Now that the Gunners have a comfortable lead over Juventus, Lyon and Benfica have prevented away goals and have a slight upper hand, and the Inter vs Villa match is evenly poised, vote for your favourite before they are thrown out!

As for odds I guess it is as follows:
1. Ars vs Juve - 3 to 1 for Ars
2. Barca vs Ben - 9 to 8 Ben
3. Milan vs Villa - 9 to 8 Villa
4. Inter vs Lyon - 5 to 4 Inter

Schevchenko will destroy everybody. Arsenal is not that good.

It'll just be BArcelona all the way!
 Competition is out of question.

Nice voting!
Both Inter and Lyon got thrown out, and nobody voted for them in spite of my prediction that they have slight edges!
But Juve and Benfica got many votes though.
Anyway, keep voting!
It's getting interesting...

Barca pwns everyone. Except for Real MAdrid. Cuz there just awesome like that.  :P

Barca should win pretty easily, but i really hope Milan can at least put up a fight. But, i hope Arsenal win.

Milan did [i]not[/i] put up a fight.
Barca is heading for Paris, and with luck the gunners should join them!

Arsenal for sure.

I mean something has happened with Arsenal this year. Usually they struggle in Europe especially away, even against far inferior opposition. And yet they had an unbeaten group stage record, spotless away record and then fail to win a single away game in the Premiership for about 6 weeks.

The two teams most voted for have gone through to the final in Paris.

Give us a final spurt of voting! Com'mon!!

only barcelona...

only ronaldinho,,,, :P

[quote="Centy"]$\text{May 17th}$[/quote]

thanks... ;)

[size=150][color=orange]BARCELONA WIN 2-1!!!!![/color][/size]

YEAHH,TAKE  that you gunners fans.

Goals scored-

Arsenal-1(sol campbell,36 mins)

bara-2(eto'o (76muin),belleti(i hink 80mins)

Match summary

Early in the match,Henry gets the ball in the 6-yard box and taps it to the keeper!!!!

18mins-Ronaldinho gives a through pas to eto'o,keeper(Lehmann) comes diving and trips eto'o!!!Lehmann shown red card!!!!!Robert Pires is sent out for replacement.Alumina replaces Lehamann.Ronaldinho misses free kick

36mins-Arsenal get a wrong free kick little to the right of penalty box.Henry takes it and sol campbell heads it in the goal.Arsenal 1-0

Ronaldinho and otherr barca forwards unable to penetrate theArsenal 7 MAN DEFENCE!!!!

Alumina is as strong as a wall,saves a wonderful try by Eto'o in the penalty box.

Henrik larsonn replaces som,eone after half time

76 min-:arsonnn gets a pass in the penalty box,does not e selfish,beautifully taps the ball to Eto'o running towards the gpoal kick line,Eto'o gets th ball and taps it past ALumina.GOALLL.1-1!!!

Henry gets a beautiful pass and he is alone with the keeprer,agaoin drives it itoi the keepre!!

4 mins later,again the reat Larsonn gets a pass little outside the pena;ty box,quickly gives a pass to Belleti through 3 defenders and belleti scores!!!!Barca 2-1!!!

And it stays like that,Barcelona are the European champions!!!!!Woohoooo.

Larsonn really passed so quickly in both the goals,the arsenal defenders were left dumbstruck.Ronaldinho could not create too much magic.

congratulations my dears.............

only barcelona :D

'scuse me shady, but pires was not sent off. he was taken off to let in almunia or whatever his name was.

anyway let us end this discussion here. thanks for your votes friends! 27 in all.

whaddyamean man? of course pires was sent off. he got a [color=red]RED CARD[/color] it's that small red strip of card paper the ref. holds up
 to a player, when the player decides he's fed up of playin' old fashioned football.

no,he was sent out so that a new keeper could come.

I bet Barcelona wins   (jajajajajaja lol)

did any one see the footbal betwin France and mexico?

what a bout Iran and Croatia?

[quote="José"][quote="Ashegh"]did any one see the footbal betwin France and mexico?[/quote]

Me. Not really interesting math really. The Iran one I didn't see.[/quote]

yes none of them where so nice.

France1  Mexico 0

Iran 2  Croatia 2

what a bout   Argentina vs Angola  ?

Argentina really played very nice.

very very good passes. ;)

I was wondering.... What are you talking about???????????????????????

About the match between Argentina (2) and Angola (0)

wonderful work - Ashegh and Jose and ragnarok23, keep it up and soon we will be discussing next year's Champions League here!


Given that b_1+b_2+....+b_n = b with each b_i a nonnegative number prove that

                 \sum (i=1,..,n-1)b_ib_i+1 \leq b 2 /4

Expanding, we have :
b 2 - 4*sumb_ib_(i+1)  \geq  sum(b_i) 2  + 2*sum( 1 \leq i<j \leq n) (-1) i+j b_i*b_j = (sum (-1) i b_i) 2   \geq 0.

Pierre.


1. Let C be a point on a semicircle of diameter AB and let D be the midpoint of the arc AC. Denote by E the projection of the point D onto the line BC and by F the intersection of the line AE with the semicircle. Prove that BF bisects the line segment DE.

Let $K$ be the intersection of $BF$ and $ED$. Then $MD^2 = MF. MB = ME^2$


algebra
function
domain
real analysis
real analysis unsolved

Let $\Omega$ be a bounded domain in $\mathbb R^n$ ($n\geq 2$), $a\in \mathbb L^q(\Omega)$ ($q\geq 1$), $a(x)>0$ in $\Omega$, $0<r<\frac{n+2}{n-2}$ and $b,\eta>0$.
Consider the set of mesurable functions $X=\{f:\Omega\times\mathbb R\to \mathbb R\mid |f(x,s)|\leq b|s|^r+a(x)\ \forall x\in\Omega,\,s\in \mathbb R\}$. For any two functions $f,g\in X$ define \[R_\eta(f,g)=\inf_{T>1}\big(||f-g||_{L^1(\Omega\times [-T,T])}+T^{-\eta}\big).\]
Prove that $(X,R_\eta)$ is a complete metric space.

I don't know right now why I need this property of $R_\eta$. Actually, I used $R_\eta$ to measure proximity of nonlinearities. And some time ago I realized that it is a real distance.
Unfortunately, class of functions I use in my work is not so simple as I described above, so I am losing completeness there :(

P.S. Everyone who read it! Please, help me by following link in my signature. :)

[quote="Myth"]P.S. Everyone who read it! Please, help me by following link in my signature. :) [/quote]
You mean they give you some benefit for each user who views their web page following the link? :? Sure, I'll click it then. I just wonder: will it help more if I do it several times or they count several same user views as one? (I hope I do not need to really download Opera to trigger the count  :) ) As to the problem you posted, it is a fairly standard one (though it takes some time to understand what is asked ;) ), so I'll leave it to someone else.

Indeed, that is why I titled it as an exercise.

You need click the link once. They count unique referrals. I don't know exactly: maybe ip-address, maybe they set cookies.
I don't get any valuable benefit from it. It is a special promotion program from Opera Software. It allows to receive a free registration if you get 250 hits.


number theory
advanced fields
advanced fields unsolved

I'm reading Ireland/Rosen's book, "A Classical introduction to Modern Number Theory," and trying to understand his proof for the following theorem, (p.41)

[quote][b]
$U(\mathbb{Z}/ p \mathbb{Z})$ is a cyclic group.

Proof.

Let $p-1=q_{1}^{e_{1}}\cdot q_{2}^{e_{2}}\cdot \cdot \cdot q_{t}^{e_{t}}$ be the prime decomposition of $p-1.$ Consider the congruences
\[(1)\hspace{.5in}x^{q_{i}^{e_{i}-1}}\equiv 1 \hspace{.5in}(p) \]

\[(2)\hspace{.5in}x^{q_{i}^{e_{i}}}\equiv 1 \hspace{.5in}(p) \]
Every solution to congruence (1) is a solution of congruence (2). blah, blah, blah, ...
[/b][/quote]

Here I don't understand what he's trying to do. Could anyone explain why "Every solution to congruence (1) is a solution of congruence (2)."?

The second congruence is the first congruence raised to the $q_{i}$ power.


Hello!
I heard that AMC have changed their format slightly by putting emphasis on logarithmic and trigonometry-based problems, and not allowing calculators. Is this true for 2008 exams?
Thanks,
Lena

I don't know about the first one, but there are definitely no longer calculators.

See the answers to Frequently Asked Questions AMC 10/12 #2, 5, 6, 7 at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

Thank you for the clarification. Why aren't calculator allowed anymore? :?:

Lots of reasons.
1. All problems are already solvable without a calculator.
2. Very little math aptitude is needed to use a calculator. The AMC is also not supposed to be a contest of writing a good program to solve a tough problem.
3. Sometimes using a calculator stops you from seeing a very nice solution to a problem.
4. You don't get a calculator on the AIME. So if you relied on the calculator to get past the AMC, you probably wouldn't do well on the AIME anyways. The cumulative change will probably be minimal.

I agree that it blocks you from seeing a good solution. In fact, I rarely see a need of a calculator for myself. However, having a calculator by your side often gives students reassurance. Like one can do some calculation by hand, but this student may need to check their answer a few times before resting assured that it is right.

Then they'll just have to trust themselves not to make idiotic calculation mistakes like me, won't they?  :wink:

Well, I've actually been making much less calculation mistakes. So I stand a decent chance of not being really affected by the change.


...what exactly is so special about nov 1?

Suspicions confirmed... it's the 3rd anniversary of Brother Bear.  :|

what a coincidence...it's the day of the dead too

what a coincidence, it's All Saint's Day, or All Hallow's Day

Ignore the unimportant ones - I was of course of what besttate said.

What a coincidence, it's the day we have to change into winter uniforms for my school.  Also....nothing.

Well, if this was a holiday that was very special, why did Dad make me do school?  :wink:

[quote="Koda"]How are you celebrating this very special day? I watched parts of Brother Bear (because I did not find enough time for the full thing). It was my tenth time  :)[/quote]
dude you've officially gone insane no offense

Of course it's cool, should be less than an hour now.


If $ x,y,z\in \mathbb{R}$ solve this equation :

$ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{z \plus{} x}} \plus{} \sqrt {\frac {z}{x \plus{} y}} \equal{} \frac {3}{2}$

hello,i think in your last term is a typo, could it be that you mean $ \sqrt{\frac{z}{x\plus{}y}}$?
Sonnhard.


I have some calculus problems where I'm asked to use a computer algebra system to find the
derivative of the function, then use the utility to graph the function and it's derivative.
What would be a good program to use to do this please?

ummm...
maybe this belongs in at least high school intermediate...

[url]http://www.numberempire.com/derivatives.php[/url] to calculate derivatives. Use a graphing calculator for graphing.

Hmmm - thanks, but I like calc101.com better to calculate derivatives, as they show all the steps.  You REALLY like
using a graphing calculator? - I prefer to use a pc tool as I can save the results,
print them out, make quick changes, have a big viewing window, use colour etc etc...

I love Mathematica, but it isn't free. If you want a free one, check out Maxima.

I second TZF. IF you can get your hands on a copy, Mathematica is by far the best option for basically anything you'll have to compute in high school. I know a few grad students who swear by Fortran coding for some esoteric project, but I think that for most intents and purposes, Mathematica will do just fine.

If you ARE going to go out and get a copy, definitely get the Student edition. It's significantly cheaper with no appreciable difference in features.

Actually I got Mathematica a while ago, but it's answers to calculus problems often
look really weird and complicated, and rarely match the book answers, unlike for instance calc101.com,
which is great as it shows all the steps leading to the answer.
Do you guys find this too, or is it likely I'm doing something wrong?  I haven't played
around with it much yet, only a rookie user...

I bought [i]Mathematica 6 for Students[/i] in May and I have a free upgrade to [i]Mathematica 7[/i] because I subscribe to Premier Service. I'll be in eight grade next year but I do higher-year math but I'm currently still in precalculus. But [i]Mathematica[/i]'s a great investment and I currently use it as a calculator. I hope I can put it to further use next time.


Prove that in a triangle $ ABC$,

$ \frac{a\minus{}b}{a\plus{}b}\equal{}\tan(\frac{A\minus{}B}{2})\cdot \tan \frac{C}{2}$.

hello, this theorem can be written in the form
$ \frac{a\minus{}b}{a\plus{}b}\equal{}\frac{\tan\left(\frac{A\minus{}B}{2}\right)}{\tan\left(\frac{A\plus{}B}{2}\right)}$.
Sonnhard.

[hide="Explanation?"]

$ a\equal{}2R\sin A,b\equal{}2R\sin B$

$ \sin A\pm\sin B\equal{}2\sin(\frac{A\pm B}{2})\cos(\frac{A\mp B}{2})$

This should do the trick

[/hide]

[hide]$ tan(\frac{A\minus{}B}{2}).tan\frac{C}{2}$

$ \equal{} \frac{{\sin \frac{{A \minus{} B}}{2}.\sin \frac{C}{2}}}{{\cos \frac{{A \minus{} B}}{2}.\cos \frac{C}{2}}}$

$ \equal{}\frac{{\cos (\frac{{A \minus{} B \minus{} C}}{2}) \minus{} \cos (\frac{{A \minus{} B \plus{} C}}{2})}}{{\cos (\frac{{A \minus{} B \minus{} C}}{2}) \plus{} \cos (\frac{{A \minus{} B \plus{} C}}{2})}}$

$ \equal{}\frac{sinA\minus{}SinB}{sinA\plus{}sinB}$

$ \equal{}\frac{a\minus{}b}{a\plus{}b}$
[/hide]


[quote="vangelismak"]Κύριε Μπάμπη μήπως υπάρχει κάποιο λάθος στη διατύπωση ? :(
Βγάζω κάποια περίεργα πράγματα ... :(  :(

Θα το ξανακοιτάξω πάντως ... το πιο πιθανό είναι να έχω κάνει λάθος εγώ :|[/quote]

 Δεν ξέρω μόνο μήπως πρέπει να προσθέσω : οξυγώνιο. Κατά τα άλλα δε βλέπω κάτι περίεργο.Είμαι σίγουρος ότι θα τη λύσετε σχετικά εύκολα , αρκεί να μην ψάχνετε για απίθανα πράγματα, αλλά να σκεφτείτε τελείως πρακτικά(preolympiad level).Αν κάτι δυσκολέψει, υπάρχει και η αναλυτική γεωμετρία ή τα διανύσματα ! Στους διαγωνισμούς όλα επιτρέπονται !
 Αυτές οι ασκήσεις δε χάνονται σε διαγωνισμούς.Εϊναι κρίμα να επιμείνετε μέχρι τέλους με συνθετική γεωμετρία , αν η άσκηση κάπου κλωτσάει.Πρώτα κοιτάζετε να λύσετε το πρόβλημα και μετά ψάχνετε τον καλύτερο τρόπο.

[b][color=red] Καλή επιτυχία σε όλους σας στον ΕΥΚΛΕΙΔΗ !!![/color][/b]

 [i]( Θα λύνω και γω μαζί σας το Σάββατο στο εξεταστικό κέντρο Εύβοιας και θα σας σκέφτομαι !)[/i]

Oπως είπα και πρίν  :|

Λάθος εκτίμηση δική μου, ...και επειδή είμαι ανυπόμονος βιάστηκα να ανεβάσω το προηγούμενο Post :(

[quote="stergiu"]Σε ισοσκελές τρίγωνο [b]ΑΒΓ(ΑΒ=ΑΓ)[/b] θεωρούμε το μέσο [b]Μ [/b]της [b]ΑΓ[/b] και σημείο [b]Δ[/b] στη βάση [b]ΒΓ[/b] ώστε [b]ΒΔ=2ΔΓ[/b]. Αν [b]Ε[/b] είναι η προβολή του [b]Δ[/b] στην [b]ΒΜ[/b], να αποδείξετε ότι οι γωνίες [b]ΑΒΓ[/b] και [b]ΜΕΓ[/b] είναι ίσες.[/quote]

Πολύ όμορφη άσκηση κ. Στεργίου :) Οπως είπατε μπορεί σχετικά εύκολα να βγει με αναλυτική και διανύσματα όμως η καθαρά γεωμετρική λύση με δυσκόλεψε πολύ  :blush:

Ευτυχώς όμως τελικά τη βρήκα:

Εστω κατ' αρχάς $ O$ το κέντρο του περιγεγραμμένου του $ ABC$. Θα θεωρήσω ένα σημείο $ E'$ τέτοιο ώστε $ \angle{MEC}=\angle{ABC}$ και θα αποδείξω ότι αυτό ταυτίζεται με το $ E$.

Για να το αποδείξω αυτό θα θεωρήσω ένα σημείο $ D'$ πάνω στη $ BC$ ώστε $ DE'\perp{BM}$ και θα αποδείξω ότι αυτό ταυτίζεται με το $ D$.

Κατ' αρχάς έχουμε $ \angle{ME'C}=MOC=B\iff$ $ MOE'C$ εγγράψιμο $ \iff \angle{OE'C}=OMC=90\iff \angle{D'E'C=OE'M=MCO=\frac{A}{2}}$

Τώρα θα βάλω και λίγο συμβολισμό για οικονομία:
$ \angle{E'BC}=E'CA=x$, $ \angle{E'CB}=E'BA=y$

Απο νόμο ημιτόνων στο $ E'D'C$ έχουμε:
$ \frac{\sin{y}}{\sin{\frac{A}{2}}}=\frac{E'D'}{D'C}$  (1)

Απο νόμο ημιτόνων στο $ E'D'B$ έχουμε:
$ \sin{x}=\frac{E'D'}{D'B}$  (2)

Me diairesi kata meli ton 1, 2 pernoume:
$ \frac{\sin{y}}{\sin{x}\cdot\sin{\frac{A}{2}}}=\frac{D'B}{D'C}$

Αρκεί λοιπόν $ D'B=2D'C$ δηλαδή:

$ \frac{\sin{y}}{\sin{x}}=2\sin{\frac{A}{2}}$ που ισχύει αφού αν θεωρήσουμε $ N$ το μέσο της $ BC$

$ \frac{\sin{y}}{\sin{x}}=\frac{BN}{MN}=\frac{BC}{AB}=2\sin{\frac{A}{2}}$

Ουφ τελείωσα :P

Η λύση μου με αναλυτική γεωμετρία  ( αν μπορεί κάποιος ας βάλει και το σχήμα  :)  )

Θεωρούμε σύστημα συντεταγμένων χοy, και δίχως βλάβη της γενικότητας ας είναι Α(0,α), Β( -1,0) και Γ(0,1). Λόγω της σχέσης ΒΔ=2ΔΓ εύκολα βρίσκουμε ότι είναι Δ(1/3,0) και επειδή το Μ είναι μέσο της ΑΓ θα είναι Μ(1/2,α/2).
Έχουμε στη συνέχεια ότι:

1) Ο συντ/στης διεύθυνσης του διανύσματος $ \vec{BM}$ είναι $ \lambda_2\equal{}\frac {a}{3}$

2) Η εξίσωση της ευθείας ΒΜ είναι  y=$ \frac {a}{3}$(x + 1),   (1)

3) Η εξίσωση της ευθείας ΔΕ είναι  y=$ \minus{} \frac {3}{a}$(x$ \minus{} \frac {1}{3}$),   (2)

Λύνοντας το σύστημα των  (1),(2) βρίσκουμε ότι οι συντεταγμένες του Ε είναι

$ \left(\frac {3 \minus{} a^2}{9 \plus{} a^2},\frac {4a}{9 \plus{} a^2}\right)$

Στη συνέχεια βρίσκουμε ότι ο συντ/στης διεύθυνσης του διανύσματος $ \vec{E\Gamma}$ είναι $ \frac { \minus{} 2a}{3 \plus{} a^2}$

Εφαρμόζοντας τώρα τον τύπο εφφ=$ \frac {\lambda_2 \minus{} \lambda_1}{1 \plus{} \lambda_2\lambda_1}$,...όπου $ \lambda_2,\lambda_1$ είναι οι συντελεστές διεύθυνσης των ευθειών ΕΜ, ΕΓ, βρίσκουμε ότι
εφφ = α. Αλλά η ευθεία ΒΑ έχει συντελεστή διεύθυνσης ίσο με α.
Άρα $ \widehat{AB\Gamma} \equal{} \widehat{ME\Gamma}$

Με αναλυτική ( ανάμικτος τρόπος) μπορούμε να απλοποιήσουμε πολύ τη λύση ως εξής :

Έστω Ζ το σσημείο που η κάθετη στη ΒΓ στο Γ τέμνει την ευθεία ΒΜ. Το ΖΕΔΓ είναι εγγράψιμο , οπότε αρκεί ΖΔ // ΑΒ. Θέτουμε ( λίγο περίεργα) Β(0,0), Γ(4α,0) , Α(2α,2β) κλπ.Η εξίσωση της ΒΜ είναι πολύ απλή και το Ζ βρίσκεται εύκολα.

[quote="vangelismak"]Πολύ ωραίος τρόπος κύριε Μπάμπη :)
Την ιδέα του εγγράψιμου σκέφτηκα αρχικά αλλά με διαφορετική κατασκευή.
Προεκτείνοντας τη ΒΜ κατα ίσο ευθύγραμμο τμήμα ΜΗ=ΒΜ προκύπτει ότι το
τετράπλευρο ΑΒΓΗ είναι παραλληλόγραμμο. Αρκεί λοιπόν να δείξουμε ότι
το ΑΗΓΕ είναι εγγράψιμο. Δυστυχώς όμως με συνθετική γεωμετρία δεν μπόρεσα να το συνεχίσω :(  Δεν ξέρω αν κάποιο από τα παιδιά μπορεί να τελειώσει αυτή την ιδέα :)   καλό θα ήταν να έχουμε και μια λύση με καθαρά συνθετικό τρόπο.
Να πώ ότι και ο τρόπος του Δημήτρη είναι πολύ όμορφος και δύσκολος...
Μπράβο Δημήτρη που τα κατάφερες :)[/quote]

 [i]Η άσκηση είναι ήδη όμορφα λυμένη και αυτό είναι το ζητούμενο.Με βασάνισε και μένα συνθετικά. Η αρχή της αναλυτικής λύσης που έγραψα πριν , με κάποιες ακόμα κινησούλες δίνει λύση καθαρά γεωμετρική( ομοιότητες κλπ).Θα σας στείλλω τη λύση μου αργότερα , για την συνεχίσετε όσο θέλετε.[/i]Μπάμπης

Θα παραθέσω μια εντελώς συνθετική λύση (βασικά έχω δύο τετοιες) .Η πιο γρήγορη είναι του Μιχάλη οπότε παρουσιάζω αυτή και θα αναφέρω απλά το σκεπτικό για τη δική μου λύση .

Έστω Κ το μέσον του BC και έστω G το κέντρο βάρους .
Τα σημεία Β, D είναι τα συζηγή αρμονικά των Μ,C (απλά εξετάστε τους λόγους) και η γωνία DEB=90 άρα τα D,B είναι τα ίχνη της εσωτερικής και εξωτερικής διχοτόμου της $ \angle{KEC}$ άρα $ \angle{MEC} \equal{} \angle{BEK}$ .
Όμως το ΕΚDG είναι εγγράψιμο και επιπλέον GD//MC .
Οπότε $ \angle{BEK} \equal{} \angle{GDK} \equal{} \angle{ACB}$

***
H δική μου προσέγγιση .
Θεωρώ τον περιγεγραμμένο κύκλο του ΜΚC και έστω ότι αυτός τέμνει την MB στο Ε' .
Τότε αν F το μέσον του τόξου KC , $ FE'\bot BM$ και έστω D' το σημείο τομής της FE me tin BC .
Τότε Ε'D' διχοτόμος της KE'C και D'E' κάθετη στην ΒΜ άρα τα σημεία D' , B είναι τα ίχνη της σωτερική και εξωτερικής διχοτόμου άρα B,K,D',C συζυγή αρμονικά άρα $ BD' \equal{} 2D'C$ άρα $ D'\equiv D$ και $ E'\equiv E$ και μετά είναι εύκολο

[quote="Yperbatikos_e"][quote="stergiu"]Σε ισοσκελές τρίγωνο [b]ΑΒΓ(ΑΒ=ΑΓ)[/b] θεωρούμε το μέσο [b]Μ [/b]της [b]ΑΓ[/b] και σημείο [b]Δ[/b] στη βάση [b]ΒΓ[/b] ώστε [b]ΒΔ=2ΔΓ[/b]. Αν [b]Ε[/b] είναι η προβολή του [b]Δ[/b] στην [b]ΒΜ[/b], να αποδείξετε ότι οι γωνίες [b]ΑΒΓ[/b] και [b]ΜΕΓ[/b] είναι ίσες.[/quote]

MR Babis. ώς μαθητης τρίτης λυκείου
φαινομαι πολύ μικρός μπροστά σε αυτά
πάλι καλά που έχω τον Δημήτρη τον Αϊβάτογλου
μαθηματικό στο φροντηστήριο των ξαδέρφων του
Σαββάλα και μου τις λύνει.....[/quote]

Λοιπόν , αφού είσαι Γ΄Λυκείου , σου ευχόμαστε όλοι μας [b]Καλή Επιτυχία [/b]!
 Αφού αγαπάς τα μαθηματικά , θα έχεις ως φοιτητής πολλές ευκαιρίες να τα γνωρίσεις βαθύτερα και να ασχολείσαι με τις ασκήσεις αυτού του φόρουμ , πότε ως αναγνώστης και πότε ως λύτης .
 Να ξέρεις ότι οι μαθητές που λύνουν σήμερα εδώ αυτές τις ασκήσεις , ξεκίνησαν νωρίς , αφιέρωσαν χιλιάδες ώρες  σκυμένοι πάνω στο χαρτί , άλλοτε με τη χαρά της επιτυχίας  και άλλοτε με τη μοναξιά της αποτυχίας - πάντα όμως με την ευχαρίστηση της δημιουργίας!

 [color=red][b]Ευφυϊα είναι η εργαστικότητα ,[/b][/color]

 έλεγε ο D. Hilbert , ένας από τιους πιο ευφιείς μαθηματικούς.Αύριο λοιπόν που θα είσαι φοιτητής να συνεχίζεις να μας επισκέπτεσαι και με το ζήλο σου να μας βοηθάς , άλλοτε λύνοντας και άλλοτε προτείνοντας ασκήσεις που σου άρεσαν!
  Μπάμπης


I want a different solution of this because my solution is ungly.

Find all the positive integers such that

$a!=b!+c!+d!$

I let you judge if this one is less ugly  ;)

wlog, $b \leq c \leq d$ then $a! \leq 3d!$

But $a > d$ or $a \geq d+1$ and $a! \geq (d+1)d!$ which is $> 3d!$ as soon as $d > 2$.

Only solution I find is $b=c=d=2$


[quote="Math^_^"]Is there an algorithm to convert the Rubix Cube back into its original position?[/quote]
By "original position," do you mean solved or something else?

And no matter what your response is, the answer is yes! Because the standard "solve" algorithm can be applied to any form you want given that you adjust values (i.e. say this red is blue to get a red on the blue side).

yeah there is. That's how people solve cube sin like 10 seconds. They learn a lot of algorithms (like 100+) and practice them

i know about twenty
im using a beginner method and part of fridrich.

I have a website with an online [url=http://www.cubetimer.com/]rubik's cube timer[/url] for timing yourself as you solve. I have a links page on there with some good resources for rubik's cube solutions.

[url]http://www.cubetimer.com/[/url]

Personally, I recommend [url]http://lar5.com/cube/[/url]. i can get 1:40 if I'm at least half lucky. It needs Java though.


calculus
integration
LaTeX
derivative
function
real analysis
real analysis unsolved

This one is prety easy but anyways
Integrate[e^(-x^2),x]
that is to say the integral of e=2.71828... to the -x^2 with respect to x

Oh and P.S how do you all type in math equations in here[/u]

The pretty math is $\LaTeX$. There are plenty of sources around here to help you learn it.

You do realize the nature of the question you ask? The derivative is not an elementary function, and there are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63767]many[/url] ways to find the definite integral over the whole real line. You could have scanned for recent threads on the subject- that one hasn't even dropped off the first page yet.


Let $ I_n$ some compact intervals of the real line having the sum of their lengths a finite number. Is it true that we can always find a positive number $x$ such that for all natural number $n$,  $nx$ does not belong to any of these intervals?

Take $k$ large enough so that the sum of the lengths of $I_{k+1}$, $I_{k+2}$, ... is less than 1/2. Let $M$ be the supremum of $I_1\cup\ldots\cup I_k$. Then in every interval $[A,A+1]$, with $A\geq M$, there is a set of measure at least 1/2 whose points have the property you are looking for.

Xamog, really, I don't get your arguments. Could you please be more explicit?

Ok, let me try to give a clearer explaination of my argument.

Assume $A$ is $\geq 100$ and $\geq M$. Take any $x\in [A,A+1]$. Assume that $nx\in I_m$ for some $n$ and $m$. This means that $m\geq k$ and $x\in (1/n)I_m$, which is an interval whose length is $\leq$ than the length of $I_m$. Now remark that, for a given $m$, there is at most one $n$ such that $(1/n)I_m$ intersects $[A,A+1]$ (this is because $A$ is large and the length of $I_m$ is small, being $m\geq k$).

So if you assume by contradicion that for any $x\in [A,A+1]$ there exist $n$ and $m$ such that $nx\in I_m$, then you can cover $[A,A+1]$ with intervals whose total length is $\leq$ than  the total length of $I_k$, $I_{k+1}$, ..., which is <1/2.

[quote="Xamog"]there is at most one $n$ such that $(1/n)I_m$ intersects $[A,A+1]$ (this is because $A$ is large and the length of $I_m$ is small, being $m\geq k$).
[/quote]

I tried to do the exact same thing, but I don't think this part is correct :?. You can easily find points $y$ as far away from $[A,A+1]$ as you want s.t. both $\frac 1n y$ and $\frac 1{n+1}y$ lie in $[A,A+1]$ for some $n$, and then you can take $I_m$ to be any interval with left endpoint $y$.


What is the sum of all the real solutions of x if $ \mid x\plus{}\sqrt{4\minus{}x}\mid\minus{}4x$=0?

I'm not sure with this

First case, $ x\plus{}\sqrt{4\minus{}x}>0$
We get $ 3x\equal{}\sqrt{4\minus{}x}$
$ 9x^2\plus{}x\minus{}4\equal{}0$
$ x\equal{}\frac{\minus{}1\pm\sqrt{145}}{18}$
The sum of the real solutions from this case is $ \frac{\minus{}1}{9}$

Second case, $ x\plus{}\sqrt{4\minus{}x}<0$
We get $ 5x\equal{}\minus{}\sqrt{4\minus{}x}$
$ 25x^2\plus{}x\minus{}4\equal{}0$
$ x\equal{}\frac{\minus{}1\pm\sqrt{401}}{50}$
The sum of the real solutions from this case is $ \frac{\minus{}1}{25}$

So the sum of all the real solutions of $ x$ is $ \frac{\minus{}34}{225}$

$ 4x \equal{} |\cdots| \Longrightarrow x \ge 0$; then $ x \plus{} \sqrt{4\minus{}x} > 0$, so there is just one solution $ x \equal{} \frac{\minus{}1 \plus{} \sqrt{145}}{18}$.


geometry
geometric transformation
set theory
number theory
prime numbers

I've been reading recently about the history of mathematics, about the axiomization of certain geometries (such as the importance of the Parallel Postulate and how changing it around can create whole new areas of work) and in doing so I have read about such people that were able to prove that certain things cannot be proven, or that all axiomization is never complete, for example.  I have been wondering, how do you go about such "impossible" proofs?

Well, as for now, you don't.  :)

Human intelligence only goes so far; surely we have to assume some things.

That is certainly true, but its the nature of what we can do with these assumptions is what I'm asking.  Being able to say that something cannot be proven is definitely one of the greatest reaches that these assumptions can make.  My question: How is it done?

[quote="math154"]Well, as for now, you don't.  :) [/quote]
I'm not sure what you mean.  There are legitimate proofs of these kinds of independence statements (for example, that the axiom of choice is independent of ZF), although I can't say I'm familiar with them.

(somehow I managed to edit this post instead of reply, so I just destroyed everything that was written here; if there's any way for a mod/admin to recover edit histories, could they do that?)

[quote="Z-coli"]Godel's incompleteness theorem..what's that?[/quote]
[quote="MellowMelon"]Proving that there is stuff that isn't provable is Godel's incompleteness theorem...[/quote]
Also see [url=http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems]here[/url].

[quote="AIME15"]	You need at least 2 givens to make a proof...[/quote]
What are you referring to?

I mean you need at least two postulates, theorems, etc.  Because if you only have one thing, you can't do anything with it.

That depends on how many things you're willing to take as definitions rather than postulates.  In any case, it's not strictly relevant to this discussion; proving that a given axiom is independent of a given system is not the same thing as proving the incompleteness theorem, which only tells you that there exists a true statement that can't be proven in a given system.

This is [b]not equivalent[/b] to independence.  Independence of a proposition $ P$ from a system $ A$ means that $ A \cup P$ and $ A \cup \neg P$ are both consistent, whereas incompleteness says that a sufficiently powerful system can't prove its own consistency.  These two statements are not contradictory because we prove the consistency of systems within [b]larger[/b] systems.

Z-coli,

First, here's some terminology for you.

A set of axioms are [b]consistent[/b] if they do not contradict each other. For example, it was shown that given certain assumptions, the axioms of arithmetic are consistent. However, if I have three axioms:

- All rich people are happy.
- All happy people are hippies.
- No hippies are rich. (No offense)

the axioms are contradictory; they are thus called inconsistent.

Given a set of axioms, we call a statement [b]independent[/b] of the axioms if it does not follow from the axioms.

For example, say I am given the following axioms:
    - All red hats are round (call this R for short.)
    - All blue hats have right angles (call this B for short.)

Now, let's say I have the statement "All green hats have circles on them" (call that statement G for shirt.) Does G follow from R or B? The answer is clearly no.

So how could we rigorously prove that a statement is independent from a set of axioms? First, consider this: if it's logically possible that something did happen, and it is also logically possible that it did not happen, we can't say anything about its truth value.

Consider this example. Say Alex and James kick two balls across a fence simultaneously. The balls are identical. One balls lands on the lawn, and another shatters a window. Is it possible that Alex kicked the ball that shattered the fence? Yes. Is it also possible that James kicked the ball that shattered the fence? Yes. So with the information we're given, we can't deduce anything about who kicked the ball that broke the fence (as it is possible that any one of us did.) Thus, the statement "Alex kicked the ball that broke the fence" is unprovable.

We apply a similar technique in mathematics. Suppose we have a statement P, and a consistent set of axioms. Call the set of axioms together A. If it is possible that the set of axioms A and P is consistent and A and ~P is consistent, P is independent from A.

As an example, consider the parallel postulate. Suppose we are given that
1. Any two points can be joined by a line.
2. Any line segment can be extended into a line.
3. Given any line segment, a circle can be drawn having the segment as radius and one endpoint as center.
4. All right angles are congruent.
(Taken from Wikipedia.)

Call these set of axioms A. A statement equivalent to the parallel postulate is that the sum of the angles in a triangle is 180 degrees. Call this statement P.

Given that A is consistent (I think it is), is it possible to deduce P? Consider Euclidean geometry. Euclidean geometry is pretty much A + P (the four axioms and the parallel postulate); Euclidean geometry is consistent (as far as I know.)

Now, consider A + ~P, that is, A along with the condition that there exist triangles whose angles don't add up to 180. Elliptic geometry satisfies A + ~P, and since it is consistent (as far as I know), we see that it is possible to have A and ~P in a set of axioms that is still consistent.

Since it is possible for a set of axioms containing A + P and A + ~P to be consistent, we cannot possibly show that P is provable/disprovable in A; if it's provable in A, then A + ~P is inconsistent (but since it's consistent, we have a contradiction.) Similarly, we cannot show that it is disprovable. Thus, P is independent.

Hope that helps.

Ohhh
thanks Zhero
By the way...Alex definitely was the one that shattered the glass
James kicked it perfectly into the yard.

Building on Zheros post, note that things get more complicated in the case of set theory:
Lets use ZF (Zermelo-Frenkel) set theory. Then it can be shown (Gödel) that there is no proof in ZF proving that ZF is consistent; so it's impossible to show that there are no contradictions (I will come back on this later). So now there axioms like the AC (Axiom of Choice) or CH (Continuum Hypothesis), that are still indepedent from ZF in an abuse of language: If we assume that ZF is consistent (so has no contradictions in itself), then AC and CH are independent, so can neither be proven nor disproven. But in case ZF is inconsistent, by the logical axiom of EFQ (Ex Falso Quodlibet, translation: from false follows everything) we get that AC, CH and even it's negations are true, as everything is true then.
Note that for all this, one still uses some axioms of logic, too. You need some information on what "implies", "and" or "false" mean. And then you also need some way of proving, thus axioms like EFQ or the classical DNA (Duplex Negatio Affirmat, translation: negation of negation of A implies A).

Now one attempt of solving the problem of ZF not being able to prove it's consistency is simple: consider ZF_Con, which we defined by the axioms of ZF and the additional axiom Con that says "ZF is consistent". But this cannot work for two reasons:
a) If ZF is inconsistent, then so is ZF_con (what can be proven in ZF can surely be proven in ZF_Con); conversely, if ZF_con is consistent, then so is ZF_Con (as the new axiom can cause no contradiction). So we get that ZF is consistent if and only if ZF_Con is. But that's just completely useless, we have shifted our problem just to another, even larger and thus more complicated, set of axioms.
b) Gödel also proved that no "reasonable" set of axioms can prove it's consistency if it is not inconsistent (and ZF is reasonable enough); this is called "Gödels second incompleteness theorem". Thus already the attempt of finding a bigger set of axioms must fail.

So we live in a world where we cannot even show that not false is different from true, and even if ist is, there are a lot of theorems that cannot be proven nor proven.

[quote="Zhero"]Given a set of axioms, we call a statement [b]independent[/b] of the axioms if it does not follow from the axioms[b] and neither does its negation.[/b][/quote]
Just a small correction.

As a note, for logicians the Euclidean axioms of geometry aren't rigorous enough. The axioms they came up with (that are basically equivalent to what Euclid had) were actually proven to form a consistent and decidable model of Euclidean geometry. Because plane geometry is not sufficiently powerful to express things like the prime numbers, Godel's theorem doesn't apply.

Of course, though, logic is certainly not rigorous either so it can't be really used for determining the validity of another system; but I get what you're saying Mellow.

[quote="Z-coli"]logic is certainly not rigorous either[/quote]
I'm not sure what you mean.


limit
geometry
geometric transformation
logarithms
calculus
calculus computations

Consider the sequence $ a_n\equal{}\sqrt[3]{n^3\plus{}3n^2\plus{}2n\plus{}1}\plus{}a\sqrt[5]{n^5\plus{}5n^4\plus{}1}\plus{}\frac{ln(e^{n^2}\plus{}n\plus{}2)}{n\plus{}2}\plus{}b$. Find $ a,b \in \mathbb{R}$ such that $ \displaystyle\lim_{n\to\infty}a_n\equal{}5$.

[url=http://www.mateforum.ro/viewtopic.php?t=2807]Here[/url] you can find a solution. And there is also a nice comment (to remember) by Virgil Nicula.

the language doesnt really make much of a barrier for the idea of the proof. subtract multiples of n to make each term have an easy limit.
the third one only converges for certain a, which gives us b

the last guy in the post seems to have given a generalized form for solving the square root limits easily..

The other post makes the argument that if you have a limit such as
\[ \displaystyle\lim_{n\to\infty} \left(an^p \plus{} bn^{p\minus{}1} \plus{}... lower order terms\right)^{1/p} \] As n gets large, we need to only consider highest order terms. we can then say that our limit is essentially \[ \displaystyle\lim_{n\to\infty} \left(an^p \plus{} bn^{p\minus{}1}\right)^{1/p} \equal{}\displaystyle\lim_{n\to\infty} n\left(a\plus{}\frac{b}{n} \right)^{1/p} \] Recall, \[ \left(a\plus{}\frac{b}{n}\right)^{1/p} \equal{} a \plus{} \frac{1}{p}\frac{b}{n} \plus{} O \left(\frac{b^2}{n^2}\right) \] So as n gets large we see that \[ \displaystyle\lim_{n\to\infty} \left(an^p \plus{} bn^{p\minus{}1} \plus{}... lower order terms\right)^{1/p} \equal{} an \plus{} \frac{b}{p}\]

[quote="moldovan"]Consider the sequence $ a_n \equal{} \sqrt [3]{n^3 \plus{} 3n^2 \plus{} 2n \plus{} 1} \plus{} a\sqrt [5]{n^5 \plus{} 5n^4 \plus{} 1} \plus{} \frac {ln(e^{n^2} \plus{} n \plus{} 2)}{n \plus{} 2} \plus{} b$. Find $ a,b \in \mathbb{R}$ such that $ \displaystyle\lim_{n\to\infty}a_n \equal{} 5$.[/quote]

Since $ \lim_{n\to\infty} a_n \equal{} 5$, we have $ \lim_{n\to\infty} a_n\cdot \frac {1}{n} \equal{} 5\cdot 0 \equal{} 0$.

$ \therefore \lim_{n\to\infty} \frac {a_n}{n} \equal{} \lim_{n\to\infty} \left( \sqrt [3]{1 \plus{} \frac {3}{n} \plus{} \frac {2}{n^2} \plus{} \frac {1}{n^3}} \plus{} a\sqrt [5]{1 \plus{} \frac {5}{n} \plus{} \frac {1}{n^5}} \plus{} \frac {\ln (e^{n^2} \plus{} n \plus{} 2)}{n(n \plus{} 2)} \plus{} \frac {b}{n}\right)$

$ \equal{} \lim_{n\to\infty} \left(1 \plus{} a \plus{} \frac {1}{n(n \plus{} 2)}\ln e^{n^2}\left(1 \plus{} \frac {n \plus{} 2}{e^{n^2}}\right)\right)$

$ \equal{} \lim_{n\to\infty} \left(1 \plus{} a \plus{} \frac {n}{n \plus{} 2} \plus{} \ln \left(1 \plus{} \frac {n \plus{} 2}{e^{n^2}}\right)\right) \equal{} 0$, yielding $ a \equal{} \minus{} 2$. Thus we have

$ a_n \equal{} \sqrt [3]{n^3 \plus{} 3n^2 \plus{} 2n \plus{} 1} \minus{} 2\sqrt [5]{n^5 \plus{} 5n^4 \plus{} 1} \plus{} \frac {\ln (e^{n^2} \plus{} n \plus{} 2)}{n \plus{} 2} \plus{} b$

$ \equal{} \sqrt [3]{n^3 \plus{} 3n^2 \plus{} 2n \plus{} 1} \minus{} (n \plus{} 1) \minus{} 2\{\sqrt [5]{n^5 \plus{} 5n^4 \plus{} 1} \minus{} (n \plus{} 1)\}$

$ \plus{} \frac {\ln (e^{n^2 \plus{} n \plus{} 2})}{n \plus{} 2} \minus{} (n \plus{} 1) \plus{} b$

$ \equiv A_n \minus{} 2B_n \plus{} C_n \plus{} b$,

$ \lim_{n\to\infty} A_n \equal{} 0,\ \lim_{n\to\infty} B_n \equal{} 0$ and $ \lim_{n\to\infty} C_n \equal{} \lim_{n\to\infty} \left( \minus{} 3 \plus{} \frac {4}{n \plus{} 2} \plus{} \frac {1}{n \plus{} 2}\ln \left(1 \plus{} \frac {n \plus{} 2}{e^{n^2}}\right)\right) \equal{} \minus{} 3$, yielding $ \lim_{n\to\infty} a_n \equal{} 5\Longleftrightarrow b \minus{} 3 \equal{} 5$.

Consequently the desired values of $ a,\ b$ are $ a \equal{} \minus{} 2,\ b \equal{} 8$.


function
integration
email
articles
LaTeX
complex numbers

What does it mean in mathematics when something is trivial, such as a trivial solution or trivial step, and what would a non-trivial solution be?

Thanks.

Usually, I take  trivial to mean that something is well-known, obvious, or easy to establish.  Check out the definition [url=http://mathworld.wolfram.com/Trivial.html]here[/url].  Pay special note to this part:

[quote]
According to the Nobel Prize-winning physicist Richard Feynman Eric Weisstein's World of Biography (Feynman 1997), mathematicians designate any theorem as "trivial" once a proof has been obtained--no matter how difficult the theorem was to prove in the first place.[/quote]

A lot of people will use this definition which is very different from the one I wrote above.  So sometimes things that are 'trivial' to one person aren't really 'trivial' (depending on the perspective).

Two trivial solutions in the question "Find and plot the solutions to the equation x :^3: + x = 0" would be +/- i.

Trivial solutions are generally considered solutions such as $x=0$.

[quote="Jimmy"]Thanks, but what does someone mean when he says a trivial solution, such as the Riemann Hypothesis "All non-trivial zeros of the zeta function have a real part of one half."[/quote]

Jimmy,

When we establish the analytic continuation for the zeta function from $\Re(s)>1$ to $\mathbb{C}\setminus\{1\}$, we use the gamma function.  Properties of the gamma function make it easy to establish that $\zeta{n}=0$ for $n=-2,-4,-6 \ldots$.  These are the so-called trivial zeros.  The nontrivial zeros are the zeros that lie in the critical strip $0<\Re(s)<1$, and the Riemann Hypothesis is that all of these zeros have a real part of $1/2$.

Here's an exercise that you might be able to appreciate:

Establish the analytic continuation (looking at the real line will suffice) of the zeta function from $\Re(s)>1$ to $\Re(s)>0$ given by

$\zeta(s)=s\int^{\infty}_1 \frac {[x]dx}{x^{s+1}}$

where $[x]$ is the greatest integer function.  Basically, you need to show that this agrees with the classical definition of the zeta function for $s>1$.  See if you can also show that the zeta function has a simple pole at $s=1$.

By the way, did you get my earlier email?

[quote="blahblahblah"]
By the way, did you get my earlier email?[/quote]

Yes I got it, thanks (sorry for not responding earlier I went out for a while), I'm going to have to re-read it a few times before I can fully understand it though :lol:

I'll give that exercise you mentiod a shot, it doesn't look too bad...

To quote two articles from dictionary.com:
[quote]
# Mathematics.
   1. Of, relating to, or being the solution of an equation in which every variable is equal to zero.
   2. Of, relating to, or being the simplest possible case; self-evident.[/quote]
and
[quote]...4. Any problem one has already solved (some claim that
hackish `trivial' usually evaluates to `I've seen it before').
Hackers' notions of triviality may be quite at variance with those
of non-hackers. See nontrivial, uninteresting.

The physicist Richard Feynman, who had the hacker nature to an
amazing degree (see his essay "Los Alamos From Below" in "Surely
You're Joking, Mr. Feynman!"), defined `trivial theorem' as "one
that has already been proved".[/quote]

in a problem, if you can see an answer in less than 15 seconds (the answer can't be some random number, something simple, like 0 or 1) the answer is probably trivial.

for fermat's last theorem: No solution exist for $x,y,z \in \Re$ such that $x^n + y^n = z^n$

But wait, (1,0,0) works! So we say  no nontrivial solutions exist

[quote="buzzer11"]trivial is connected to the word trivia, which is an easy quiz. the word trivial means easy or well known.[/quote]Let me work my latin magic here.. by which I mean look it up in my latin dictionary. The latin trivium means a crossroads or public place (literally "three roads"). Since people hang out at the crossroads, trivialis means commonplace, which is like "common place" but also means well-known or in this case trivial. Interesting.

[quote="blahblahblah"]$\mathbb{R}$ not $\Re$[/quote]I would say $\mathbf{Z}$ here..

$\mathbb{Z}$?

Anyways, the german-ish R is pretty much only used to refer to real parts of complex numbers (functions, whatever).

He was giving the example of Fermat's Last Theorem which of course involves the set of integers $\mathbf{Z}$. The font you're using ($\mathbb{Z}$) is what people write on blackboards to represent boldface since that's hard with chalk. But I guess people thought that looked nicer so they adapted it to computer font. I prefer the traditional boldface though. Let me know if I don't have the story straight.

[quote="blahblahblah"]Trust me, I know what fermat's last theorem states, I know that what he really meant was $\mathbb{N}$, and I know that $(1,0,0)$ is not actually a counterexample.  I was just stating that, in any event, $\Re$ does not refer to a set of numbers, particularly not the set of real numbers.  Some people new to latex do not know the commands - I remember a night where I couldn't find this symbol when I needed to hand in an assignment ($\mathcal{R}$) so I was just trying to help.[/quote]

my bad....I am not great at $\LaTeX$

No need to apologize - a few weeks ago I realized that every tex'd document I handed in this year had a rather ugly and easily fixed typesetting flaw.

To prevent any further discussions on higher things like $\Re$ and $\mathbb{R}$, I'll lock the topic since Jimmy seems to get the question.

Silverfalcon


How big can the ratio of the momento of the photoelectrons coming from aluminium and the momento of the incoming photon be?

Let $p_{e}$ be the momentum of the photoelectrons and $p_{f}$ the momentum of the photons. They are related as
\[p_{f}c = A+\frac{p_{e}^{2}}{2m},\]
and we need to minimize the function $f(p_{f}, p_{e}) = p_{f}/p_{e}$. We'll use Lagrange multipliers; let
\[F(p_{f}, p_{e}) \equiv f(p_{f}, p_{e})-\frac{1}{\alpha}\left(p_{f}c-A-\frac{p_{e}^{2}}{2m}\right),\]
where $\alpha$ is a constant multiplier. The condition for the extremum is \[\frac{\partial F}{\partial p_{f}}= 0 \Rightarrow p_{f}m \alpha = p_{e}^{3},\] \[\frac{\partial F}{\partial p_{e}}= 0 \Rightarrow \alpha = cp_{e}.\]

Now it is trivial to find
\[\frac{p_{e}}{p_{f}}= c\sqrt{\frac{m}{2A}}.\]


suppose I gave you one dollar and your friend gave you one dollar. Would you get richer or would we just get more poor?

because, the somatory of dollar before is equal to the somatory of dollars after the transaction. If you got richer at same time we got poor, it would have been 4 dollars, because $2+2 = 4$. So, how to explain that? did we get more poor or did you get richer? it can't be both...

How interesting problem!!!!

I remember that when I was Young I Saw the same problem in a Calculus book of one of my friend!!!

And the calculus book was given by, or to, dont know, etc...

I think You explode, because the paradox itself.

See Zenos paradox in wikipedia.
It has a good aprox of this

The two people giving dollars each get 50 cents poorer, while the person receiving gets 1 dollar richer.

but, I like pie, imagine that, instead of 1 dollar, it was 55 cents? how could one get 27,5 cents poorer?

and, treething, who is back?

[quote="i_like_pie"]In that case, my idea doesn't really make sense... At all...

I think Treething meant I was back.[/quote]
no, he's referring to fambiombr, who is a SPAMMMER

and no, we don't want to buy a calculus book

[quote="kohjhsd"]Um, why is this a paradox?

And I think Zeno's Paradox is different...[/quote]

Hey everybody.

This post is just a trick.

There is no paradox.
This post is in Games and Fun.

I sayed that it's like zeno's paradox, because zeno's paradox isnt a paradox!
It just a trick in the thougth.

I have a book that say something about paradoxes just like that, a pseudo-paradox
but those paradoxes are much more well write. hehe

what is spammer? SPA? "Systems and Process Assurance", "Sanitas Per Aqua", "Society of Performing Arts", "SOUTHWARK PASTORAL AUXILIARY", "Societa per Azioni", "Software Publishing Association" ?

I dont really know

maybe it is "Solving of Problems Art", an anagram of AoPS

and, even if you wanted, I wouldnt give you my book, I love it

hahaaha

I can't get the paradox :(
If all that in your book are these "paradox", i'm definitely not interested in it

I don't understand how this is a paradox.
Don't you just each get a dollar poorer and I two dollars richer?


ratio
geometric sequence
algebra unsolved
algebra

1.Let ${a_n}$ be the sequence such that $a_{n+1}=2^n-3a_n$
Find all $a_o$ such that $a_0<a_1<a_2<...$
2.$a\in Z+$ and $a_0=0$, $a_{n+1}=(a_n+1)a+(a+1)a_n+2\sqrt{a(a+1)a_n(a_n+1)}$
Prove that $a_n\in N \forall n\ge 0$

1. It is obviosly
$a_{n+1}=2^n-3a_n=2^n-3*2^{n-1}+3^2a_{n-1}=\dots=2^n-\dots+(-3)^n+(-3)^{n+1}a_0$
It give $a_n=\frac{2^n+(5a_0-1)3^n}{5}$ Sequence increase only if $a_0=\frac 15$.
2. This problem about solutions Pelle equations. In this case we have recurent equation for $a_n$:
$a_{n+2}=(4a+2)a_{n+1}-a_n+2a$. It give
$a_n=(q^n+q^{-n}-2)/4=[-1+\sum_{k=0}^{[n/2]} C_n^{2k}(2a+1)^{n-2k}2^ka^k(a+1)^k ]/2$. $q^{\pm1}=2a+1\pm2\sqrt{a(a+1)}.$
Polinom $a_n(a)\in Z[a]$.


$A, B, T \in M_n(C), AB - BA = T, AT = TA. \Rightarrow  det(T) = 0$

It follows from this [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=84219[/url]

 :)


Let $ H$ and $ O$ are othocenter and circumcenter of $ \triangle{ABC}$ respectively
,$ O_A,O_B$ and $ O_C$ be the mirror image of $ O$ across $ BC,AC$ and $ AB$ respectively.
,$ HO_A \cap BC \equal{} D,HO_B \cap AC \equal{} E$ and $ HO_C \cap AB \equal{} F$
Prove that $ AD,BE,CF$ concurrent.

Let Ha be the mirror image of H across BC, then Ha D O on a line
Easy to gain that BD/DC=S△HaBG/ S△HaCG=BG*BHa/CG*Cha=CA*BH/BA*CH
The same way we can calculate  CE/EA and AF/BF
BD/DC* CE/EA *AF/BF=1
Thus we are done

[quote="mr.danh"]A consequence of the problem: AD,BE,CF are symmetric to OA,OB,OC wrt the medians of triangle ABC respectively.[/quote]

Nope. You probably meant: The points $ D$, $ E$, $ F$ are symmetric to the points $ OA\cap BC$, $ OB\cap CA$, $ OC\cap AB$ with respect to the midpoints of $ BC$, $ CA$, $ AB$, respectively.

  darij


function
calculus
quadratics
geometry
geometric transformation
reflection
algebra

[quote="mattzeiler"]... but I just can't find the inverse. I tried so many things... [/quote]
mattzeiler, can you show us what you did try? If we see that, perhaps we can see where you went wrong.

I'm also moving this problem to the "Calculus calculations and tutorials" forum, even though it's not really calculus at all.

I tried multplying the x by x to get y=(x^2+1)/x and then I moved the x to the other side and set the whole thing equal to zero as:   0=x^2-xy+1  The I tried to do the quadratic and got lost from their.

Just write $x+\frac{1}{x}=y$ $x\geq 1$ and $y\geq 2$.Then solve a quadratic equation respected to $x$.That would be your inverse function.

$x^2-xy+1=0$

$x=\frac{y\pm\sqrt{y^2-4}}2.$

If $y\ge2$ this quadratic has two roots. After all, the original unrestricted function wasn't one-to-one. But if we're dealing with the restricted version, then we need $x\ge 1$ and that mean we take the plus sign out of the two possible signs.

So $f^{-1}(x)=\frac{x+\sqrt{x^2-4}}2$ for $x\ge2.$

[quote="mattzeiler"]Hi, I had a question on an assignment that I couldn't get, can you help me out.  The ask you to plot f(x)=x+(1/x) on a graph.  That was fine.  Then they said is this an inverse function.  It isn't on the whole interval because it isn't one to one.  The next step said, what is the inverse of the function on the interval from [1,infinity).  This would make it one to one but I just can't find the inverse.  I tried so many things and the professor says it works.  Can someone help me?[/quote]
Doesn't inverse functions symmetrically reflect about in the y=x..?


It's a long story but...
simply put, I thought it was due on tuesday like last time, until I realized today was the 23rd. So I went to the post office - but basically no one was there. I'm not sure what it will be postmarked as, but if it will be postmarked as the 24th, is there anything I can do so that it would be "received" on the 23rd? Or do I get a 0/25 for this contest - despite all my work and effort?

This makes me really annoyed for assuming it would be like last month.

 :(

Should I e-mail them? I'm seriously panicking and I can't seem to be able to do anything else. Does anyone have any advice? I am really beginning to think that it'll be postmarked tomorrow.
I'm going to have a rough time until I get some advice. So much stress:(

So you gave it to the post office, but they haven't taken it out of the thing yet?

If you email them, be sure to include your ID number and name.

So would the email be usamts@usamts.org?

Thanks. I'll include my id number, name, and an explanation.

The same thing happened to me. I decided just not to mail it because my mom was sick and I was unsure about my solutions. Good thing too, because I would have scored like a 10, judging by comparing my solutions to the official ones.  :(

So how did your situation turn out?


trigonometry
function
limit
calculus
calculus computations

Hello everyone,

For the following problem, I am attempting to complete it using a different method which seems reasonable to me. However, I do not get the same answer as if I had used my workbook's solution.

Could anyone please justify my method?

Thank you.

---

1. Calculate the value of $ k$ that makes $ h(x)$ continuous over the interval $ ( \minus{} \frac {\pi}{2}, \frac {pi}{2})$.

$ f(x) \equal{} \left\{\begin{array}{cc}2(\tan x \plus{} 1), & \mbox x\leq \minus{} \frac {\pi}{4} \\
\tan(kx \minus{} 1), & \mbox x > \minus{} \frac {\pi}{4}\end{array}\right.$

---

[b]My Method :[/b]

I found the limit of the first piecewise-defined function and then set the second to the limit.

$ \lim_{x \to \frac {pi}{4}^ \minus{} } 2(\tan x \plus{} 1) \equal{} 2(tan(\frac {\pi}{4} \plus{} 1)) \equal{} 4$

Therefore:

$ \tan(kx \minus{} 1) \equal{} 4$

$ kx \minus{} 1 \equal{} \arctan(4)$

$ k\frac {\pi}{4} \equal{} 2.326$

$ k \equal{} 2.961$

[b]Workbook's Answer:[/b]

[hide]Basically, the workbook sets both piecewise-defined function equal and solves for $ k$ from there. Their answer is $ k \equal{} \minus{} \frac {4}{\pi}$.[/hide]

[quote="catalan"]

$ \lim_{x \to \frac {pi}{4}^ \minus{} } 2(\tan x \plus{} 1) \equal{} 2(tan(\frac {\pi}{4} \plus{} 1)) \equal{} 4$

[/quote]

I think you meant to set $ x\equal{}\minus{}\pi/4$ rather than $ x\equal{}\pi/4$. This changes the value to 0, instead of 4. You also have an error with the parentheses here, it's $ 2(\tan(\minus{}\pi/4)\plus{}1)$ rather than $ 2\tan(\pi/4\plus{}1)$.

Both you and the workbook are wrong.

Your ideas are fine- this is about the limits from the two sides agreeing. Both formulas are naturally continuous (except at the poles $ \frac{\pi}{2}\plus{}n\pi$ of $ \tan$), so these limits can be found by evaluating the formulas at that point.
Where did you go wrong? First, you used the wrong point. You found a limit at $ \frac{\pi}{4}$, not $ \minus{}\frac{\pi}{4}$. Second, you didn't check for continuity on the rest of the interval- your expression blows up at $ x\equal{}\frac1k\left(\frac{\pi}{2}\plus{}1\right)\approx .868$. There are other arctangents of $ 4$, and the first negative one would actually work here.

Now, what if we actually use the correct point $ \minus{}\frac{\pi}{4}$? The first formula gives $ 2(\minus{}1\plus{}1)\equal{}0$, so the second must be zero as well. That means that $ \minus{}\frac{\pi}{4}k\minus{}1$ is an arctangent of $ 0$, so it's a multiple of $ \pi$. If we use the standard arctangent, this is zero, and we get $ k\equal{}\minus{}\frac{4}{\pi}$. The other options are of the form $ \minus{}\frac{4}{\pi}\plus{}4n$.
There's still one more thing to check, and this is where the workbook fails. Is this formula actually continuous? $ \minus{}\frac{4}{\pi}x\minus{}1$ varies between $ 0$ and $ \minus{}3$ on the interval $ (\minus{}\frac{\pi}{4},\frac{\pi}{2})$, crossing one of the poles. It's not continuous. None of the other options are continuous either- they lead to much larger $ k$, and have even more points of discontinuity.

The correct answer: there is no such $ k$.

@ catalan:

Can I ask for the name of your workbook ?  It looks like an interesting one.
Thanks in advance.


calculus
integration
logarithms
calculus computations

For what value of $ a > 1$ is
\[ \int_a^{a^2} \frac{1}{x} \log \frac{x \minus{} 1}{32} \,dx
\] a minimum?

If your $ \log$ is natural logarithm, I get...

$ \int_a^{a^2} \frac {1}{x} \log \frac {x \minus{} 1}{32} \,dx \equal{} \frac {3}{2}\log^{2}a \minus{} 5\log 2\log a \minus{} S\left(\frac {1}{a^{2}}\right) \plus{} S\left(\frac {1}{a}\right)$ where $ S(\alpha) \equal{} \sum_{k \equal{} 1}^{\infty}\frac {\alpha^{k}}{k^{2}}$.

But the only thing I know about $ S$ is $ \forall\alpha\in (0,1)$ we have $ S(\alpha) \plus{} S(1 \minus{} \alpha) \equal{} \frac {\pi^{2}}{6} \minus{} \ln\alpha\ln(1 \minus{} \alpha)$.

And my solution is very long so I think either I went through the wrong way or I've made some mistake. :P

hello, Maple 12 gives
$ {\int_{a}{}^a{}^2\frac{1}{x}\log\left(\frac{x-1}{32}\right)\,dx=-{\it dilog} \left( a \right) -\ln  \left( 1/32\,a-1/32 \right) \ln \left( a \right) +{\it dilog} \left( {a}^{2} \right) +\ln  \left( 1/32\,{a}^{2}-1/32 \right) \ln  \left( {a}^{2} \right)}$
Let $ {f(a)=-{\it dilog} \left( a \right) -\ln  \left( 1/32\,a-1/32 \right) \ln
 \left( a \right) +{\it dilog} \left( {a}^{2} \right) +\ln  \left( 1/
32\,{a}^{2}-1/32 \right) \ln  \left( {a}^{2} \right)}$, then i have got
$ f'(a)=\frac{-5\ln(2)+\ln(a-1)-2\ln((a-1)(a+1))}{a}$ and $ f'(a)=0$ gives $ a=3$.
Sonnhard.

The log there is base 10.. how does that change the answer? This is a competition problem so we can't use Maple  :)

$ f(a) \equal{} \int_a^{a^2} \frac {1}{x} \log \frac {x \minus{} 1}{32}dx\ (a > 1)$,

$ f'(a) \equal{} \left(\frac{1}{a^2}\ln \frac {a^2 \minus{} 1}{32}\right)*2a \minus{} \frac {1}{a}\ln \frac {a \minus{} 1}{32}$

$ \equal{} \frac {1}{a}\ln \frac {(a \plus{} 1)^2(a \minus{} 1)}{32}$ yielding $ f'(a) \equal{} 0\ (a > 1)\Longleftrightarrow (a \plus{} 1)^2(a \minus{} 1) \equal{} 32$.

$ \therefore (a \minus{} 3)(a^2 \plus{} 4a \plus{} 11) \equal{} 0$, yielding $ a \equal{} \boxed{3}$.

Ohhh  :blush:  I got it. Thank you very much kunny. :lol:


function
real analysis
real analysis unsolved

Show that there does not exist a continuous function $ f$ such that $ f(\beta)$ is rational for all irrational $ \beta$ and $ f(\alpha)$ is irrational for all rational $ \alpha$.

[hide="Solution"]The preimage of any point under such $ f$ would have to be closed, with empty interior. Therefore, the set $ f^{-1}(\mathbb Q)$ is of the first category. [/hide]
How can one describe the sets $ A\subset \mathbb R$ for which there exists $ f\in C(\mathbb R)$ such that $ f(A)\subset \mathbb R\setminus A$ and $ f(\mathbb R\setminus A)\subset A$?

I'll try to think about your question although I don't see me making much progress :oops:. (I take it $ C(\mathbb R)$ denotes the space of continuous functions?)
All countable sets fail to have this property since $ f(\mathbb R\setminus A)\subset A$ and $ f(A)\subset \mathbb R\setminus A$ implies that $ f(\mathbb R)$ is countable, however a continuous (non constant) function has an uncountable range by the intermediate value theorem. It follows that all sets $ B$ with $ \mathbb R\setminus B$ countable also fail to have this property - however I highly doubt that all uncountable sets having uncountable complement relative to $ \mathbb R$ have this property - so I've lived up to my prediction and not made much progress.
P.S. If your solution was intended to point me in the way of the solution to the question beneath it then just ignore this post. I know what category is and I have a shaky understanding of Baire's theorem but I really amn't that comfortable with  ideas to do with category - sorry if this is the case.
[b]Edit[/b]. @JBL - I had a feeling I completely made that up. Thanks to your comment I now feel that I  have to make an effort to use impeccable English throughout the remainder of my posts! (I've spent the last year studying only maths and physics at school and can't remember anything from previous years English classes - English is my first language by the way :blush: ).

[quote="EBz89"]...I really amn't...[/quote]

This is entirely off-topic, but the contraction "am not" $ \mapsto$ "amn't" really caught my attention.  It is not standard English: if you wanted to contract "I am not," the only option is "I'm not," even though "it is not" can be contracted in two different ways.

Yes, by $ C(\mathbb R)$ I mean the set of all continuous functions from $ \mathbb R$ to $ \mathbb R$. And no, I don't have an answer myself. The property we are interested in can be restated in a better form: $ f^{-1}(A)=\mathbb R\setminus A$. Such a set $ A$ must be neither open nor closed ... so much for standard English.  :P

Also, taking preimages (with continuous $ f$) preserves the classes $ G_{\delta}$ and $ F_{\sigma}$. Which implies that $ A$ must be either
1) both $ G_{\delta}$ and $ F_{\sigma}$
or
2) neither $ G_{\delta}$ nor $ F_{\sigma}$.
Both cases can indeed occur. Choose a set $ B\subset [0,1)$ and let $ A=\bigcup_{n\in\mathbb Z}(B+2n)\cup \bigcup_{n\in\mathbb Z}([0,1)\setminus B+2n+1)$. This set satisfies $ f^{-1}(A)=\mathbb R\setminus A$ with $ f(x)=x+1$. Since the choice of $ B$ is completely arbitrary, the set $ A$ may be even nonmeasurable. So it looks like my question does not have any sensible answer. Oh well.


If AD,BQ,CP are cevians concurring at a point 'o' ,
then prove that
$ \frac {OD}{AD} + \frac {OQ}{BQ} + \frac {OP}{PC} = 1$
i think it is some known theorem plz provide the name if u kno
PLZ HIDE SOLUTIONS
also tell me if wrong [hide="this"]
Draw $ RS \parallel{} BC$ extend,
$ DP$ to meet $ RS$ at $ X$,
similarly, $ CP$ to $ H$,
              $ DQ$ to $ Y$,
              $ BQ$ to $ G$,
so now consider, $ \bigtriangleup AOH$
$ XD$ - transversal,
therefore by menelaus theorem,
$ \frac {AD}{OD} \cdot \frac {OP}{PH} \cdot \frac {HX}{XA} = 1$
$ \Rightarrow$
$ \frac {OD}{AD} = \frac {OP}{PH} \cdot \frac {HX}{XA}$ - $ (1)$
also,
$ \bigtriangleup PXH \sim \bigtriangleup PDC (RS \parallel{} BC)$
$ Rightarrow$ $ \frac {PH}{PC} = \frac {XH}{DC}$
$ \Rightarrow$ $ \frac {1}{PH} = \frac {DC}{PC \cdot XH$
replacing in $ (1)$,
we get, $ \frac {OD}{AD} = \frac {OP \cdot DC}{PC \cdot XA}$ - $ (2)$
also,
$ \bigtriangleup DBP \sim \bigtriangleup XAP$
$ \Rightarrow$ $ \frac {XA}{DB} = \frac {AP}{BP}$
again substituting similarly in $ (1)$,
we get,$ \frac {OD}{AD} = \frac {OP \cdot DC}{PC \cdot YA}$ - $ (3)$
from, $ (2)$ and$ (3)$ we can conclude that $ XA = YA$
so is wat i have done right?[/hide]

Actually it is too simple. So for the sake of posting a solution:
$ \frac{OD}{AD}\equal{} \frac{[OBC]}{[ABC]}$
$ \frac{OQ}{BQ}\equal{} \frac{[OAC]}{[ABC]}$
$ \frac{OP}{CP}\equal{} \frac{[OBA]}{[ABC]}$ whihc gives you the result.
I suppose you meant $ PC$ insteadof $ PF$ ......


Let $ S_n \equal{} \sum_{k\equal{}1}^n \frac{4k\minus{}1}{2^k}$.
Please calculate $ S_n \minus{} \frac{1}{2}S_n$.

hello, prove by induction that
$ \sum_{k\equal{}1}^n\frac{4k\minus{}1}{2^k}\equal{}\minus{}6\left(\frac{1}{2}\right)^{n\plus{}1}\minus{}8\left(\frac{1}{2}\right)^{n\plus{}1}(n\plus{}1)\plus{}7$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, prove by induction that
$ \sum_{k \equal{} 1}^n\frac {4k \minus{} 1}{2^k} \equal{} \minus{} 6\left(\frac {1}{2}\right)^{n \plus{} 1} \minus{} 8\left(\frac {1}{2}\right)^{n \plus{} 1}(n \plus{} 1) \plus{} 7$.
Sonnhard.[/quote]

$ \Rightarrow \lim  \equal{} 7$

hello, yes, it is $ \sum_{k\equal{}1}^\infty\frac{4k\minus{}1}{2^k}\equal{}7$.
Sonnhard.

Can you use the derivation to prove that $ \lim \sum\limits_{i \equal{} 1}^n {\frac{i}{{2^i }}}  \equal{} 2$?

hello, prove by induction that
$ \sum_{i\equal{}1}^n\frac{i}{2^i}\equal{}\minus{}2\left(\frac{1}{2}\right)^{n\plus{}1}(n\plus{}1)\minus{}2\left(\frac{1}{2}\right)^{n\plus{}1}\plus{}2$.
Sonnhard.

I have nice solution: use derivation! :)  :)

[quote="Poincare"][quote="L_Euler"]i don't really understand the first equation...[/quote][/quote]

Do you see that form: $ {\frac{{4n \minus{} 1}}{{2^n }}}$?  :)  :)

[quote="Dr Sonnhard Graubner"]hello, prove by induction that
$ \sum_{k \equal{} 1}^n\frac {4k \minus{} 1}{2^k} \equal{} \minus{} 6\left(\frac {1}{2}\right)^{n \plus{} 1} \minus{} 8\left(\frac {1}{2}\right)^{n \plus{} 1}(n \plus{} 1) \plus{} 7$.
Sonnhard.[/quote]

[quote="Dr Sonnhard Graubner"]hello, prove by induction that
$ \sum_{i \equal{} 1}^n\frac {i}{2^i} \equal{} \minus{} 2\left(\frac {1}{2}\right)^{n \plus{} 1}(n \plus{} 1) \minus{} 2\left(\frac {1}{2}\right)^{n \plus{} 1} \plus{} 2$.
Sonnhard.[/quote]

How are you getting the above identities?


trigonometry
calculus
calculus computations

The numerator is artificial. Split it into $\sum x^2e^{-(2n+1)x}+\sum\frac{(-1)^n}{2n+1}e^{-(2n+1)x}.$
Standard methods work here- remember that these are power series in $e^{-x}.$

[hide="Answer"]$\frac{x^2}{2\sinh x}+\tan^{-1}(e^{-x})$[/hide]

Standard series in exp(-x)? never heard of those... but seems quite interesting.

do you know by accidence a good online paper on these? [ just the basics to learn ]

I think (or better am sure) that he means standard power series in a variable $z$, where $z$ is set to be $z=e^{-x}$, not more.


superior algebra
superior algebra unsolved

Prove that the Galois group of x^6+24x-20 is A_6.

Thanks.

Use Maple:
infolevel[galois]:=2:
> p:=x^6+24*x-20:
galois(p,x);
     There exist 16 Galois groups in $S_{6}$.
i) $discrim(p)=864000^{2}$ is a square over $\mathbb{Q}$; this result eliminates 11 candidates; it remains 5 candidates ($A_{6},...$).
ii) Study $p$ in $\mathbb{Z}_{7}$; it remains 2 candidates.
iii) Study $p$ in $\mathbb{Z}_{23}$; it remains $A_{6}$.
   You can work according these lines. :P


On a large, flat field, $n$ people ($n>1$) are positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires and hits the person who is closest. When $n$ is odd, show that there is at least one person left dry. Is this always true when $n$ is even?

This uses the extremal principle.  This is a part of a problem shown in the extremal principle chapter of Problem-Solving Strategies.

I think I've solved this problem. Here are my thoughts.

[hide="Solution"]First consider when $n$ is odd. Let the positions where the people are standing be labelled as $P_{1}, P_{2}, \ldots, P_{n}.$ Now, note that if one of the $n$ people is squirted by at least two others, we are done (a total of $n$ squirts distributed).

WLOG, say $P_{1}P_{2}$ is the segment of shortest length of out all possible segments that can be formed by the $n$ points. It follows that $P_{1}$ and $P_{2}$ each squirt each other. Now, denote the set containing the remaining $n-2$ points as $S.$ We have two cases:

Case I: $P_{1}$ and/or $P_{2}$ is squirted at least twice

If one of the $n-2$ points is closer to $P_{1}$ or $P_{2}$ than any of the other points in $S,$ we are done.

Case II: $P_{1}$ and $P_{2}$ are squirted once

This means that every point in $S$ is closer to some other point in $S$ than the points $P_{1},P_{2}.$ Hence, the segment $P_{1}P_{2}$ is effectively isolated from $S,$ and the problem reduces to a $n-2$ point case (still odd).

Continuing this pattern, it is clear that we will end up with a single person, who must squirt one of the other $n-1$ people (who have already been squirted once), so in bith cases, one person is left dry.

Note that when $n$ is even, it's possible for everyone to get wet. Just consider $n/2$ isolated chords.
[/hide]

My solution seems quite unformal. Is there a way to fix it up?

Edit: Hmm, is this how they used the extreme principle in PSS?  :D

[quote="tjhance"]what's the extremal principle?[/quote]Here's how to use the extreme principle. Assume the given information in the problem is in some type of order from "largest" to "smallest." Then focus on the largest or smallest elements to shed light on possible solutions.

4everwise, your solution seems fine. I have a slightly different solution.

Assert that everyone gets shot. Note that because n is odd we can not pair people up as one person will be left out. Also, everyone has to shoot exactly one person and be shot by exactly one person.

Label all the people $1$ to $n$. Further let $2$ be the person shot by $1$, $3$ be the person shot by $2$, all the way up till $n-1$ shoots $n$ and finally $n$ shoots $1$. Also let $d_{i,j}=d_{j,i}$ be the distance between two people $i$ and $j$.

Now $d_{1,2}$ is the smallest distance from $1$ to any other person. For $2$ to shoot $3$, $d_{1,2}>d_{2,3}$ because otherwise $2$ would should $1$. Similarly $d_{2,3}>d_{3,4}$ and eventually $d_{n-1,n}>d_{n,1}$. But here is a contradiction as $d_{n,1}=d_{1,n}<d_{1,2}$ implies $1$ would have shot $n$. Hence proved.

Also note that this has been proved without taking odd numbers into concern. What it does take into concern is that people are not paired up, which has to be the case when $n$ is odd. Therefore as long as people are not paired together, someone is going to be dry even if $n$ is even.

What do you think? :)


trigonometry
quadratics
analytic geometry
complex numbers

simple trig problem, but my answer simply doesnt match the answer key:

Suppose that  :theta:  is an angle in the first quadrant of the Cartesian plane, such that cos(3 :theta: ) = 11/16.  It is know that cos  :theta:  is of the form  (a :sqrt:b - c )/d  If this fraction is completely simplified and the above symbols represent positive integers, what is the value of a+b+c+d?

thanks in advance.

Did you get [hide]a = 3, b = 5, c = 1, and d = 8[/hide]?

I agree with the posted solution except for the sign of c (if c = -1, then a+b+c+d would equal 15, not 17).  But then there appears to be an error in the statement of the problem.

If the required form of cos(theta) does include that subtraction symbol, then c should be -1, not 1.  But the problem stipulated that a, b, c, and d are positive integers.  As written, no solution exists.  Simplest fix: change the subtraction to addition, and then the answer indeed is 17.

lol, ok thank god, i thought i was going crazy...yea, that's exactly what i figured...one of the numbers is negative...

thanks, just making sure i wasnt doing something really dumb..

Pardon me, but what exactly are you talking about?

ahhh.... i can't seem to solve this.  i get an ugly cubic that i am unable to factor.  is there another way to go about this instead of putting cos(3:theta:) in terms of cos(:theta:)?

First, addressing what you wrote -- the cubic does factor, with some cleverness.

Here's how I did it:

[hide]Put it in the cubic equation form by turning cos(3 :theta: ) into cos( :theta: )'s.  You should end up getting the equation 64x^3 - 48x - 11 = 0, which we can also write as (4x)^3 - 12(4x) - 11 = 0, which shows us the obvious linear factor so that we can divide and solve a quadratic.

Other ways of getting to the linear factor --

1) Note the form of the answer you're looking for.  It is the solution to a quadratic.  Write out the quadratic that it must be a solution to and express the cubic as the product of a linear factor with the quadratic.

2) Note the form of the answer you're looking for, and that it must be the root of a quadratic.  Figure that the third root has to be rational.  Use the rational root theorem.[/hide]

How did you even get that cubic?  Is there a formula I am missing?

Expand out Cos(3:theta:) either by taking
Cos(3:theta:) = Re( (Cos(:theta:) + i*Sin(:theta:)):^3:)  (using complex numbers) or by writing
3:theta: = 2:theta: + :theta: and using the cosine sum formula, then the double-angle formulae, then converting into all cosines.


trigonometry
algebra
polynomial
Euler
calculus
calculus computations

Arne, I was wondering if there is a nice elementary solution. :?

Hm. I guess you know there isn't a really elementary way to find $\zeta(2)$?

So, why would there be an easy way to find $\zeta(4)$? ;)

Just for one of the various ways - admittedly, none of them elementary - let's use Euler's original method. (And let's be Eulerian - don't invest much energy into justifying some of these manipulations.)

Start with $\frac{\sin x}x=\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)$

Now multiply out the "polynomial" on the right, collecting like coefficients as in Viete's relations, and expand the left hand side as a power series:

$1-\frac{x^2}6+\frac{x^4}{120}+\dots=1- \left(\sum_{k=1}^{\infty}\frac1{k^2\pi^2}\right) x^2+\left(\sum_{1\le j<k}\frac1{j^2k^2\pi^4}\right)x^4+\cdots$

Equating the $x^2$ terms, we get that $\sum_{k=1}^{\infty}\frac1{k^2\pi^2}=\frac16.$ We'll need that in what follows.

Now, look at the $x^4$ terms. We have a sum over the "upper triangle". Add in a second copy of that for the "lower triangle", and then add in the "main diagonal" terms, and we'll have a sum over the "whole square." (I put all that in quotes becuase each piece is infinite.)

$2\sum_{1\le j<k}\frac1{j^2k^2\pi^4}+\sum_{k=1}^{\infty}\frac1{k^2k^2\pi^4} =\left(\sum_{j=1}^{\infty}\frac1{j^2\pi^2}\right) \left(\sum_{k=1}^{\infty}\frac1{k^2\pi^2}\right)$

$\frac2{120}+\sum_{k=1}^{\infty}\frac1{k^4\pi^4}=\frac16\cdot\frac16$

$\sum_{k=1}^{\infty}\frac1{k^4\pi^4}=\frac1{36}-\frac2{120}=\frac1{90}$

So $\sum_{k=1}^{\infty}\frac1{k^4}=\frac{\pi^4}{90}$

You can keep doing this style of argument for term after term. Euler himself pushed this calculation out to $\zeta(26),$ which for hand calculation is a prodigious feat.


if $ a_1\geq a_2\geq ....\geq a_n>0$ and $ \prod_{i\equal{}1}^nb_i\geq\prod_{i\equal{}1}^na_i$ then $ \sum_{i\equal{}1}^nb_i\geq\sum_{i\equal{}1}^na_i$

[quote="toanIneq"]if $ a_1\geq a_2\geq ....\geq a_n > 0$ and $ \prod_{i \equal{} 1}^nb_i\geq\prod_{i \equal{} 1}^na_i$ then $ \sum_{i \equal{} 1}^nb_i\geq\sum_{i \equal{} 1}^na_i$[/quote]
If i am not mistaken,this problem is from some Bulgarina MO,and it was posted here before. :wink:

What happens if $ n\equal{}2, a_1\equal{}8,a_2\equal{}2,b_1\equal{}4,b_2\equal{}5$?

[quote="nayel"]What happens if $ n \equal{} 2, a_1 \equal{} 8,a_2 \equal{} 2,b_1 \equal{} 4,b_2 \equal{} 5$?[/quote]
I think that there must be one more condition,inequality
$ \prod_{i \equal{} 1}^{j} b_i\geq\prod_{i \equal{} 1}^{j} a_i$,holds for all $ j\in\{1,2,\dots,n\}$,not for only $ j \equal{} n$.

oh,i'm sorry.I have edited it:
[u]PROBLEM[/u]:If $ a_1\geq a_2\geq...\geq a_n>0$ and $ b_1\geq a_1,b_1b_2\geq a_1a_2,....\prod_{i\equal{}1}^nb_i\geq\prod_{i\equal{}1}^na_i$ then we have
$ \sum_{i\equal{}1}^nb_i\geq\sum_{i\equal{}1}^na_i$

[quote="toanIneq"]oh,i'm sorry.I have edited it:
[u]PROBLEM[/u]:If $ a_1\geq a_2\geq...\geq a_n > 0$ and $ b_1\geq a_1,b_1b_2\geq a_1a_2,....\prod_{i \equal{} 1}^nb_i\geq\prod_{i \equal{} 1}^na_i$ then we have
$ \sum_{i \equal{} 1}^nb_i\geq\sum_{i \equal{} 1}^na_i$[/quote]
We have
$ \sum_{i \equal{} 1}^nb_i \minus{} \sum_{i \equal{} 1}^na_i \equal{} \sum_{i \equal{} 1}^{n\minus{}1}(a_i\minus{}a_{i\plus{}1})(\sum_{j \equal{} 1}^i \frac{a_j}{b_j}\minus{}i) \plus{}a_n(\sum_{i\equal{} 1}^n \frac{a_i}{b_i}\minus{}n) \ge 0$
by AM-GM.


Did you need to post this stupid thread in the Olympiad section? :mad:

conjurer is right this is nothing but just a spam
 :mad:

Hey, this is not spam. I just posted it here by mistake. Sorry, I will never do it again. Is there any way to delete the thread?

Mod is too busy than trying to sort out this childish probs.
Just tell the user the info.
@Lakshya: Post these in the General discussion

@chemrock->he wanted to delete the thread and not just create a new one anywhere else
by the way your suggestion is right :)


integration
calculus
calculus computations

Let $l,m$ be positive integers.
Evaluate

\[\int^2_0 x^l(2-x)^m dx\]

The first thing I'd do here is make a modest change of variables: $x=2u,$ from which $dx=2du$ and $2-x=2(1-u).$ The integral then becomes

$2^{l+m+1}\int_0^1u^l(1-u)^mdu.$

This puts it into the well-known family of beta integrals. For nonnegative integer values of $l$ and $m,$ a recursion derived from integration by parts (very similar to what liyi just did) suffices for computing it, but there's another calculation that works even when the exponents are not integers.

Suppose $\alpha,\beta>0.$ Lets look at the following double integral

$I=\int_0^{\infty}\int_0^{\infty}x^{\alpha-1}y^{\beta-1}e^{-(x+y)}dy\,dx.$

On the one hand, we can just do this as an iterated integral, getting

$I=\left(\int_0^{\infty}x^{\alpha-1}e^{-x}dx\right) \left(\int_0^{\infty} y^{\beta-1}e^{-y}dy\right)=\Gamma(\alpha)\Gamma(\beta).$

On the other hand, we can go back to the original integral and make the change of variables $x=uv,\,y=(1-u)v$ for $0\le u\le1,\,0<v<\infty.$ The Jacobian of this change of variables is $\frac{\partial(x,y)}{\partial(u,v)}=v$ so $dx\,dy=v\,du\,dv.$

$I=\int_0^{\infty}\int_0^1v^{(\alpha-1)+(\beta-1)+1}u^{\alpha-1} (1-u)^{\beta-1}e^{-v}\,du\,dv$

$=\left(\int_0^{\infty}v^{\alpha+\beta-1}e^{-v}dv\right) \left(\int_0^1u^{\alpha-1}(1-u)^{\beta-1}du\right)$

$=\Gamma(\alpha+\beta)\int_0^1u^{\alpha-1}(1-u)^{\beta-1}du.$

Set this equal to our other equation for $I$ and we get

$\int_0^1u^{\alpha-1}(1-u)^{\beta-1}du= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.$

If we return to the case of nonnegative integer exponents, we get

$\int_0^1u^{m}(1-u)^{n}du=\frac{m!n!}{(m+n+1)!}.$

Thank you for your detail explanation,Kent Merryfield.

Anbody could show simple usage calculation to insist an accuracy of the above last answer?

Mathematics shown dominent than other subject, may be the random property and accuracy meet in some formula .

Phongphan Porana.


number theory
relatively prime
number theory unsolved

There's one more solution (239,13). There no more solutions.
[quote="Baronov"]THe only solutioin is (1,1).
Using Gaussian integers we get :
(x+i)(x-i)=$2y^{4}$. Since x+i and x-i are relatively prime in Z[i] we get WLOG x+i=ef$(a+ib)^{4}$ where e=1,-1,i,-i;f is either 1 or 1+i and a,b are integers. From here checking all the cases gives us desired result[/quote]
For x odd $(x+i,x-i)=(1+i)$

This 's a special case of Lagrange 's theorem .
That sum is :Find all($x,y,z\in{N}$) such that:
$x^{2}+y^{4}=2z^{4}$
It 's solved by Lagrange in 1777 .However.His solution isn't so clear as i want :)


There is a plane flying above the sea at a distance of 20 km from an antena radio situated on the shore at a hight of 200 m.Fiind the heights of the plane from the sea so that the pilot hears the best radio signal.

I must also say that it's a old plane(second world war type)

Can you post the numerical value of the answer?I got something,but I'm hesitating about that.

Please post the solution.I couldn't get the same answer. :(

[quote="radac_mail"]250*k+125 ,k is from N[/quote]
What $N$ are you talking about?


[hide="{If we neglect the decrease of the rocket mass and the change of its velocity compared to 1500 m/s,}"] we can say that it gives the thrust-out fuel an impulse of 1500 m/s * 20 kg/s * 4 s = 120000 kg.m/s, which must be equal to the change of momentum of the rocket, which is $M.\Delta v$. Therefore the y-axis-velocity change is 4 m/s.
Total velocity (using pythagorean theorem) is 5 m/s.
[/hide]

[quote="warut_suk"]A $30000$-kg rocket is going forward at $3$ m/s. It releases gas to its right, with velocity $1500$ m/s and amount $20$ kg/s. After four seconds of gas releasing, what is the velocity of the rocket?[/quote]

From the conservation of momentum:

$(3\times 10^4 \text{ kg })(3 \text{ m/s}) \hat{i} = \vec{p} + (20 \text{ kg/s })(4 \text{ s })(1500 \text{ m/s } \hat{j})$

$\vec{p} = 9 \times 10^4 \text{ kg m/s } \hat{i} - 1.2 \times 10^5 \text{ kg m/s }\hat{j}$

$\vec{v} =\frac{9 \times 10^4 \text{ kg m/s } \hat{i} - 1.2 \times 10^5 \text{ kg m/s }\hat{j}}{(3\times 10^4 - 80) \text{ kg}} = 3.008 \hat{i} - 4.0107 \hat{j} \text{ m/s }$

[b]The rocket's velocity after the gas discharge is 3.008 m/s forward, 4.0107 m/s left.[/b]

That's right, I think you forgot about it Dr. No. It should read

$(3\times10^4\text{ kg})(3\text{ m/s})\hat i=\vec p+(20\text{ kg/s})(4\text{ s})(1500\text{ m/s }\hat j)+(20\text{ kg/s})(4\text{ s})(3\text{ m/s }\hat i)$.

And you also forgot that the rocket releases the gas as it accelerates - so the gas released after 4th second will really have velocity $3\text{ m/s }\hat i+1500\text{ m/s }\hat j-v_j\hat j$ where $v_j$ is rocket's speed in j direction after 4 seconds, so the equation won't hold anyway. You'll have to write down some diff equations to solve this properly.

Not that it really matters :) ...

[quote="kubus"][quote="Dr. No"][b]The rocket's velocity after the gas discharge is 3.008 m/s forward[/b][/quote]

Wait. I know that my answer did some approximation in the Y-direction, but please explain me how can the X-component of velocity change? I understood that the rocket thrusts the fuel to ITS right, with respect to the rocket, regardless of its velocity. Therefore the fuel has also the X-component of velocity (3 m/s) and the X-component of the velocity of the rocket does not change.[/quote]kubus,

bus made the proper correction in post #5. Including the forward drift of the propellant gas leaves the forward component of the rocket's velocity unchanged.

[quote="bus"]That's right, I think you forgot about it Dr. No. ...

Not that it really matters :) ...[/quote]I did indeed. I often make those kinds of errors. (Character flaw, I suppose)  :oops:


How many positive integers are less than $ 200\pi$?

$ 200\pi \equal{} 628.31...$ (This can seen by calculator or by memorizing the first few digits of $ \pi$). So we are looking at the number of positive integers from $ 1$-$ 628$. There are $ \boxed{628}$ integers.


trigonometry
LaTeX
logarithms
function
complex numbers

I was just taking a look at some $\sin$ and $\cos$ curves after avoiding it for as long as possible and concentrate on something else. But can anyone confirm the results of this equation doesn't exist:

$\cos x=1.4$

oh I get it.

[hide="here is my explanation"]The cosine has to be less than one because

$\cos= \frac{adj}{hyp}$ Ane the hypotenuse, by definition is the longest side of a triangle, so there is no way that the adjacent side of some angle of a right triangle is longer than the hypotenuse.

So $\cos x=p$        $\text{where p>1$ has no solutions.[/hide]

[quote="Zenith Xan"][quote="ccy"]It says $\cos x=1.4$.

In $\LaTeX$ you must have a space between \cos and x in \cosx.

If you have a rt. triangle, $\frac{opp}{hyp}<1$..[/quote]

Almost. $\frac{opp}{hyp}\le1$[/quote]

Nondegenerate right triangle.   ;)

The equation has no real solutions.  However,

$\cos x = \frac{e^{ix} + e^{-ix}}{2} = 1.4$

Has complex solutions.  Specifically:

$u = e^{ix}$
$u + \frac{1}{u} = 2.8$
$u^2 - 2.8u + 1 = 0$
$u = \frac{2.8 \pm \sqrt{7.84 - 4}}{2} = 1.4 \pm \sqrt{0.96}$
$ix = \ln (1.4 \pm \sqrt{0.96})$
$x = \boxed{ - i \ln (1.4 \pm \sqrt{0.96} )}$.

[quote="t0rajir0u"]
$ix = \ln (1.4 \pm \sqrt{0.96})$
$x = \boxed{ - i \ln (1.4 \pm \sqrt{0.96} )}$.[/quote]

Is this step correct???

[quote="deej21"]Yes, dividing by $i$ is the same as multiplying by $-i$, because $\frac 1 i = \frac i {i^2} = -i$[/quote]

You're right... thanks...

[quote="Centy"]Oh yes, don't forget those glorious complex angles :!:[/quote]

What would a complex angle look like?

[quote="Sporally"]Only problem is that now i've gone back in the book, i am working with $\cos$ and $\sin$ before we have been introduced to $e$. So i guess i will just write that $x$ doesn't exist in this case.[/quote]Before you write up your homework solution, make sure what values you are working with for the variable $x$. If you are working with real numbers, $\cos x=1.4$ is an impossible situation because $|\cos x|\le1$. I'll avoid complex numbers, weird things can happen with complex numbers. :P

That's what both my calculator and me figured, but i just got confused that the book has now given me a couple of tasks where $\cos x$ is greater than 1. I simply got worried that i had missed a point :D

Both sine and cosine are less than or equal to one since they are circle functions, and the maximum they can acquire in a unit circle is 1.  The minimum is -1.

Masoud Zargar

P.S.  This proves that I a am a horrible teacher.  :blush:

EDIT: If we have $\sin x=3$, then we do the following:

$\\sin x=\frac{e^{ix}-e^{-ix}}{2i}=3\\u-\frac{1}{u}=6i\\u^2-6ui-1=0\\u=\frac{6i\pm\sqrt{-36+4}}{2}\\u=3i\pm i\sqrt{8}=i(3\pm 2\sqrt{2})\\\\e^{ix}=i(3\pm 2\sqrt{2})\\x=-i\ln i(3\pm 2\sqrt{2})\\ x=-i(\ln e^{i\frac{\pi}{2}}(3\pm 2\sqrt{2}))\\x=-i(\ln e^{i\frac{\pi}{2}}+\ln (3\pm 2\sqrt{2}))\\ x=\frac{\pi}{2}-i\ln (3\pm 2\sqrt{2})$

Masoud Zargar

[quote="boxedexe"]$x=\frac{\pi}{2}-i\ln (3\pm 2\sqrt{2})$

Masoud Zargar[/quote]
You guys are forgetting the fact that $e^{ix}$ is periodic!  The actual answer would be

$x=\frac{\pi}{2}+2 \pi n -i\ln (3\pm 2\sqrt{2}), n \in \mathbb{Z}$  :roll:

Guess we got it all settled for those who wanted the complex answer aswell...


Consider $a, b, c$ three integrer numbers such that $a+b+c=0$. If $a^{47}+b^{47}+c^{47}=d$,

a) Can $d$ be equal to $2$?

b) Can $d$ be a prime number?

I think we can't just state $47|a+b$, just because $47$ is prime, $47|d$ and $a+b|d$.

Anyway, you state if, but don't state another possibilit[b]ies[/b] such as $a+b=1$
and you said it is proven?

isn't it enough to say that $n^{47}\equiv n \mod 47$ for any $n$, so that $a^{47}+b^{47}+c^{47}\equiv a+b+c = 0 \mod 47$?
moreover, $n^{47}\equiv n \mod 2$, so $94|a^{47}+b^{47}+c^{47}$...


inequalities
topology
inequalities unsolved

Let $a,b,c,d$ are positive numbers.Prove the following inequality.

\[(ac^2+bd^2)^3\leqq(a^3+b^3)(c^3+d^3)^2\]

This is the special case (with c = e and d = f) of the following inequality:

$\left(ace+bdf\right)^3 \leq \left(a^3+b^3\right)\left(c^3+d^3\right)\left(e^3+f^3\right)$

for any six nonnegative numbers a, b, c, d, e, f. Now it's almost clear to me that this one must be well-known, but I don't remember where exactly it can be found, so I post here a solution (quite an idiot solution, I admit):

Since

$\left(ace+bdf\right)^3 = a^3c^3e^3+3a^2c^2e^2bdf+3aceb^2d^2f^2+b^3d^3f^3$

and

$\left(a^3+b^3\right)\left(c^3+d^3\right)\left(e^3+f^3\right) = a^3c^3e^3+a^3c^3f^3+a^3d^3e^3+a^3d^3f^3+b^3c^3e^3+b^3c^3f^3+b^3d^3e^3+b^3d^3f^3$.

These sums have the expressions $a^3c^3e^3$ and $b^3d^3f^3$ in common, so we can subtract these expressions, and it remains to show that

$3a^2c^2e^2bdf+3aceb^2d^2f^2 \leq a^3c^3f^3+a^3d^3e^3+a^3d^3f^3+b^3c^3e^3+b^3c^3f^3+b^3d^3e^3$.

But this can be splitted into

$3a^2c^2e^2bdf \leq a^3c^3f^3+a^3d^3e^3+b^3c^3e^3$

and

$3aceb^2d^2f^2 \leq a^3d^3f^3+b^3c^3f^3+b^3d^3e^3$,

both of which follow from AM-GM.

  Darij

Thank you for your reply, darij.

I never thought the first inequality.This problem is made from [b]tangent line[/b], by accident.I will post my solution.Plese wait for a while.

kunny

Actually, I believe that I posted a proof of Darij's inequality a while back.

It follows immediately from Holder, because 1/3+1/3+1/3=1, and so, if we write:

P(a,b,3) as (a^3+b^3)^(1/3), then:

(abc+def) <= P(a,b,3)*P(c,d,3)*P(e,f,3), which is the desired result.

You are right, of course. Even without applying Hölder (which is one of the inequalities I always have to look up in some book since I can't remember them), we can state a more general fact: If $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ are 2n nonnegative numbers, then

$\left(a_1a_2...a_n+b_1b_2...b_n\right)^n \leq \left(a_1^n+b_1^n\right)\left(a_2^n+b_2^n\right)...\left(a_n^n+b_n^n\right)$.

In fact, denoting $p_i=\frac{a_i}{b_i}$ for every i, this is easily seen to be equivalent to

$\left(p_1p_2...p_n+1\right)^n \leq \left(p_1^n+1\right)\left(p_2^n+1\right)...\left(p_n^n+1\right)$.

And this is the Huygens inequality for the numbers $\sqrt[n]{p_1}$, $\sqrt[n]{p_2}$, ..., $\sqrt[n]{p_n}$. You can prove this inequality using convexity, but you can also show it by expanding and applying AM-GM with the help of a simple counting.

  Darij

I think I've seen this inequality:
${
(a_{1,1}...a_{1,n}+a_{2,1}...a_{2,n}+...+a_{m,1}...a_{m,n}})^n\leq(a_{1,1}^n+...+a_{m,1}^n)(a_{1,2}^n+...+a_{m,2}^n)...(a_{1,n}^n+...+^a_{m,n}^n)
$
I remember heard it called general cauchy's inequality. I don't know it is really called that or my friends jsut called it that because it's like cauchy.
The inequality can be proved, as darij said, by expanding and applying AM-GM with the help of a simple counting.

[quote="darij grinberg"]You are right, of course. Even without applying Hlder (which is one of the inequalities I always have to look up in some book since I can't remember them), we can state a more general fact: If $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ are 2n nonnegative numbers, then

$\left(a_1a_2...a_n+b_1b_2...b_n\right)^n \leq \left(a_1^n+b_1^n\right)\left(a_2^n+b_2^n\right)...\left(a_n^n+b_n^n\right)$.

In fact, denoting $p_i=\frac{a_i}{b_i}$ for every i, this is easily seen to be equivalent to

$\left(p_1p_2...p_n+1\right)^n \leq \left(p_1^n+1\right)\left(p_2^n+1\right)...\left(p_n^n+1\right)$.

And this is the Huygens inequality for the numbers $\sqrt[n]{p_1}$, $\sqrt[n]{p_2}$, ..., $\sqrt[n]{p_n}$. You can prove this inequality using convexity, but you can also show it by expanding and applying AM-GM with the help of a simple counting.

  Darij[/quote]

hi, may i know what's huygens inequality?

The Huygens (or Huyghens?) inequality states that

$\left(1+x_1\right)\left(1+x_2\right)...\left(1+x_n\right) \geq \left(1+\sqrt[n]{x_1x_2...x_n}\right)^n$

for any n nonnegative numbers $x_1$, $x_2$, ..., $x_n$.

  Darij

I'm sorry to have kept you waiting for, darij.

I will show my solution as follows.

For $t>0$, $t^3\geqq3t-2$, because $y=3t-2 $ is the [b]tangent line [/b] of $ y=t^3 $ at the point  $(1,1)$, setting $t=\frac{p}{q}( p>0,q>0)$,

we have $p^3\geqq{3pq^2}-2q^3 \cdot\cdot\cdot$ (☆)　,i.e. $pq^2\leqq{\frac{p^3+2q^3}{3}}$.

Denoting $x=\frac{a}{\sqrt[3]{a^3+b^3}},y=\frac{b}{\sqrt[3]{a^3+b^3}},z=\frac{c}{\sqrt[3]{c^3+d^3}},w=\frac{d}{\sqrt[3]{c^3+d^3}}\cdot\cdot\cdot$(☆☆),

we obtain $x^3+y^3=z^3+w^3=1$.so in using (☆),$xz^2+yw^2\leqq\frac{1}{3}(x^3+y^3)+\frac{2}{3}(z^3+w^3)=1$.

Substituting (☆☆)　for　this　inequality,and cubing the both sides,we are done.

kunny

This is a nice solution, Kunny. Just wanted to note that the inequality

[quote="kunny"]$pq^2\leqq{\frac{p^3+2q^3}{3}}$[/quote]

follows from AM-GM applied to the numbers $p^3$, $q^3$ and again $q^3$. You don't need tangent lines for that. But still, tangent lines are a quite useful method. Some minutes ago, before reading your message, I solved [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=126794]a problem from Stun[/url] using the tangent line method (actually, I used it to find the inequality $\frac{a^3}{a^2+b^2}\geq \frac12\left(2a-b\right)$ which I needed in the solution).

  Darij

[quote="darij grinberg"]You are right, of course. Even without applying Hlder (which is one of the inequalities I always have to look up in some book since I can't remember them)

  Darij[/quote]

I find it so much easier to remember inequalities when you have applications for them.  In fact, Holder and Minkowski are both used to prove that $\left(\displaystyle\sum^n_{i=1} a_i^n\right)^{\frac{1}{n}}$ is a norm (essentially a measure of distance in a metric space or something similar).

Oh, and I have to say that I was very, very impressed by your proof of Stun's inequality in that other thread.


I am really weak with inequalities and I have seen many problems that involve them.  Could someone post the basics about the relationships between AM-GM-HM?  I would really appreciate it.

Prove that $(1+a)(1+b)(1+c) \ge 8$ if $a, b, c > 0$ and $abc = 1$.  Also (unrelated) I'd like to point out that GM-HM is really just AM-GM.  So you'll almost never use GM-HM, although you might use AM-HM.  The overwhelming majority of inequalities using RMS-AM-GM-HM uses AM-GM.

[hide]

Knowing , we move on to  and find it's value(s) using AM-GM.

Moving on to the last part, we use AM-GM again:

Adding the previous two inequalities, we have

Add two to both sides (since 1=1 and abc=1 in the original expression), and we have

which is the desired relation.

[/hide]

Was that incorrect?

That's right.  New problem: In triangle ABC with area 2, let the sides be a, b, and c and the altitudes d, e, and f.  Prove that $12 \le a + b + c + d + e + f$.

I think one of the altitude should be $d$. In that case:
We have $ad=be=cf=4$. Thus by AM-GM inequality
\[
a+b+c+d+e+f\geq2\sqrt{ad}+2\sqrt{be}+2\sqrt{cf}=12.
\]

[quote="probability1.01"]That's right.  New problem: In triangle ABC with area 2, let the sides be a, b, and c and the altitudes e, f, and g.  Prove that $12 \le a + b + c + d + e + f$.[/quote]

Since the area of the triangle is 2,

 ad = be = cf = 4

  So a + b+ c+d+e+f = a + b+ c+4/a+4/b+4/c

= (a+4/a) +(a+4/b)+(a+4/c)

>= 2*sqrt(a*4/a) + 2*sqrt(b*4/b) + 2*sqrt(c*4/c)

= 4 + 4 + 4

= 12

Errr... that was meant for Euler1001 but that's ok :).  I'll have to think a little longer to come up with some more RMS-AM-GM-HM problems.

RMS=Root Mean Squared.

For any sequence $x_1,x_2,\ldots,x_i$, the RMS is equal to:

$\displaystyle\sqrt{\frac{\displaystyle\sum_{i=1}^nx_i^2}{n}}$

So... since someone said RMS-AM-GM, I'm assuming the inequality would be
$RMS \ge AM \ge GM$?

Yes.  You could also add in the harmonic mean:

\[\text{RMS}\ge \text{AM} \ge \text{GM} \ge \text{HM}.\]

The harmonic mean of the nonnegative numbers $\{a_1,a_2,\ldots,a_n\}$ looks like:

\[\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}.\]

There is an even further generalization called the POWER MEAN inequality which says:

\[\sqrt[m]{\frac{a_1^m+a_2^m+\cdots+a_n^m}{n}}\ge \sqrt[k]{\frac{a_1^k+a_2^k+\cdots+a_n^k}{n}} \ \ \text{for} \ m>k.\]

An other AM-GM problem:
$\text{find the minimum of:} y=\sin^2{x} \cos{x}$

(this is closely related to physics)


Sometimes I have difficulty finding what I want in the search.
Is there any way AoPS can use a more Google style search?
I've been to several websites where the search is powered by Google.

would google search still work for egoseach?

You can search for "keyword site:artofproblemsolving.com" if you really want to.

PS And, no, I cannot design a search engine "just like" Google. If I could, that would mean that all those thousands of employees at Google are pathetic or I'm the new Einstein :D

[quote="Temperal"]I thought it was [quote]site:http://artofproblemsolving.com "keyword"[/quote]

A google search looks less professional, since you have to put the Google logo there or it's illegal... plus the results show up on a google page. Plus, member searches wouldn't be as accurate, and as Klebian said, egosearch (That's looking for your own posts) would not work.[/quote]This is a good answer.

[quote="Temperal"][quote="AMC37.5"][quote="Temperal"]

A google search looks less professional, since you have to put the Google logo there or it's illegal... plus the results show up on a google page.[/quote]

Google (here: http://www.google.com/coop/cse/) lets you create a custom search engine with your own logo. (Google's "Powered by Google" logo is relatively unobtrusive, on the side.) I think you can also change the colors so it looks like the same site.[/quote]

"We're sorry.

We were unable to access the page that you requested.

Some suggestions:

    * Check the spelling of the page address you are trying to reach.
    * Try accessing the page you are looking for from the Google Co-op homepage.
    * If this problem continues, the page you are trying to reach may be unavailable.
    * Go back to the previous page.

"[/quote]

http://www.google.com/coop/cse/

Oh, I see, the parentheses.  :blush:

Well, what I said still holds true. It would have less features, and the Google logo is still very visible.


calculus
integration
calculus computations

A sprinkler distributes water in a circular pattern, supplying water to a depth of $ e^{\minus{}r}$ feet per hour at a distance of $ r$ feet from the sprinkler.

What's the total amount of water supplied per hour inside of a circle of radius $ 13$ in $ ft^3/h$ ?
What's the total amount of water that goes through the sprinkler per hour in $ ft^3/h$?

[quote="qwesl"]A sprinkler distributes water in a circular pattern, supplying water to a depth of $ e^{ \minus{} r}$ feet per hour at a distance of $ r$ feet from the sprinkler.

What's the total amount of water supplied per hour inside of a circle of radius $ 13$ in $ ft^3/h$ ?
What's the total amount of water that goes through the sprinkler per hour in $ ft^3/h$?[/quote]

Dimensional analysis (a.k.a. "looking at the units") will help you here.  The sprinkler supplies water to a depth of $ e^{ \minus{} r}\,$ $ \, ft/h$ at a distance of $ r\,$ $ \, ft$.  I'll assume third quarter calculus here: Let $ W(R)$ denote the total amount of water supplied per hour inside a circle of radius R. Then

\[ dW \equal{} e^{\minus{}r}\, dA \equal{} e^{\minus{}r}r\, dr\, d\theta\]

the units confirm this

\[ \frac{ft^3}{h}\equal{}\frac{ft}{h}\cdot ft^2 \equal{} \frac{ft}{h}\cdot ft\cdot ft\]

recall that radians are unit-less.  Evaluate the double integral in polar coords

\[ W(R)\equal{}\iint e^{\minus{}r}r\, dr\, d\theta\]

then consider $ W(13)$ and $ W(R)$ as $ R\rightarrow\infty$.