_n}<\frac{1}{a_1}\plus{}\frac{2}{a_1\plus{}a_2}\plus{}...\plus{}\frac{n}{a_1\plus{}a_2\plus{}..\plus{}a_n}$ where $ b_i$ is permutation of $ a_i$ and$ a_1\le a_2 \le . . . \le a_n$ because
$ a_1\le b_1,
a_1\plus{}a_2\le b_1\plus{}b_2,
 . . .
a_1\plus{} . . . \plus{}a_n\equal{}b_1\plus{} . . . \plus{}b_n$

[quote="dima ukraine"]Alternate $ a_i$ so $ a_1\leq\ a_2\leq\ . . . \leq\ a_n$ then left part of ineq will increase then ineq follow from
$ \frac {2}{a_1 \plus{} a_2}\leq\ \frac {1}{a_1}$
$ \frac {2k \minus{} 1}{a_1 \plus{} . . . a_{2k \minus{} 1}}\leq\frac {2k \minus{} 1}{a_k \plus{} . . . a_{2k \minus{} 1}}\leq\frac {2k \minus{} 1}{ka_{k \minus{} 1}} < \frac {2}{a_{k \minus{} 1}}$ for $ 2\leq k\leq (n \plus{} 1)/2$
$ \frac {2k}{a_1 \plus{} . . . a_{2k}}\leq\frac {2k}{a_{k \plus{} 1} \plus{} . . . a_{2k}} < \frac {2}{a_{k}}$ for $ 2\leq k\leq n/2$[/quote]

 But $ \sum {\frac{i}{a_1\plus{}a_2\plus{}....\plus{}a_i}}<\sum_{k\equal{}2}^{\frac{n\plus{}1}{2}}{\frac{2}{a_{k\minus{}1}}\plus{}\sum_{k\equal{}2}^{\frac{n}{2}}{\frac{2}{a_k}}}$    :|

sorry i didnt see $ 2\sum_{i\equal{}1}^{n}\frac{1}{a_n}$ its for $ 4\sum_{i\equal{}1}^{n}\frac{1}{a_n}$ :wallbash_red:

Sorry,I am not able to understand how can you get $ \frac{1}{a_n}$ on the right hand sight.
In your opinion,I think the left hand sight must have something like $ a_{2n}$?
Sorry for my poor ability.
and Hoping to see some answer

See

[url]http://www.mathlinks.ro/viewtopic.php?p=834159#834159[/url]

 :wink:

Well done!
Actually I think you would solve it with some 'well-know' inequalities which i had never heard of..
It seems natural,really good

[quote="wxydx00"]$ a_1,a_2,...a_n > 0$,prove:
$ \frac {1}{a_1} \plus{} \frac {2}{a_1 \plus{} a_2} \plus{} ... \plus{} \frac {n}{a_1 \plus{} a_2 \plus{} .. \plus{} a_n} < 2\sum_{i \equal{} 1}^{n}\frac {1}{a_n}$[/quote]

It's the special case for Hardy's inequality.  :)


can anyone explain to me what quicktime 7 is about?

i tried to play one of my dvd movies but it would not open.

I doubt quicktime can play the DVD videos {generally VOB files}..Quicktime is generally used to play the .mov files.... :?
If you want to play a DVD , I guess Windows Media Plaer is sufficient....
However, $Quictime \neq Useless.$ Plenty of content on the web can be played with quicktime. :)

i got a Mac, so can't use Windows media player
so quicktime is mostly for the web?

Plenty of content on the web is in the ".mov" format ---the format which quicktime mainly supports. :)


Find all integers $ x,y,z \in \mathbb{Z}$ such that $ x\plus{}y\plus{}z \equal{} 1$ and $ x^{3}\plus{}y^{3}\plus{}z^{3}\equal{}1$.

[hide="Possible Hint"]
$ x(x \plus{} 1)(x \minus{} 1) \plus{} y(y \plus{} 1)(y \minus{} 1) \plus{} z(z \plus{} 1)(z \minus{} 1) \equal{} 0$ is a necessary condition for the problem, but not sufficient. In other words, $ x,y,z$ is a solution only if the above equation is true. I haven't done this problem, so I don't know whether this will help or not.[/hide]

$ (x\plus{}y\plus{}z)^3\minus{}3(xy\plus{}xz\plus{}yz)(x\plus{}y\plus{}z)\plus{}3xyz\equal{}1$

$ xy\plus{}xz\plus{}yz\equal{}xyz\equal{}k$ for some integer k.

Hence consider the polynomial $ (t\minus{}x)(t\minus{}y)(t\minus{}z)\equal{}t^3\minus{}t^2\plus{}kt\minus{}k\equal{}(t\minus{}1)(t^2\plus{}k)$

Hence, $ t\equal{}1$ is going to be one of x,y,z, and the other 2 will be found as $ \pm\sqrt{\minus{}k}\equal{}\pm a$, where obviously we must take k to be non-positive, so $ a\equal{}\sqrt{\minus{}k}$ will be non-negative.

Hence, the complete solution set will be $ (x,y,z)\equal{}(1,a,\minus{}a)$ and all permutations thereof for all non-negative integer values of $ a$.


[quote="nguoivn"]Given $ a, b, c > 0$. Find the best k for this inequality is trues:

$ \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} k(ab \plus{} bc \plus{} ca)}{2(k \plus{} 1)} \geq\ \frac {ab^2}{a \plus{} b} \plus{} \frac {bc^2}{b \plus{} c} \plus{} \frac {ca^2}{c \plus{} a}$

I obtain $ 3 \leq\ k < 4$  :maybe:[/quote]
$ k_{max}\equal{}\sqrt{9\plus{}6\sqrt{3}}\minus{}1 \equal{} 3.403669475\cdots.$

[quote="Guest"][quote="nguoivn"]Given $ a, b, c > 0$. Find the best k for this inequality is trues:

$ \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} k(ab \plus{} bc \plus{} ca)}{2(k \plus{} 1)} \geq\ \frac {ab^2}{a \plus{} b} \plus{} \frac {bc^2}{b \plus{} c} \plus{} \frac {ca^2}{c \plus{} a}$

I obtain $ 3 \leq\ k < 4$  :maybe:[/quote]
$ k_{max} \equal{} \sqrt {9 \plus{} 6\sqrt {3}} \minus{} 1 \equal{} 3.403669475\cdots.$[/quote]

How did you get it ??  :|


A car has the speed 120 km/h on going to a place then it returns to cut the same distance at the speed of 80 km/h. What is the average speed on both trips?

I'm not completely sure I understand the question, but it seems like it's just the average of the two speeds, so it would be 100 km/h.

Edit: Ignore this, I misinterpreted.

[hide]
Let d = the distance in one direction

Let $ t_1$ = the distance traveled for 120 km/hr
Let $ t_2$ = the distance traveled for 80 km/hr

average speed = the total distance divided by the total time

$ \frac {120t_1 \plus{} 80t_2}{t_1 \plus{} t_2}$

$ 120t_1 \equal{} d$ and $ 80t_2 \equal{} d$.

$ \frac {2d}{(d/120) \plus{} (d/80)}$

$ \frac {480d}{2d \plus{} 3d}$

$ \frac {480d}{5d}$

$ 96$

$ 96$ km/hr

[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

Hello,
let $k$ be an algebraically closed field and consider two coprime polynomials $f,g$ in $k[x,y]$
I want to prove that the varieties they define only have a finite number of points in common.

My book takes this approach: work in $k(x)[y]$ to manipulate the equations to get a polynomial in $x$, which tells us only a finite number of values is possible for $x$
Switching $x$ and $y$ will give us the same for $y$, and thus the number of points $(x,y)$ satisfying both equations is finite

But they use this apparently straightforward property that if $f$ and $g$ are coprime in $k[x,y]$, they will also be coprime in $k(x)[y]$.
Is this true and/or obvious?   I mean, how can I manipulate "noncoprimeness" in this ring to get a common factor in $k[x,y]$ as well?

All hints are welcome  :blush: , thank you, fredbel6

[quote="harazi"]This apparently obvious fact is just Gauss's lemma disguised.[/quote]
I guess it is, I'm sorry, I didn't look at it that way. :)  :blush: Thank you


modular arithmetic
quadratics
Gauss
number theory
number theory unsolved

The following statement is true or false?

If $GCD(2^{n}+1,n!-1) > 1$ then $GCD(2^{n}+1,n!-1) = 2n+1$.

It is verified for $n$ up to $45,000$.

Can you prove that there are infinite many $n$ such that $GCD(2^{n}+1,n!-1)>1$?I think this one should be consider first,though I have no idea about both these two problems.... :o
It is too hard for me  :blush:

If $p=2n+1$ is prime then Wilson's theorem implies
\[(n!)^{2}\equiv (-1)^{n}(2n)! \equiv (-1)^{n+1}\pmod{p}.\]
Then for $n=4k+1$ we have $n! = \pm 1\pmod{p}$ and $2^{n}\equiv-1\pmod{p}.$
If $n! = 1\pmod{p}$ then $p$ divides $\gcd(2^{n}+1,n!-1).$

Call prime p special if
\[ p|gcd(2^{(p \minus{} 1)/2} \plus{} 1,n! \minus{} 1)
\]
and exclusive if exist $ n\not \frac {p \minus{} 1}{2}$, suth that $ p|gcd(2^n \plus{} 1,n! \minus{} 1).$
It is easy to prove, that $ p$ is special if and only if $ p \equal{} 3\mod 8$ and $ \frac {h \plus{} 1}{2}$ is even, were $ h$ is number of classes $ Q(\sqrt{\minus{}p})$ (or number of quadratic residius less than $ p/2$ - odd, or there are odd number of odd quadratic residius).
There are 3 exclusive primes, less than 5 000 000:
$ p \equal{} 521881, \ n \equal{} 221799, \ n \equal{} 300081,$
$ p \equal{} 774593, \ n \equal{} 338884,$
$ p \equal{} 4327489, \ n \equal{} 3652374.$

[quote="heartwork"]The following statement is true or false?

If $ GCD(2^{n} \plus{} 1,n! \minus{} 1) > 1$ then $ GCD(2^{n} \plus{} 1,n! \minus{} 1) \equal{} 2n \plus{} 1$.

It is verified for $ n$ up to $ 45,000$.[/quote]
Sorry I don't read above solution . Here is mysolution :
The $ n\in N$ satisfy the condition then $ 2n \plus{} 1$ is a prime in the form $ 8k \plus{} 3,8k \plus{} 5$
[u]Proof [/u] $ 2n \plus{} 1|n! \minus{} 1$
If $ 2n \plus{} 1$ is a composite then it can represent in the form $ ab$ where $ a,b\geq 2$
Imply that $ \min\{a,b\} < n$ so $ \min\{a,b\}\not |n! \minus{} 1$
So $ 2n \plus{} 1 \equal{} p$
$ p|2^{\frac {p \minus{} 1}{2}} \plus{} 1 \Longleftrightarrow p\equiv 3,5 (\mod 8)$
So the statement true for n in this form.

[quote="TTsphn"]So $ 2n \plus{} 1 \equal{} p$
$ p|2^{\frac {p \minus{} 1}{2}} \plus{} 1 \Longleftrightarrow p\equiv 3,5 (\mod 8)$
So the statement true for n in this form.[/quote]
The statment equavalent, that there are not exclusive pairs (p,n). I found exclusive pairs, therefore statment is not true.
If $ p\equal{}1\mod 4$, then $ (\frac{p\minus{}1}{2}!)^2\equal{}\minus{}1\mod p$, therefore $ p\not |\frac{p\minus{}1}{2}!\minus{}1$ or p is not special.

Oh I am sorry have a mistake .
$ p|(\frac{p\minus{}1}{2})!\minus{}1$
Imply that $ p\equiv 3 (\mod 4)$ so it has form $ 8k\plus{}3$
But it is only necessary.

[quote="Rust"]Call prime p special if
\[ p|gcd(2^{(p \minus{} 1)/2} \plus{} 1,n! \minus{} 1)
\]
and exclusive if exist $ n\not \frac {p \minus{} 1}{2}$, suth that $ p|gcd(2^n \plus{} 1,n! \minus{} 1).$
It is easy to prove, that $ p$ is special if and only if $ p \equal{} 3\mod 8$ and $ \frac {h \plus{} 1}{2}$ is even, were $ h$ is number of classes $ Q(\sqrt { \minus{} p})$ (or number of quadratic residius less than $ p/2$ - odd, or there are odd number of odd quadratic residius).
There are 3 exclusive primes, less than 5 000 000:
$ p \equal{} 521881, \ n \equal{} 221799, \ n \equal{} 300081,$
$ p \equal{} 774593, \ n \equal{} 338884,$
$ p \equal{} 4327489, \ n \equal{} 3652374.$[/quote]
Your condition  $ \Longleftrightarrow p|(\frac{p\minus{}1}{2})!\minus{}1$ and it is not clear for n large enoght.

If $ p\equal{}3\mod 8$, then $ p|2^{(p\minus{}1)/2}\plus{}1$ and $ n\equal{}(p\minus{}1)/2, \ n!\equal{}\pm 1\mod p$ (because for $ n\equal{}(p\minus{}1)/2$ we get $ \minus{}1\equiv (2n)!\equiv (p\minus{}1)...(p\minus{}n)n!\equiv (\minus{}1)^nn!n!$, it give $ n!\equal{}\pm 1\mod p$).
If we have odd number quadratic residius in set $ \{1,2,...,n\}$ mod p, then $ n!$ is quadratic residius mod p. But for $ p\equal{}3\mod 8$ -1 is not quadratic residius. Therefore in these case must be $ n!\equal{}1\mod p$.

[quote="Rust"]If $ p \equal{} 3\mod 8$, then $ p|2^{(p \minus{} 1)/2} \plus{} 1$ and $ n \equal{} (p \minus{} 1)/2, \ n! \equal{} \pm 1\mod p$ (because for $ n \equal{} (p \minus{} 1)/2$ we get $ \minus{} 1\equiv (2n)!\equiv (p \minus{} 1)...(p \minus{} n)n!\equiv ( \minus{} 1)^nn!n!$, it give $ n! \equal{} \pm 1\mod p$).
If we have odd number quadratic residius in set $ \{1,2,...,n\}$ mod p, then $ n!$ is quadratic residius mod p. But for $ p \equal{} 3\mod 8$ -1 is not quadratic residius. Therefore in these case must be $ n! \equal{} 1\mod p$.[/quote]
Yes it is quadric reside with Gauss Lemma .


Let $ a_1 \le a_2 \le \cdots \le a_n$ be real numbers with sum $ 0$. Prove that
\[ a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n \plus{} 2na_1a_n \le 0.
\]
:)

We will use that $ a_1.a_n\leq a_i.a_j$ for all $ i$ and $ j$.

$ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2\equal{}\minus{}2.\displaystyle\sum_{i<j}^{n} a_i.a_j$

If  $ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2\plus{}2n.a_1.a_n>0$ then  $ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2\plus{}2n.a_i.a_j>0$ for all $ i$ and $ j$.

If we sum these inequalities for all $ i<j$

$ 2.\dbinom{n\plus{}1}{2}.\displaystyle\sum_{i<j}^{n} a_i.a_j>0$

But $ 2.\displaystyle\sum_{i<j}^{n} a_i.a_j\equal{}\minus{}(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)\leq0$. It's a contradiction.

Why inequality $ a_1a_n<\equal{}a_ia_j$ holds for all $ i,j$? I think it's not true for example for $ i\equal{}1,j\equal{}2$

[quote="crazyfehmy"] If we sum these inequalities for all $ i < j$

$ 2.\dbinom{n \plus{} 1}{2}.\displaystyle\sum_{i < j}^{n} a_i.a_j > 0$.[/quote]
I think this is wrong. :D

On the other hand, I think the following stronger result holds:

If $ a_1 \le a_2 \le \cdots \le a_n$ are real numbers with sum $ 0$, then

\[ a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n \plus{} na_1a_n \le 0.
\]


geometry
analytic geometry
graphing lines
slope
rotation
geometry unsolved

Prove that the equation of the circle of which the points $ (x_1, y_1)$ and $ (x_2, y_2)$ are the ends of a chord of segment containing an angle $ \theta$ is
$ (x \minus{} x_1)(x \minus{} x_2) \plus{} (y \minus{} y_1)(y \minus{} y_2) \pm \cot \theta [(x \minus{} x_1)(y \minus{} y_2) \minus{} (x \minus{} x_2)(y \minus{} y_1)] \equal{} 0$

I think your problem have error.
Because if $ \theta \equal{} 180^0 \Longrightarrow cot \theta \equal{} \infty \Longrightarrow$ the equation of the circle is not exist.

isnt that obvious? How can a chord/segment subtend 180 degree at the centre???  :|

Let $ O = (x_O, y_O)$ be the center the circle through the points $ A_1 = (x_1, y_1)$ $ A_2 = (x_2, y_2)$. Let $ M = (x_M, y_M) = \left(\frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2}\right)$ be the midpoint of the segment $ (A_1A_2)$, and $ m = \frac {y_2 - y_1}{x_2 - x_1}$ the slope of the line $ A_1A_2.$ Since $ OM \perp A_1A_2,$ slope of the line $ OM$ is $ - \frac {1}{m}$. Assume that $ \theta$ is the smaller angle subtending the longer circle arc, $ \angle MOA_1 = \angle MOA_2 = \theta \le 90^\circ,$

$ \cot \theta = \frac {(OM)}{(MA_1)} = 2 \cdot \frac {(OM)}{(A_1A_2)} \ge 0$

$ \frac {\cot^2 \theta}{4} = \frac {(x_O - x_M)^2 + (y_O - y_M)^2}{(x_2 - x_1)^2 + (y_2 - y_1)^2} =$

$ = \frac {(x_O - x_M)^2 \left(1 + \frac {1}{m^2}\right)}{(y_2 - y_1)^2 \left(1 + \frac {1}{m^2}\right)} = \frac {(x_O - x_M)^2}{(y_2 - y_1)^2}$

$ = \frac {(y_O - y_M)^2 \left(m^2 + 1\right)}{(x_2 - x_1)^2 \left(m^2 + 1\right)} = \frac {(y_O - y_M)^2}{(x_2 - x_1)^2}$

To find the sign of the 1st RHS fraction, consider special case of $ A_1, A_2$ on the y-axis symmetrical with respect to the origin and $ O$ on the positive x-axis. Then $ x_1 = x_2 = x_M = y_M = y_O = 0$, $ y_2 > 0 > y_1$, $ |y_1| = |y_2|$, $ x_O \ge 0.$ Then $ \frac {\cot\theta}{2} = + \frac {x_O - x_M}{y_2 - y_1} \ge 0$.

Likewise, to find the sign of the 2nd RHS fraction, consider the previous special case rotated by $ 90^\circ,$ to get the same circle. Then $ y_1 = y_2 = y_M = x_M = x_O = 0$, $ x_2 < 0 < x_1$, $ |x_1| = |x_2|$, $ y_O \ge 0.$ Then $ \frac {\cot\theta}{2} = - \frac {y_O - y_M}{x_2 - x_1} \ge 0$.

In conclusion, $ \frac {\cot \theta}{2} = \pm \frac {x_O - x_M}{y_2 - y_1} = \mp \frac {y_O - y_M}{x_2 - x_1}.$

The given equation can be written as

$ x^2 - [(x_1 + x_2) \pm \cot \theta (y_2 - y_1)]x + y^2 - [(y_1 + y_2) \mp \cot \theta (x_2 - x_1)] y + C = 0$

where $ C$ is some constant independent of $ x, y$. Substituting $ x_1 + x_2 = 2x_M$, $ y_1 + y_2 = 2y_M$ and $ \cot \theta = \pm 2\ \frac {x_O - x_M}{y_2 - y_1} = \mp 2\ \frac {y_O - y_M}{x_2 - x_1}$,

$ x^2 - 2xx_O + y^2 - 2yy_O + C = 0$

$ (x - x_O)^2 + (y - y_O)^2 = R$

where $ R$ is some other constant. All that remains is to show $ R > 0$, so that we can put $ r^2 = R,$ where $ r$ is the circle radius. But the given equation is obviously satisfied for 2 real points $ A_1, A_2$, therefore it is a real curve, a circle passing through $ A_1, A_2$ and $ R > 0.$

Thank Yetti. Your solution was very careful.

My solution same Yetti:
The equation of circle $ (x - x_O)^2 + (y - y_O)^2 = R^2$ with $ R = OA_1$ = $ \frac {A_1M}{sin \frac {\theta}{2}}$= $ \frac {A{}_1A{}_2}{2sin \frac {\theta}{2}}$ = $ \frac {\sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}}{2sin \frac {\theta}{2}}$
We have equation of line $ OH$: $ (x{}_A{}_2 - x{}_A{}_1)(x - x_M) + (y{}_A{}_2 - y{}_A{}_1)(y - y_M)$  $ \Longleftrightarrow$ $ (x_2 - x_1)(x - \frac {x_1 + x_2}{2}) + (y_2 - y_1)(y - \frac {y_1 + y_2}{2})$.
We use $ O \in OH$ and $ \Vec {OM}. \Vec{AB} = 0$ $ \Longrightarrow$ $ x_O, y_O$.
From that and $ (x - x_O)^2 + (y - y_O)^2 = R^2$, we have the equation is $ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) \pm \cot \theta [(x - x_1)(y - y_2) - (x - x_2)(y - y_1)] = 0$

I think you must limit $ \theta$ for $ A_1A_2$ is real chord, (not is diameter of circle).
So: $ 0^0 < \theta < 180^0$.

This is picture of two circle:

[/img][b]consider the point P (h,k) as the point where angle t(for theta) is subtended

also let

A - (x1 , y1)

B - (x2 , y2)

slope of AP = m1 = (k - y1) / (h - x1)

slope of BP = m2 = (k - y2) / (h - x2)

tan t = (+-) [m1 - m2]/[1 + m1m2]

on putting values

[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/b/9/2/b922a0f84193caed6c47a3cd52129078c1dcf18f.gif[/img]

in the above it should be tant = (+-)....

or

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/3/8/638afa49d8f9b6c0cd665b098ff353fb00bdec8e.gif[/img]

now replace h,k with x,y and bring right hand side to left hand side to get the desired answer
now it should be easy


Find the value of $ x$ to make $ (x, 3)$ a solution of $ 2x \plus{} 5y \equal{} 21$.

we'll plug in the $ y$ value and solve for $ y$

$ 2x\plus{}5y\equal{}21$
$ 2x\plus{}5(3)\equal{}21$
$ 2x\plus{}15\equal{}21$
$ 2x\equal{}6$
$ x\equal{}3$

our value of $ x$ is $ \boxed{3}$


Can you show me the steps of how you got it? :D

[quote="Ani"]punch - 20% mixed fruit, 80% lime perrier
you add 15 mL of the previous punch solution to 10 mL of 30% fruit and 70% perrier.
How many mL of the new punch is actually fruit?

Is the answer 6?[/quote]

15 mL of this solution is:
3 mL mixed fruit
12 mL lime perrier

10 mL of the other solution is:
3 mL mixed fruit
7 mL lime perrier

3 mL + 3 mL = 6 mL

That is so much punch -_-;'

You can use ca=t formula...

(Concentration)(amount)=total

Or make a chart, but this way is the easiest to show on a forum.


limit
trigonometry
calculus
calculus computations

If a, b, c, d are constants such that lim x to 0, (ax^2 + sinbx + sincx + sindx) / (2x^2+3x^3+4x^4) = 3, evaluate a, b, c, d.

$ \lim_{ x\to 0} \frac{ax^2 \plus{} \sin(bx) \plus{} \sin(cx) \plus{} \sin(dx)} { 2x^2\plus{}3x^3\plus{}4x^4} \equal{}$
$ \lim_{ x\to 0} \frac{2ax \plus{} b \cos(bx) \plus{} c \cos(cx) \plus{} d \cos (dx)} { 4x\plus{}9x^2\plus{}16x^3}\equal{}$
note this $ \to \pm \infty$ unless $ b \plus{} c  \plus{} d \equal{}0 , (*)$
$ \lim_{ x\to 0} \frac{2a \minus{} b^2 \sin (bx) \minus{} c^2 \sin(cx) \minus{} d^2 \sin(dx)} { 4\plus{}18x\plus{}48x^2}\equal{} \frac{2a}{4}\equal{}\frac{a}{2} \to a\equal{}6$

so, with (*) and a=6, we have the complete set of solutions.

$ (6, \minus{}\alpha, \beta, \alpha \minus{}\beta )$


calculus
geometry
geometric transformation
reflection
calculus computations

I was wondering if you guys had any suggestions for a good(friendly) calculus book. If this is the wrong forum, I apologize - I wasn't sure where to place this. Thanks in advance!

"Friendly" could mean a lot of things, and I'm not going to try to guess what you mean. Some of the main categories of books:

** Books written with an eye to theory:

Spivak
Apostol

(Both of these strongly reflect the personalities of their authors, which is not the case in the next category.)

** "Mainstream" calculus books:

Stewart
Edwards & Penney
Thomas & Finney
- and several others.

These books have each appeared in many different editions, with only incremental changes between editions. (We're currently using Stewart's "Essential Calculus" which is a mainstream book shrunk down to fewer pages.) They read as if written by committee and focus group. The information is there, of course. These books are not completely interchangeable, but they're more alike than they are different.

** Books intended for calculus-for-business and other short calculus courses. (I'd stay away from those if you want to understand anything.)

** "Reform" calculus texts: Hughes/Hallett and imitators. These books tend to downplay algebra and algebraic manipulation - which isn't my reading of the history of 17th century mathematics, but so it is.

** Popularizations and "made easy" books: I'm not that familiar with this category, but such things do exist.

Speaking of books, Spivak finally released a 4th edition of his book after 14 years.


geometry
3D geometry
function
algorithm
group theory
combinatorics open
combinatorics

Do there exist some way solve rubik using only elemental methods(without group theory as other people did)? I am thinking to design a function that represents the disorder of a cube, in such way that when cube is solved the function have value of 0. and then design a algoritm that makes that function always decrease, then because a cube have only finite states. then it will reach the absolute minimun with will solve the cube.
   Is this method possible? how can I define a function that represents the disorder of the cube??

How can you do that? If that algorithm existed, when u had the funcion equal to 0, and applied the algorithm the funcion would become negative and because of that you cannot use it to solve it. You have to be sure that the function takes only non-negative values and if that is so, there can't be an algorithm that always decreases the function...

For each combination of moves you apply to a rubik's cube, if you apply it enough times, the cube will return to the same state he was before, so you can't define if that combination decreases or increases the value of the function.

Tell me if I'm wrong, but I believe that the method as you described it is not possible.

I don't know how is the function like.
But may be we can make a non-increasing function that leads any state of the cube to the solved one.
But I think the function has many cases that you have to consider it first before beginning the algorithm.
If it's that kind of function, I think it'd better not use the function since it is nothing but harder than the nowaday solution.
 :D


[quote="shortlist"]I  think we will prove When c=0 , With c=min{a,b,c}  :roll:[/quote]

can you upload about your solution?

thanks a lot.

Let $ min(a,b,c)\equal{}c$ and $ a\equal{}c\plus{}x , b\equal{}c\plus{}y$

Inequality equivalents to $ kc^2\plus{}lc\plus{}m \geq 3\plus{}4\sqrt{2}$ where

$ k\equal{}\frac{4}{x^2}\plus{}\frac{4}{y^2}\plus{}\frac{4}{(x\minus{}y)^2}$

$ l\equal{}\frac{2x\plus{}4y}{y^2}\plus{}\frac{2y\plus{}4x}{x^2}\plus{}\frac{2x\plus{}2y}{(x\minus{}y)^2}$

$ m\equal{}\frac{x(x\plus{}y)}{y^2}\plus{}\frac{y(x\plus{}y)}{x^2}\plus{}\frac{xy}{(x\minus{}y)^2}$

Let $ f(c)\equal{}kc^2\plus{}lc\plus{}m$

Since $ k>0 , l>0$

$ f(c) \geq f(0)\equal{}m$

Let's prove that $ m \geq 3\plus{}4\sqrt{2}$

$ m \geq 3\plus{}4\sqrt{2} \Longleftrightarrow$

$ u^3\plus{}5u^2\minus{}(4\sqrt{2}\minus{}1)u\plus{}1 \geq 0$ where $ u\equal{}\frac{x}{y}\plus{}\frac{y}{x}\minus{}2$

$ u^3\plus{}5u^2\minus{}(4\sqrt{2}\minus{}1)u\plus{}1 \geq 0 \Longleftrightarrow$

$ (u\minus{}\sqrt{2}\plus{}1)^2(u\plus{}3\plus{}2\sqrt{2}) \geq 0$ which is obviously true.

Inequality holds if and only if

Let $ (x_1,x_2,x_3)$ is a permutation of $ (a,b,c)$ where $ x_1 < x_2 < x_3$

$ x_1\equal{}0$

$ \frac{x_3}{x_2}\equal{}\frac{\sqrt{2}\plus{}1\plus{}\sqrt{2\sqrt{2}\minus{}1}}{2}$

[quote=hedeng123]Let a,b,c be positive real numbers .Prove that:

$ \frac {(a \plus{} b)(a \plus{} c)}{(b \minus{} c)^2} \plus{} \frac {(b \plus{} c)(b \plus{} a)}{(c \minus{} a)^2} \plus{} \frac {(c \plus{} a)(c \plus{} b)}{(a \minus{} b)^2}\ge 3 \plus{} 4\sqrt {2}.$[/quote]

[img]https://i.imgur.com/3IVebNj.png[/img]


Prove that if three real numbers $ a$, $ b$, $ c$ lie in the interval $ \left[0,1\right]$, then

$ \frac {a}{bc \plus{} 1} \plus{} \frac {b}{ca \plus{} 1} \plus{} \frac {c}{ab \plus{} 1}\le 2$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=15431 .

  Darij


inequalities
irrational number
real analysis
real analysis unsolved

Let the sequence $ (a_n)_{n\in{N^{*}}}$, defined by $ a_n\equal{}\frac{1}{3}\plus{}\frac{1}{3^{2^{2}}}\plus{}...\plus{}\frac{1}{3^{n^{2}}}$, for every $ n\in{N^*}$.
Prove that the sequence converges to an irrational number

Use the inequality:

$ \sum_{i\equal{}n\plus{}1}^{m} \frac{1}{3^{i^2}} \le \frac{1}{n 3^{n^2}} \forall m>n \in N$


Prove that:with each $ n$ is positive integer,exits unique a real  $ x_n$  so that  $ \frac {1}{2008^{x_n}} \minus{} x_n \plus{} n \equal{} 0$
Find  $ lim(x_{n \plus{} 1} \minus{} x_n)$

Let $ f_n(x)\equal{}2008^{\minus{}x}\minus{}x\plus{}n$, then $ f_n(n)>0,f_n(n\plus{}2008^{\minus{}n})<0$, therefore $ n<x_n<n\plus{}2008^{\minus{}n}$.

[quote="Rust"] therefore $ n < x_n < n \plus{} 2008^{ \minus{} n}$.[/quote]
thus $ 1 \minus{} \frac {1}{2008^n}\le x_{n \plus{} 1} \minus{} x_n\le 1 \plus{} \frac {1}{2008^{n \plus{} 1}}$
when $ n$ tends $ \infty$ we get $ lim x_{n \plus{} 1} \minus{} x_n \equal{} 1$
Thank Rust for the first idea


calculus
derivative
integration
geometry
logarithms
function
complex analysis

If $ B$ is a convergent infinite Blaschke product, then $ \int_{\mathbb{D}} |B'|^2 \, dA \equal{} \infty$.

After much thought I found a solution using Frostman's theorem. I believe there should be a much more elementary solution, but I can't find it.

You cover the disk infinitely many times, so what do you expect the total area to be? :P

But why do I do that without frostman? Also, can't a Blaschke product actually omit uncountably many points?

OK, I see your point. :) You can circumvent Frostman, but in the process you get a different proof that is rather short and uses only that $ |B|\to 1$ radially a.e. on the boundary. Start with Green's formula
\[ \int_{\mathbb T}u(z)|B_n'(z)|\,|dz|\equal{}\int_{\mathbb D}\Delta u(z) \log\frac 1{|B_n(z)|}\,dA(z)\plus{}2\pi\sum_{j\equal{}1}^n u(z_j)\]
valid for all finite Blaschke products $ B_n$ with zeroes $ z_1,\dots,z_n$ (counted with multiplicities) and all reasonable functions $ u$. Now apply it with $ u\equal{}|B|^2$ (which is reasonable enough to justify the validity of Green's formula by the DCT) and $ B_n$ that is the $ n$-th partial product of $ B$. You'll get
\[ 2\pi n\equal{}\int_{\mathbb T}|B_n'|\,|dz|\equal{}16\int_{\mathbb D}|B'(z)|^2 |B(z)|^2\log\frac 1{|B_n(z)|}\,dA(z)\]
and it remains to note that $ x^2\log \frac 1x$ is bounded on $ (0,1]$ and $ |B|\le |B_n|$ everywhere.

This argument works equally well for arbitrary $ H^\infty$-functions with infinitely many zeroes in the unit disk whose boundary values are uniformly bounded from below by some positive constant. I believe that $ H^\infty$ here can be replaced by $ H^2$ or even Nevanlinna but do not see how to do it at the moment. On the other hand, the boundedness from below is crucial: one can construct a non-zero bounded function $ f$ with infinitely many zeroes such that $ \int_{\mathbb D}|f'(z)|^2\,dA(z)<\plus{}\infty$.


algebra
polynomial
number theory unsolved
number theory

well,i don't think so.
for example :
Let $ p\equal{}5$ and consider this polynomial:
$ f(x)\equal{}x^8\minus{}x^4\plus{}5$
u can easily see that for all $ x \in Z$ we have $ 5 | f(x)$

[quote="mkn19"]Is the following statement true: f(x) is a polynomial with integer coefficents and p (which is a prime bigger than three) divides f(x) for all integer x. Should the coefficents of f(x) be divisible by p?[/quote]

As long as you do not put any restrictions on the degree of your polynomial,
the answer is no: The little Fermat tells you that $ x^p\equal{}x\bmod p$, and this yields
plenty of examples.

But if you restrict the degree to $ \le p\minus{}1$, then indeed all coeffcients must be divisible by $ p$.


geometry rectangle trigonometry circumcircle

[quote="Silverfalcon"]If the area of a rectangle is 1/2 of the area of its circumscribed circle, compute the acute angle between the diagonals of the rectangle to the nearest degree.  [We will accept any integer answer that is no more than 4 degrees away from the correct answer to this problem.]  [Reminder: Calculators may NOT be used!][/quote]
Not so sure about this:
[hide="Solution?"]
Let the desired angle be $ \alpha \equal{} \angle BOC\implies [BOC] \equal{} \frac {r^2\sin\alpha}{2}$ where $ r$ is the circumradius of the rectangle.  Notice that $ [ABCD] \equal{} 4[BOC]$.  Hence, $ \frac {\pi r^2}{2} \equal{} [ABCD] \equal{} 2r^2\sin \alpha$ which yields that $ \frac {\pi}{4} \equal{} \sin \alpha\implies \sin \alpha\approx .78$.  Now, $ \sin 45\approx .7$, I guess that we could estimate that the angle is $ \boxed{50}$ (we can be a 4 degrees off, we we should be fine.)

Remark: The actual angle is $ 52$ degrees, so the estimate is quite close. [/hide]

Nice job. I personally didn't get this problem on my own so I looked into yours and understood it. Silly me on not able to use classic sine area formula.  :blush: I guess I have more work to do!

Anyway, I think your answer is right, and I'm not sure if there's any other solution.  Even if there is, I don't think it's any simpler than yours.


Hi!
I want to write some formulas, that include greek chars, in italic and bold style. I can do that in LaTex, but need to make somehow to be translated in MathType to be used in Microsoft Office. How can do that?
Mathtype don't recognize some commands from LaTex.
Bellow i put an example of what i want to do:

[code]
$\frac{\beta }{2} = \bm{\beta} $
[/code]
$ \frac{\beta }{2} = \bm{\beta}$
Or, if someone can suggest any possibilities to write bold and italic greek chars and export to mathtype, will be great!

MathType will accept your code and create an identical equation to what you've shown in your post -- with one exception. It doesn't recognize the command \bm, so that forces you to do 2 things -- delete the red [color=red]\bm[/color] in the MathType window, and change the beta to bold manually.

For someone with a little programming expertise, MathType's translators are editable, and pretty well commented. Adding this functionality to the TeX input translator may not be a huge hurdle. I haven't looked at it myself though, so I can't say for sure.

In mathType, if symbols cannot be set individual. If i set, greek char to bold (from Style->Define ), all greek's in my formula are marked bold. I don't want that.
If i make all bold, and i set translate to Tex - Plain text, i don't get any command to make bold. For my formula, this is what i have from clipboard even greek it's bold or not.
[code]
$$\beta  = {\beta  \over 2}$$
[/code]
I try'it to use MathMl with MathCast (an very interesting math editor), but with same results.
So problem it's with greek chars.

[quote="vlad2005"]In mathType, if symbols cannot be set individual. If i set, greek char to bold (from Style->Define ), all greek's in my formula are marked bold. I don't want that.[/quote]

No, in MathType, symbols [b]can[/b] be individually set to bold. It's just that you don't do it through Style>Define>Symbol. There are 2 ways you can do it. One way is to define a User1 style in Style>Define>Advanced. Define it the same as Symbol style, except set the Bold property. Then when you create your equation, choose User1 style for the one or two Greek letters you want to be bold.

The second way to do it is to use Style>Other, but I suggest using this method only as a last resort.

Please note also, if you use the first method (User1), [b]it will mess up your translation to TeX[/b], so if you need to translate the equation, use the second method. Still though, [i][b]neither method[/b][/i] will allow you to translate bold characters into bold TeX characters.

[quote]If i make all bold, and i set translate to Tex - Plain text, i don't get any command to make bold. For my formula, this is what i have from clipboard even greek it's bold or not.[/quote]

That's right. Like I said in my first post, MathType doesn't recognize TeX's styling commands, whether it's for Bold characters, for a particular font size, or whatever. That's true whether you're copying TeX to paste into MathType, or whether you're translating a MathType equation into TeX.

Yes, seam that work, using what u suggest. Thanks!!


function
geometry
3D geometry
sphere
topology
ratio
real analysis

a nice one:
Let $f:B->S$ be a continuous function from $B$, the unit
ball of $\mathbb{C}$, towards $S$, the unit sphere of $\mathbb{C}$.
Prove that there exists $z \in S$ such that $f(z)=-z$

I can prove it using the other well-known lemma which comes in handy when we want to prove Browuer's Theorem, namely that $S$ is not a retract of $B$, but if we assume that lemma, then there's only one tiny step to take towards the proof of the theorem, so we wouldn't need this lemma right here anymore.. :?

The \C code generates a gigantic blank box- about a screen wide and tall. Please edit your post to fix this.
Is $\mathbb{C}$ intended?

there is a solution with winding numbers .

Ok, let's give it a try (but we're still proving that lemma I mentioned, even though we don't have to mention it :)).

Assume there is such a function for which $f(x)\ne x,\ \forall x\in S^1$. We can define a homotopy $H:S^1\times[0,1]\to S^1$, s.t. $H(x,t)$ equals the point dividing the small arc $(x,f(x))$ in the ratio $1-t:t$ (it's easy to see that from the condition that $f(x)\ne -x,\ \forall x\in S^1$ we get the continuity of $H$). This means that $f$ is homotopic to the identity map on $S^1$.

On the other hand, there is a homotopy $H':S^1\times[0,1]\to S^1$ defined by $H'(x,t)=f(x(1-t))$. $H'$ is a homotopy from $f$ to the constant map $f(0)$.

The two result above yield a contradiction, since they have different winding numbers ($1$ and $0$, respectively).


[quote="EUCLID 3.14"]Happy new year everyone!!!!!!!!!!![/quote]

Thank!  :D

EUCLID 3.14 wrote:
Happy new year everyone!!!!!!!!!!!

Thank!
 so am i


Let $P = \{P_1, P_2, ..., P_{1997}\}$ be a set of $1997$ points in the interior of a circle of radius 1, where $P_1$ is the center of the circle. For each $k=1.\ldots,1997$, let $x_k$ be the distance of $P_k$ to the point of $P$ closer to $P_k$, but different from it. Show that $(x_1)^2 + (x_2)^2 + ... + (x_{1997})^2 \le 9.$

Draw circles $C_i$ with centers in $P_i$ and radius $\frac{x_i} 2$. It is easy to see that these circles are disjoint. Assume the contrary: $C_i\cap C_j\neq \Phi$. Hence $P_iP_j < \frac{x_i+x_j} 2 \le \max {x_i,x_j }$. Contradiction with the minimality of $x_i$.
Since each circle have radius at most $1/2$, because $PP_i\le 1$, all these disjoint sircles are contained in a circle of center P and radius $3/2$. Hence the sum of areas of $C_i$ is less than the area of this circle, hence $\sum \pi\frac{x_i^2} 4\ge\pi \frac 9 4$. The conclusion follows.

Note that the number 1997 have no relevance for the problem.


calculus
integration
trigonometry
logarithms
calculus computations

This may be simple to you, but i'm only in 4th grade. Please show me the solution!  :D

1 $ \int_{\minus{}\infty}^{\plus{}\infty}e^{tx}e^{\minus{}e^{\minus{}x}}e^{\minus{}x}dx$
2 $ \int_{\minus{}\infty}^{\plus{}\infty}e^{tx}\frac{e^{\minus{}x}}{(1\plus{}e^{\minus{}x})^2}dx$
3 $ \int_{\minus{}\infty}^{\plus{}\infty}e^{tx}\frac{1\minus{}cos(ax)}{\pi ax^2}dx$

for 1, put $ e^{\minus{}x}\equal{}y$ to get $ \int_0^\infty\;y^{\minus{}t}\;e^{\minus{}y}\;dy\equal{}\Gamma{(1\minus{}t)}\equal{}\frac{\pi}{\sin(\pi\;t)\;\Gamma{(t)}}$

for 2, put $ x\equal{}2\ln(\tan\;y)$ to get $ 2\int_0^{\frac{\pi}{2}}\;\sin^{2t\plus{}1}y\;\cos^{\minus{}2t\plus{}1}y\;dy\;\equal{}\;\mathrm B{(1\plus{}t,1\minus{}t)}$
$ \equal{}\Gamma{(1\plus{}t)}\;\Gamma{(1\minus{}t)}\equal{}\frac{\pi\;t}{\sin(\pi\;t)}$
(hope they taught you in the 4th grade what the [b]beta [/b]integral is)

for 3, i should leave it to you, all it requires is a trivial trigonometric substitution  ( you should eventually get expressions of $ \tan^{\minus{}1}\left(\frac{a}{t}\right)$ and $ \frac{t}{a}\ln\sqrt{\;1\plus{}\frac{a^2}{t^2}\;}$ in your answer )


algebra
polynomial
function
topology
linear algebra
matrix
linear algebra unsolved

Here is a quite difficult problem: (at least to me)
  Let $P$ be a complex polynomial and $A$ the set of those complex  matrices $A$ such that $P(A)=0$. What is the closure of this set?

Maybe there's something missing? :?

On the one hand, the set is included in its closure. On the other hand, if $A_n$ is a sequence of matrices in this set and $A_n\to A$, then for all $m$, we have $A_n^m\to A^m$, so $P(A)=O$, meaning that $A$ also belongs to the set. These two show that the closure of the set is the set itself. :?

Or to put it differently: the polynomial function is a continuous function from $\mathcal M_n(\mathbb C)$ to itself, and this set is the preimage of $O_n$, which is a closed set, so it must be a closed set.

The set of $A$ with $P(A)=0$ is already closed, in any reasonable topology where polynomials are continuous.  Maybe you intend some other question?

[b]Edit: [/b] I see grobber has written the same thing.  Maybe the intended question is to explicitly describe the set of $A$ given $P$, or given $A$ to describe the set of $P$ that annul it.

Excuse, I was stupid, find all isolated points of the set.

First of all, we prove that all the isolated points must be scalar matrices $\lambda I_n$, where $\lambda$ is a simple root of $P$:

If a matrix belongs to our set, then all its conjugates belong to the set. If we assume that our matrix is not diagonal, then we can conjugate it with a sequence of invertible matrices $A_k\to I_n$, where, for instance, if $a_{12}\ne 0$, then $A_k$ are diagonal matrices for which the first and second diagonal entry are different. This shows that $A$ must be diagonal. Now, if, say, the first and second diagonal entries are not equal, then we can conjugate it with matrices $A_k\to I_n$, where $A_k$ has $1$ on the main diagonal, and $\frac 1k$ as its $(2,1)$ entry (and $0$ everywhere else). The conclusion is: $A$ is an isolated point $\Rightarrow\ A$ is a scalar matrix $\lambda I_n$, where $\lambda$ must, of course, be a root of $P$.

If $(x-\lambda)^2|P$, then $A_k\to A$, where $A_k$ has $\lambda$ on the main diagonal and $\frac 1k$ in position $(2,1)$ (and $0$ everywhere else), so $A$ is not isolated.

Conversely, suppose $A=\lambda I_n$, with $\lambda$ a simple root of $P$. All the matrices from our set in a small enough neighbourhood of $A$ must have all eigenvalues equal to $\lambda$. If they are different from $A$, then their Jordan forms have some $1$'s below the main diagonal, and their minimal polynomials have $\lambda$ as a root with multiplicity $>1$, so they do not belong to our set, contradiction. This means that $A$ is isolated in this case.

So the matrices are precisely those of the form $\lambda I_n$ with $\lambda$ a simple root of $P$.

Is this correct?


This is a really nice problem with pure intristical value, altough the results I used can be shown by elementary means I hesitated to put the problem here. On a second thought they are well-known by most high school students so I decided to post it here in order to see the brilliant heads of this forum derive a even more elementary solution.

Define a function $\mu$ over the rationals by $\mu(n/m) = nm$ when $(n,m)=1$. Show that

\[\sum_{r \in \mathbf{Q^+}} \frac{1}{\mu(r)^2} = \frac{5}{2}\]
and derive other similar identities for integers higher than $2$.

Hmmmm....if the reference is good, it may be too early to post this problem here.

Pierre.


Determine(with proof)whether there exists a set $X$ of integers s.t.for any integer $n$ there is exactly one solution of $a+2b=n$,where $a,b$ are elements of $X$(not necessarily different).

you find a solution? see please!  :wink:

[url=http://www.kalva.demon.co.uk/usa/usoln/usol966.html]USA MO[/url]


Given that integers a and b satisfy the equality
a^2-(b^2-4b+1)a-(b^4-2b^3)=0,   (*)
i) prove that b^2+a is a square of an integer.
ii) do there exist an infinitely many of pairs (a,b) satisfying (*)?