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So..

It seems like this forum is overflowing with people who are really on top of their math.  But such is not the case with anyone at my school.  It's really inspiring to see so many people who actually spend time and are smart enough to call the AMC or harder competitions easy, and ace them.  Good freaking job, guys.

Now me..  I haven't ever scored high on math competitions.  I mean sure, 99th percentile or something, but that means zilch.  I scored in the mid 90s on the AMC 10B.  :(:(  I'm a Junior, taking Calc this year.  Yes, I know it's late.  But the way I see it is: come on.  These are just standardized tests, damnit.  And my life will be completely committed to school this year, I'm going to have zero life heh.  Bio, Calc, US Hist, Lang, Stats, Germo 4, Sci Research, robotics, engy club.  I'm fine with it.  My weakness is the standardized test, however.  SO, now, this brings me to my question(s).

Please, AoPS forumgoers - what would you recommend I do, what is the the best way for me to ultimately score 125+ on the AMC, or whatever the benchmark is to pass on to the AIME?  What have you guys done to reach the point where you can ace the AMC, & etc?

Thank you in advance, so much.

Brian

fake edit:  Oh, and does anyone recommend I buy the 2 AoPS books?  Are they worth it?  Or are they really just a waste of cash?  Please cut the BS, I know it's kind of risky asking the barber if I should get my hair cut, but.. yeah.  And if they are good, are they enough on its own?  Probably not, when is anything enough on its own ever?  But okay.  That's all for now.  Promise.

Standardized test?  I wouldn't consider AMC's to be standardized tests.... :?:

But yes, I would definitely recommend the AOPS books.  Since you got 90 something on AMC10 you would probably learn a lot from Volume 1.  Just try to do lots of problems on the forums and past math competitions.

what is your goal? just to qualify past the AMC? if you're looking [b]just[/b] make it past the AMC, aops vol. 1 should be enough. if you're looking to go further, and be successful on the AIME, then you'll want to pick up vol. 2.
and i will repeat what eryaman said, it's not really like a standardized test at all.

Does AoPS 1 cover everything on the AMC?

Where can I find past tests?

Get some old AMC tests. You can buy them at AMC website (just search google for American Math Competition).

As for AoPS book, you can get them if you want (they're very good) but don't rush into it --- I guarantee that the materials in there won't be in your brain after reading the chapter once. This takes a lot of time and you have to be patient.

This seems simple but work on your algebra and geometry book that you used on the school. Borrow them from school and look at every single of them and see if you can do ALL of them. If not, do them and complete the basics to full extent. AMC doesn't require you to have extraordinary skills in math (at least for AMC-10 and AMC-12) so once you have solid base, you'll be easily qualified to AIME.

Of course, I won't guarantee it.

Why darktreb? Please explain the reasoning.

Isn't it obvious? :huh: I have unreasonably high goals! (See sig. :D )

[quote="sr6622"]Yup I have heard that one before....but to go from non-AIME qualifier to MOP is impossible.[/quote]

No it's not...especially with freshman MOP.  I'm betting at least one person has gone from non-AIME to MOP this way.  I know beta went from non-AIME to coming really, really close to MOP (only a few points on the USAMO).

I seem to remember that white_horse_king88 made MOP on his first year (this last year), but I could be completely off. Maybe I misread something.

[quote="Pakman2012"]Shoot for the moon, even if you miss, you'll still wind up among the stars  :P[/quote]

This is a common misconception.  The moon is actually much closer than the stars, so if you shoot for the moon and miss, chances are you did not put on enough velocity to reach the stars, so it just means you fall back down and the Earth's atmosphere burns you up the end yay.

I didn't know you were also a physicist, Stephen...  :P

[quote="JesusFreak197"][quote="probability1.01"]This is a common misconception.  The moon is actually much closer than the stars, so if you shoot for the moon and miss, chances are you did not put on enough velocity to reach the stars, so it just means you fall back down and the Earth's atmosphere burns you up the end yay.[/quote]

Okay... shoot for the stars, and you'll end up at the moon. :P[/quote]

How about..

Just do your math to go IMO and make a lot of money and buy the spaceship to go to the moon?

This seems like a good plan.. :roll:

I don't see what's so unrealistic about getting into MOP as a ninth grader, even if you just started. Nowadays you learn a lot of math, see an easy USAMO (like this year's), and do #1 and #4 and get into MOP on a freebie. Really not that unrealistic ... getting 1 and 4 this year required little skill beyond MathCounts.

I remember my ninth grade year, it was a 30 person MOP. Believe me, that makes a big difference.

But then to go from a low USAMO score to regular MOP(aka blue MOP) seems impossible also.

[quote="darktreb"]If I could give only one piece of advice to everyone, this would be it.

1. Set [i]unreasonably[/i] high goals, and believe you can reach them.[/quote]

I guess his goal was to get a perfect score on IMO.
This is why he ended up becoming only one of the best math students in USA.

OH yeah...I guess you guys were right about the non-AIME freshaman to MOP  :D , but a rising junior non-AIME qualifier to MOP....that seems impossible to me.

Well...out of all the olympiads, I would say the math olympiad IS the hardest one to pull off...

Good luck still though.


$ \frac {2}{1! \cdot 9!} \plus{} \frac {2}{3! \cdot 7!} \plus{} \frac {2}{5! \cdot 5!} \equal{} \frac {(2^{m} \minus{} 1)}{n}$
Actually i think denominator shd be n! but anyways it doent matter  :)
Hint
[hide]
Multiply and divide by 10! And write them as $ \binom{10}{1}$ and $ \binom{10}{3}$ and $ \binom{10}{5}$  :D
After that trivial
[/hide]

The value of m is 10 and n  is 2*fact(10)
where fact(n) represents n factorial..........

[hide]PS.sorry i don know to use LATEX[/hide]

@madness  please latexify ur posts ur three chances are already over :)
[size=200]
[color=red]NO POSTING TILL THEN[/color][/size]

Wait a sec man, how do you post factorial  :huh: . Wth man, its too much. If the author of the prob didnt post in latex, then how am i supposed to knw?  :huh:  :huh:

use 1  da on ur keyboard :huh:
he will latexify it i have pmed him :P

WTH MAN MOD YOU ARE REALLY STUPID, ITS REALLY IRRITATING TO LATEXIFY UNLESS THE PROBLEM IS COMPLEX  :mad:  :mad:   :huh:

MAdness ,u have taken the question wrong
Chappi, your answer,(part of it atleast is wrong)
valerimaxi, ur m is correct, n is wrong

is n fact(10) i.e factorial of 10.........

@pardesi--how to latex factorial....

i think i told that to Rohit three posts above...well anyways use shift+1(together)

Can you post the question correctly?  :huh:  :huh:  Wth i copied ur question and latexified the question. Anyways nitish, i gave the method which im pretty sure its right  :D . Please explain the question in words or short form math terms if u want to be more clear and dont know latex, otw use latex

@nitish: Wth? Do you have any problem with my method?  :huh:  Forget the answer

Valerimaxi, ur answer is actually correct, but if there is a 2 in the denominator, wont u take it intro the numerator. u will lose it out if it is an objective question

tht is what i told u at first that n is 2* fact(10)....

My dear baby,ur method is obviously correct.  :!:  :!: . wanted to check the answer.

Answers dont matter in these kinda questions, only the method matters as it helps to solve faster  :D . And ofc if u had latexified ur question (Sorry pardesi  :D ) it would have been easier to answer but i think you might have made a typo somewhere in the ques because i tried to latexify whatever u had posted that's all. Even if u hadnt used latex, you could have mentioned like 2^(n-1) or something  :)

BABY,the question is (1 DIVIDED BY 5!.5!) AND NOT 2 DIVIDED BY 5!.5!

hey!!!!!!!!!!!!!!!!!!!!!!! :mad:  :mad:
will any of u check my answers.....

okay da big deal, you split the other terms as also $ \frac {1}{1! \cdot 9!} \plus{} \frac {1}{9! \cdot 1!}$ and so on making it $ \binom {10}{1}$ as well as $ \binom {10}{9}$ i.e $ \binom {10}{r}$ where r is odd thus givivng the result  :huh:
@ranga: Dont cry da, as i already said, the ans isnt inportant for this sum at all  :P

My method:
[hide="method"]
multiply and divide by 10!
Then $ \frac {1}{10!}\times(\binom{10}{1} \plus{} \binom{10}{3} \plus{} \binom{10}{5} \plus{} \binom{10}{7} \plus{} \binom{10}{9}) \equal{} \frac {2^{9}}{10!}$
n=10
and m=10
I think the original sum goes like $ \frac {1}{5!\times5!}$ and not $ \frac {2}{5!\times5!}$ and $ 2^{m \minus{} 1}$
[/hide]

Ok............ all of u are correct. Check Rakesh's for final answer


inequalities
trigonometry
inequalities proposed

[color=purple] Let ABC is an acute triangle.Prove that
  $\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}+\frac{1}{4}[(\cos\frac{A}{2}-\cos\frac{B}{2})^{2}+(\cos\frac{B}{2}-\cos\frac{C}{2})^{2}+(\cos\frac{C}{2}-\cos\frac{A}{2})^{2}]\le \frac{3}{2}$
  Have fun ! :D [/color]

Hey I posted it here http://www.mathlinks.ro/Forum/viewtopic.php?t=108951

The proposed inequality holds true in any triangle, as the above link shows and the following proof ...

[quote]$\triangle\, ABC\ \implies\ \boxed{\ \sin\frac A2+\sin\frac B2+\sin\frac C2+\frac 14\cdot\left[\left(\cos\frac A2-\cos\frac B2\right)^2+\left(\cos\frac B2-\cos\frac C2\right)^2+\left(\cos\frac C2-\cos\frac A2\right)^2\right]\, \le\, \frac 32\ }$[/quote]
[u][b]Proof [/b](sketch)[/u] The conclusion reduces to proving that the inequality $\boxed{\ \sum\, \cos A+\frac 14\sum\, \left(\sin B-\sin C\right)^2\, \le\, \frac 32\ }$ holds true

in any [b]acute-angled[/b] triangle. Denoting $R,r,s$ the elements of this acute triangle and using well known identities, the latter inequality

is equivalent to: $1+\frac rR+\frac {s^2-12Rr-3r^2}{8R^2}\, \le\, \frac 32\iff\boxed{\ s^2\, \le\, 4R^2+4Rr+3r^2\ }$ , which is just [u][b]Gerretsen[/b]'s inequality[/u].


When you type a URL into the message body, it automatically becomes clickable.

For example, http://www.artofproblemsolving.com/Forum/index.php becomes clickable

However, when you stick a comma in front of it, it is also accepted as part of the URL. Is this a bug or something?

The website http://artofproblemsolving.com, is a great website.

Also happens with http://www.artofproblemsolving.com, a great website.

Could someone fix this, so it is not requred to put a space between the website and the comma.

Thank you,
~anirudh

[quote="anirudh"]Yes. But sometimes I just type the URL and not put [url][/url] tags in.[/quote]
Copy and paste work...

[quote="t0rajir0u"][quote="anirudh"]Is this a bug or something?[/quote]

I wouldn't call it a bug as much as the lack of a feature.  The site's code most likely detects links by searching for strings beginning with "http://" or "www." and then cuts off the end of those strings at the next space, or whatever.[/quote]Some links contain commas in their body, so that's why I didn't bother to search for that special case :)

But it is known that spaces aren't, so could perhaps you fix it so that dot-space, comma-space, dash-space, i.e. punctuation-not-slash-space doesn't get made into part of the link?

Ok I think it is fixed for comma and dot-space. No need for further code complications I belive :)

Hmm just testing: http://just-testing.com. It worked :)


if the edges of the $ K_{n}$are colored red or blue,and $ r_{i}$,i=1,2...,n denote the number of red edges with vertex $ i$ as an endpoint,and if$ T$ denotes the number of monochromatic triangles then
T=n(n-1)(n-2)/6-1/2 sum[1-->n]$ r_{i}(n\minus{}1\minus{}r_{i})$

Please actually learn Latex.

[quote]If the edges of the complete graph $ K_{n}$ are colored red and blue, $ r_{i}$ denotes the number of red edges with vertex $ i$ as an endpoint, and if $ T$ denotes the number of monochromatic triangles, prove
$ T \equal{}\frac{n(n\minus{}1)(n\minus{}2)}{6}\minus{}\frac{1}{2}\sum_{i \equal{} 1}^{n}r_{i}(n\minus{}1\minus{}r_{i})$.[/quote]

It's nearly trivial: every non-monochromatic triangle has two vertices at which one of its red edges and one of its blue edges meet.  Thus, the second term counts the number of non-monochromatic triangles, while the first term is the total number of triangles.


You have three fair icosahedral (20-faced) die, with faces numbered 1 to 20. The die are red, green, and blue. Let the number that shows up when you roll the red die to be $ r$, the number that shows up when you roll the green die to be $ g$, and the number that shows up when you roll the blue die to be $ b$. What is the probability that $ r^{2}\plus{}b^{2}\plus{}g^{2}\minus{}rb\minus{}bg\minus{}gr$ is positive?

[hide]
$ r^{2}\plus{}b^{2}\plus{}g^{2}\minus{}rb\minus{}gb\minus{}gr\not>0$ when $ r\equal{}b\equal{}g$.

The probability that $ r\equal{}b\equal{}g$ is $ \frac{1}{400}$.

Therefore, the probability that $ r^{2}\plus{}b^{2}\plus{}g^{2}\minus{}rb\minus{}gb\minus{}gr$ is positive is $ \frac{399}{400}$[/hide]

[hide]its only untrue if theyre all the same

thats a possibility: 20/20x20x20=1/400

probablitly=1-1/400=399/400[/hide]

[quote="13375P34K43V312"][quote="Kevin K."]
$ r^{2}\plus{}b^{2}\plus{}g^{2}\minus{}rb\minus{}gb\minus{}gr\not > 0$ when $ r \equal{} b \equal{} g$.
[/quote]

[quote="mz94"]its only untrue if theyre all the same[/quote]

You haven't told me why...[/quote]

if r=b=g, that makes the equation (in terms of any color, im using r)

r*r+r*r+r*r-r*r-r*r-r*r=0

which is not greater than 0

[quote="mihail911"]are there any cases where the expression $ r^{2}\plus{}b^{2}...$ is negative?[/quote]

That's exactly what I meant, mz94 and Kevin K. are asserting this but have not told me why.

yeah..i see what u mean, 1337.
i mean, assuming $ r<b<g\Longrightarrow r^{2}<rb$ and $ b^{2}<gb$ and i don't know about the rest. what else does this imply?

[hide="hint for that"]We want to show that something is nonnegative. Well, what kinds of expressions are always nonnegative?[/hide]

[hide="answer to question in hint"]squares[/hide]

If no one's solved it yet...

[hide]$ r^{2}\plus{}b^{2}\plus{}g^{2}\minus{}rb\minus{}bg\minus{}gr\equal{}\frac{1}{2}[(r\minus{}b)^{2}\plus{}(b\minus{}g)^{2}\plus{}(g\minus{}b)^{2}]\ge0$, with equality if and only if all the squares are equal to 0, that is, $ r\equal{}b,b\equal{}g,g\equal{}r$, or simply $ r\equal{}b\equal{}g$.[/hide]


ΕΔΩ ΘΑ ΓΡΑΦΟΥΜΕ ΠΡΟΒΛΗΜΑΤΑ ΣΥΝΔΥΑΣΤΙΚΗΣ

Νίκο αυτό που θέλεις να γράψεις στο τόπικ  δεν φαίνεται καθόλου κάνεις κάτι για να διορθωθεί το πρόβλημα σε παρακαλώ???

[color=red]ΣΥΝΔΥΑΣΤΙΚΗ 1-ΕΚΦΩΝΗΣΗ[/color]
Πόσοι θετικοί ακέραιοι  από το 1 μέχρι το 1000 μπορούν να γραφούν στη μορφή $2^{a}+2^{b}+2^{c}$ όπου $a,b,c$ μη αρνητικοί ακέραιοι ?

[b][color=darkblue]ΣΥΝΔΙΑΣΤΙΚΗ 1-ΛΥΣΗ[/color][/b]

Τους ζητούμενους αριθμούς που είναι μεταξύ του $1$ και του $1000$ θα τους χωρίσουμε σε τρεις κατηγορίες, αυτούς που γράφονται στη μορφή $2^{k}=2^{k-1}+2^{k-2}+2^{k-2}$ με $k\geq 2$, αυτούς που γράφονται στη μορφή $2^{k}+2^{l}=2^{k}+2^{l-1}+2^{l-1}$ με $k<l$ και αυτούς που γράφονται στη μορφή $2^{k}+2^{l}+2^{m}$ με $k<l<m$, όπου $k,l,m \in \mathbb N$.
Οι αριθμοί της πρώτης κατηγορίας είναι οι $2^{2},2^{3},...,2^{9}$ δηλαδή συνολικά είναι $\binom{8}{1}$ αριθμοί.
Το πλήθος των αριθμών της δεύτερης κατηγορίας είναι οι συνδυασμοί ανά δύο των αριθμών $2^{0},2^{1},...,2^{9}$ ,δηλαδή $\binom{10}{2}$.
Τέλος, το πλήθος των αριθμών της τρίτης κατηγορίας είναι οι συνδυασμοί ανά τρία των αριθμών $2^{0},2^{1},...,2^{9}$ ,δηλαδή $\binom{10}{3}$.

Συνεπώς, πλήθος ζητούμενων αριθμών $=\binom{8}{1}+\binom{10}{2}+\binom{10}{3}=173$.

Στέλιος...

Πεντε διαφορετικοι αριθμοι εκλεγονται τυχαια απο το συνολο {$1,2,3...n$}
ενας καθε φορα,χωρις επανατοποθετηση.Να δειξετε οτι η πιθανοτητα <οι τρεις πρωτοι επιλεγμενοι αριθμοι,οπως επισης και οι πεντε επιλεγμενοι αριθμοι να σχηματιζουν αριθμητικη προοδο(αν διαταχτουν καταλληλα...)>ειναι μεγαλυτερη του $6(n-2)^{-3}$

+ το english version (οπως το βρηκα)

[hide]Five different numbers are drawn at random from {$1,2,3...n$} ,one at a time,without replacement.Show that the probability that the first three numbers drawn,as well as all the five numbers,can be arranged to form an arithmetic progression is greater than $6(n-2)^{-3}$[/hide]

 :D

Λοιπόν παιδιά συγχωρήστε την πολυκαιρη απουσία μου λόγω ανωτέρας βιας μια ασκηση λοιπόν για να επανέλθω

Έστω ένα πολύγωνο με ν>=6 Βρείτε τον αριθμό των τριγώνων που σχηματίζονται άπό 3 κορυφές του πολυγώνου τέτοια ώστε ανα δυο να είναι μη διαδοχικά στο πολύγωνο
Καλή διασκέδαση

Κανείς για αυτήν ???? Γιατί την σνομπάρετε επειδικτικά?????
 :mad:  :mad:
Επειδή ο Σιλουανός κάνει πλακίτσες σνομπάρετε τις ασκήσεις μου¨???
 :rotfl:  :mad:

RAKAR και ambro τι θα λέγατε να βάζετε και τίτλους στις ασκήσεις
π.χ. [b]ΣΥΝΔΥΑΣΤΙΚΗ 2 , ΣΥΝΔΥΑΣΤΙΚΗ 3[/b]  ??  :D

Παιδιά κανε΄ςι δεν ενδιαφέρθηκε για την άσκηση ????

Κρίμα
γιατί είναι ωραία κι όχι τόσο δύσκολη περιμένω λύσεις σας
σε 2-3 μέρες θα δημοσιεύσω λύση

Όταν λες διαδοχικά δύο τρίγωνα τι εννοείς? να μην έχουν κοινές κορυφές?

Ακριβώς ανα 2 να μην έχουν κοινές κορυφές

Aν τα  τρίγωνα δεν έχουν κοινές κορυφές μεταξύ τους και αν κατάλαβα καλά το πρόβλημα (γιατί πολύ εύκολο μου φαίνεται έτσι όπως το κατάλαβα) τότε ο αριθμός είναι

$\left[\frac{n}{3}\right]$

sorry ala den ginete edit....(apetite paremvasi moderator...)
kala min pnigomaste se mia kutalia nero...ksexorizi oti ine askisi sindiastikis+i deyteri sti sira...
pantos ine arketa kalo 8ema....
ambro:i ekfonisi tis askisis su,ine akrivos etsi pos tin grafis sta ipogramismena??8elo na po oti tin exis metaferi aytoleksi apo eki pu tin vrikes,sosta??oxi tipota alo ,apla giati mpori h paramikri alagi leksis(i siras lekseon.... )na alazi entelos tin askisi...

 :D

κοίταξε φίλε μου είναι πιστή μετάφραση μιας άσκησης στα Αγγλικά
το ξέρω αυτό που λές για τις λέξεις κι την σειρά τους αν θές κάποια στιγμή δημοσιεύω κ ιτο πρόβλημα στα αγγλικά αλλά επειδη το μετέφρασα με φίλο μου του ΕΜΠ έτσι το είδαμε κι τό έλυσα

Αμβρόσιε δώσε ένα παράδειγμα για ένα οκτάγωνο ας πούμε .Ποια είναι και πόσα τα ζητούμενα τρίγωνα ? :maybe:

Σιλουανέ απορώ τι ακριβώς δεν καταλαβαίνεις από την έκφραση ανα 2 σημεια να μην είναι  διαδοχικά σημεία του πολυγώνου ?? ή από την έκφραση τρίγωνο με τρείς κορυφες του πολυγώνου ???
αυτά δεν χρειάζονατι διευκρίνιση κι παράδειγμα για παιδιά σαν κι εσένα
περιμένω λύσεις σας

[quote="ambros"]Φιλικά το προβλημα το κατάλαβες αλλά θα το επαναλάβω με υπογράμμιση
[u]τον αριθμό των τριγώνων που δημιουργούνται από 3 κορυφές του πολυγώνου τέτοια ώστε ανα δυο να μην έιναι διαδοχικές κορυφές στο πολύγωνο[/u] η απάντηση είναι λάθος ξαναπροσπάθησε[/quote]

ambro ...λεσ απο τρεισ κορυφεσ ....τετοια ωστε.....

η εννοεισ τετοιεσ ωστε (τισ κορυφεσ δηλαδη..)
η αλλιως εννοεισ τα τριγωνα(ειναι το μονο ουδετερο στο πληθυντικο που χρησιμοποιεισ...)αλλα τοτε δεν εχει νοημα η φραση <να μην ειναι διαδοχικεσ κορυφεσ στο πολυγωνο>...
τα τριγωνα να μην ειναι διαδοχικεσ κορυφεσ στο πολυγωνο??
γι αυτο σε ρωτιησα για την ορθοτητα της εκφρασησ σοθ στο προηγουμενο post...καλ λεει ο siluan...δωσε ενα παραδειγμα για ν=6 (για ν=8 μπορει να εχει πολλα τριγωνα...)
[b]αυτα δεν χρειαζονται διευκρινηση και παραδειγμα για παιδια σαν κι εσενα...[/b]
φιλε ambro σε πληροφορω οτι στην ΙΜΟ ,ο κινεζος με 42/42!!!,ζητησε διευκρινηση στο 6 θεμα...

 :D

Αμβρόσιε, έχω δηλώσει ότι δεν είμαι καλός σ'αυτά. Μήπως όμως πρέπει να είναι $n>6$ και όχι $n\geq6$ $?$

Αν είναι έτσι, έχω βρει ότι για $n =$ περιττός $\Longrightarrow$ $S_{t}= \frac{n\cdot(n-5)}{2}$ και για $n =$ άρτιος $\Longrightarrow$ $S_{t}= \frac{n\cdot(n-4)}{2}$ $,$ όπου $S_{t},$ ο αριθμός
των τριγώνων όπως (κατάλαβα ότι) ορίζει το πρόβλημα και συγχωρέστε με αν κάνω λάθος.

Κώστας Βήττας.

Αμβρόσιε, μήπως η απάντηση είναι $\frac{n(n-4)(n-5)}{6}$ ;
 :roll:
Αν κατάλαβα καλά το πρόβλημα και είναι πράγματι αυτή τότε μπορώ να βάλω και τη λύση μου.  :wink:

Τρίτη διαφορετική άποψη .Μάλλον ο καθένας μας έλυσε διαφορετικό πρόβλημα  :D .
Στέλιο δώσε ένα παράδειγμα για το οκτάγωνο $A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}A_{8}$

Ποια είναι τα ζητούμενα τρίγωνα ?

Βάζω τα δικά μου τρίγωνα στο οκτάγωνο, που συμφωνούν με του stedes στον αριθμό.

$A_{1}A_{3}A_{7},$ $A_{1}A_{4}A_{6},$ $A_{2}A_{4}A_{8},$ $A_{2}A_{5}A_{7},$ $A_{3}A_{1}A_{5},$ $A_{3}A_{6}A_{8},$ $A_{4}A_{1}A_{6},$ $A_{4}A_{7}A_{1}.$

$A_{5}A_{3}A_{7},$ $A_{5}A_{2}A_{8},$ $A_{6}A_{4}A_{8},$ $A_{6}A_{5}A_{1},$ $A_{7}A_{1}A_{5},$ $A_{7}A_{2}A_{4},$ $A_{8}A_{2}A_{6},$ $A_{8}A_{3}A_{5}.$

Με τον stedes συμφωνούμε και στο επτάγωνο. Από το εννεάγωνο και πέρα υπάρχει μεγάλη διαφορά και βλέπω ότι εγώ έιμαι λάθος, γιατί τώρα βρίσκω τρίγωνα, επί πλέον αυτών που μου δίνει ο τύπος.

Ο τύπος του stedes μου αρέσει γιατί δουλεύει και στό εξάγωνο και επί πλέον προκύπτει μία ακολουθία αριθμών προβλέψιμη (μπορώ να βρίσκω τον επόμενο όρο κάθε φορά αρκεί να ξέρω τους προηγούμενους).

$2,$ $7,$ $16,$ $30,$ $50,$ $77,$ $112,$ $...$

Κώστας Βήττας.

Φίλε Ράλλη αδιαφορώ για τον Κινέζο φίλο μου έτσι κ ιαλλιώς αυτός δεν ΄΄ελυσε τέτοιας σδυσκολίας θέμα :wink:
τέλος πάντων

τα τρίγωνα είναι τέτοια ώστε οι κορυφές του ανα δυο να μην ειναι διαδοχικές στο πολύγωνο

τώρα είμαι σαφής????
 :wink:

Για ευκολία, νομίζω ότι μπορούμε να αναδιατυπώσουμε το πρόβλημα ως εξής:

Από το σύνολο των θετικών ακεραίων αριθμών $1,$ $2,$ $3,$ $...$ $,n$  να ευρεθεί το πλήθος των δυνατών συνδυασμών ανά τρεις , έτσι ώστε σε  κάθε συνδυασμό, να μην υπάρχουν διαδοχικοί αριθμοί και ακόμα
να μην υπάρχουν δύο συνδυασμοί, που να περιέχουν τους ίδιους αριθμούς.

(Πως ξεθωριάζουν μερικά πράγματα (άμα τ' αφήσεις) μετά το λύκειο και το πολυτεχνείο...).

Κώστας Βήττας.

Αμβρόσιε άλλο αυτό που λες τ΄'ωρα. Τόση ώρα ασχολούμουνα με λάθος πρόβλημα . :mad:

Πρόσθετη διευκρίνηση:

Οι αριθμοί $1,$ $2,$ $3,$ $...$ $,n$ έιναι σημεία πάνω σε κύκλο, οπότε οι $1$ και $n,$ θα πρέπει να θεωρούνται διαδοχικοί (ως διαδοχικά σημεία).

Προσπαθώντας να προσεγγίσω το σκεπτικό του srtedes, από το σύνολο των δυνατών, εν γένει, συνδυασμών ανά τρείς (ευκαιρία να ξεσκονίσω τον Δασκαλόπουλο...), αρκεί να αφαιρέσουμε αυτούς που περιέχουν δύο διαδοχικούς αριθμούς καθώς επίσης και τους συνδυασμούς που περιέχουν τρεις. Αυτός είναι ο αριθμητής στον τύπο που ζητάμε.

Σε κάθε έναν από αυτoύς που μένουν (που δεν περιέχουν διαδοχικούς αριθμούς) μπορεί να γίνει αναδιάταξη των αριθμών, κατά έξι δυνατούς τρόπους. Έτσι έχουμε τον αριθμό $6$ στον παρανομαστή.

Πως όμως βρήσκουμε τους συνδυασμούς με δύο και με τρεις διαδοχικούς αριθμούς; Πως επηρεάζεται το αποτέλεσμα από το ότι ο$1$ και ο $n$ θεωρούνται (ως κορυφές πολυγώνου) διαδοχικοί; Δηλαδή πόσοι είναι οι παραπάνω συνδυασμοί που αφαιρούνται, σε κάθε περίπτωση;

Επικαλούμαι τη βοήθειά σας (αν βέβαια, αυτό το σκεπτικό είναι σωστό).

Κώστας Βήττας.

Sorry ρε παιδια αλλα εχω πραγματικα μπερδευτει..... :o [quote="ambros"]
Έστω ένα πολύγωνο με ν>=6 Βρείτε τον αριθμό [b]των τριγ[/b][b]ώνων[/b] που σχηματίζονται άπό 3 κορυφές του πολυγώνου τέτοια ώστε [b]ανα δυο να είναι μη διαδοχικά[/b] στο πολύγωνο
Καλή διασκέδαση[/quote]
pos orizonte ta diadoxika trigona....???
[quote="ambros"]Ακριβώς...([b]τα τριγωνα  [/b])ανα 2 να μην έχουν [b]κοινές [/b][b]κορυφές[/b][/quote]
dekto...
[quote="ambros"][b]τα τρίγωνα [/b]είναι [b]τέτοια ώστε[/b] οι [b]κορυφές [/b]του ανα δυο [b]να μην ειναι [/b][b]διαδοχικές[/b] στο πολύγωνο [/quote]
alo pragma i korifes tus na min ine diadoxikes(δεν μπορο να φανταστο καν ποσ οριζονται οι διαδοχικεσ κορυφες ενοσ τριγωνου...) kai alo na min exun kines korifes....
[quote="ambros"]τον [b]αριθμό των τριγώνων[/b] που δημιουργούνται από 3 κορυφές του πολυγώνου [b]τέτοια ώστε ανα δυο να μην έιναι διαδοχικές κορυφές [/b]στο πολύγωνο[/quote]
epidi pragmatika den exo katalavi ti zitai i askisi,ambro se rotao,mipos enois telika:Τον αριθμο  των τριγωνων,που εχουν ως κορυφες,μη διαδοχικεσ κορυφες του πολυγωνου,και ανα δυο (τα τριγωνα) να μην εχουν κοινες πλευρες???....

 :D

μάλλον αυτό εννοεί  :mad: (Αν κατάλαβα από αυτά που γράφει ο Στέλιος....)
Ρε Αμβρόσιε ξεκαθάρισε επιτέλους τι θες....... :maybe:

Nai tsakothite tora na gelasoume....
Re paidia-ligo iremia sto forum.Telospanton ego me aformi auto to skiniko thelo na episimano gia akomi mia fora tin anagaiotita yparksis moderator sto elliniko forum.

Parakalo na epanelthei i taksi!

P.s.Re ambrosie ksereis poso se ektimo-alla den katalavaino giati den deineis tis dieukriniseis pou sou zitane o Rakar kai o Silouan.Giati antidras toso perierga?Exeis tpt prosopiko mazi tous?

Λοιπόν, από τις παρατηρήσεις του Κώστα νομίζω ότι κατάλαβα τι ζητάει το πρόβλημα:[quote]Έστω ένα πολύγωνο με ν>=6 Βρείτε τον αριθμό των τριγώνων που σχηματίζονται άπό 3 κορυφές του πολυγώνου [u]τέτοιες[/u] ώστε ανα δυο να είναι [u]μη διαδοχικές[/u] στο πολύγωνο [/quote]
Δηλαδή πρέπει να επιλέξουμε 3 κορυφές του πολυγώνου, έτσι ώστε ανα δύο να μην είναι διαδοχικές

Αυτό που βρήκα είναι ότι ισχύει ο τύπος που δίνει ο Στέλιος

[hide][color=brown][b][u]Λήμμα[/u][/b] Με $n$ διαδοχικούς ακεραίους μπορούμε να φτιάξουμε $\frac{(n-1)(n-2)}{2}$ ζευγάρια αριθμών μη διαδοχικών.[/color]

[color=darkblue]Πχ αν οι αριθμοί είναι οι $1,2,...,n$, τότε από τα $\binom{n}{2}$ ζευγάρια που υπάρχουν συνολικά, τα μόνα που αποτελούνται από διαδοχικούς αριθμούς είναι τα $(1,2),(2,3),...,(n-1,n)$, το πλήθος $n-1$.

Άρα επιθυμητά ζευγάρια  $\binom{n}{2}-(n-1) = (n-1)\frac{n}{2}-(n-1) = (n-1)\frac{n-2}{2}$

Η απόδειξη δεν αλλάζει αν θεωρήσουμε τους $x+1,...,x+n$[/color]

Έστω $a_{n}$ των τριγώνων στο n-γωνο

Ας υποθέσουμε ότι γνωρίζουμε το $a_{n}$ και θέλουμε να βρούμε το $a_{n+1}.$
Προσθέτουμε λοιπόν μία κορυφή, ας τη συμβολίζουμε $n+1$ αντί για $A_{n+1}.$
Προφανώς όλες οι υπάρχουσες τριάδες διατηρούνται. Θα βρούμε τις επιπλέον.

Η κορυφή $n+1$ μπορεί να συνδυαστεί μόνο με ζευγάρια από την περιοχή $\{2,...,n-1\}$
Σε πλήθος $n-2$ διαδοχικών αριθμών έχουμε $(n-3)\frac{n-4}{2}$ το πλήθος επιθυμητά ζευγάρια, επομένως αποκτούμε άλλες τόσες (καινούριες) τριάδες.

Επίσης, θα μπορούμε πλέον να συνδυάσουμε το $1$ με το $n$, κάτι που στην προηγούμενη περίπτωση απαγορευόταν επειδή αυτά θεωρήθηκαν διαδοχικά.
Το ζεύγος $(1,n)$ μπορεί να συνδυαστεί με οποιοδήποτε από τα $3,4,...,n-2$
Έτσι αποκτούμε ακόμη $n-4$ τριάδες

Συνολικά το "κέρδος" είναι
$n-4+(n-3)\frac{n-4}{2}=$
$=(n-3)+(n-3)\frac{n-4}{2}-1 =$
$=(n-3)\frac{n-2}{2}-1=$

$=\frac{(n-2)(n-3)-2}{2}$

Άρα $a_{n+1}=a_{n}+\frac{(n-2)(n-3)-2}{2}$

Θα αποδείξουμε ότι $a_{n}= \frac{n(n-4)(n-5)}{6}, \forall n\geq 6$

Για $n=6$ ισχύει, αφού $a_{6}=2$
(έχουμε τις τριάδες $1-3-5$ και $2-4-6$)

Υποθέτουμε ότι ισχύει για το φυσικό αριθμό $n$.

Μετά από πράξεις καταλήγουμε ότι $\frac{n(n-4)(n-5)}{6}+\frac{(n-2)(n-3)-2}{2}= \frac{(n+1)(n+1-4)(n+1-5)}{6}$
δηλαδή ισχύει και για το $n+1$[/hide]

Σας παρακαλώ πάρα πολύ με κανλενα σας δεν έχω προσωπικά κι το ξέρετε
οκ περιμενετε σε πολύ λίογ θα σας βάλω την αγγλικά έκδοση όπως την έχω από την πηγή μου
αλλά μην ζητήσει κανένας καμια άλλη διευκρίηνση :mad:
δικαιούμαι κ ιεγώ έτσι δεν είναι  :rotfl:  :D

Let P be a convex n- gon where n>=6  Find the number of triangles formed by any three vetrices of P that are pairwise nonadjacent in P
 :D  :D

[quote="ambros"]
αλλά μην ζητήσει κανένας καμια άλλη διευκρίηνση :mad:
δικαιούμαι κ ιεγώ έτσι δεν είναι  :rotfl:  :D[/quote]

Τι θα πει δε δίνεις διευκρίνηση ?? :maybe:
Α και κάτι άλλο η άσκηση έχει λυθεί ήδη  :D .

Με την πολύτιμη βοήθεια των Stedes και Pontios, βρήκα την άκρη και μπορώ τώρα να ολοκληρώσω, το σκεπτικό που ξεκίνησα σε προηγούμενο μήνυμα.

1) – Από το σύνολο των θετικών ακεραίων αριθμών $1,$ $2,$ $3,$ $...$ $,n$ θεωρουμένων ως διαδοχικών σημείων πάνω σε κύκλο, το πλήθος των συνδυασμών τους ανά τρεις, δίνεται από τη σχέση

$\Sigma_{1}= \binom{n}{3}= \frac{(n-3)!(n-2)\cdot(n-1)\cdot n}{3!(n-3)!}= \frac{n\cdot(n-1)\cdot(n-2)}{6}$ $,(1)$

2) – Το πλήθος $\Sigma_{2},$ των συνδυασμών που περιέχουν δύο μόνο διαδοχικούς αριθμούς, το υπολογίζουμε ως εξής:

a) – Συνολικά μπορούμε να συγκροτήσουμε $n$ ζεύγη διαδοχικών αριθμών,

τα $(1, 2),$ $(2, 3),$ $(3, 4),$ $...$ $,((n-1), n),$ $(n, 1).$

b) – Το κάθε ζεύγος συγκροτεί τριάδα, με δύο διαδοχικούς αριθμούς, με οποιονδήποτε αριθμό του συνόλου από τους υπόλοιπους ($n-2$ το πλήθος). Εξαιρούνται οι αριθμοί προηγούμενος και επόμενος του ζεύγους, με τους οποίους συγκροτούνται $2$ τριάδες με τρεις διαδοχικούς αριθμούς που τις υπολογίζουμε ξεχωριστά. Άρα οι συνολικές τριάδες με δύο μόνο διαδοχικούς αριθμούς, είναι

$\Sigma_{2}= n\cdot (n-2-2) = n\cdot (n-4)$ $,(2)$

3) – Το πλήθος $\Sigma_{3},$ των συνδυασμών με τρεις διαδοχικούς αριθμούς, ισούται με $n$ και είναι οι

$(1, 2, 3),$ $(2, 3, 4),$ $(3, 4, 5),$ $...$ $,((n-2), (n-1), n),$ $((n-1), n, 1),$ $(n, 1, 2)$ $,(3)$

Από $(1),$ $(2),$ $(3),$ προκύπτει ότι οι τριάδες $\Sigma$ που ζητάμε συνολικά, όπως τις ορίζει το πρόβλημα (τρίγωνα από μη διαδοχικές κορυφές δοθέντος κυρτού πολυγώνου), δίνεται από τη σχέση:

$\Sigma = \Sigma_{1}-\Sigma_{2}-\Sigma_{3}= \frac{n\cdot(n-1)\cdot(n-2)}{6}$ $-n\cdot(n-4)$ $-n$

$\Sigma = \frac{n\cdot(n^{2}-3n+2)-6n\cdot(n-3)}{6}= \frac{n\cdot(n^{2}-9n+20)}{6}= \frac{n\cdot(n-4)\cdot(n-5)}{6}$ και το πρόβλημα έχει λυθεί.

Κώστας Βήττας.

:coolspeak:
Ας γράψω και τη δική μου λύση:

Από την κορυφή $A_{i}$ ξεκινάν τα $2\frac{(n-5)}{2}$ τρίγωνα με κορυφή μία τουλάχιστον από τις $A_{i-2},A_{i+2}$ και τα $(n-6)\frac{(n-5)}{2}$ τρίγωνα με κορυφή μία τουλάχιστον από τις $A_{1},A_{2},...,A_{i-3},A_{i+3},...,A_{n}$.

 Άρα, κάθε κορυφή του πολυγώνου είναι κορυφή σε $x=(n-5)+\frac{(n-5)(n-6)}{2}=\frac{(n-5)(n-4)}{2}$ τρίγωνα.

Συνεπώς όλα τα ζητούμενα τρίγωνα είναι στο πλήθος $\frac{nx}{3}$ (...διαιρούμε με το τρία γιατί μετρήσαμε κάθε τρίγωνο τρεις φορές, μία για κάθε κορυφή του), δηλαδή $\frac{n(n-5)(n-4)}{6}$.

P.S. To $x$ θα μπορούσαμε να το υπολογίσουμε κατευθείαν χρησιμοποιώντας το λήμμα του pontiou για τις κορυφές $A_{1},A_{2},...,A_{i-2},A_{i+2},...,A_{n}$ ($(n-3)$ στο πλήθος).

Στέλιος...

sindiastiki 5

Exoyme $a_{1}$ xehorista antikeimena toy $w_{1}$  eidous,
 $a_{2}$  xehorista antikeimena toy $w_{2}$  eidous kai $a_{n}$ xehorista antikeimena  toy $w_{n}$ eidous.
Kai ikanopoiountai oi exis sheseis :
$a_{2}\leq a_{1}+1$, $a_{3}\leq a_{2}+a_{1}+1$,...,
$a_{n}\leq \sum_{i=1}^{n-1}a_{i}+1$

An theloume na ta topothetisoume se mia seira opou kanena
antikeimeno tou $w_{i}$ eidoys na min brisketai dipla se ena antikeimeno
toy idiou eidoys me $i=2, 3, ... , n$ me posous tropous ginetai
ayti i topothetisi


modular arithmetic
number theory
divisibility tests

Let $S(n)$ denote the sum of the digits of $n$.  Prove that $S(n)\equiv n\pmod9$.

this is also the same reason for the divisibility rule of 3.

extention:
find the divisibility rule for 99

Divisibility by 99 is just the divisibility by 11 and 9. Alternating sum of the digits must be divisible by 11, and the sum of the digits must be divisible by 9.


algebra
polynomial
number theory
superior algebra
superior algebra unsolved

Some nice corollary I encountered while doing some number theory:
Note the well known Sophie Germain identity $ x^4\plus{}4a^4 \equal{} (x^2\plus{}2ax\plus{}2a^2)(x^2\minus{}2ax\plus{}2a^2)$. Claim: this is the only "unexpected" factorisation of $ x^n\minus{}c$. Formally:

Let $ K$ ba a field, $ c \in K$. Then the polynomial $ x^n\minus{}c$ is irreducible iff the following holds:
- for all primes $ p|n$, $ c$ isn't an $ p$-th power in $ K$.
- if $ 4|n$, then $ c$ isn't of the form $ c\equal{}\minus{}4a^4$ with $ a \in K$.

Additional question: can you describe the factorisation of $ x^n\minus{}c$ in general¿ The result is quite nice, too.

I try to show that the general statement is the following
                                                       $ x^n \plus{} \frac {n^2}{4}a^n$
but i believe that i lose something.

That's precisely Theorem 9.1 in Lang's Algebra (3ed), Chapter VI, Section 9, p. 297 (Galois theory of Kummer extensions).


trigonometry
algebra proposed
algebra
trig identities

$ \sum\limits_{k \equal{} 1}^{p \minus{} 1} {\left( {\frac{k}{p}} \right)} e^{2\pi i\frac{{ak}}{p}}  \equal{} \left( {\frac{a}{p}} \right)i\sqrt p$ for prime $ p \equiv 1\left( {\bmod 4} \right),\left( {a,p} \right) \equal{} 1$
and
$ \sum\limits_{k \equal{} 1}^{p \minus{} 1} {\sin \frac{{k\pi }}{p}}  \equal{} \tan \frac{{\left( {\frac{{p \minus{} 1}}{2}} \right)\pi }}{p}$ for odd $ p$

wow, the first one is very nice... i haven't tried the one with the multiples of $ \pi/23$, but it seems nice too... let $ \omega\equal{}\dfrac{\pi}{19}$....

multiplying by $ \cos 9\omega$ and applying the prod-to-sum formulas it follows that the lhs equals

$ \sin 9\omega\minus{}2(\cos 6\omega\minus{}\cos 12\omega\minus{}(\cos 5\omega\minus{}\cos 13\omega)\plus{}\cos 4\omega\minus{}\cos 14\omega\minus{}(\cos 2\omega\minus{}\cos 16\omega))$

it is easy to note that this equals

$ \sin 9\omega\plus{}2(\sin 3\omega\minus{}\sin 4\omega\plus{}\sin 5\omega\minus{}\sin 7\omega)$

squaring both sides it follows that we have to prove that

$ \sin^29\omega\plus{}4\sin9\omega(\sin 3\omega\minus{}\sin 4\omega\plus{}\sin 5\omega\minus{}\sin 7\omega)\plus{}4(\sin 3\omega\minus{}\sin 4\omega\plus{}\sin 5\omega\minus{}\sin 7\omega)^2$

again, from the prod-to-sum formulas we have that the "second" term equals

$ 2(\cos 6\omega\minus{}\cos 12\omega\minus{}(\cos5\omega\minus{}\cos13\omega)\plus{}\cos4\omega\minus{}\cos14\omega\minus{}(\cos2\omega\minus{}\cos16\omega)$

which equals $ 2(\cos 7\omega\plus{}\cos4\omega\minus{}\cos 3\omega\minus{}\cos2\omega)$...

applying prod-to-sum formulas on the "non-squares terms" of the third term it follows that it equals

$ 4(\sin^2 3\omega\plus{}\sin^2 4\omega\plus{}\sin^2 5\omega\plus{}\sin^2 7\omega)\plus{}4(\minus{}(\cos\omega\minus{}\cos7\omega)\plus{}\cos2\omega\minus{}\cos8\omega\minus{}(\cos4\omega\minus{}\cos10\omega)\minus{}(\cos\omega\minus{}\cos9\omega)\plus{}\cos3\omega\minus{}\cos11\omega\minus{}(\cos2\omega\minus{}\cos12\omega))$

which equals

$ 4(\sin^2 3\omega\plus{}\sin^2 4\omega\plus{}\sin^2 5\omega\plus{}\sin^2 7\omega)\plus{}4(\minus{}2\cos\omega\minus{}\cos4\omega\plus{}\cos3\omega)$...

then, we have to prove that

$ \sin^29\omega\plus{}2(\cos 7\omega\plus{}\cos4\omega\minus{}\cos 3\omega\minus{}\cos2\omega)\plus{}4(\sin^2 3\omega\plus{}\sin^2 4\omega\plus{}\sin^2 5\omega\plus{}\sin^2 7\omega)\plus{}4(\minus{}2\cos\omega\minus{}\cos4\omega\plus{}\cos3\omega)\equal{}19\cos^29\omega$

substracting $ \sin^29\omega$ and using that $ \minus{}\cos\omega\equal{}\cos18\omega\equal{}(2\cos^29\omega\minus{}1)$ it follows that we have to prove that

$ 2\cos7\omega\minus{}2\cos4\omega\plus{}2\cos3\omega\minus{}2\cos2\omega\plus{}2\cos\omega\plus{}4(\sin^2 3\omega\plus{}\sin^2 4\omega\plus{}\sin^2 5\omega\plus{}\sin^2 7\omega)\equal{}9$

using the identity $ 2\sum_{k\equal{}1}^9\cos\left(\dfrac{(2k\minus{}1)\pi}{19}\right)\equal{}1$ it follows that we have to prove that

$ 4(\sin^2 3\omega\plus{}\sin^2 4\omega\plus{}\sin^2 5\omega\plus{}\sin^2 7\omega)\equal{}8\plus{}2(\cos5\omega\plus{}\cos9\omega\plus{}\cos11\omega\plus{}\cos13\omega)$

the result follows easily from the fact that $ 1\minus{}\cos 2x\equal{}2\sin^2x$ and some basic algebraic manipulation... and we're done  :D

My post on  "Tue Feb 26, 2008 10:57 am"  -   is a simple solution of all such  equalities


find all function $ f;\mathbb{R} \to \mathbb{R}$
such that

$ fof(x)\equal{}f(x)\plus{}6x$

[quote="mathmen"]find all function $ f;\mathbb{R} \to \mathbb{R}$
such that

$ fof(x) \equal{} f(x) \plus{} 6x$[/quote]
Are you sure there is not a continuity constraint ?


[quote="Nebraska boy"]In the triangle ABC , the incircle with center I touches AB,AC,BC at D,E,F. Let M be the midpoint of BC. Prove that ED,AM,IF concur[/quote]

here is a figure

[geogebra]b6b75f4d578c8fece848977e007312d7d9f959b4[/geogebra]

Dear Mathlinkers,
see for example
http://perso.orange.fr/jl.ayme  vol. 3  The cross-cevian point  P. 16-18
THis problem has been posted several times. Any reference?
Sincerely
Jean-Louis

Dear Mathlinkers
for first references:
http://www.mathlinks.ro/Forum/viewtopic.php?t=152959
Sincerely
Jean-Louis


calculus
integration
limit
logarithms
calculus computations

For $a>1$, let $f(a)=\frac{1}{2}\int_0^1 |ax^n-1|dx+\frac{1}{2}\ (n=1,2,\cdots)$ and let $b_n$ be the minimum value of $f(a)$ at $a>1$.

Evaluate

\[\lim_{m\to\infty} b_m\cdot b_{m+1}\cdot \cdots\cdots b_{2m}\ (m=1,2,3,\cdots)\]

[quote="kunny"]For $a>1$, let $f(a)=\frac{1}{2}\int_0^1 |ax^n-1|dx+\frac{1}{2}\ (n=1,2,\cdots)$ and let $b_n$ be the minimum value of $f(a)$ at $a>1$.

Evaluate

\[\lim_{n\to\infty} b_m\cdot b_{m+1}\cdot \cdots\cdots b_{2m}\ (m=1,2,3,\cdots)\][/quote]

I think you mean: \[\lim_{m\to\infty} b_m\cdot b_{m+1}\cdot \cdots\cdots b_{2m}\ (m=1,2,3,\cdots)\] ?

Anyway...Let me try ...

Use the fact that \[| \int  f(x) dx| \leq \int |f(x) dx|\] Hence
\begin{eqnarray*} f(a) &\geq& \frac{1}{2}(  | \int_0^1 ax^n - 1 dx | + 1)  \\
&=& \frac{1}{2} \left( 1 + |\frac{a}{n+1} - 1 | \right)
\end{eqnarray*}
So $b_n = 1/2( |a/(n+1) - 1| +1 )$. We then have $\lim_{n\rightarrow \infty} b_n = 1$, that is, the limit exists.
since $\lim_{n\rightarrow \infty} b_n = \cdots = \lim_{n\rightarrow \infty} b_{2n} = 1$, it follows that the required limit is just $1$.

Sorry for that it was typo.

Regrettably your answer is incorrect.

kunny

[quote="kunny"]Hello!  Thank you for your reply,fskilnik.

You are wrong in calculation for $\frac{df_n}{da}(a)$. :(

kunny[/quote]
You´re right, kunny!  I typed it wrong!  :oops:  (I corrected it above.)

All the rest seems right to me, anyway. Thank you for the nice problem.  :)

fskilnik

[quote="skanderbeg"]hi kunny. Can you point out where the proof is wrong?[/quote]

$\left|\int  f(x)dx\right|\leq \int |f(x)|dx$, when does the equality hold?

[quote="fskilnik"][quote="kunny"]You are wrong in the result of $b_n$, fskilnic. :([/quote]
Please help me then, kunny.    :blush:

It seems to me that $b_n  = 2^{n\left( {n + 1} \right)^{ - 1} }  = 2^{\frac{n}{{n + 1}}} \, $ is right , isn´t it?    :(

fskilnik[/quote]

Regrettably you are wrong. :(

kunny

Sure, kunny!  Stupid mistake of mine...

I´ve got (say) $ a_{n}$ , not $ b_{n} \, \, $ !

Therefore $
b_n  = f_n \left( {a_n } \right) = 2^{ - \,\frac{1}{{n + 1}}}
\, \, \, $   and the rest is easy. We get the answer 1, right?!

[quote="kunny"]Your answer for $b_n$ is right, but the desired answer is wrong.
kunny[/quote]
Yes, it´s wrong.        :o

From $
\left( {m + 1} \right)\frac{1}{{2m + 1}} < \frac{1}{{m + 1}} + \frac{1}{{m + 2}} +  \ldots  + \frac{1}{{2m + 1}} < \left( {m + 1} \right)\frac{1}{{m + 1}}
\, \, \, $

I realize that $\, \, \,
\frac{1}{2} \le \,\,\,L = \mathop {\lim }\limits_{m \to  + \infty } \left( {\frac{1}{{m + 1}} + \frac{1}{{m + 2}} +  \ldots  + \frac{1}{{2m + 1}}} \right)\,\,\,\, \le 1
$

The answer seems to me to be  $\, \,  2^{ - L} \, \,$ , but... what is the value of $L$  ?             :?

[quote="kunny"][quote="fskilnik"][quote="Moubinool"]$L=\ln 2$[/quote]
Thanks, Moubinool.  Could you please give me a hint on how to prove that?

By the way, kunny, is the answer correct now?  (Finally?)   :lol:[/quote]

Yes, this answer is right. You can use the limit of Riemman's sum. :)[/quote]

Thanks, kunny! (I will look other problems posted by you, for sure.)  :)


induction
combinatorics unsolved
combinatorics

I have a question on graph theory. How can we show that for every simple graph we can orient it's edges in such a way that the indegree and the outdegree of each vertex differ by at most one?

Use induction on the number $ e$ of edges :
The result is trivial for $ e \geq 2$.

Let $ n >0$ be an integer. Assume the result holds for $ e\equal{}k$ for any $ k \leq n$.

Wlog, we may assume that each vertex has degree at least $ 1$.

- If there is a vertex, say $ M$, with degree $ 1$ (before the orientation of the edges).
Then omit $ M$ and its edge $ MA$, and use the induction hypothesis for the resulting graph. Then, choose the orientation of the edge $ MA$ according to the values of the indegree and outdegree of $ A$.

- If there each vertex has degree at least $ 2$, then the graph has a cycle $ C$ (just start from any vertex and travel along the edges, none being used more than once. Since there is a finite number of vertices, you will eventually reach an already visited vertex).
Now, omit $ C$, and use the induction hypothesis on the resulting graph (if when omiting $ C$, we erase all the edges, then go to the next step...). Then choose the orientation of the edges of $ C$ as if you travel around $ C$ (this gives one more for the indegree and outdegree of each of the vertices in $ C$). And we are done.

Pierre.


Let $1=d_1<d_2<....<d_k=n$ be all different divisors of positive integer n written in ascending order. Determine all n such that:
i)$d_4= 4d_3 - 5d_2$
ii)$d_5=2d_2d_3d_4 - 1$
iii)$n$min

Since $ d_5$ is coprime to $ d_2, d_3, d_4$, it must be prime.

We also have $ d_3<d_4\equal{} 4d_3 \minus{} 5d_2$, implying that $ 5 d_2 < 3d_3$.

The smallest $ n\equal{}1990$ is delivered by $ d_2\equal{}2$, $ d_3\equal{}5$, $ d_4\equal{}10$, and $ d_5\equal{}199$.


not really diffcult but i have thought a lot
let 0<x<3 \pi /2
prove that
(sinx/x)^3<cosx

[quote="darkmaster"]This inequality is wrong! The right one is :
     (sin(x)/x) ^3  > cos (x) with 0<x< pi /2
But my computer gives me the answer that it is right for x \in ]0;(3*p)i/2[.
I'll try to solve this problem.[/quote]
No one should do this problem because:
1) it is already a duplicate!!!! check PLEASE
2) this problem is WRONG
your computer did get the right answer, you just have to spot for the right detail. to see this, let f(x)=(sin(x)/x)3  - cos(x). Try finding a root and you will see that the root of this f(x) is very close to (3\pi)/2. Another way quick check is to use the intermediate value: f(\pi)>0 and f((3\pi)/2)<0, then there must be a root between \pi and (3\pi)/2.

[quote="darkmaster"]This inequality is wrong! The right one is :
     (sin(x)/x) ^3  > cos (x) with 0<x< pi /2
But my computer gives me the answer that it is right for x \in ]0;(3*p)i/2[.
I'll try to solve this problem.[/quote]
yes you are right darkmaster
(sin(x)/x)^3>cos(x)
thanks a lot :(


I dont have time to work on this but it seems related to [url=http://en.wikipedia.org/wiki/Wallis_product]this.[/url]

[hide="In more elementary terms"] $\prod_{k=1}^{\infty}(1+a_{k})$ converges iff $\sum_{k=1}^{\infty}a_{k}$ converges.  In this case, it's clear that for $a_{k}= \frac{1}{2m-1}$ we have divergence. [/hide]

Thank you. I asked this because I studied the character of the series $\sum_{n=1}^\infty (-1)^{n}\log(n)=\log(\prod_{m=1}^\infty \frac{2m}{2m-1})$. So it diverges.

[hide]Suppose that the product converges to S, if it does converge. For $m>1$, $\frac{2m}{2m-1}>\frac{2m+1}{2m}$.

Then, $S^{2}> S \prod_{m=1}^\infty \frac{2m+1}{2m}=\prod_{m=1}^\infty \frac{m+1}{m}$, which very clearly diverges.[/hide]

Got it. Another fine solution. Anyway, this equality holds $\sum_{n=1}^\infty (-1)^{n}\log(n)=\log(\prod_{m=1}^\infty \frac{2m}{2m-1})$, doesn't it?

Yes it does. If I may ask, what about the character of that sum were you analyzing? Just curious.

[quote="Stewie"]Anyway, this equality holds $\sum_{n=1}^\infty (-1)^{n}\log(n)=\log(\prod_{m=1}^\infty \frac{2m}{2m-1})$, doesn't it?[/quote]

On the left you have a sum that diverges and on the right you have a product that diverges to infinity, so it doesn't make much sense to call them equal.  What you mean is that \[\sum_{m=1}^{2n}(-1)^{m}\log(m)=\log\left(\prod_{m=1}^{n}\frac{2m}{2m-1}\right),\] and this just follows from rules for logarithms.

If we take logs, then we get $\sum (-1)^{n}\log n$...so that series diverges because the sequence diverges.

You're right, JBL, but what I meant is that the 2 sides of that equality have the same character to the infinity.

Zuton Force, I see now that your first inequality is wrong..

[quote="Stewie"]You're right, JBL, but what I meant is that the 2 sides of that equality have the same character to the infinity.[/quote]

As you've written it, they don't: you need something to make clear that you are only taking the even partial sums of the series on the left, since the odd partial sums diverge to $-\infty$.

Got it. So what should it be said in this case? I mean, should I say that the sum doesn't diverge neither to $\infty$ nor to $\infty$.

Your sum still diverges to $\infty$, because the product $\prod_{k=1}^{N}\ln \frac{2k}{2k-1}$ does diverge (we can put a lower bound on it on the order of $\sqrt{N}$). However, you can't use the sum $\sum_{k=1}^\infty (-1)^{k}\ln k$ to prove this, because you're only considering even partial sums.

For a clearer example, we know that $\sum_{k=1}^\infty \frac{1}{2^{k}}$ converges to 1. On the other hand, we have
\[\frac{1}{2^{k}}=2^{k}-\frac{4^{k}-1}{2^{k}}\]
and the sum
\[2-\frac{3}{2}+4-\frac{15}{4}+8-\frac{63}{8}+\cdots\]
diverges, since the odd partial sums tend to $\infty$.


Here are two oral questions i'm having doubts in......

1)in the figure,the surface is rough only between block [b]A[/b] and block [b]B[/b].In equilibrium,the frictional force by A on B is in which direction?

2)If we have a rectangular brick first pushed along its longer face then on its shorter face(Making it stand) on a rough table,in which case will the frictional force be more(I know this is pretty dumb,just want to clear some stupid doubts :lol: )

1.) Is there some gravitational force? What is F?

2.) We usually consider the friction to be area-independent so in that case the forces are the same. In real life, however, there might be slight differences between the two.

1)of course there is gravitational force,F is just given like that,it's not specified what it is exactly,though i have second thoughts that the questuon depends on F

2)I though there might be something of pressure and stuff....according to the formaula it won't matter but when you hold the brick in your hand according to the conditions in the question,won't the second seem tougher?

I just thought that gravitational force would point downwards... In this case, the frictional force should be zero, don't you think? For 2.), no, the force will be about the same.

I still haven't got a perfect answer for the 1st one,Please explain the method you use and bus,i don't quite get you :huh:

The friction will act directly upwards, and it will be equal to F times the coefficient of static friction, however if F is insufficient to keep the block A up, the friction will then be equal to F times the coefficient of dynamic friction. But in either case, the friction acts upwards because this opposes the motion the block A would otherwise have.

Oh, I'm sorry, it seems that I misunderstood the problem statement... degreestudent is right. The only thing I have to add is that the friction force is the [i]lesser[/i] of $m\cdot g$ and $f\cdot F$.

As object $A$ is in equilibrium then the vectoral sum of forces acting on the $A$ will be zero. They are first gravitational force $m \vec{g}$ then friction force $\vec{S}$ normal force $\vec{N}$ and at last the force that we are applying on the object $\vec{F}$. As you understood $m\vec{g}+\vec{F}+\vec{N}+\vec{S}=0$. The only vertical forces acting on the object are the gravitational force and force of friction thus $\vec{S}+m\vec{g}=0$ so the force of friction is upward and equal to the force of gravitational pull. This is the force acting on object $A$ by object $B$, so the force acting on object $B$ by object $A$ will be the $-\vec{S}$ the same $m\vec{g}$.

I knew that the questoin depended on [b]F.[/b]Degreestudent,if the block satrts slipping,then you can't say that the frictional force acts upwards since the condition given in the question that the blocks are in equilibrim would be wrong and the answer would be that the block cannot remain in equilibrium.

The reason i asked this question was that in the book,the answer was that [hide]block cannot remain in equilibrium[/hide].

the conditions i mention in my reply were the only possible conditions for equilibrium.

I didn't understand exactly what you said. But the direction of friction does not depend on F (only requires it to be non-negative). Also, the block [i]can[/i] be slipping and still remain in equilibrium

Let's start with #1.  Now go on and answer what will the force direction be even if they are are not in equilibrum.

If there is no friction and no force opposing the gravity force, both blocks are accelerating at g.  (Remember the cannonballs and the leaning tower. . . )  This acceleration is irrelevant to the direction of the force between A and B, which is in equilibrum in the direction of F.

If this helps, the situation is similar to a case of objects in orbit, with a -F force on the other side of B.  B has to push A back with F to keep from changing orbits (moving through the solid wall).

Relative to the blocks, the force is normal to the interface.

Enought there I think for you all to see the light.

On to #2.  The real world questions are if the brick will tumble, exceed the force on the interface necessary for a different coefficient of friction (different deformation of the two surfaces).  If one assumes no and that the friction force is area indepedent and that g changes over the height difference are neglible, the previous answers seem to be correct to me.

"Relative to the blocks, the force is normal to the interface."

I don't understand what this means at all..

For the question 2 the answer is equal. You can understand it from my previous reply.

[quote="bus"]Of course, the objects can't be in equlibrium if there is no friction between B and the wall.[/quote]

But there is friction between A and B. This frictional force can prevent A from moving downwards if F is large enough.

Now I get your point. I was a fool saying that. :(

No problem, really. I think that the whole task is very questionable... the people who write these books should give besides answers at least some kind of hints as to why is the answer what it is.

Hey varun, you joined MathLinks exactly one day before me! :)

bus, i actually joined exactly two days before you.  :D

I loved doing physics until i explored the excitement in maths. since then i'm a lover of maths. And had not done physics questions for a long time(say 2 months). I would be happy if we have more and more mechanics question as I liked it very much. :ninja:

[quote="varun"]Actually F can be.
But to stop them from falling an external force must act.
Here frictional force is internal, thus cannot prevent the fall.

OK shady...[/quote]

The force F is external,see the diagram.

The force F is balanced by the normal reaction.
The weight of the bodies is not balanced by any external force. Thus, they may be together, but they will fall down.
The same explaination has been given by someone else also.

[quote="bus"][quote="shadysaysurspammed"][quote="bus"]A and B won't move relative to each other but they will both fall down if there's no friction between B and the wall..... That's the way I see it.[/quote]

Why can't F be large enough to prevent the blocks from slipping?[/quote]

How can F prevent the blocks from accelerating downwards if it's a horizontal force??[/quote]

Well there is friction between the blocks so maybe [b]F[/b] can create frictional force greater than the weight of the bigger block and keep it up.

Well bus, that's the way small students are scolded and told to believe what the teachers believe. ;)
I too don't have a very good answer for shady...'s doubt. So try to answer him. It is not necessary that what we have never seen can never happen.
I've tried by saying there is no external force balancing the weights. So how about that?

[quote="bus"]Right. The friction is [i]between[/i] the blocks. But this frictional force cannot hold [i]both[/i] of the blocks in the air. What holds block A in the air? Block B does. What holds block B in the air? Block A does. Doesn't it sound strange? Have you ever noticed that the wall plays no part in this whole thing? The wall itself creates no vertical force because it's frictionless. Therefore the wall cannot support any of the two blocks, and the blocks can't support each other only by themselves. They will fall.[/quote]

so then isnt the answer that the system cant remain in equilibrium?

Yes, according to me it should be the answer.
 :P

[quote="bus"]Right. The friction is [i]between[/i] the blocks. But this frictional force cannot hold [i]both[/i] of the blocks in the air. What holds block A in the air? Block B does. What holds block B in the air? Block A does. Doesn't it sound strange? Have you ever noticed that the wall plays no part in this whole thing? The wall itself creates no vertical force because it's frictionless. Therefore the wall cannot support any of the two blocks, and the blocks can't support each other only by themselves. They will fall.[/quote]

This is what i supposed to be the answer but i started thinking about it as someone pulling a block on a frictionless surface with a block over it with acceleration g and someone puts his hand on top of the block A and presses it down with a force F such that it stops by creating sufficient normal force between the blocks and thus increasiong friction to a great amount.I think this isn't right,please correct me and i won't post anymore nonsense :D  :blush:

In order to be consistent with your original problem, You would have to have strings, say, attached to both blocks, one on each, and pulling them both along with an acceleration of g.
Then if you pressed down on Block A, all you would do is increase the normal force between the two blocks.
However, since there is [i]no relative movement[/i] between the two blocks, then there can be no friction.
Since they are both moving in the same direction with the same acceleration, g, that's why there is no relative movement.

Hence no friction.

I think i get the whole thing now....conceptual problem,sorry guys :D


Solve this Maths Problems within 60 Secs with Tricks

Plz provide your solution alongwith

[b][color=#FF0000]
1) 68% of 595 - 43% of 372 = ?

2) (96)² + (63)² = (?)² - (111)² - 8350

3) [(135)² ÷ 15 * 32] ÷ ? = 45 * 24

4) (739% of 383) ÷ 628 = ?

5) 814296 * 36 = ? * 96324

6) 628.306 + 6.1325 * 44.0268 = ?

7) (935.82)² = ?

8 ) 92.5% of 550 = ?

[/color][/b]

[color=#0000FF][b]Solve above problems within 60 secs..Without use of calculator (calc)

Use Paper, Pen & Brain only

Plz provide ur solution alongwith time u consumed[/b] [/color]

Posting this once was more than enough:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308060

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308060[/url]


Please answer the poll and explain why you think so.

Could you at least add a 3rd option (other?) Nothing is black and white you know.

Iraq is a country for Iraqis not a country to the Americans
America must be administered only the same

Well I'm not the person to discuss all this war and politics, but I think that we should just stop!!!! Because we already have oil!!! But I remember from class one day that we get our oil from some Middle Eastern Organization called.... OPEC????  Please do not insult me for getting this wrong. But Saudi Arabia is like... helping us right now so we have enough oil. The major problem is that Venezuela, a major  part of the OPEC situation is wanting to become a communist country. :(  So if they become a communist country, then America is screwed because they might cut off America from the OPEC imported oil. :(

My opinion is that the war in Iraq was a war over oil, and nothing else. Democracy is a non sequitur from the actions of the American army in Iraq because of its inept handling of the transition of power. This shows that either a) Bush did not really want democracy (actually would be Cheney in this case) and used it as a cover, which seems most likely, b) that the US government really wants democracy to be established but doesn't know how (unlikely, as we've already replaced totalitarianism with democracy in Grenada), or c) that there are conflicting forces in Congress/the executive branch, or both, which is hindering efforts to institute democracy (possible but there is no motive). The other reason I do not believe the US government really wants democracy and stability, etc., etc. is that when the war started, it was all "WMDs, WMDs, WMDs," with no mention of democracy whatsoever. Then, when it was realized that there were no WMDs, the focus suddenly shifted to democracy. I was still in elementary school at the time, but I found that very, [i]very[/i] fishy.
Now, the reasons why I believe oil is the cause of the war:
1. The Saudis wanted the war (and they supply 90% of OPEC oil, which accounts for over half of our energy supply). It makes sense. They are bordered by Iraq, and they wanted to draw attention away from the fact that they, too, harbor terrorists (it's widely known).
2. Iraq is still pretty oil-rich. When the war started, the idea of the US was probably "let's take down Iraq, a country that is perfect because it has oil but doesn't constitute much of OPEC's exports, and has a dictator who might have WMDs to cover up the effort." The idea probably was: After the 21-day drive, we would have brushed up insurgencies, then instituted a democratic republic in Iraq, and started training Iraqis at a leisurely pace to be security forces. In the meantime, we would have used the presence of American troops and American-provided security against Iran, Israel, and Saudi Arabia to coerce oil contracts out of the Iraqis, cutting our dependence on oil from the Saudis, allowing us to attack the Saudi monarchy later on and then take [i]their[/i] oil.

Now, however, it's shown that we can't get a good government in Iraq or take their oil. So as such, we should withdraw, because there's no point to staying in Iraq except pissing off the world and getting ourselves killed.

Whoa.. You can become the addresser for like.. politics!!!! How do you know all this??? All I know is that America is screwed right now, but I hope Obama will bring it up. He's also a good speaker too! :D

I believe that the US shouldn't be in Iraq, we really have no business there.  As for Bush/Cheney lies, WMD's, false statements, getting us into the war, fighting for oil, etc. I really have no opinion.  I have not done enough research to make an informed decision.

The good thing is that Americans are using less and less gas now. I see more people on bikes now. :)

Zeitgeist??? Wth... is that spam or a thing about the war in iraq?

No it's a controversial documentary that is based off of some truths and mostly uncertified assumptions said as fact.

right now, the economy is not so good and the war is really costly, which is not helping our debt. Also, the war is not creating jobs like WW2 did, we no longer need people to make guns and such. The only reason we entered iraq was for 1. 911 and war on terror and 2. oil. thats really it and we should just tighten up homeland security and get out of iraq.

Right. The war is taking out a lot of our money that the governement uses. Is Obama planning to take out the troops?  :o

In 2011. That was previously agreed upon by the Bush administration.

I'm pretty sure they are ending major combat operations in 2010 but I may be wrong.

[quote="Ihatepie"]Well it all started in the eighties. The United States saw Saddam is a strong voice against communism, which they were ultimately against. So we gave money, supplies, and basically anything Saddam wanted to fight back the Soviets and resist their takeover.[/quote]  You appear to be very confused.

[quote="Ihatepie"][quote="boxedexe"][quote="jonathanchou711"] ...Venezuela, a major part of the OPEC situation is wanting to become a communist country. :([/quote] Why is it that you express sadness that a country is nationalizing its resources?  In fact, misery and poverty in this world is largely due to capitalism.  Needless to say, socialism has not been understood very well by many, including myself.[/quote] There are two parts to communism. There's the economic and the social side. On the economic side, I don't care whether we are capitalist or socialist or communist, all I care is what the best government is in that situation. But I seriously disagree with the social side of communism or socialism for that matter. People should be able to think what they want, do what they want, and voice what they want unless it harms another person.[/quote]
From Wikipedia: Socialism refers to a broad set of economic theories of social organization advocating public or state ownership and administration of the means of production and distribution of goods, and a society characterized by equality for all individuals, with a fair or egalitarian method of compensation.

Socialism is not introduced properly in Capitalistic societies for obvious reasons.

[quote="Ihatepie"][quote]Regarding another post of yours, I don't think that Obama is any different (in terms of policy) than any other American president.  I view America as one system and Obama as the voice of this system.[/quote]Then you view it wrong ;). The majority of presidents we've had have had very different outlooks. For example, Clinton encouraged extra regulating, abortion rights, gay rights, and was against religion as a major role in government. On the other hand, Bush had the opposite view on all of these things.[/quote]
The politicians may have different views on different issues, but their ultimate goals are the same and their essential principles are roughly the same.

[quote="Ihatepie"]
[quote][quote="jain_harshi"]Um, I think people will hate me for this, but no. Iraq needs those troops to restore order. Without troops, insurgents will be back with suicidal bombs, rockets, and what not. Troops are also needed to keep peace in the country until democracy has been established. After that, troops can be withdrawn but for now, no.[/quote]
America is not there to restore order; no country spends this amount of money to give peace to another country.  Also, there is no peace in Iraq to be kept.  In fact, I believe that Iraq was a much safer place during Saddam's time in comparison to now.  Finally, I would like to point out that Iraq did not threaten America in terms of terrorism during Saddam's time.  I believe that America thought of Saddam as a person who was unyielding.
[/quote] Well it all started in the eighties. The United States saw Saddam is a strong voice against communism, which they were ultimately against. So we gave money, supplies, and basically anything Saddam wanted to fight back the Soviets and resist their takeover. However, when Iraq took control of the oil-rich-Kuwait in the early 90s, the United States got angry. We invaded and encouraged Iraqi civilians to rise up against their leader. And that's why we were angry with Saddam.[/quote]
What JBL said.

Ok here's what I meant to say:

The US saw Saddam Hussein as a voice against communism (which the United States were against.)
So we (the united states) gave him (saddam hussein and iraq) weapons, money, and other stuff to defend his country from the Soviets. When Iraq invaded Kuwait, we got angry at Iraq and Saddam Hussein

You appear to be confusing Iraq and Afghanistan -- the Soviet Union never engaged in any military campaign in Iraq.  [url=http://www.gwu.edu/~nsarchiv/NSAEBB/NSAEBB82/]This[/url] seems like a good history of the US role in the Iran-Iraq war (1980-88), and the Wikipedia article on Saddam Hussein isn't bad, either.  Read up ;)

I know they didn't. But the US remained scared that they might.

The U.S. should not only withdraw all troops  but also apologize and pay reparations for damages done and crimes commited by the U.S. government  during the past thirty years. This includes the support for Saddam Hussein's in the 1980's when he commited his worst atrocites, the brutal sanctions that starved the Iraqi population while strengthened Saddam, and the 2003 invasion/current occupation that killed probably at least one million innocent civilians directly (millions indirectly), left millions homeless and a society that's completely devastated.

Similarly, Obama really has no right to put more troops in Afghanistan; that's up to the people of Afghanistan to decide.

First, let me make it clear that I am trying to answer many different posts.

Beta is right-absolutely right. The U.S. did a bad job in the Middle East, violating many ethics. A third-party observer would definitely agree with you. But remember that the world is "survival of the fittest". Executives of nations have left massive amounts of appalling and unjust decisions in their wake. But ultimately, they all benefited the leader. Hitler and Stalin are two great examples-and they were only revealed because they were defeated. If the U.S. wins (near impossible, I know) then in 20 years it will blow over and the world will be none the wiser. This was the attitude of Bush in 2001. The US is smart enough to realize helping other countries for no benefit weakens it. I am sorry if I seem harsh or cynical, but I truly believe that's how the world works.

Ihatepie-after 1985 there was no such thing as the Soviet threat. Gorbachev wrecked the economy with perestroika, the August Coup helped even more, and the U.S. saw that coming. Even before that the Soviets were on the decline after OPEC ended the embargo, because that wrecked the Russian barrel.

We did aid Iraq, but that was in the Iran-Iraq war. When this occurred, the Western world saw that the Middle East was gaining independence. This would have led to disastrous infighting between Shiites and Sunnis, Arabs and Persian, and Jews and Muslims, cutting off petroleum resources. Hence Western intervention. So it did start in the 1980's. Another factor was the Iranian revolution, which ousted the last Western puppet state in the Middle East.

And in response to boxedexe's latest post, socialism is (as my Social Studies teacher said-I still have the notes from last year) where the central government provides work, healthcare, welfare, economic regulation, etc. to the people, but does not fully control the economy. It has nothing to do with social issues. Capitalism, socialism, and communism are only economic structures! The "political" and "social" aspects are completely different. They have to do with government systems and social hierarchies.

[quote="boxedexe"]That is certainly a plausible reason; however, I think that another reason is that the Middle East has always been a very important area for which many countries have fought with one another. [/quote]

I'd like to address this. The Middle East has historically not been
an important resource region. Before the 20th century (and thus, the need for oil), countries conquered the Middle East because of proximity. The Soviets bordered the Middle East on half their southern edge. The British? Right next to Egypt and Pakistan. The Ottomans? On their front steps. Same with the Islamics, the Egyptians and the Persians. The Romans just wanted land. The land of the Middle East is not especially important barring oil, it just connects Europe, Africa, Russia, and eastern Asia, where most of our superpowers have come from. The Middle East is just a hub through which they passed.

[quote="evilhamster"]Gorbachev wrecked the economy with perestroika[/quote]  Why is it that this thread can't go two posts in a row without someone saying something completely nonsensical?

I'll clarify what I meant. Perestroika blew open the USSR economy and suddenly overhauled the economic system without social stability, and created a vacuum in the economy with the sudden withdrawal of the government, as well as sending inflation and national debt through the roof. This followed on the heels of failed reform, the war in Afghanistan, and the stalling Soviet economy. These factors also made the status of "superpower" untenable for the Soviets, because unlike the Americans, who could sap wealth from rich Western European allies, the Soviets had no such banks of wealth in Eastern Europe, further exacerbating the problem.

[quote="evilhamster"]And in response to boxedexe's latest post, socialism is... [/quote]
I think that this section of your post is more of a response to one of Ihatepie's posts.  I would like to note that with the growing power of China, Afghanistan seems to be a strategic land that may aid the United States in the control of China to a certain degree.

the reason i dont want our troops to withdraw is simple: if we do, al-Qaeda and/or other terrorist groups will not hesitate to make 9/11/09 a terrible day.

Are you serious? Do you even know that Iraq was not linked to 9/11 in any way shape or form?

And: what do you mean by making it a horrible day¿

Hi folks -- we've obviously been infested by a troll with a high assertion-to-information ratio.  Aside from rating its posts appropriately, try not to feed it [i]too[/i] much.

It's possible Iraq could become a base for Al Qaeda but I don't think Obama will let it happen. The next 9/11 doesn't have to happen on U.S. soil for it to be pretty devastating to us.

I'm not too happy about redirecting 20,000 additional troops to Afghanistan. It's a worthless piece of land hardly worth dying over. At least not until the Talibahn or Al Qaeda can fully reconstitute themselves again and are harder to hide. Let's bring them home.


algebra
system of equations
algebra unsolved

[quote="FelixD"]Let $ A$ be an integer not equal to $ 0$. Solve the following system of equations in $ \mathbb{Z}^3$.
$ x \plus{} y^2 \plus{} z^3 \equal{} A$
$ \frac {1}{x} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^3} \equal{} \frac {1}{A}$
$ xy^2z^3 \equal{} A^2$[/quote]

$ xy^2\plus{}y^2z^3\plus{}z^3x\equal{}xy^2z^3 ( \frac {1}{x} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^3} )\equal{}A$

$ \{ x,y^2,z^3 \}$ is the solution on t:
$ t^3\minus{}At^2\plus{}At\minus{}A^2\equal{}0$

$ (t\minus{}A)(t\minus{}(\minus{}A)^{\frac{1}{2}})(t\plus{}(\minus{}A)^{\frac{1}{2}})\equal{}0$

$ \{ x,y^2,z^3 \} \equal{} \{ A,(\minus{}A)^{\frac{1}{2}},\minus{}(\minus{}A)^{\frac{1}{2}} \}$

$ A < 0$ is clear. And we have

$ y^2\equal{}(\minus{}A)^{\frac{1}{2}}$

$ x\equal{}A,z^3\equal{}\minus{}(\minus{}A)^{\frac{1}{2}}$ or $ x\equal{}\minus{}(\minus{}A)^{\frac{1}{2}},z^3\equal{}A$.

So if $ A\equal{}\minus{}k^{12}$ for an integer $ k$ then we can take $ ( x,y,z ) \equal{} ( \minus{}k^{12}, \pm k^3, \minus{}k^2 )$ or $ ( x,y,z ) \equal{} ( \minus{}k^3, \pm k^3, \minus{}k^4 )$.

I solved and found no solution for the system. A cannot be negative.

Oops! I used the problem description from a different website and it's not the same as this one. So my solution above does not apply here.


absolute value
number theory unsolved
number theory

let P is a prime number such that   P>lal+1  prove that

x^n+ax+P is irreducible.

Since $P>|a|+1$ we conclude that $|z^n+az+P|>0$ for $|z|\leq1+\varepsilon$, where $\varepsilon>0$. So $z^n+az+P$ has not root in $|z|\leq 1+\varepsilon$. Suppose $z^n+az+P=f(z)g(z)$, where $f,g\in\mathbb{Z}[x]$, then $P=f(0)g(0)$, so either $f(0)=\pm 1$ or $g(0)=\pm 1$, but $f(0)$ is a product of some zeros of $z^n+az+P$,  which are greater than $1$ at absolute value, so $|f(0)|>1$. In the same way $|g(0)|>1$. Contradiction.


How many of the 4 digit integers have four digits that are either non-decreasing or non-increasing?

[hide="hint"]If I give you the 4 digits of such a hypothetical number, in how many ways can you rearrange them to meet the condition?  For example, if I told you the number has digits 3,5,1, and 7, how many "desirable" numbers could you find fitting this criteria?[/hide]

[hide="hint_2"]Do what b-flat suggested, also keep in mind complementary counting might be useful.[/hide]

[quote="pascal12"][quote="mathepic"]The question is confusing.

If it has to be either non-decreasing OR non-increasing, and no 4 digits can be both, then therefore all 4 digit numbers fit the criteria.

If they are increasing, they are clearly not decreasing, and vice versa.[/quote]

Not necessarily. For example, 1243 is neither increasing nor decreasing. It has to be all 4 numbers.[/quote] And And the question states that.

Choose four digits between the ten available. You can choose digits with repetition such as (2122). We have exactly two ways to put the digits in non-decreasing or non-increasing order (2221, 1222). The problem is that we have included three digits integers in the counting. We need to get rid of the four digits that start with zero. Starting with 0 we only have non-decreasing order (except for 0000 wich is non-increasing too)

$ 2\text{CR}^4_{10}\minus{}\text{CR}^3_{10}\equal{}2\text{C}^4_{13}\minus{}\text{C}^3_{12}\equal{}1210$ where $ \text{CR}^p_{n}\equal{}\text{C}^p_{n\plus{}p\minus{}1}$


An equilateral triangle  $ABC$  is inscribed in a circle. Let $BD$ a diameter of this circle. On sides $AB , CB$  we take points $M , N$ respectively,   such that $MN = AM + CN$ . Prove that $\angle MDN = 60$

Babis

It's obvious that $AD=CD$.

Now, construct $G$ on $AB$ so that $AG=CN$. It's trivial that $\Delta DAG \cong \Delta DCN$. Now it follows that $\Delta DMN \cong \Delta DMG$, so $\angle NDM = \frac{\angle CDA}{2} = \frac{\pi}{3}$. $\mathbb{QED}!$ :D

[quote="JBL"][quote="Jan"] Now it follows that $\Delta DMN \cong \Delta DMG$[/quote]

This can't possibly be what you mean.[/quote]

 I do not see any problem. The solution seems right.Point $G$ is on the extension of $BA$ . Probably your figure does not help.

 [u]Babis[/u]

Ah, $G$ is on the other side of $A$ from $B$!  It is not clear in the solution.  Thanks.

Yes you are right, I should have mentioned that.


Prove that there exists a set A with 1905 elements included in the set
{1,2,3...,1993} such that A does not contain any arithmetic sequence with 41 terms

Ive found a solution for this problem(it belongs to teacher Mihai Onucu Drambe), but its quite long, and i havent translated it into english.If you want it in romanian email me at xxxxtt@math.com

I think the problem had a set {1,2,3,...1992} and number of elements of A = 1904.  Try to prove that the fianl set is
{1,2,3,...1992} \ ({41,41*2,...,41*48}U{1641,1642,...,1680}). Which indeed has 1904 elements.


search
function
number theory open
number theory

Is there any way to prove that:
If between n+2 and 2n is no prime numbers for any integer n>2, then n=3 or n=5

Well, I don't quite see another way then try to search the net for Chebyshev's theorem and it's strengthenings. :wink:
A quote from Wikipedia that will help:
"there always exists at least one prime number p with n < p < 2n − 2" :wink:

I though that theorem said that there is at least one prime between n and 2n

Well, there are many more strengthenings for this theorem, but the form I gave you is the original postulate :wink: You can search google for them :wink:
Plus the form I gave you is for n>3 :wink:

Thank you.
I tried with n>p>2n but didn't get anything useful out of it.
But it might work with the original postulate.

I think I got one:
Maximum nuberm of primes on interval (n,2n)-(n+2,2n) for n>2 is 1: {n+1 or n+2, both cant be at the same time}

pi(x)- pi(x/2)<=1 (pi of x minus pi of x halfes is less or equal to 1) ( pi(x) is function that generetes number of primes that does not exceede x)
Therefore x<11 (x is less then 11) ---> Ramanujan primes, since for every x=>11 (x equeal or bigger then 11) there are at least 2 primes between x and x/2.

From here you can check every number and you get 3 and 5 are the only ones that have no primes between n+2 and 2n.
10: between 12 and 20 ---> 17
9 : between 11 and 18 ---> 17
8 : between 10 and 16 ---> 13
7 : between 9 and 13   ---> 13
6 : between 8 and 12   ---> 11
5 : between 7 and 10  ---> none
4 : between 6 and 8    ---> 7
3 : between 5 and 6    --->none
for 2 and 1 it doesnt make sense because 2*2=2+2 and 1+2<2*1

Therefore 3 and 5 are only numbers with such restrictions.

This is what I came up with.


I cannot convert the file [i]solution.tex [/i]into PDF file.
When I try converting, it appears at the end:
LaTeX-result: 0 error(s), 0 page(s)

There was also a line: No pages of output.

Have you put any text into it? Nothing will be output unless you add some text.

Oh, thank you very much.
I have just downloaded the file and haven't put any information.


[quote="lillyiggie"]log base 4 (5x-4) = 3[/quote]

$log_{4} {5x-4}=3$ $\rightarrow 4^3=5x-4$
$\boxed {x=\frac{68}{5}}$

[quote="seesaw"][hide] $log_{4}{5x-4}\: =\: 3$ can be converted into $4^3\: =\: 5x-4$
Thus, $64\: =\: 5x-4$,
so  $68\: =\: 5x$
and $68/5\: =\: x$ [/hide][/quote]
ne` ong noi gi` vay cha noi!
hhahahha!

[quote="nguoichuyentoan"]
ne` ong noi gi` vay cha noi!
hhahahha![/quote]Use English please  :roll:

[hide]$\log_4 (5x-4)=3\implies 4^3=5x-4$
$\implies 64=5x-4$
$\implies x=68/5$[/hide]

EDITed typo

[quote="lillyiggie"]pianoforte your wrong[/quote]

Yeah, I think you typed wrong, because the steps are correct, only the result is wrong....

[hide]
log4(5x-4)=3

5x-4=4^3
5x-4=64
5x=68
x=68/5[/hide]

Come on. I'm stupid and can even do that.


=================================
Let $ n$ be a positive integer and let $ A_1, A_2,\ldots,A_{2n\plus{}1}$ be subsets of a set $ B$.
Suppose that:
(i)  Each $ A_i$ has exactly $ 2n$ elements,
(ii)  The intersection of every two distinct $ A_i$ contains exactly one element, and
(iii)  Every element of $ B$ belongs to at least two of the $ A_i$.

For which values of $ n$ can one assign to every element of $ B$ one of the numbers 0 and 1
in such a way that each $ A_i$ has 0 assigned to exactly $ n$ of its elements?
=================================

This was the second problem posed at IMO'1988, and it has been proposed by Czechoslovakia.

Does anybody in this forum know the name of the composer of this problem?
It should be added to
http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions


analytic geometry
LaTeX
topology
set theory
combinatorics solved
combinatorics

Can the plane be coloured in 2000 colours so that any nondegenerate circle contains points of all 2000 colors?

Such a coloring does exist :
Let $c_1,c_2,...,c_{2000}$ be the colors. In th following, subscripts are considered modulo 2000.
Let's give an orthonormal system of coordinates.
Now, for any given real number $a$, let $D_a$ be the line with equation $x=a$.

Let $a$ be a decimal number, which means that its decimal expansion has a finite number of non-zero digits after the decimal point. Let $n = n(a)$ be the rank of its last nonzero digit, which will be called the length of $a$
Then color all the line $D_a$ with color $c_n$. We then color each point of the plane with decimal x-coordinate. Now, give any of the colors to all the other points to construct a coloring of the entire plane.

Now, let $C$ be a circle with center $M(x_M,y_M)$ and radius $r$. Since the interval $[x_M - r ; x_M + r]$ contains decimal points with all lengths sufficientely large, we deduce that for each $i \in \{1,2,...,2000 \}$ the circle $C$ intersect a line $D_a$ which has been colored with $c_i$, which proves the desired property of the coloring.

Pierre.

Yes, the plane can be coloured in the required manner.

Label the colours $C_i, 1 \leq i \leq n$.  Colour the lines $L_i = \{y= \frac{a_i}{b_i}: a_i, b_i \in Z, b_i == i (mod n)\}$ colour $C_i$, and the lines $y=r$, $r$ is 0 or irrational, arbitrarily.

Lemma: There exist $a,b \in Z$ such that for any $L<U$,  $L,U \in R$, $L <\frac{a}{nb+i}<U$.
Proof: The difference $U(nb+i)-L(nb+i)=(U-L)(nb+i)$ can be made larger than 1 since $b$ can be made arbitrarily large.  Therefore, there exists $a \in Z$ such that $U(nb+i)>a>L(nb+i)$ or equivalently, $U>\frac{a}{nb+i}>L$, proving the lemma.

It follows, by the lemma, that any circle with maximum $y$ coordinate $U$ and minimum $y$ coordinate $L$ intersects some member of $L_i$ which, by construction, is coloured $C_i$, proving that every circle contains a point coloured $C_i$ for all $1 \leq i \leq n$.

Sorry for reviving this very, very old topic, but I'm struck by it. We want to show the points of the plane can be colored with $ n$ colors, so that any non-degenerate circle contains points of all colors.

1. The set $ D \equal{} \mathbb{Q}^2$ is dense in $ \mathbb{R}^2$;
2. There exist $ n$ real numbers $ x_i$, linearly independent over $ \mathbb{Q}$ (take them from a Hamel basis);
3. The sets $ D_i \equal{} (x_i, x_i) \plus{} D$ are disjoint, and each still dense in $ \mathbb{R}^2$;
4. Color $ D_i$ using color $ C_i$, and the rest of the points of the plane (randomly) at will.

So why all the fuss?

There exist circles that do not pass through any point of $ {\bf Q}^2$ (e.g., center $ (0, 0)$, radius $ \pi$), and translates of these circles avoid some of your $ D_i$.  (Actually, we can probably avoid all of your $ D_i$ simultaneously.)  Density is not sufficient, as a circle does not contain an open set in $ {\bf R}^2$.

Ah, the old confusion between circle and disk (I probably got confused by the word "contain", as it seemed to point to the interior of the circle (i.e. disk), rather than something like "lie on"...).

Q is also dense in R, why not just paint the lines x=q*(pi^i) for all y q in Q (Pi isn't algebric so all are different) and a circle projection contains an open segment therefore containig an x of all types

Your post is incomprehensible -- please use LaTeX (see e.g. http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:LaTeX_on_AoPS ), fix your various typos, etc.

There is a standard technique from set theory that solves problems like this one in an almost mindless fashion.  Namely, fix a 'well-ordering' over the set of all circles in the plane (of which there are continuum many).  Go through these circles in order; for each circle C in turn, color 2000 uncolored points on C in rainbow fashion.

This can be done because

(a) for each circle C, there are only fewer-than-continuum many circles before C in the well-ordering;

(b) each (nondegenerate) circle has continuum many points, so there will be enough points 'left over' on any circle C we reach in this inductive process.

Many more interesting constructions are possible in this way; one of my favorites: prove that 3-dimensional space can be partitioned into circles of unit radius.

The following solution shows how we can strengthen the problem quite a bit:

Let the colors be $\{1,2,3,\dots ,k\}$. Take a surjective function $f:\mathbb R \rightarrow \{1,2,3\dots k\}$.

We color point $(x,y)$ with color $f(C(y))$. Where $C$ is the conway base $13$ function.

Why does it work? because if we have a circle then there is an open interval $I$ so that for every $a$ in $I$ there is a point $x,y$ on the circle with $y\in I$. The Conway function satisfies that for every interval $I$ and real number $x$ there is $i\in I$ with $f(i)=x$.


Find non-consecutive integer solutions to
\[ 1992^{50}\plus{}x^2\plus{}1\equal{}1992^{26}\plus{}1992^{24}\plus{}y^2.\]

[quote="Cassi"]Find non-consecutive integer solutions to
\[ 1992^{50} \plus{} x^2 \plus{} 1 \equal{} 1992^{26} \plus{} 1992^{24} \plus{} y^2.
\]
[/quote]

$ (y\minus{}x)(y\plus{}x) \equal{} 1992^{50} \minus{} 1992^{26} \minus{} 1992^{24} \plus{} 1992^0$

which is odd.

meaning that every single pair of factors of that ridiculous number corresponds to a set of solutions (except the pair where $ y\minus{}x\equal{}1$, as that'd be consecutive).

so let's try to break that monstrosity into factors

$ (1992^{24}\minus{}1)(1992^{26}\minus{}1)$

$ \equal{}(1992^{12}\minus{}1)(1992^{12}\plus{}1)(1992^{13}\minus{}1)(1992^{13}\plus{}1)$

$ \equal{}(1992^6\minus{}1)(1992^6\plus{}1)((1992^4)^3\plus{}1)(1992^{13}\minus{}1)(1992^{13}\plus{}1)$

$ \equal{}(1992^3\minus{}1)(1992^3\plus{}1)((1992^2)^3\plus{}1)((1992^4)^3\plus{}1)(1992^{13}\minus{}1)(1992^{13}\plus{}1)$

i don't really feel like doing the cube factoring out

and some of those constituent factors might not even be prime

this isn't gonna end nicely at all.

You can even reduce $ 1992^{13}\minus{}1 \rightarrow (1992\minus{}1)(1992^{12}\plus{}1992^{11}\plus{}...1992\plus{}1)$.  Just to make the solutions longer.

I'm guessing the problem is just asking for one solution, not all of them.

Let $ 1992^{24}\equal{}a$ and $ 1992^{26}\equal{}b$.

Then $ ab\plus{}1\plus{}x^2\equal{}a\plus{}b\plus{}y^2\iff (y\minus{}x)(y\plus{}x)\equal{}(a\minus{}1)(b\minus{}1)$.

Now let $ y\minus{}x\equal{}a\minus{}1$ and $ y\plus{}x\equal{}b\minus{}1$. We have that $ y\equal{}\frac{a\plus{}b}{2}\minus{}1$ and $ x\equal{}\frac{b\minus{}a}{2}$.

It is easy to verify that these satisfy the requirements and the original equation.


My problem set features the following line: [img]http://komrades.org/screenshots/s3.png[/img]
I'm basically just wondering what the LaTeX code to render that would be. I tried making a simple array with the text in it, but the spacing was awkwardly large.

Anyone have any clue how he typeset this?

Thanks :)

You can use the following

$ \left(\begin{array}{c}\left(\vec x_{k}\right)_{k \equal{} 1}^{\infty}\,\text{is Cauchy}\\ \text{in }\mathbb R^{n}\end{array}\right)$

[color=blue]\left(\begin{array}{c}\left(\vec x_k\right)_{k=1}^\infty\,\text{is Cauchy}\\
\text{in }\mathbb R^n\end{array}\right)[/color]

Hope it helps.

Hm, you're probably right, that does look like what he used. I guess I was thinking there was something more elegant for that sort of situation. Thanks, Hetidemek :)

On an aside, how about something like [img]http://komrades.org/screenshots/s4.png[/img] Surely there's something in LaTeX for set notation like that?

$ B=\left\{\begin{array}{c}\vec x\in\mathbb R^{n}\end{array}\Bigg|\right.\left.\begin{array}{c}&\exists\text{ a sequence }\left(\vec x_{k}\right)_{k=1}^{\infty}\text{ in }A\\ &\text{such that }\left(\vec x_{k}\right)_{k=1}^{\infty}\text{converges to }\vec x\end{array}\right\}$

[color=blue]B=\left\{\begin{array}{c}\vec x\in\mathbb R^n\end{array}\Bigg|\right.\left.\begin{array}{c}&\exists\text{ a sequence }\left(\vec x_k\right)_{k=1}^\infty\text{ in }A\\
&\text{such that }\left(\vec x_k\right)_{k=1}^\infty\text{converges to }\vec x\end{array}\right\}[/color]

I couldn't set it equal like the image.

Hope stevem may give you a better answer.

The array package allows one to centre text both vertically and horizontally in a cell. So with
\usepackage{array}
you can write (in a document, not the forum)
[code]\[
B:=\left\{
\begin{array}{c|>{\centering}m{6cm}}
 \vec{x} \in \mathbb{R}^n & $\exists$ a sequence $(\vec{x}_k)_{k=1}^\infty$ in A \newline
 such that  $(\vec {x}_k)_{k=1}^\infty$ converges to $\vec{x}$
\end{array}
\right\}
\][/code]The great thing here is there's no need to use \text and you get
[img]http://sixthform.info/gblog/pictures/778b02fb1757691ef9f524f485bd85fc.gif[/img]

The example in the first post is harder because of the 2 () but they are best done using \pmatrix (which means using \text unfortunately) as in
[code]\[
\begin{array}{>{\centering}m{4cm}c>{\centering}m{7cm}}
 $\begin{pmatrix}
   (\vec{x}_k)_{k=1}^\infty \text{ is Cauchy} \\
   \text{ in }\mathbb{R}^n
  \end{pmatrix}$
 & \Longleftrightarrow
 & $\begin{pmatrix}
     \text{ each of the component sequences }\\
     (x_k^{(1)})_{k=1}^\infty, \dots, (x_k^{(n)})_{k=1}^\infty \\
     \text{ is Cauchy in } \mathbb{R}^n
    \end{pmatrix}$
\end{array}
\][/code] which looks like
[img]http://sixthform.info/gblog/pictures/fedf6249f12a433ca4dab7c0878cde8b.gif[/img]

The array package documentation is [url=http://tug.ctan.org/tex-archive/macros/latex/required/tools/array.pdf]here[/url].

[i]Wow[/i], stevem, thanks! Talk about a thorough answer, this was exactly what I was hoping for :)

I'm taking a look now at the lengthy array package documentation. There's a lot more meat to this thing than I first suspected!


geometry
geometric transformation
rotation
counting
distinguishability

There are 3 angels and 6 devils. The angels and devils are having a conference, so they are seating on equally-distanced chairs around the circular table.
Assuming that rotations of an arrangement count as the same, in how many ways can 3 angels and 6 devils be seated around the circular table?

EDIT:

oh didnt mention that angels and devils are undistinguishable

Fix a devil at the top (to account for rotations).  Then there are $ 8!$ ways to arrange the rest, assuming the angels and devils are distinguishable.

hmm
wrong (i have the answer, but dunno the solution)

Wow. Blanked out.

Let the number of devils in between each pair of angels be $ 0\leq a,b,c$ such that $ a\plus{}b\plus{}c\equal{}6$.

However, for every $ (a,b,c)$, $ (b,c,a)$, and $ (c,a,b)$, we can only count one.

Listing, we can have

$ (0,0,6)$
$ (0,1,5)$ and $ (0,5,1)$
$ (0,2,4)$ and $ (0,4,2)$
$ (0,3,3)$
$ (1,1,4)$
$ (1,2,3)$ and $ (1,3,2)$
$ (2,2,2)$

Thus, there are $ \boxed{10}$ ways total.

is it 11? I just did cases with an angel at the top and do some cases.  If it is correct I will post a more detailed solution.

math154 is correct

thanks for the solution


function
real analysis
real analysis unsolved

hey,I'm confused with this exercice that I found in a oral examination!

[color=darkblue]Find all functions $ f$ such that differential equation $ y" \plus{} y' \plus{} fy \equal{} 0$ has a pair of solutions linearly independent as the forme $ (g,g^{2})$.[/color]

It appears that they intend to have you substitute $ g, g^2$ into the given equation, giving you two differential equations in terms of $ f, g$.  Then you can eliminate $ f$, solve for $ g$ and substitute back in to find all possible values of $ f$.  (The "linearly independent" condition is just to eliminate $ y \equal{} 0$ as a "relevant" solution -- otherwise, every $ f$ works.)

So, we have $ g'' \plus{} g' \plus{} fg \equal{} 0$ and $ 2gg'' \plus{} 2g'^2 \plus{} 2gg' \plus{} g^2f \equal{} 0$, and subtracting $ g$ times the first equation from the second gives $ gg'' \plus{} 2g'^2 \plus{} gg' \equal{} 0$.  Solving this gives something like $ g \equal{} \left(c_1 \minus{} e^{\minus{}t \plus{} c_2}\right)^{1/3}$, and then you can plug this back in to acquire $ f$.


linear algebra
matrix
function
calculus
linear algebra unsolved

Find all extreme points of $ X^{T}AX$ on the set $ \{X^{T}X \equal{} I\}$,  where $ A$-given nxn matrix and $ I$ is unite matrix.

[Topic moved from analysis to linear algebra.]

Two clarifications:

1. Are we restricting all of these to being matrices with real entries? (I would guess that we are making that restriction.)

2. Nothing is being said about $ A$ being symmetric, right?

[quote="Ibrogimov"]Find all extreme points of $ X^{T}AX$ on the set $ \{X^{T}X = I\}$,  where $ A$-given nxn matrix and $ I$ is unite matrix.[/quote]That's a function from $ O(n)$ to ${ \mathbb{R}^{n^2}}$. What do you mean by extreme points?

Hello!
Thanks!

$ X,A$ - nxn real matrices.

We should find all(local, global) extreme points(max(min)), may be it is useful to use Lagranges multiplier method.

P.S. This problem is from Variational calculus and Optimization methods book.

Maximum and minimum values only make sense for a function to $ \mathbb{R}$. We don't have one of those.

Oh, Sorry!
I mean there $ max(min)$ of $ det(X^{T}AX)$....

It's a constant. The maximum and minimum are both $ \det(A)$.

Thank you jmerry.

I'm sorry.
Yes, $ det(A^{T}AX)\equal{}det(A)$, but in this case there is no meaning of problem.

But, the problem the teacher gave me was:
Find all extreme points of  $ X^{T}AX$on the set  $ \{X^{T}X\equal{}I\}$, where $ A$-given nxn matrix and $ I$ is unite matrix.


The planning committes at school has 10 members.Exactly four of these memebers are teachers. A four-person subcommitte with at least one member who is a teacher must be formed from the members of the planning committe. How many distinct subcommittes are possible.

[hide="solution"]Ok, so there are 10c4 total committees that can be formed. However, this includes committees that cannot be formed since those committees don't have a teacher. There are 6c4 committees without a teacher. So there are 10c4 - 6c4 = 210-15=[b]195[/b][/hide]


geometry
rectangle
linear algebra
matrix
calculus
integration
function

Let $ g: \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation. Prove that $ | \det{g} | \cdot v(U)$ is the volume of $ g(U)$ for any rectangle $ U$.

What is $ v(U)$?  For any $ \mathbb{R}^{n}$, the determinant of the matrix will equal the "volume" of the enclosed space.  Most linear algebra textbooks should have a proof of this (at least mine does but I'm too lazy to re-type the whole thing :lol: ).

One approach is to write $ g$ as a composition of particularly simple linear transformations. If you want to do it in detail it's going to be a huge mess. One can invoke Fubini's theorem (jmerry calls it Tonelli?) and it follows directly, though. But Fubini's theorem is also either a huge mess or very abstract.

Um, I don't think Fubini's Theorem is what you want here.  Fubini's Theorem concerns reversing the order of double and triple integrals.

I know what I am talking about...

Fubini's theorem let's you express the measure of $ g(U)$ as $ n$ iterated single-variable integrals. Fubini's theorem is more than what you said. Not only is the order of the iterated integrals unimportant, but all orderings also give the integral in the corresponding product measure-space

Kalle can u give more details for proof using Fubini's theorem?

[quote="Kalle"]One can invoke Fubini's theorem (jmerry calls it Tonelli?) and it follows directly, though.[/quote]

you use Fubini when the function is in $ L^1(A\times B)$, you use Tonelli if it is merely nonnegative + measurable

The question of how to prove the formula depends on what your definition of the volume of a parallelepiped is.

One standard definition of volume is using integration:  The volume of an object $ A$ in $ R^n$ is just the integral of the constant function $ 1$ over the set $ A$, or equivalently, the integral of the characteristic function of $ A$ over $ R^n$.  Using this definition, your formula follows easily from the change of variables formula for multivariable integrals.  But this is a stupid proof, first because the proof of the change of variables formula is complicated, but more importantly, because the change of variables formula itself is motivated by the very formula that you want to prove!

Here's another bad answer to your question:  One totally reasonable definition of the volume of a parallelepiped is to USE your formula.  In other words, your formula may be taken as a definition rather than a theorem.  But then you would still have to prove that this definition of volume does not depend on choice of $ g$ or $ U$.  One way to do it would be to prove that it matches the calculus definition of volume (see previous paragraph).

Here's my real answer to your question:  What you should really care about here is not what the proof is (since that depends on pesky definitions), but rather intuition about WHY this formula is true.  Here it is, and it works for any $ U$ (not just a rectangle) for which any intuitive notion of volume makes any sense:

Hopefully, it is intuitively clear that if det $ g$ is zero, then the volume of $ g(U)$ should be zero (since $ g$ crushes $ R^n$ down to a lower dimensional space).  If det $ g$ is not zero, then $ g$ can be row-reduced to the identity, which is equivalent to saying that $ g$ can be written as a product of so-called elementary linear transformations.  Now you need only verify that applying any of these elementary linear transformations $ E$ has the effect of multiplying the volume of any object by (the absolute value of) det $ E$.  There are three kinds of $ E$'s.  (1) Swaps any two components, e.g. $ (x,y)\mapsto(y,x)$--leaves volume fixed and has det $ \equal{}\minus{}1$.  (2) Multiply any component by a constant $ c$, e.g. $ (x,y)\mapsto(cx,y)$--multiplies volume by $ |c|$ and has det equal to $ c$.  (3)  Adding a multiple of one component to another component, e.g. $ (x,y)\mapsto(x\plus{}cy,y)$--leaves volume unchanged and has det $ \equal{}1$ (think of a paralellogram and a square with the same base and height).

This intuitive proof can then be souped up a bit to rigorously prove the formula, using the calculus definition of volume.  You just have to verify that (1), (2), and (3) do the right thing to the calculus definition of volume.  (1) and (2) obviously do.  In order to prove (3), you can invoke Fubini's Theorem (just like using Fubini to calculate the area of a parallelogram).


I was doing problems from 2003 putnam.  I did A1, A2, A3 and A4, and I was able to do all of them except for A3!  This is completely different what have happened on 2006.  Besides the first one about torus, which I happen to have memorized the equation of a torus before, I wasn't able to get any other question right.

  But by my opinion, putnam 2003, at least the part A's, are at least easier than 2006.  AG-AM is used twice, and A4 is just AIME level problem.

  Do any of you guys agree with me or what?  I only started to take Putnam since last year so I am not sure how this works, but sometimes I have the feeling that questions over years are not really on the same level.  For example, 2006 A6 was extremely difficult to me, I think it was much harder than any other problems on 2006.

   So I have this feeling of this test being haphazardly put together by a bunch of lunatics that have no consistency at all, either in topics, or in level of difficulty.

Look on the 2006 results topic (the sticky at the top of the forum) and look at the table I posted about halfway down the first page. The 2003 Putnam was the easiest one of the last nine years for avoiding a score of zero, but by several other measures, including the mean score and the score needed to rank in the top 500, it was one of the harder years.

On the other hand, 2006 A6 was just off-scale difficult. I am quite confident that it was the hardest question of the last decade.

Also: 2006 does rank as a fairly difficult year, and the majority of those who took it got zeros.

If you want to see what an "easy" Putnam looks like, try 1998.