allright, well i just graduated high school about a week or so ago

i'm going to miami of ohio with a full ride cause of my "credentials" [not that great in comparison to ppl here and my close friends, thus quotations] in mathematics...

but im drawn as to what to do... i really do love math, im not that great at it as most people here are, but i need help in determining my future, haha, so i turn to these math forums which i have been an idle member of

allright simply put: i dont know whether to pursue math or just go pre-med cause of money issues upcoming in my life... my original goal was to double major in chemistry and math and then have the option of med-school or not, but now im tempted to go all out w/ bio/chem major

what would people who know more than me reccomend?  if i had it my way and didnt have to worry about making money in my future i'd go all out w/ a math/stat major and just study math [haha, my dorkness is showing]

i have orientation on the 21st/22nd and need to come go there with my mind just about made up w/ what i want to do... so the college wants me to major in math for the putnam but im pretty sure i wouldnt measure up and im just more or less concerned with my future

is there much of a career in majoring in math and a pre-med science? of course diversity helps, plus it would help me stay more motivated cause im pretty sure i  have a deep passion for math [i do it in my free time]

help a hormone driven teenager who is confused as all hell about his future out here!

any suggestions/experiences are welcome here, and i would respect any opinion or comment

ps: it's 4:45 am local time here, thus the random breakdown/rambling

-Hessam

You don't have to decide yet. How much difference would deciding now make to your first-semester freshman schedule? For later, changing majors is something that can be done and is often done.

Depending on where your AP courses and/or other preparation place you, take the appropriate calculus course and the appropriate chemistry course. That's already a big hunk of your schedule - a physics course may be a possibility but might not be something you have to do in the very first semester. Some medical schools don't trust AP credits - if there's a health professions advising office, you might want to call them and ask. And then fill your schedule with all the other things that freshmen are expected to do.

You can probably spend at least a year with a schedule that marks you as a "math or science or something like that" major without getting too specific. Now, once you're into it - what subjects do you really like? Will you be able to make good enough grades to make medical school a realistic possibility? Do you even want to go to medical school? (Pre-meds need to do some volunteering in hospitals or other health care facilities - that advising office I referred to above, if it exists, could tell you more. One reason is to weed out the people who don't really want to be there.) There are other things in the world - as a chem/biochem/biology major, perhaps you'll get attracted into scientific research rather than medicine.

As for the math - for as long as you look forward to your next math course, keep taking them. Don't weigh every single course against a career - rest assured that knowing more math isn't going to hurt you. As for the Putnam: I don't know how it is there. I run the Putnam program at my school. It isn't really worth my time to try very hard to recruit outside of math majors, but if someone came to me and expressed interest, I'd say, "Sure, give it a try. Let's see if you like it." And if that someone were telling me about AMC and USAMTS and so on, I'd certainly listen.

I think Kent Merryfield's advice is really good.

Enjoying your work is probably much more important than making a lot of money.  For one thing, the amount that you enjoy what you do makes a huge difference in determining how good you are at it.  It may be dangerous to become a doctor for the money, if you'd rather be doing something else.  It's possible to get paid pretty well doing math.

Don't be a doctor just for the money - you'll be a terrible doctor, then.

If you must do something 'just for the money', become an actuary - you may find you like it.  They get paid well, have consistently extraordinarily high job satisfaction ratings, and get to do lots of stats/math.

However, it's way too early to worry about it.  Most of the jobs I've had since leaving college didn't exist when I started college.

[quote="Hessm"]i'm going to miami of ohio...
i have orientation on the 21st/22nd and need to come go there with my mind just about made up w/ what i want to do... [/quote]

I checked with someone in the Physics Department at MUOhio who said that this is not actually true, despite the way it may seem.  There's a point during orientation when students divide up by intended major; then they go to different rooms and receive advice for registration. Physical sciences, math, and statistics are all together.  My source says "But the largest room in the building is reserved for "undecided,"..." so it seems like you could just go there, or listen to the spiel in one of the major-advice rooms and then do what you want.

I have a Math/Stat degree from Miami (1980) and it served me reasonably well in graduate school.  When I was at Miami the Math/Stat major contained plenty of electives and required a couple of areas of specialization outside math/stat.   It would have been relatively easy to set up a "double major" with Zoology (pre-med/pre-vet at Miami in the those days) or a math/pre-med 4-yr program. Here's the page with the Miami math/stat requirements

[url]http://unixgen.muohio.edu/%7EMathStat/undergraduate/InfoMSM_May2006.htm#_MATHEMATICS_AND_STATISTICS[/url]

Looks like requirements are still pretty flexible.

You might look at some med school admissions requirements.  I'm not sure they are all that particular about your major program - they're mainly interested a good background in the physical and life sciences, good GPA, and demonstrated committment to health care (volunteering in hospitals,...).  While most universities have pre-med programs, I'm not sure that they are the only way into med school.

Take a couple of computer programming courses while you're there no matter what else you plan to do.

Hopefully, you'll place into the honors accelerated calculus sequence (MTH 155H/255H) or above.  It was difficult to place out of this sequence when I was there.  Like most folks, I needed some relatively easy courses when I was a freshman.

In my day, the high demand subjects restricted enrollments to students who were listed as majors.  Find out if this is still the case for zoology or whatever passes for pre-med these days.  You might need to sign up in the high-demand major just to get classes in that area.

Anyone should be able to take the math/stat classes - they've never been in particularly high demand.  The Putnam shouldn't be restricted to math majors.  I think it's open to all majors.

Good Luck.

Greg

Yes, it is definitely possible for math majors to be admitted to medical school, so this is not an either/or decision. I agree with rrusczyk's advice that you shouldn't go to medical school because someone else thinks it's a good idea, but only because you desire to be a good doctor. Check requirements for a math major, and check requirements for (possible) future medical school admission, and take the courses that give you a "both-and" path. Being an actuary is a good career, and so is being a biomedical engineer, and such a path leaves both careers open to you.

A major in math will not preclude a career in medicine, if indeed that is appropriate for you.  Math, humanities and physical science majors all outscore biological science majors on the MCAT and in general do well in medical school.  See [url]http://www.aamc.org/students/mcat/examineedata/pubs.htm[/url]

If earning potential is of primary concern, medicine is unlikely to be your best choice -- ask any doctor in private practice.

well thanks everyone for your input

it's true that i dont need to declare it "officially'now, but with all the credit's i have the classes i registered for are junior and senior level courses

i decided that for now: going for a double major, getting two bachelors of science degrees in mathematics and chemistry, along with meetin proper pre-med requirements [difference of one class to go pre-med]

we'll see what happens in time


integration
calculus
function
real analysis
real analysis solved

2. If F is a primitive of f:R->R, f(x) = ex^2 for every x \in R, compute lim{x->oo}(xF(x))/f(x).

F(x)=   \int _[1,x]exp(t^2)dt integration by part gives
F(x)= exp(x^2)/2x - e/2 +   \int _[1,x]exp(t^2)/(2t^2) dt
this leads to
xF(x)/f(x) = 1/2 - x/exp(x^2).e/2 +
x/exp(x^2).1/2. \int _[1,x]exp(t^2)/t^2 dt ---> 1/2 when x-->+oo

To see that the last term in the previous some tend to 0 use
  \int _[1,x]exp(t^2)/t^2 dt =   \int _[1,x] (2t.exp(t^2))/(2t^3)dt =
exp(x^2)/(2x^3) - e/2 + 3/2. \int _[1,x]exp(t^2)/t^3dt

Remark: D(x) = F(x)/f(x) = e^(-x^2).\int _[1,x]exp(t^2)dt is called
Dawson' integral
Spanier, Oldham " Dawson's Integral " Ch.42 in Atlas of Functions.
Washington, DC:Hemisphere, pp.405-410, 1987.

i think i didn't get smth
F(x)= \int _[1,x] f(x)???
ain't there F(x) = \int _[0,x]f(x)???

you can take F(x) = \int x a f(t)dt for any a, that is because a function has an infinity of primitives.

i have a question(maybe i'm wrong) but isn't it easier if we use l'Hospital??


If you are going to IMO or know who goes please post in this thread here!  :)

Hi guys, you can answer that question. As I heard some countries already have chosen their teams. One of those countries is Molova! :D

Also Vietnam had chose team for IMO 2008 in Spain  :lol:

[quote="mornik"]Croatia also chose the team for IMO. I think two of the people in the team are on Mathlinks (including me), but we post very little.[/quote]

I think I am the second... :)

We will be the [b]most beautiful[/b] team in Spain, we have [b]3 girls[/b]!

[quote="rachid"]Results are more important :lol: .[/quote]There have been very powerful contestants in the past IMOs that were girls (a few examples Eugenia Malinnikova 3 times gold 41, 42, 42!, Ana Caraiani 1 silver, 2 gold, and many others.)

[quote="FOURRIER"]Our country's (Morocco) TST final round is not done yet unfortuntaly.

We have TST  1 week before Baccaleaurate exam :mad:  :mad: . I don't think this exists in other countries.[/quote]

Yes, it does :P
But I have no problem with that.

[quote="TTsphn"]Vietnam team has a girl .All of them post quite few in Mathlinks  :)[/quote]
why they post few???

[quote="Valentin Vornicu"][quote="FOURRIER"]We have TST  1 week before Baccaleaurate exam :mad:  :mad: . I don't think this exists in other countries.[/quote]When I was in my last year of highschool, this was our program for the IMO TST: Saturday TST, Sunday last TST (the one which I aced and qualified :)), Monday first day of Baccalaureate, Tuesday the 2nd day of Bacc., Wednesday the 3rd day, Thursday the 4th day, Friday the 5th day, Saturday the 6th day.

So you have it easy my friend :)[/quote]
but did you pass to imo :huh: ,i think it's difficult.

[quote="Valentin Vornicu"][quote="FOURRIER"]We have TST  1 week before Baccaleaurate exam :mad:  :mad: . I don't think this exists in other countries.[/quote]When I was in my last year of highschool, this was our program for the IMO TST: Saturday TST, Sunday last TST (the one which I aced and qualified :)), Monday first day of Baccalaureate, Tuesday the 2nd day of Bacc., Wednesday the 3rd day, Thursday the 4th day, Friday the 5th day, Saturday the 6th day.

So you have it easy my friend :)[/quote]

WOW :D  .

After the bac you should have had at least one weak of rest :P

[quote="rachid"][quote="Valentin Vornicu"]When I was in my last year of highschool, this was our program for the IMO TST: Saturday TST, Sunday last TST ([u]the one which I aced and qualified[/u] :))
[/quote]
but did you pass to imo :huh: ,i think it's difficult.[/quote]
You seem weak at english :lol:

[quote="FOURRIER"][quote="Valentin Vornicu"][quote="FOURRIER"]We have TST  1 week before Baccaleaurate exam :mad:  :mad: . I don't think this exists in other countries.[/quote]When I was in my last year of highschool, this was our program for the IMO TST: Saturday TST, Sunday last TST (the one which I aced and qualified :)), Monday first day of Baccalaureate, Tuesday the 2nd day of Bacc., Wednesday the 3rd day, Thursday the 4th day, Friday the 5th day, Saturday the 6th day.

So you have it easy my friend :)[/quote]

WOW :D  .

After the bac you should have had at least one weak of rest :P
[/quote]Actually right afterwards we went in Hungary for IMO training.

sad that germany didn't win^^ and we have the IMO in 2009, so i guess, the geman women team will be world champion or whatever comes in this year^^

Naphthalin

Nothing comes in 2009. It's an odd number :) The next world cup is in South Africa in 2010. Sorry, Ballack will lose yet one more final :)

I am going to my second IMO.

So that's why greece won the final against portugal  :roll:

The portuguese team will be:
POR 1 - Daniel Vaz
POR 2 - Eloísa Pires
POR 3 - Filipe Valeriano (me)
POR 4 - Jorge Miranda (also has a mathlinks account)
POR 5 - Pedro Vieira
POR 6 - Ricardo Campos

See you all in Spain.

IMO is coming soooooooooooooon  :D
and the moroccan team is well trained as usual  :P  :P

Greek team is again going to win the laugh game ?  :P  (if you don't know what is the laugh game then just ask the German team from Vietnam ) .
BTW , we will have again a mathlinks meeting ? And does anyone know about the videos of the ML meeting last year ?

The problem was most of the videos were much more bigger than Youtube's resolution. Now they have upgraded, so this year we should have more videos.

[quote="Valentin Vornicu"]I remember that the Moroccan team is always trained for swimming and water polo also ;) :D[/quote]

Well, it depends on the year, some years were very good for morocco:
1991, 1995, 2004.

In fact, the country doesn't offer good conditions, mathematical olympiad is not given any importance.

I'll be there (Trinidad and Tobago team). By the way, which other Mathlinker/Aopser IMO-ers will be going to Stanford in September? I'd like to meet you in person.

And this is the AZE team in IMO 2008

1.RUSLAN MUSLUMOV (Bronze medal in IMO2007,IZHO 2008,BMO 2008)
2.ELYAR GULAMHUSEYNLI(Bronza medal in IMO 2007)
3.SERKHAN BEDIRLI(Silver medal in IZHO 2008)
4.FERID MMEMEDOV(Bronze medal in BMO 2008)
5.ZULFU ASLANLI(Silver medal in I_SWEEEP 2008)...(me)
6.SHAHRIYAR GARALOGLU

This year our team will be great...6 medals.....(INSHALLAH)

[quote="FOURRIER"]In fact, the country doesn't offer good conditions, mathematical olympiad is not given any importance.[/quote]Many places don't, but that's why we're all here ;)

Hello!!
I too am going to the IMO.I from Peru,
the peruvian team is:
PER 1.-JESUS RAMOS
PER 2.-CESAR CUENCA
PER 3.-FERNANDO MANRIQUE
$ PER 4. \minus{} AMILCAR VELEZ..(IO !!)(Me)$
PER 5.-TOMAS ANGLES
PER 6.-IVAN MUÑOS.

OK people!..See you all in España!!(spain :P)..
PD. In mathlinks.
$ PER 4 \minus{} \minus{} > Aev5peru$., $ PER1$ AND $ PER3 \minus{} \minus{} > Feeurl1$, $ PER 2 \minus{} \minus{} > CUENK$
                                                                                Bye
http://www.mathlinks.ro/viewtopic.php?t=213355

GIL Publishing House will be in Madrid during 17.07.2008 (Thursday) - 21.07.2008 (Monday). Our publications in English will be available for sale and you can save the shipping charges. Any of you interested in acquiring books please contact us using this username or using imo@gil.ro e-mail address.

Thank you!

[quote="Armenia"]I am Tigran Hakobyan from Armenia(ARM1).
I will happy to meet with you.[/quote]

I think everybody knows you from the famous sleepy ML gallery picture here:

- http://www.mathlinks.ro/album.php?mode=pic&id=5744

But let us have a more formal picture from this year's IMO 2008 web page as well:

- http://www.imo-2008.es/paises/ARM.html


If $ \log_{k}{x}\cdot \log_{5}{k} \equal{} 3$, then $ x$ equals:

$ \textbf{(A)}\ k^6\qquad
\textbf{(B)}\ 5k^3\qquad
\textbf{(C)}\ k^3\qquad
\textbf{(D)}\ 243\qquad
\textbf{(E)}\ 125$

[hide="Solution"]
Using the rule $ \log_{a}{x}\cdot \log_{b}{y} \equal{} \log_{a}{y}\cdot \log_{b}{x}$

The expression $ \log_{k}{x}\cdot\log_{5}{k}\equal{} 3$ becomes $ \log_{k}{k} \cdot \log_{5}{x} \equal{} 3\implies 5^3 \equal{} x \implies \text{ Ans. } \{\mathbf{(E)}\}$
[/hide]

[hide="Solution"]
$ \log_kx\cdot\log_5k\equal{}3$

$ \log_kx\equal{}\frac{3}{\log_5k}\equal{}3\log_k5$ using the rule $ \log_ba\equal{}\frac{1}{\log_ab}$.

$ \log_kx\equal{}3\log_k5\equal{}\log_k5^3\implies x\equal{}\boxed{\textbf{(E)}\ 125}$.
[/hide]

[hide="Cheap"]
Assume $ k\equal{}25$ for the sole reason that it makes all the answer choices different and is a power of $ 5$.

$ 2\log_{25}x\equal{}3\implies x\equal{}25^{3/2}\equal{}\boxed{\textbf{(E)}\ 125}$.
[/hide]

In a lot of the AIME questions there are questions similar to this but since there are only integer combinations people are able to plug in numbers.

AIME questions aren't this trivial. There is actually an identity that leads to this answer right away. I have yet to see an AIME problem that can be solved in one step using a relatively simple theorem/formula.

$ \log_{k}x \cdot \log_{5}k \equal{} 3$

$ \log_{5}k \equal{} \frac {1}{\log_{k}5} \implies \log_{k}{x} \equal{} 3\log_{k}{5} \equal{} \log_{k}5^3$

Taking $ \log_{k}{\frac {x}{5^3}}$ to other side we have

 $ \log_{k}{\frac {x}{5^3}} \equal{} 0 \implies x \equal{} 5^3$ or $ \boxed {E}$


[quote="tdl"]The real sequence is defined by $ x_0 \equal{} 1, x_1 \equal{} k, x_{n \plus{} 2} \equal{} x_n \minus{} x_{n \plus{} 1}$. Show that there is only one value of $ k$ for which all the terms are positive.[/quote]

The solution of this induction is :
$ x_n\equal{}\alpha\left(\frac{\minus{}1\plus{}\sqrt{5}}{2}\right)^n\plus{}\beta\left(\frac{\minus{}1\minus{}\sqrt{5}}{2}\right)^n$ with $ \alpha\plus{}\beta\equal{}1$ and $ \alpha\frac{\minus{}1\plus{}\sqrt{5}}{2}\plus{}\beta\frac{\minus{}1\minus{}\sqrt{5}}{2}\equal{}k$

$ x_n\geq 0$ $ \forall n$ implies $ \beta\equal{}0$, and so $ \alpha\equal{}1$ and so $ \boxed{k\equal{}\frac{\minus{}1\plus{}\sqrt{5}}{2}}$


Να αποδειξετε οτι υπαρχουν δυο υποσυνολα των αρρητων τετοια ωστε να ειναι ξενα, υπεραριθμησιμα και το καθενα πυκνο στους πραγματικους και η ενωση τους ειναι ολοι οι αρρητοι. Σκεφτηκα αυτο το προβλημα αλλα δεν μπορω να βρω παραδειγμα.

@NickNafplio
[hide]Γραψε την ιδεα που μου ειπες προφορικα αν μπορεις[/hide]

Το πρόβλημα αυτό, όπως το θέτεις, έχει τετριμμένη λύση. Για παράδειγμα πάρε:
\[ A \equal{} \sqrt {2} \plus{} \mathbb{Q}\quad B \equal{} \sqrt {3} \plus{} \mathbb{Q}\quad C \equal{} (\mathbb{R}\setminus\mathbb{Q})\setminus (A\cup B)\]
Μετά θέσε: $ C_ \plus{} \equal{} C\cap (0, \plus{} \infty)\quad C_ \minus{} \equal{} C\cap( \minus{} \infty,0)$.

Τα δύο σου σύνολα είναι τα $ B\cup C_ \plus{}$  και $ A\cup C_ \minus{}$.

Θα ήταν πολύ καλύτερο να απαιτήσεις κάθε σύνολο να έχει υπεραρίθμησιμη τομή με οποιοδήποτε ανοικτό διάστημα. Τότε, ένας τρόπος να το καταφέρεις αυτό είναι να θεωρήσεις τα σύμπλοκα του $ \mathbb{R}$ ως προς το         $ \mathbb{Q}$ και να πάρεις έναν αντιπρόσωπο από κάθε σύμπλοκο στο διάστημα $ (0,1)$. Ονόμασέ το το σύνολο αυτό $ C$. Είναι υπεραριθμήσιμο. Χώρισέ το σε δύο υπεραρίθμησιμα σύνολα $ C_1$ και  $ C_2$ και τότε τα σύνολα με την ισχυρότερη ιδιότητα είναι τα:
\[ C_1 \plus{} \mathbb{Q},C_2 \plus{} \mathbb{Q}\]

Η δικια μου ιδεα (η οποια αρχικα εχανε σε ενα μικρο σημειο αλλα τελικα την διορθωσε ο Αλεξανδρος) ηταν να δουλεψουμε με τη δεκαδικη μορφη.

Παιρνουμε το συνολο των αρρητων τα οποια εχουν πεπερασμενο πληθος αρτιων ψηφιων στη δεκαδικη τους αναπαρασταση, τοτε αυτο και το συμπληρωμα του στους αρρητους ειναι προφανως πυκνα στους πραγματικους, υπεραριθμισημα και ξενα μεταξυ τους, ενω η ενωση τους δινει ολους τους αρρητους.

Θέλουμε τα δυο υποσύνολα να τέμνουν κάθε ανοικτό διάστημα. Αρκεί να τέμνουν κάθε διάστημα με ρητό κέντρο και ρητή ακτίνα. Τώρα έχουμε αριθμήσιμες συνθήκες να ικανοποιήσουμε (αντί υπεραριθμήσιμες) και μπορούμε να τις ικανοποιούμε μία προς μία...

Με παρόμοιο τρόπο (αλλά τεχνικά αρκετά πιο δύσκολο) μπορεί να λυθεί και το πιο δύσκολο πρόβλημα που έβαλε  ο/η Jondrette.

[quote="Demetres"]Με παρόμοιο τρόπο (αλλά τεχνικά αρκετά πιο δύσκολο) μπορεί να λυθεί και το πιο δύσκολο πρόβλημα που έβαλε  ο/η Jondrette.[/quote]

[hide]Με μια επιπόλαια σκέψη θα πρότεινα just do it με transfinite induction, αλλά δεν έχω κάτσει να το δουλέψω με λεπτομέρεια. Κάτι τέτοιο έχεις κι εσύ στο μυαλό σου?[/hide]

Cheerio,

Durandal 1707

Edit: ερώτηση ορολογίας (γιατί δεν το'χω καθόλου με τη μαθηματική ορολογία στα ελληνικά :blush:): σύμπλοκα λέμε τα cosets? Ρωτάω επειδή με μπέρδεψε λίγο το post του Jondrette...

Ναι, τα σύμπλοκα είναι τα cosets. Γενικά μπορούν να οριστούν σαν κλάσεις ισοδυναμίας. Με άλλα λόγια μπορεί να αποφύγει κανείς τη θεωρία ομάδων. Στην προκειμένη περίπτωση μπορούμε να χωρίσουμε το $ \mathbb{R}$ σε κλάσεις ισοδυναμίας με βάση τη σχέση $ a\sim b\iff a \minus{} b\in\mathbb{Q}$.
Οι κλάσεις ισοδυναμίας που προκύπτουν είναι όλες σύνολα της μορφής $ \mathbb{Q} \plus{} a$ για καποιο πραγματικό αριθμό $ a$ που δεν είναι μοναδικά ορισμένος. Τα σύμπλοκα λοιπόν είναι αριθμήσιμα πυκνά και ξένα ανά δύο. Και πόσα τέτοια έχουμε; Προφανώς υπεραριθμήσιμο πλήθος από τέτοια σύμπλοκα. Είναι σα να κόβουμε το $ \mathbb{R}$ σε φέτες που είναι "παράλληλες" με το $ \mathbb{Q}$ (παράβαλε με την περίπτωση που κόβουμε το $ \mathbb{R}^2$
σε φέτες που είναι ευθείες παράλληλες με δοσμένη ευθεία που διέρχεται από την αρχή των αξόνων: ομαδοθεωρητικά οι δύο κατασκευές είναι ταυτόσημες).

Τέτοιες ιδέες χρησιμοποιούνται εκτεταμένα στη θεωρία μέτρου (βλέπε απόδειξη ύπαρξης μη μετρήσιμου συνόλου). Μάλιστα ένα από τα κλασσικά σύνολα που κατασκευάζει κανείς στη θεωρία μέτρου δίνει και αντιπαράδειγμα σαν αυτό που χρειαζόμαστε: Αρκεί να θεωρήσουμε ένα σύνολο το οποίο είναι τέτοιο ώστε τόσο αυτό όσο και το συμπλήρωμά του να έχουν τομή θετικού μέτρου με κάθε ανοικτό διάστημα. Για μια τέτοια κατασκευή παίζει κανείς με σύνολα τύπου Cantor, με επαγωγή (απλή - όχι υπερπεπερασμένη) και με το θεώρημα Baire. Όταν πάρουμε ένα τέτοιο σύνολο τότε αυτό και το συμπλήρωμά του (ή για την ακρίβεια οι τομές τους με το $ \mathbb{R}\setminus\mathbb{Q}$) αποτελούν λύση του προβλήματος.

Όλα αυτά τα λέω σαν μια επεξήγηση της προηγούμενης ανάρτησης μου και απαντώντας στο φίλο Durandal. Παρ' όλ' αυτά η λύση που έδωσε ο NickNafplio είναι καταπληκτική και καθιστά τη χρήση των παραπάνω  εργαλείων περιττή. Μπορεί, δε, κανείς να παίξει με την ιδέα του και να βρεί παρόμοιες λύσεις. Για παράδειγμα μπορούμε να θεωρήσουμε από τη μιά τους άρρητους που το ανάπτυγμα τους καταλήγει σε μια ουρά από άπειρα τριάρια και άπειρα τεσσάρια, και από την άλλη τους υπόλοιπους άρρητους.

[quote="Durandal"]
[hide]Με μια επιπόλαια σκέψη θα πρότεινα just do it με transfinite induction, αλλά δεν έχω κάτσει να το δουλέψω με λεπτομέρεια. Κάτι τέτοιο έχεις κι εσύ στο μυαλό σου?[/hide]
[/quote]

Ναι αυτό έχω υπόψη. Για κάποιον εξοικοιωμένο με την μέθοδο η απόδειξη δεν παρουσιάζει ιδιαίτερα προβλήματα.

[quote="Jondrette"]Ναι, τα σύμπλοκα είναι τα cosets.[/quote]

Thanks. Αυτό φαντάστηκα ότι είναι, απλώς ήθελα να βεβαιωθώ :oops:

[quote]Παρ' όλ' αυτά η λύση που έδωσε ο NickNafplio είναι καταπληκτική και καθιστά τη χρήση των παραπάνω  εργαλείων περιττή. Μπορεί, δε, κανείς να παίξει με την ιδέα του και να βρεί παρόμοιες λύσεις. Για παράδειγμα μπορούμε να θεωρήσουμε από τη μιά τους άρρητους που το ανάπτυγμα τους καταλήγει σε μια ουρά από άπειρα τριάρια και άπειρα τεσσάρια, και από την άλλη τους υπόλοιπους άρρητους.[/quote]

Συμφωνώ και επαυξάνω, η λύση του Nick είναι πολύ όμορφη και, όπως λες κι εσύ, οδηγεί σε πολλές άλλες λύσεις (π.χ. μια παρόμοια ιδέα είναι να πάρεις από τη μία τους άρρητους που έχουν μόνο περιττά ψηφία, από την άλλη τους άρρητους που έχουν μόνο άρτια ψηφία και να κολλήσεις τους υπόλοιπους όπου και όπως θέλεις).

Nick, και πάλι well done :)

Cheerio,

Durandal 1707

[quote="Jondrette"]Όλα ωραία και καλά ως τώρα αλλά ας το χοντρύνουμε λίγο το παιχνίδι.

Να δειχτεί ότι υπάρχει υπεραριθμήσιμη(!) οικογένεια ξένων ανά δύο υποσυνόλων των αρρήτων έτσι ώστε καθένα από αυτά να έχει υπεραριθμήσιμη τομή με κάθε ανοικτό διάστημα.

(Μέχρι στιγμής το κάναμε για δύο. Ισχύει όμως για υπεραριθμήσιμα...)[/quote]

Η δεύτερη λύση που έδωσες γενικεύεται. (Χρειαζόμαστε μόνο να δείξουμε ότι κάθε σύνολο με πληθικότητα $ |\mathbb{R}|$ μπορεί να χωριστεί σε $ |\mathbb{R}|$ σύνολα με πληθικότητα $ |\mathbb{R}|$.

Ακριβώς!

Η επόμενη κατεύθυνση στην οποία θα μπορούσαμε να θεωρήσουμε το πρόβλημα αυτό είναι σε γενικούς μετρικούς χώρους. Ας πούμε λοιπόν ότι έχουμε ένα διαχωρίσιμο μετρικό χώρο $ X$ και ένα πυκνό υποσύνολο $ A$ το οποίο έχει την ιδιότητα ότι η τομή του με κάθε ανοικτό υποσύνολο του χώρου είναι υπεραριθμήσιμη. Μπορούμε να το χωρίσουμε σε α) δύο β) απειρα αριθμήσιμα γ) υπεραριθμήσιμα σύνολα με τις ίδιες ιδιότητες;

Δεν το έχω σκεφτεί αλλά οι λύσεις που δόθηκαν παραπάνω δε γενικεύονται.

Δεν ξέρω την απάντηση και το αφήνω ανοικτό για οποιονδήποτε ενδιαφέρεται αλλά θα επιχειρήσω να το λύσω αν βρω ώρα να σκεφτώ.

Το transfinite induction γενικεύεται.

Θα το αποδείξω για κάθε τοπολογικό χώρο Χ ο οποίος έχει αριθμήσιμη βάση. Ας αρχίσουμε από το (α). Παίρνουμε μια αριθμήσιμη βάση $ \mathcal{B}$ του Χ και μια καλή διάταξη του $ \mathcal{B} \times \omega_1$. Κοιτάμε το πρώτο στοιχείο του $ \mathcal{B} \times \omega_1$, έστω $ (U_1,\alpha_1)$. Υπάρχουν υπεραριθμήσιμα στοιχεία του $ Α$ στο $ U_1$. Αφαιρούμε δύο και τα βάζουμε στα δύο υπο κατασκευή σύνολα. Μετά προχωράμε στο δεύτερο στοιχείο, τρίτο, κτλ. Αν είμαστε στο $ \beta$ στοιχείο, επειδή $ |\mathcal{B} \times \omega_1| = |\omega_1|$, έχουμε μέχρι στιγμής αφαιρέσει από το $ A$ το πολύ αριθμήσιμο αριθμό στοιχεία. Άρα η επαγωγή μπορεί να συνεχιστεί. Επειδή κάθε $ U \in \mathcal{B}$ εμφανίζεται υπεραριθμήσιμες φορές σαν πρώτο στοιχείο του $ \mathcal{B} \times \omega_1$ κάθε ένα από τα δυο υποσύνολα του Α έχουν υπεραριθμήσιμη τομή με το $ U$.

Η ίδια απόδειξη δουλεύει και για το (β), αλλά το (γ) με δυσκόλεψε περισσότερο. Για το (γ) θα χρησιμοποιήσω ξανά transfinite induction και αυτό που ήδη απέδειξα για τα δύο σύνολα.

Στο βήμα $ \alpha$, θα διαμερίζουμε το "υπάρχον" σύνολο στα 3. Στο $ A_\alpha,B_\alpha$ και $ C_\alpha$ τα οποία θα έχουν τις ζητούμενες ιδιότητες. Τα $ A_\alpha$ δεν θα πειραχθούν ποτέ και θα μας δώσουν την ζητούμενη ιδιότητα. Το πως θα διαμερίσουμε εξαρτάται αν το $ \alpha$ είναι successor ή limit ordinal.
Αν είναι successor ordinal, π.χ. $ \alpha = \beta + 1$ χωρίζουμε το $ B_\beta$ στα τρία. Μένει να δούμε τι γίνεται αν το $ \alpha$ είναι limit ordinal. Τα $ A_\beta$ και $ B_\beta$ τα έχουμε ήδη χρησιμοποιήσει. Στο πρώτο limit ordinal (το $ \omega$) θα χρησιμοποιήσουμε το $ C_1$ χωρίζοντάς το στα $ A_\omega,B_\omega,C_\omega$. Στο δεύτερο limit ordinal θα χρησιμοποιήσουμε το $ C_2$ κτλ.

Όταν άρχισα να γράφω αυτή τη λύση νόμιζα πως όλα θα ήταν εντάξει επειδή πάντα θα είχα περίσσευμα κάποιο $ C$ να χρησιμοποιήσω. Επειδή όμως $ \omega \epsilon_0 = \epsilon_0$ μάλλον υπάρχει πρόβλημα.  :wallbash:

Πολύ ωραία. Θα χρησιμοποιήσω την ιδέα της κατασκευής σου για να δώσω τη λύση στην περίπτωση γ). Έχει ως εξής:

Ας υποθέσουμε ότι  $ X$ είναι ένας τοπολογικός χώρος με μια αριθμήσιμη βάση $ \mathcal{B}$, και  $ A$ είναι ένα πυκνό σύνολο του
οποίου η τομή με κάθε σύνολο της $ \mathcal{B}$ είναι υπεραριθμήσιμη. Θα δείξουμε ότι υπάρχει υπεραρίθμησιμη οικογένεια $ A_i$ από
ξένα ανά δύο υποσύνολα του $ A$ τα οποία έχουν την ίδια ιδιότητα με το $ A$.

Η ιδέα είναι η εξής: Αρκεί να βρούμε μια υπεραριθμήσιμη οικογένεια ξένων ανά δύο υποσυνόλων του τα οποία έχουν μη κενή τομή (απλά
αυτό!) με κάθε σύνολο της  $ \mathcal{B}$. Μπορείτε να δείτε γιατί; (υπόδειξη: κάθε άπειρος πληθάριθμος είναι ισοδύναμος με το τετράγωνό του...)

Έστω $ I$ ένα σύνολο πληθικότητας $ c_1$ (πρώτος υπεραριθμήσιμος πληθάριθμος). Διατάσσουμε καλώς το σύνολο  $ J \equal{} \mathcal{B}\times I$
έτσι ώστε να έχει ισομορφική διάταξη με τον πρώτο υπεραριθμήσιμο διατακτικό. Κατασκευάζουμε τα σύνολα
 $ A_i$ ως εξής. Αν $ (U,i)$ ένα στοιχείο του $ J$, τότε επιλέγουμε στοιχείο του  $ U$ που δεν έχει επιλεγεί ως τώρα και το βάζουμε
στο  $ A_i$. Μέχρι στιγμής έχουμε κάνει μόνο αριθμήσιμες επιλογές οπότε η επιλογή αυτή είναι δυνατή. Με την ολοκλήρωση της
υπερπερασμένης αυτής επαγωγής λαμβάνουμε μια υπεραριθμήσιμη οικογένεια ξένων ανά δύο υποσυνόλων of $ A$ καθένα  από τα οποία έχει
μη κενή τομή με οποιοδήποτε σύνολο της βάσης $ \mathcal{B}$, με άλλα λόγια αυτό που ήταν ανάγκη να δείξουμε.

[quote="Jondrette"]
Η ιδέα είναι η εξής: Αρκεί να βρούμε μια υπεραριθμήσιμη οικογένεια ξένων ανά δύο υποσυνόλων του τα οποία έχουν μη κενή τομή (απλά
αυτό!) με κάθε σύνολο της  $ \mathcal{B}$. Μπορείτε να δείτε γιατί; (πλάκα δεν έχει...)
[/quote]

Έσπαγα το μυαλό μου χθες βράδυ να καταλάβω γιατί αυτό είναι αρκετό. Σήμερα πρόσεξα πως και εγώ είχα προτείνει κάτι παρόμοιο λίγο πιο πάνω.  :rotfl:

[quote="Jondrette"]Διατάσσουμε καλώς το σύνολο  $ J \equal{} \mathcal{B}\times I$
έτσι ώστε να έχει ισομορφική διάταξη με τον πρώτο υπεραριθμήσιμο διατακτιό. [/quote]

Κάτι το οποίο χρησιμοποίησα και εγώ αλλά παρέλειψα να το γράψω.


Several drugstores received a shipment of 10 bottles, each containing 1000 pills. After shelving them, each pharmacist received a fax saying the pills in one bottle were 10 mg too heavy.

Knowing what a good pill should weigh, what is the fewest number of weighings necessary to guarantee determination (exclude luck) of the heavy bottle at every pharmacy? Explain the process used to arrive at your solution.

hold on. When you 'weigh' does that mean wieighing the wieght of one bottle, or two together?

can you explain this problem? it is hard and i vaguely remember seeing before that the answer was much less than i expected :)

What are you weighing the bottles with? If you used a scale, you'd only need a single weighing.

From what I think that you are trying to say, you need 2 weighings.
1. Two bottles of the same weight.
2. The heavier bottle with one of the bottles previously weighed.

It's kind of like the rigged coin problem where 11 coins are normal and one is light or heavy.  You have to figure the bogus in 3 weighs.  But here it's easier because it gives that it's heavier.
1.So weigh 333 vs 333 and if they weigh not equal the heavy one gets done into step two.  If they weigh same the other 334 not weighed are done into step two
2.Do the same thing with these only 111 and take the 111 or 112 into step three.
3.Same thing with 37, leaves37 or 38
4.Same thing with 12 leaves 12or ,or13, or14

6 steps for a bottle but first you have to weigh 3 steps for the 10 bottles
9 steps?  Am i close?

we're using a scale that tells us a weight, not a balance scale. Therefore to do it your way, you'd have to do it 12 times, which is obvioulsy worse than the 10 brute forcing it, however the actually solution is much more clever. (i'm not going to give it away)

[hide="Answer"]  The minimum weighings that will garuntee our finding of the heavier bottle is [b] 3 [/b].

There are 10 bottles.  Divide them into two equal groups of 5.  Then sub-divide each group into the following portions: 2, 2 and 1.  It will look like this:

                                                    10
                                                   /  \
                                                  /    \
                                                 /      \
                                                /        \
                                              5          5
                                             /|\        /|\
                                            / | \      / | \
                                           /  |  \    /  |  \
                                          2  2  1   2  2  1
                        Pool:           A  B      C  D

  First we compare Pool A with Pool B.  Say Pool A is heavier.  Now we divide Pool A into two individual bottles, comparing them 1 on 1, which ever is heavier is obviously the faulty bottle.  But suppose Pool A and pool B are the same, we then compare Pool C and Pool D.  If 1 of these is heavier than we repeat the process mentioned earlier.  If their the same, we go on and measure the two individual pills and which ever is heavier must be the faulty pill.  The worst case scenario is that in which pool A and B are the same, Pool C and D are the same.  And the faulty bottle is among one of the individual bottles.  Meaning the minimum weighing to garuntee us the faulty pill is  [b] 3[/b]

[/hide]

EDIT:
[b] TO READ THIS, YOU MUST QUOTE IT, AND THEN READ IT BECAUSE OF THE DIAGRAM.[/b]

[quote]we're using a scale that tells us a weight, not a balance scale.[/quote]
So your way doesn't apply, G-Unit.
Do you think you could just reclarify the problem? Nobody seems to know what you're talking about.  :?

I think we're using a balance scale, so G-UNIT's solution seems correct to me (exactly what i did too) .

[hide]

There are 5 bottles on one side, and 5 on the other.
You take the heaviest one and weigh 2 bottles and 2 bottles.
Then, since if you aren't lucky, you take the last one and wiegh it with one of the 4 you already made; since you know that they all weigh the same.

The least amount of weighing is $\framebox{3}$[/hide]


Find \[ \lim_{n \to \infty} n! \cdot \left( e - \sum_{k=0}^n \frac{1}{k!} \right) .  \] (hopefully, it exists)

(edit: he fixed it. ;) )

When I use his expression, I get the same recurrence as you had, and the term $a_1 = e-2 < 1$, so it is not at all obvious that the sequence is increasing. In fact, it seems like it will converge to 0, but I'm not certain. This is just based on the first few terms,

$a_0 = 1.72$
$a_1 = 0.72$
$a_2 = 0.44$
$a_3 = 0.32$
$a_4 = 0.28$

I really have no idea how you'd show it gets all the way to 0 though. Also it is not entirely obvious to me how to show that it won't start increasing again, though I suspect it won't.

Maple won't evaluate the limit, and it won't compute terms in the series past 11. $a_{11} = 0.0906...$ according to Maple. Anyone have Mathematica? I heard that's better. Or a proof would be good too...   ;)

Edit: I just realized that rounding error actually makes a huge contribution in calculations here. I tried starting with $e$ and just doing the recursive operations, and in that way Maple gave me $a_{11} = 0.0719...$, but this is obviously bad because it makes $a_{12}$ negative, and we know that's impossible.

It seems like guessing by calculation isn't an effective approach for this problem...  :blush:

Edit again :oops: : I forgot to mention, the first value for $a_{11}$ results in the sequence starting to increase a few terms later (though I had to proceed using the recursive relation, not using the explicit definition of the term). I'm not sure what to make of that (I still think the limit should be zero, but I've been known to say totally moronic stuff in the past...).

[quote="oooooh ya"]i hope this isnt dumb,

because $e = \sum_{n=0}^\infty \frac{1}{n!}$ would that make the limit 0?

$\sum_{n=1}^\infty \frac{1}{n!} = e - 1$[/quote]

Yes, but just because the limit of the expression in the inside goes to $0$ doesn't mean the limit of the expression on the inside [b]multiplied by $n!$[/b] will also go to $0$.

Read it again: he bounds $a_n$ from above by a nice expression.  That nice expression isn't equal to $a_n$.

Xev: Mathematica does indeed evaluate the limit correctly.

I really should get that -- Maple just doesn't cut it much of the time.

[quote="maokid7"]I understand all of the work its just that i do not see where

$\displaystyle a_n=\sum_{k=1}^{\infty}\frac{n!}{(n+k)!}$[/quote]

$e = \sum_{k=0}^{\infty} \frac{1}{k!}$
$e - \sum_{k=0}^{n} \frac{1}{k!} = \sum_{k=n+1}^{\infty} \frac{1}{k!} = \sum_{k=1}^{\infty} \frac{1}{(n+k)!}$
$a_n = n! \left( e - \sum_{k=0}^{n} \frac{1}{k!} \right) = \sum_{k=1}^{\infty} \frac{n!}{(n+k)!}$

I suppose I should've put this in the solution, but I thought it was obvious. :blush:

I know that the limit is indeed 0 from your proof but you did not prove that the limit could not be negative.
all you did was
$a_n \le b_n$
and $\lim_{n\to \infty}b_n=0$
so $a_{\infty} \le 0$ which could mean any negative #.
please tell me if i am missing something.

EDIT: o woops I just noticed that all of the terms in the sequence are always positive...silly me   :blush:  :oops:


integration
calculus
function
complex numbers
complex analysis

Prove that if $ a,b$ lies inside $ C(r)$ then

\[ \int _{C(r)} \sqrt{(z\minus{}a)(z\minus{}b)} dz \equal{} \minus{}\frac{\pi \cdot i (a\minus{}b)^2}{4}\]

for branch of square root which is equal to $ z\plus{}O(1)$ for big $ z$.

P.S. I now it's probably entirely trivial, one has to count residue at infinity, but I don't know how  :blush: , any explanation is welcomed.

Hey Megus,  I got so many holes in my understanding of this, I've been hesitant to reply even though I find it very interesting.  Here goes anyway:  If I consider a large circular contour around $ a$ and $ b$ as $ C(r)$ and then parameterize $ z$ as $ z(u)\equal{}1/u$, then the contour becomes $ H(r)$ with the reverse direction with radius $ 1/r$.  That is:

$ \oint_{C(r)}\equal{}\minus{}\oint_{H(r)}$

The integral then becomes:

$ \oint_{C(r)}\sqrt{(z\minus{}a)(z\minus{}b)}dz\equal{}
\minus{}\oint_{H(r)}\frac{\sqrt{(1\minus{}au)(1\minus{}bu)}}{u}(\minus{}\frac{1}{u^2})du$

or:

$ \oint_{H(r)}\frac{\sqrt{(1\minus{}au)(1\minus{}bu)}}{u^3}du$

I can now adjust $ r$ so that the contour, $ H(u)$ encircles only the pole at the origin and then use the Residue Theorem (it's a big now considering I haven't shown what I'm integrating over is analytic except at the origin but just for now, let's assume it is):

$ \oint_{H(r)}\frac{\sqrt{(1\minus{}au)(1\minus{}bu)}}{u^3}du\equal{}2\pi i\mathop\text{Res}\limits_{z\equal{}0}\left\{\frac{\sqrt{(1\minus{}au)(1\minus{}bu)}}{z^3}\right\}$

which reduces to:

$ 2\pi i\left(1/8(\minus{}a^2\plus{}2ab\minus{}b^2)\right)\equal{}\minus{}\frac{\pi i}{4}(a\minus{}b)^2$

Still doesn't address the question of just where on the square root multi-surface the integration is taking place, the necessity of $ a$ and $ b$ lying in the contour, and just what residue of the multi-residue square root function is being taken . . . see, holes.

Thanks for an answer, it sheds some light on the question. I'll try to ask some of my teachers about it.
And don't you have $ \int _{C(r)} \equal{} \int _H(r)$ ? (no minus)

I'm pretty sure it's a minus sign since if $ z \equal{} e^{it}$ with $ t$ going from $ 0$ to $ 2\pi$, then $ 1/z$ is $ e^{ \minus{} it}$ going in the opposite direction.  Therefore, in terms of the Residue Theorem, need the negative sign to indicate reverse direction.  Also, I can understand somewhat the surface being integrated over.  The first plot below is the imaginary multi-sheet for $ \sqrt {(z \minus{} 1)(z \plus{} 1)}$ and the yellow contour, where the integration (imaginary part) is taking place for this problem (I think).  The second plot is the chosen (imaginary) branch analytic except along the branch cut between $ \minus{} 1$ and $ 1$.  Real component is analogous to this.

[img]http://img480.imageshack.us/img480/9039/sqrtmultisheetvi8.jpg[/img]

Here's the Mathematica code to generate these plots:

[code]
a = 1;
b = -1;
f1[z_] := Sqrt[(z - a)*(z - b)];
f2[z_] := -Sqrt[(z - a)*(z - b)];
p1 = Plot3D[Evaluate[
       Im[f1[z] /. z -> x + I*y]], {x, 0, 5},
     {y, -5, 5}, PlotRange ->
       {{-3, 3}, {-3, 3}, {-5, 5}}]
p2 = Plot3D[Evaluate[
       Im[f2[z] /. z -> x + I*y]], {x, -5, 0},
     {y, -5, 5}, PlotRange ->
       {{-3, 3}, {-3, 3}, {-5, 5}}]
trace1 = ParametricPlot3D[
       {Re[z], Im[z], Im[f1[z]]} /.
         z -> 2*Exp[I*t], {t, -Pi/2, Pi/2},
       PlotStyle -> {Yellow, Thickness[
             0.02]}];
trace2 = ParametricPlot3D[
       {Re[z], Im[z], Im[f2[z]]} /.
         z -> 2*Exp[I*t], {t, Pi/2, 3*(Pi/2)},
       PlotStyle -> {Yellow, Thickness[
             0.02]}];
pic1 = Show[{p1, p2, trace1, trace2},
     BoxRatios -> {1, 1, 1}]
p1 = Plot3D[Evaluate[
       Im[f1[z] /. z -> x + I*y]], {x, -5, 5},
     {y, -5, 5}, PlotRange ->
       {{-3, 3}, {-3, 3}, {-5, 5}}]
p2 = Plot3D[Evaluate[
       Im[f2[z] /. z -> x + I*y]], {x, -5, 5},
     {y, -5, 5}, PlotRange ->
       {{-3, 3}, {-3, 3}, {-5, 5}}]
trace1 = ParametricPlot3D[
       {Re[z], Im[z], Im[f1[z]]} /.
         z -> 2*Exp[I*t], {t, -Pi/2, Pi/2},
       PlotStyle -> {Yellow, Thickness[
             0.02]}];
trace2 = ParametricPlot3D[
       {Re[z], Im[z], Im[f2[z]]} /.
         z -> 2*Exp[I*t], {t, Pi/2, 3*(Pi/2)},
       PlotStyle -> {Yellow, Thickness[
             0.02]}];
pic2 = Show[{p1, p2, trace1, trace2},
     BoxRatios -> {1, 1, 1}]
[/code]


linear algebra
matrix
superior algebra
superior algebra solved

We have $A,B \in M_{2}(R)$ 2 invertible matrixs.
Prove that:
          $Tr^2(A)+Tr^2(B)+Tr^2(AB)=Tr(A)Tr(B)Tr(AB)+Tr(ABA^{-1}B^{-1}).$

Consider $A=I_2$ and $B$ any invertible matrix

we get

$(1+1)^2+(trB)^2+(trB)^2=(1+1)tr(B)tr(B)+tr(I_2)$

$4+2(trB)^2=2(trB)^2+2$

wich is false


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry proposed

Two circles $ (O_1)$ and $ (O_2)$ touch externally at the point $C$ and internally at the points $A$ and $B$ respectively with another circle $(O)$. Suppose that the common tangent of $ (O_1)$ and $ (O_2)$ at $C$ meets $(O)$ at $P$ such that $PA=PB$. Prove that $PO$ is perpendicular to $AB$.

PA=PB  PA*PG=PH*PB  =>  PG=PH   => AG=BH
FG/OP=AG/AP=BH/BP  =>  FG=EH  => ⊙E ⊙F are congruence
⊙F is the reflection of ⊙E  W.R.T OP => AB ⊥ OP

Let the line $POC$ meet $(O)$ at $S$. From an inversion with centre $S$ and appropriate radius, we get two tangential circles $(K_1), (K_2)$(The centres need not map to the centre of the image circle) both homothetic to the original circles and their tangency point is $C'$. $(O)$ maps to a common tangent $l$ of $(K_1)$ and $(K_2)$ and $l$ intersects $(K_1)$ and $(K_2)$ at $A'$ and $B'$ respectively. $P'$ lies in the image of the line $SCP$ which means $SC'$ intersects the line $l$ at $P'$ and also, point $O'$ lies on $SC'P'$. It suffices to prove that $SP'$ is perpendicular to $A'B'$. Consider quadrilateral $K_1C'P'A'$ where $K_1C'=K_1A'$ and $\angle K_1A'P'=\angle K_1C'P'=90^{\circ}\Longrightarrow K_1C'P'A'$ is a square$\Longrightarrow SP'\parallel C'P'\parallel K_1A'\perp A'B'$ as required.

[color=#FF0000][Edit: Wrong as pointed out][/color]

Maybe I misunderstood something, but PA = PB and OA = OB mean that PO is the perpendicular bisector of AB, which would finish the problem.

gouthamphilomath, I actually don't understand your solution. It seems to me that you didn't use $PA = PB$ at all. Also, I don't understand your proof that that quadrilateral is a square. Finally, notice that $AB \perp OP$ in general is not equivalent to $A'B' \perp O'P'$, because lines may invert into circles (that is, $A'B'$ is not the inverse of $AB$)
Anyway, the problem as it is sounds weird because all the tangent circles turn out to be useless to the solution.

$\triangle PAB$ is isosceles, its circumcenter lies on the angle bisector of $\angle APB$, done.

Best regards,
sunken rock


Let f(x) = x^4 + ax^3 + bx^2 + cx + d. If the graph of y=f(x) intersects the graph of y=2x-1 at x=1,2,3, then the value of f(0) + f(4) is?

I know the answer is 30, but I don't know how to get there. Any tips on how to improve number theory skills, since I am poor at it? Thanks in advance!

f-(2x-1) is a polynomial with roots 1,2,3, so we can write

f-2x+1=(x-1)(x-2)(x-3)(x+b)

(since f is of degree 4 and has the x^4 term is 1)

Plugging in 0,4, b cancels and you get your answer.

I don't understand where you got f - (2x-1) from. Is this supposed to be f(x) - (2x-1)? Or is it a new polynomial f(2x-1)? Also, how can you assume that the intersections between f(x) and f - (2x-1) gives you the roots?

This one was posted recently, too -- try a search.  It's meant to be $ f(x) \minus{} (2x \minus{} 1)$: a value of $ x$ such that $ y \equal{} f(x)$ and $ y \equal{} g(x)$ intersect is the same thing as a zero of $ f(x) \minus{} g(x)$, and the latter object is easier to work with than the former.

Thank you very much JBL for clearing up the confusion. I understand how to work the problem perfectly, but whats the theorem or proof that states that a value of x where f(x) and g(x) intersect is the same as a zero of f(x) - g(x)?

Let $ x_0$ be a real number such that $ f(x_0)\equal{}g(x_0)$. Then $ f(x_0)\minus{}g(x_0)\equal{}0$, so $ x_0$ is a root $ f(x)\minus{}g(x)$ by definition. (When $ f(x)\minus{}g(x)$ is evaluated at $ x\equal{}x_0$, it equals $ 0$.)


algebra
polynomial
logarithms
algebra proposed

The denominator is natural logarithm of $ n\plus{}1$.

it looks like the converse of an IMC problem

[quote="Shishkin"]It is known that $ k$ and $ n$ are positive integers and \[ k \plus{} 1\leq\sqrt {\frac {n \plus{} 1}{\ln(n \plus{} 1)}}.\] Prove that there exists a polynomial $ P(x)$ of degree $ n$ with coefficients in the set $ \{0,1, \minus{} 1\}$ such that $ (x \minus{} 1)^{k}$ divides $ P(x)$.[/quote]
how to change the In(n+1) into a useful situation?


inequalities
inequalities proposed
algebra

Let $a$, $b$, $c$ be positive real numbers with $abc \leq a+b+c$. Show that \[ a^2 + b^2 + c^2 \geq \sqrt 3 abc. \]
[i]Cristinel Mortici, Romania[/i]

Why does it follows that $(ab+bc+ca)^2\geq 3abc(a+b+c)$

Because it is equivalent with $\frac 12 ( (ab-bc)^2 + (bc-ca)^2 + (ca-ab)^2 ) \geq 0$.

Or because from this well-know inequality \[ (x+y+z)^{2}\geq 3(xy+yz+zx), \] by putting $x=bc, y=ca, z=ab$ we get \[ ab+bc+ca\geq\sqrt{3abc(a+b+c)}. \]

[b] Problem. [/b]

       Let $a, b, c$ three non-negative real numbers such that $a+b+c\geq abc$. Prove that \[ a^{2}+b^{2}+c^{2}\geq abc\sqrt{3}. \]

 [i] Solution 1. [/i]

     Just like fuzzylogic said we have the followong inequalities \[ a^{2}+b^{2}+c^{2}\geq ab+bc+ca\geq\sqrt{3abc(a+b+c)}\geq abc\sqrt{3}. \]

  [i] Solution 2. [/i]

       Let's assume by contradiction that \[ a^{2}+b^{2}+c^{2} <abc\sqrt{3}. \] By applying Cauchy-Schwarz inequality, $3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}$ and the hipothesys $a+b+c\geq abc$ we get $abc<3\sqrt{3}$.

     On the other hand , by Am-Gm we have \[ abc\sqrt{3}>a^{2}+b^{2}+c^{2}\geq 3\sqrt [3] {a^{2}b^{2}c^{2}}. \]
      We get from here $abc>3\sqrt{3}$. We have a contradiction. :)

  [i] Solution 3. [/i]

     We have \[ a+b+c\geq abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 1. \] We shall prove a stronger inequality \[ ab+bc+ca\geq abc\sqrt{3}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\sqrt{3}. \]
     Now, let us denote $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$ and the problems becomes:

      "If $x, y, z$ are three non-negative real numbers such that $xy+yz+zx\geq 1$. Then the following holds:

                                                      \[ x+y+z\geq\sqrt{3}. \]
      But, this last problem follows immediately from this inequality \[ (x+y+z)^{2}\geq 3(xy+yz+zx). \]

$({a+b+c \over 3})^3 \geq abc$ and $a+b+c\geq abc$ $\rightarrow$ $(a+b+c)^4\geq 27a^2b^2c^2$. Also $\sqrt{{a^2+b^2+c^2\over 3}}\geq {a+b+c\over 3}$. Now the given inequlaity straightly folows from last two.

Hello. This is one of the first inequalities that I have ever proven. I think that my method is unnecessarily long. Would anyone mind giving some feedback as to how I can improve my ability in proving olympiad inequalities.

[b]Solution[/b]

First assume that $a \ge b \ge c$. Then we have also that $1/bc \ge 1/ac \ge 1/ab$. Hence by Chebyshev's inequality,

$\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac{(a+b+c)(1/bc + 1/ac +1/ab)}{3} \ge \frac{abc(1/bc + 1/ac +1/ab)}{3}$

This is equivalent to

$\frac{a^2 + b^2 + c^2}{a+b+c} \ge \frac{abc}{3}$ [b](1)[/b]

Now note the following. By AM-GM,

$(a^2 + b^2 + c^2)(a+b+c) \ge 9abc \Rightarrow \frac{a^2 + b^2 + c^2}{9abc} \ge \frac{1}{a+b+c}$ [b](2)[/b]

Combining (1) and (2) yields that

$\frac{(a^2 + b^2 +c^2)^2}{9abc} \ge \frac{a^2 + b^2 + c^2}{a+b+c} \ge \frac{abc}{3}$

$\Rightarrow (a^2 + b^2 +c^2)^2 \ge 3(abc)^2 \Rightarrow a^2 + b^2 +c^2 \ge \sqrt{3}abc$

EDIT: Woops. My proof can be much improved by using AM - QM instead of Chebyshev to show (1).

let $ab=\frac{1}{x},bc=\frac{1}{y},ca=\frac{1}{z}$
then $x+y+z=1$
it suffices to prove
$\sum x^2\ge\sqrt{3xyz}$
or $(\sum x^2)^2\ge 3xyz(x+y+z)$
trivialized by Muirhead.

i did not found muirhead  inequality anywhere.i know that it can be obatain by applying MG-MA,but not more.could you help me please?

1.Generalization
Let $ a,b,c,k,r>0$ be positive real numbers with \[{a+b+c}\geq{kabc}\] and
$r\leq\sqrt{3k}.$ Show that \[a^{2}+b^{2}+c^{2}\geq{rabc}.\]
C.M.A,Ion Bursuc,2006

2.Generalization
Let $ a,b,c,k,r>0$ be positive real numbers ,p,q,n positive integer ,$3q>n$,$pn>q$,
$\left(\frac{r}{3}\right)^{3p-1}\leq\left(\frac{k}{3}\right)^{3q-n} $with \[{a+b+c}\geq{k(abc)^p}.\] Show that \[a^{n}+b^{n}+c^{n}\geq{r(abc)^q}.\]
C.M.A,Ion Bursuc,2006

The following generalized inequality holds:
$$\sum_{i=1}^n a_i^2\geq\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i$$
For $n\geq2; a_1,\dots,a_n\ni\mathbb{R_+}$ and $\sum_{i=1}^n a_i\geq\prod_{i=1}^n a_i$

Proof:

Using a known inequality that follows from CBS or Chebyshev, Rearrange or other inequalities: $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2$
we get that $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2\geq\left(\prod_{i=1}^n a_i\right)^2$ so we need $\left(\prod_{i=1}^n a_i\right)^2\geq n\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i\Leftrightarrow\prod_{i=1}^n a_i\geq n\sqrt[n-1]{n}$

so we break the problem into two parts:

1) $\prod_{i=1}^n a_i\geq n\sqrt[n-1]{n}$

We can see that inequality holds and one can check that equality holds when $a_i=\sqrt[n-1]{n},\forall i=\overline{1,n}$

2)$\prod_{i=1}^n a_i<n\sqrt[n-1]{n}$
$$\prod_{i=1}^n a_i<n\sqrt[n-1]{n}=\sqrt[n-1]{n^n}=\left(\sqrt[n-1]{n}\right)^n\Leftrightarrow\sqrt[n]{\prod_{i=1}^n a_i}<\sqrt[n-1]{n}$$
We know that $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2; \sqrt[n]{\prod_{i=1}^n a_i}<\sqrt[n-1]{n}$ and by AM-GM: $\sum_{i=1}^n a_i\geq n\cdot\sqrt[n]{\prod_{i=1}^n a_i}$ so applying these three we get $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2\geq\left(n\cdot\sqrt[n]{\prod_{i=1}^n a_i}\right)^2=n^2\cdot\sqrt[n]{\left(\prod_{i=1}^n a_i\right)^2}=n\sqrt[n-1]{n^{n-1}}\cdot\frac{\prod_{i=1}^n a_i}{\sqrt[n]{\left(\prod_{i=1}^n a_i\right)^{n-2}}}>$

$>n\cdot\sqrt[n-1]{\frac{n^{n-1}}{n^{n-2}}}\prod_{i=1}^n a_i=n\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i\Rightarrow\sum_{i=1}^n a_i^2>\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i.$

This solution might seem hard to guess, but if you firstly do the case for $n=3$ as in the problem with this method, you can understand how the general method and form of the inequality was supposed, because the number of $a_i$'s and the coefficient in front of $\prod_{i=1}^n a_i$ depend on each other and all the inequalities used in the case $n=3$ are true in general case.

Edit: [b]WRONG SOLUTION[/b]
Solution by Muirhead, it is suffices to prove that\[a^2+b^2+c^2\geq \sqrt{3}(a+b+c)\] so this is equivalent to proving that \[(a+b+c)(a^2+b^2+c^2)^2\geq 3abc(a+b+c)^2\iff \sum_{cyc}a^5+\sum_{sym}a^4b+2\sum_{sym}a^3b^2+2\sum_{cyc}a^2b^2c\geq 3\sum_{cyc}a^3bc+6\sum_{cyc}a^2b^2c\] which follows from $2\sum_{sym}a^3b^2\geq 4\sum_{cyc}a^2b^2c$ and $\sum_{cyc}a^5+\sum_{sym}a^4b\geq \sum_{cyc}a^3bc+2\sum_{cyc}a^3bc=3\sum_{cyc}a^3bc$.

[color=#0ff][b]HERE's THE CORRECT ONE:[/b][/color]
\[a^2+b^2+c^2\ge ab+bc+ca \ge \sqrt{3abc (a+b+c)}\ge \sqrt{3} abc\] by the well-known inequality $(x+y+z)^2\ge 3(xy+yz+zx)$.

From C-S we know that $a^{2}+b^{2}+c^{2}\geq\frac{(a+b+c)^{2}}{3}.$ From $a+b+c\geq abc$ we have$\frac{(a+b+c)^{2}}{3}\geq\frac{(abc)^{2}}{3}$ $\frac{(abc)^{2}}{3}\geq  \sqrt{3}abc\Longleftrightarrow abc\geq3\sqrt{3}$ from AM-GM we have $\frac{a+b+c}{3}\geq\frac{abc}{3}\geq\sqrt[3]{abc}\implies abc\geq 3\sqrt{3}.$ Thus our proof is complete.

@Above
$abc\geq 3\sqrt3$ is not correct. Take $a=b=c=1$.
There is a mistake in $\frac{a+b+c}{3}\geq\frac{abc}{3}\geq\sqrt[3]{abc}\implies abc\geq 3\sqrt{3}$. How did you get it?

If $abc\geq 3\sqrt3$
\[\frac{a^2+b^2+c^2}{abc}\geq \frac{(a+b+c)^2}{3abc}\geq \frac{(abc)^2}{3abc}=\frac{abc}{3}\geq \sqrt 3\]
If $abc\leq 3\sqrt 3$
\[\frac{a^2+b^2+c^2}{abc}\geq \frac{3\sqrt[3]{a^2b^2c^2}}{abc}=\frac{3}{\sqrt[3]{abc}}\geq \frac{3}{\sqrt3}=\sqrt3\]


calculus
integration
logarithms
calculus computations

Find the following integral
$\int {\frac{x^{\frac{1}{2}}}{x(1+x^{\frac{1}{2}})}dx}$

Make the variable change $x = t^2$
:cool:


function
number theory
greatest common divisor
algebra unsolved
algebra

Find all positive integers $k$ for which there exists a function $f : N \longrightarrow Z$
satisfying $f(1997) = 1998$ and, for all $a, b$,
$f(ab) = f(a) + f(b) + kf(gcd(a, b))$.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=taiwan+function&t=18691

Pierre.


geometry algebra polynomial function Vieta

1) Find the shortest distance between the line y = 2/3x + 11 and the circle x^2 + y^2 - 4x +10y + 16 = 0

2) Let a, b, and c be the solutions of x^3 + 2x^2 - 6x +9 = 0.
    Find 1/a + 1/b + 1/c

Happy New Year!

basically how you do them...

[hide="1"]complete the square to get the circle in a good form, find the intersection of the line and a perpendicular line that goes from the center of the circle, and find center to intersection distance, and subtract the radius[/hide]

[hide="2"]say you have a new function that is f(1/x), then that is not a polynomial, but if you multiply by x^3, it would be, and this has roots that are reciprocals of the others, and use vieta's sums to find the answer from that poly[/hide]

[hide="No. 2"]Or you can use Vieta's sums directly: ${1 \over a}+{1 \over b}+{1 \over c}={{ab+ac+bc} \over {abc}}$.[/hide]

[quote="Hellstar"]So the answer for the second one $x^3 + 2x^2 - 6x +9$ would be $6$? (since the polynomial with roots $1/a, 1/b, 1/c$ is $9x^3-6x^2+2x+1$)[/quote]

that would be the correct polynomial, but you have to remember to divide by the lead coefficient, giving: [hide="answer"]$\frac{6}{9}=\frac{2}{3}$[/hide]


Cool!! -

I wanna see what a "math guy" announcing on ESPN looks like!

[quote="rrusczyk"]ESPN noon EST - the Countdown round of Nationals.

(Disclaimer: yours truly is an 'analyst' for the broadcast.)[/quote]

And was that you writing the solutions? I saw it, and I got on! :D (for a moment, about 10 seconds on there. I have approximately 14 minutes and 50 seconds of fame left, yay!)

Yah, that was my scrawl on the solutions (and me babbling during the contest).  They told me beforehand I'd get 15 seconds per problem.  Amazing how long some 15 seconds are and how short others are...

i'm proud to say that i know 3 national competitors. York Wu (11th place. so close to making it on countdown)  Erik Chen (8th place) and Andrew Chien (2nd place) i can't believe he didn't get 1st. owel. GO CALIFORNIA

[quote="jli"]is it still on?[/quote]

This happened in may of 2003. It's already 2005.


Let a,b,c>0 such that $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\equal{}1$  Prove that
$ (a^2\plus{}b^2\plus{}c^2)^2\plus{}abc(\sqrt{a^2\plus{}b^2\plus{}c^2})^3 \geq 4$

Let a,b,c>0 Prove that:
$ (a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)^2\plus{}(ab\plus{}bc\plus{}ca)^3 \geq 4abc(a\plus{}b\plus{}c)^3$

[quote="tran300491"]Let a,b,c>0 Prove that:
$ (a \plus{} b \plus{} c)^2(ab \plus{} bc \plus{} ca)^2 \plus{} (ab \plus{} bc \plus{} ca)^3 \geq 4abc(a \plus{} b \plus{} c)^3$[/quote]
It's equivalent to $ \sum_{cyc}(a^4b^2\plus{}a^4c^2\plus{}3a^3b^3\minus{}2a^4bc\minus{}a^3b^2c\minus{}a^3c^2b\minus{}a^2b^2c^2)\geq0,$ which obvious.

Let a,b,c>0 Prove that
$ 5(ab\plus{}bc\plus{}ca)^3 \plus{}27(abc)^2 \geq 18abc(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$

[quote="tran300491"]Let a,b,c>0 Prove that
$ 5(ab \plus{} bc \plus{} ca)^3 \plus{} 27(abc)^2 \geq 18abc(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$[/quote]
$ 5(ab\plus{}ac\plus{}bc)^3\plus{}45a^2b^2c^2\geq20abc(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)$ better. :wink:

Why
[quote="arqady"][quote="tran300491"]Let a,b,c>0 Prove that:
$ (a \plus{} b \plus{} c)^2(ab \plus{} bc \plus{} ca)^2 \plus{} (ab \plus{} bc \plus{} ca)^3 \geq 4abc(a \plus{} b \plus{} c)^3$[/quote]
It's equivalent to $ \sum_{cyc}(a^4b^2 \plus{} a^4c^2 \plus{} 3a^3b^3 \minus{} 2a^4bc \minus{} a^3b^2c \minus{} a^3c^2b \minus{} a^2b^2c^2)\geq0,$ which obvious.[/quote]
I don't use Muihard and Karamata inequality

[quote="arqady"]

$ 5(ab \plus{} ac \plus{} bc)^3 \plus{} 45a^2b^2c^2\geq20abc(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)$ better. :wink:[/quote]

Well Let $ ab\equal{}x$and $ bc\equal{}y$and $ ca\equal{}z$ so the problem is
\[ 5(x\plus{}y\plus{}z)^3 \plus{}45xyz \geq 20(xy\plus{}yz\plus{}zx)(x\plus{}y\plus{}z)\]
or
\[ 5\sum x^3 \plus{}15xyz \geq 5\sum x^2y \plus{}x^2z\]
wich is schur


algebra
polynomial
function
algebra unsolved

if p(2x,2y)=p(x,y) prove that polynomial is constant

The definition of a homogeneous function of two variables states that $p(x, y)$ is homogeneous with degree n if $p(tx, ty) = t^n p(x, y)$ for all x and y.

In this case, p is a polynomial and is homogenous with degree 0, which means that the sum of the exponents of each term must equal zero, which implies that the each exponent equals zero.

For, let $p(x, y) = \Sigma _{i, j = 0}^m c_{ij} x^i y^j$.

$p(tx, ty) = \Sigma _{i, j = 0}^m c_{ij} (tx)^i (ty)^j = \Sigma _{i, j = 0}^m c_{ij} t^{i+j}x^i y^j$.

For the above expression to have t^0 as a factor, i+j = 0, etc.

Except for when $x=0,y\neq0$ every point $(x,y)$ can be represented by $(x,mx)$.
Let $f_m(x)$ be the single-variable polynomial $p(x,mx)$, $m$ being constant for each polynomial.  So, $f_m(x)=f_m(2x)=f_m(4x)=f_m(8x)=...f_m(2^kx)=...$.
Because $f_m$, a single-variable polynomial, is constant an infinite number of times, it is always constant (for a particular value of m).  Therefore, except for $x=0,y\neq0$, $p(x,y)=p(x,mx)=f_m(x)=f_m(0)=p(0,0)$. Because $p$ is continuous, $p(x,y)=p(0,0)$ can be extended to all $x$ and $y$.


What's your favorite fruit?

See Fav. fruit 2 and Fav. fruit 3 if your favorite fruit is not here. The "other" option is in Fav. fruit 3. Please only vote in one of the polls.

This poll, Fav. fruit 1, is the thread for discussion.

Sorry if your favorite fruit is bunched up in a category and you don't like the other fruits in that category.

what is the point of making these polls? They're semi-useless and they don't even raise your post count.

Strawberry FTW. Shortcake, jam, jelly, plain, smoothie, syrup, YUM YUM!

ORANGE FTW!!! :icecream:

And, where is peach! I also love it!!!

uhhh... mandarin is a language. The right name is tangerine :wink:

[quote="pacman2812"]
uhhh... mandarin is a language. The right name is tangerine :wink:[/quote]

You sure mandarin isn't another kind of fruit?  :maybe: ...

[quote]And, where is peach! I also love it!!! [/quote]

Peach is on poll number 2. But if you've voted here, don't vote there.

[quote="leoxnlin"][quote="pacman2812"]
uhhh... mandarin is a language. The right name is tangerine :wink:[/quote]

You sure mandarin isn't another kind of fruit?  :maybe: ...[/quote]

[url=http://en.wikipedia.org/wiki/Mandarin_orange]Wikipedia has all of the answers...[/url]

...what a huge poll... I voted banana

lychee!!!!! they're yummy :D  lychee popsicles r amazing too!! (hmm me thinks there is some in the freezer right now...) :P

I like so many fruits it isn't funny...So I can't recall all of them.

I like Many fruits. hm.. kiwi, banana, pear, ... but My favourite  fruit is apple . so I put apple :)

i like pretty much all fruits. but my fav is blueberry, and it's been so for as long as i can remember.


Hi everybody,

I started to use bibtex to organize the references in my dissertation today.

I really like the standard natbib style that looks a little like this:

A. Alesina, ... Journal of Public Economics ..., 2004. [7, 24]

BUT: Is there any way that the pages are NOT given at the end of the list entry? I'd like to see this:

A. Alesina, ... Journal of Public Economics ..., 2004. [7, 24]

Best regards,

Christian

[quote="McLaine"]I really like the standard natbib style that looks a little like this:
A. Alesina, ... Journal of Public Economics ..., 2004. [7, 24][/quote]
[quote="McLaine"]
BUT: Is there any way that the pages are NOT given at the end of the list entry? I'd like to see this:

A. Alesina, ... Journal of Public Economics ..., 2004. [7, 24][/quote]They look exactly the same.
Where do the page numbers come from? What "standard" style? You will need to give a [b]complete[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] to show what you get.

Sorry, for the confusion:

I want to see

A. Alesina, ... Journal of Public Economics ..., 2004.

instead of

A. Alesina, ... Journal of Public Economics ..., 2004. [7, 24],

i.e., i do not want the pages on which I cited Alesina ... to appear in the references.

My settings are:

\usepackage[sectionbib,round]{natbib}

\begin{document}
\bibliographystyle{plainnat}

...

\citet{Akerlof1990}

.....

\bibliography{LiteraturDatenbank}

\end{document}

You don't seem to have read link in my request for a [b][color=red]complete[/color][/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url]. The code you have posted is not compilable and even refers to an unknown database [i]LiteraturDatenbank[/i].
natbib doesn't normally put page numbers in so unless you give a proper compilable example there's no way to analyse the problem.


At a holiday camp there were 38 holiday makers. Prove that either there must be atleast 19 people who are mutual friends, or atleast 19 people who are mutual strangers. :wink:

:huh:

Wouldn't this imply $ R(19,19) \leq 38$? (see http://mathworld.wolfram.com/RamseyNumber.html) This is very not true. Maybe you meant $ 6$ and $ 3$ instead of $ 38$ and $ 19$?

Do not use the concept of Ramsey numbers, as this is simplified case where m=n.

I was trying to point out that your problem is not correct.

Ok, Assume that 18 people know each other. This would mean that the number of people who dont know each other would be 20.
Conversely, assume 18 people do not know each other. Then there must be 20 people who would know each other.
Either way you can always have atleast 19 people who either know each other or don't.

We have different definitions of mutual friends/mutual strangers, apparently. When I read $ k$ mutual friends I mean that any two people from that group of $ k$ are friends.

Your argument can be applied for $ 8$ people and there being $ 4$ mutual friends or $ 4$ mutual strangers. Here's a counterexample to that.

A knows B C D H
B knows A D F G
C knows A F G
D knows A B E
E knows D G H
F knows B C H
G knows B C E H
H knows A E F G

It's not almost impossible, it's completely impossible because it [i]isn't true.[/i]

You still haven't provided your definition of "mutual friends," which looks like it's something like "connected subgraph."  Usually this is taken to mean [i]complete[/i] subgraph, or [url=http://en.wikipedia.org/wiki/Clique_(graph_theory)]clique[/url].

As I see it there are two definitions for mutual friends (the opposite applies for mutual strangers);
[b]Definition One:[/b] Of the $ 38$ holiday makers, any two who both consider eachother to be freinds are mutual friends. We must show that there are $ 19$ people who have/dont have a friends.

The question is true for this definition because there are $ \binom{38}{2}=703$ pairs of people.
Each pair forms an independant group, that must be either friends of strangers.
So there must be atleat $ \frac{703+1}{2}=352$ pairs of either mutual friends or stangers.
The smallest number of people that could make 352 pairs of friends/stangers is $ 28$ (because $ \binom{27}{2} = 351)$

[b]Definition Two:[/b] There is a group of $ k$ holiday makers $ (k\ge 2)$ that each consider everybody else in the group a friend. We must show that there must be a group of atleast $ 19$

The question is not true for this definition. We can split the $ 38$ holiday makers into $ 6$ families of $ 6$ people (all of whom are friends) and a couple of two.
The largest group of mutual stangers is made by taking one person from each family\couple to make a group of $ 7$ mutual stangers.

[quote="KRIS17"]It would be almost impossible to solve my problem with your interpretation. I wouldn't ask such problems here.[/quote]

his is the default interpretation if you have any reasonable sort of math background, sorry


Princeton
college
calculus
analytic geometry
graphing lines
slope
Harvard

I'm a senior, and I will be attending Princeton University this fall.  I have signed up to take economics (I am interested in becoming a business major, so this might be a good starting block), but this year, I am having a lot of trouble with Calculus.  Obviously, since economics involves Calculus, I see this as a problem.

I don't want to be completely stranded when I get to college, so I was wondering if you guys could suggest any books that would help me with Calculus.  I have tried to read through my textbook, but unfortunately, I still have trouble graping concepts.  My friends say that this might be because I'm not doing enough practice problems.  Are there any books with specific practice problems in them that could help?  Any other tips on what I can do to become strong in math this summer?

Thanks!

I used Edwards and Penney, which I liked a bit more than Stewart.  It's pretty good as far as the typical plug n chug calculus textbook is concerned.

Welcome Justice_Scalia. I think that you should have a look in this forum: [url=http://www.mathlinks.ro/Forum/index.php?f=66]Calculus Forum[/url] and also maybe you might want to have a look in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20875]princeton vs harvard[/url] topic, since you'll be going to Princeton :P :)

What is the name of the calculus textbook you are using now? What are the economics textbooks you have at hand?

[quote="hello"]There isnt any calculus in introductory level classes like the first courses in microeconomics and macroeconomics. I am taking first course in micro and all you need is slope and basic algebra 1. I think econ mainly uses statistical ideas and calculus is needed only in significantly more advanced econ classes I believe.[/quote]

Uhh, I suspect that this is at least partially dependent on what university you go to, as most universities try to teach to the level of the students in their classes, and I suspect that the vast majority of incoming freshmen at a school like Princeton have some experience with calclulus.

It's worth noting that at my school, the first courses in economics (micro/macro) require no mathematical background, whereas the second courses in micro/macro do have a calculus prerequisite.

[quote="blahblahblah"]Uhh, I suspect that this is at least partially dependent on what university you go to, as most universities try to teach to the level of the students in their classes, and I suspect that the vast majority of incoming freshmen at a school like Princeton have some experience with calclulus.[/quote]

I've never heard of an intro ec. class that required knowledge of calculus.  They usually are "intro theory" courses, big on ideas but low on calculations.  Harvard's situation is the same as yours: the intro ec class doesn't require any calc, but the second-level course does.  If you want to major in economics or take many classes, mathematical capacity will become significant, but not if you're only planning on taking one class.

Well, we use Larson, Hostetler, Edwards.  I'm not sure how well-written it is compared to other textbooks, but it seems to lack applications of concepts.  For Macro/Micro, we use McConnell.  I am self-studying both Micro/Macro, so I have not gotten far enough to know how much Calculus is needed.  Up till now we have only learned basic concepts and graphs.  A finance professor at the local university told me that for finance, calculus is a foundation.  I'm not sure about the specific differences between economics and finance (Princeton only offers an economics concentration in any case), but I am pretty sure Calculus will be useful.  Right now, besides looking for other books, I am trying to complete all the odd problems in my Calculus book.

[quote="Justice_Scalia"]Right now, besides looking for other books, I am trying to complete all the odd problems in my Calculus book.[/quote]I suggested a forum you should take a look into, don't forget about it ;)

JBL:

[quote]I've never heard of an intro ec. class that required knowledge of calculus.[/quote]

Several decades ago, Harvard's Ec 10 intro class had about 30 sections each year.  28 of those sections did not use calculus.  The other two sections, however, DID use calculus.   They were known as the "math sections" of Ec 10,   Math 21 or higher was a corequisite for enrollment in a math section of ec 10.  (Larry Summers taught one of those, by the way.)

Some Harvard  teaching fellows had experience teaching both regular and math sections of ec 10.   Some of those TFs  felt that even mathematically sophisticated students were better off learning the basic concepts of economics in a nonmathematical way.  The idea was that students who relied on mechnanical manipulation of mathematical formulas to solve problems in introductory economics often didn't think hard enough about the underlying concepts and just "plugged and chugged" their way to the answers.

Of course, some of the most mathematically sophisticated Harvard undergrads in those days just directly took the graduate economic theory sequence for economics grad students (2010) bypassing the intro undergraduate economics class altogether.   That course heavily relied on multivariate calculus, of course.   I've heard that in some years the strongest students in that class were undergrads rather than the econ grad students for whom it was designed.  From reading the current Harvard catalog, it looks like this route is no longer accessible to undergrads.  (Cynics might say that this is to keep the undergrads from embarrassing the grad students!)

Caltech's intro economics class routinely uses calculus, because the Caltech core requirements are such that the professor can expect a high level of mathematical competence from all students.

I don't know about MIT's intro economics class, but it would seem reasonable to me that their instructors might well do the same.

Undergraduates can still bypass Ec 10 and take Ec 1010 or 1011 or something like that, and those courses use calculus.  Ec 10 is officially a pre-req for those courses, however.  At MIT, I was under the impression that their intro economics course counted as a humanities requirement, making me doubt that they would require knowledge of calculus.

[quote]At MIT, I was under the impression that their intro economics course counted as a humanities requirement, making me doubt that they would require knowledge of calculus.[/quote]

I'm not so sure I'd buy this exact line of reasoning, since MIT's higher level economics courses also count towards humanities requirements (actually, there is no "humanities requirement," per se at MIT; it is, more properly, "humanities & social sciences" requirements--HASS in MIT-lingo.)   And I'm sure that some of those higher level econ courses use calculus, so the fact that a course counts towards the HASS requirement doesn't preclude that it might require the use of calculus.

Leaving that line of reasoning aside, MIT's website for their introductory microeconomics class 14.01 states that the textbook they use is Pindyck and Rubinfeld, which is a non-calculus approach to economics.

So it appears that MIT's intro economics class indeed does not take advantage of the fact that all students can be assumed to have calculus as (at least) a co-requisite.

Then again, the choice of textbook can be deceptive--Caltech's syllabus for intro econ indicates that they use a nonmathematical text but that lectures are conducted at a higher level of mathematical sophistication.

http://www.hss.caltech.edu/~wilkie/ec11/syllabus.pdf

I wrote:

[quote]MIT's website for their introductory microeconomics class 14.01 states that the textbook they use is Pindyck and Rubinfeld, which is a non-calculus approach to economics.

So it appears that MIT's intro economics class indeed does not take advantage of the fact that all students can be assumed to have calculus as (at least) a co-requisite. [/quote]

Actuallly, on further investigation, it appears that the Pindyck and Rubinfeld textbook DOES use calculus in the footnotes, though not in the body of the text itself.

(The economics of textbook publishing likely requires authors to keep calculus to the footnotes in order to broaden the appeal.  This gives faculty at different schools the flexibility to use calculus or not, as they see fit. )

It's also interesting to note that Pindyck & Rubinfeld is an intermediate textbook, though MIT uses it in its INTRO course.  (Pindyck is an MIT prof, by the way.)


An acute-angled triangle with sides $a, b, c$ is given.
Show that
$\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\geq\frac{a+b+c}{abc}$

 …!!!!

 :)

multiply by 2abc both sides

it remains to show $\sum \frac{a}{\cos A} \ge 2(a+b+c)$

The "larger side contains the larger angle", and the cosine function on [0, pi/2] is decreasing, so that $(a,b,c)$ and $(\frac{1}{\cos A} \dots)$ are similarily sorted.  By chebyshev, it remains to prove $\sum \frac{1}{\cos A} \ge 6$.

The function $f(x) = \frac{1}{\cos x}$ is increasing and concave up on the interval $[0 , \pi /2]$.  By Jensen, the result is proved.


am gresit mai devreme enuntul.incearca acum +problema asta:

Fiind the primitive of ln(1+sqrt(3)*tg x) supposing this function is well defined.


Let $x_1,...,x_n$ be positive reals such that $\frac{1}{1+x_1}+...+\frac{1}{1+x_n}=\frac{n}{2}$. Find all possible values of the expression $\sum_{i,j=1,n}{\frac{1}{x_i+x_j}}$.

can this useful?
let x_{i}=(tan a_{i}) 2
so far I think it quite hard

____________
I will be out in next ten days

I don't think you are right,because S <= 2 for n=2
can you inspect for \displaystyle sum(1+a_i)(1+a_j)(1-a_ia_j)=n^2-\sum{a_i^2a_j^2}\le n^2

Why $a_i$ lies in  $[-1,+1]$ ?

Cannot it be more negative as $x_i$ is positive ?

I am so stupid :(  but now your proof is clear.

A very nice proof. Congratulations. :)

I have never solved a Harazi inequality in my life.

I will be happy if a day I will try to solve one, only one!!!! :blush:


function
real analysis
real analysis unsolved

$F$ a finite subset of $\mathbb{Z}^d$, $B$ the boundary of $F$, that is the set of $x \in \mathbb{Z}^d- \{ F \}$ such that there is some $y \in F$ satisfying $|x-y|=1$. A function $f$ defined over $F \cup B$ is harmonic if for every $x \in F$, $f(x)$ is equal to the mean of $f$ on the $2d$ neighbors of $x$.
Prove that there exists a unique harmonic function defined over $F \cup B$ if the values of $f$ are imposed on $B$.

One could say that the existence part implies the uniqueness part, [i]and vice-versa[/i] :), even though that doesn't make that much sense.

Write down the conditions for a function to be harminic in every point of $F$. You'll get a system $Ax=b\ (*)$ with $|F|$ equations and unknowns, and with the components of $b$ equal to various linear combinations of the values of $f$ in the points of $B$ (as we change the values we assign to $f$ on $B$, only $b$ changes in $(*)$). If we prove that $A$ is invertible, we're done. On the other hand, the invertibility of $A$ follows if we prove that when $(*)$ has a solution, it is unique.

We can take $f|_B\equiv 0$. Of course, $b=0$ in this case, so $(*)$ now has the trivial solution $x=0$. All we want now is to prove that it's unique. This is done pretty much like we proved the uniqueness of the function in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=468314#p468314]this other problem[/url]. It's even easier in this case, since we don't need any graph theoretic considerations.


[b]11th class variant:[/b]

Let a, b, c be three positive numbers; prove the inequality

$a^{3}+b^{3}+c^{3}\geq a^{2}b+b^{2}c+c^{2}a$.

When does equality hold?

[b]12th/13th class variant:[/b]

Let a, b, c, d be four positive numbers; prove the inequality

$a^{3}+b^{3}+c^{3}+d^{3}\geq a^{2}b+b^{2}c+c^{2}d+d^{2}a$

When does equality hold?

[b]My addition:[/b]

Generalize: For any n positive numbers $a_{1},$ $a_{2},$ $...,$ $a_{n}$, the inequality

$a_{1}^{3}+a_{2}^{3}+...+a_{n}^{3}\geq
a_{1}^{2}a_{2}+a_{2}^{2}a_{3}+...+a_{n}^{2}a_{1}$

holds with equality only for $a_{1}=a_{2}=...=a_{n}$.

Prove this without using Rearrangement.

  Darij

This one can be done by Cauchy.
[tex] 2a_1^3+a_2^3 \geq 3a_1^2a_2 [/tex]
[tex] 2a_2^3+a_3^3 \geq 3a_2^2a_3 [/tex]
[tex] ... [/tex]
[tex] 2a_n^3+a_1^3 \geq 3a_n^2a_1 [/tex]
Plus them together and we have done.The equality occurs when $ a_1=...=a_n $ :)

I don't know why there is sometimes added constraints as "Solve it without rearrangement"...
These problems are typical rearrangement inequalities...

Pierre.

[quote="pbornsztein"]I don't know why there is sometimes added constraints as "Solve it without rearrangement"...[/quote]

Well, this was added by me, not by the problem proposers at the olympiad. The problem is really trivial with rearrangement, but I wanted to hear another proof. And Anh Cuong has found a very nice one. It is always better to have two proofs than one!

By the way, there were some people on the olympiad who solved the inequality using rearrangement, but most of them didn't get the complete credit because they hadn't written down the equality case properly.

  Darij


Anyone have any cool false proofs they would like to share?

Fierytycoon

[quote="Rep123max"]Am I the only person who hates AXIOMMSs? We are accepting something that could definetly be wrong. Our Counting System is an AXIOM (postulate)!![/quote]

Think of your basic beliefs as your base.  If you don't have any beliefs, you can't build anything off those (non)beliefs, just like you can't build a tower without a base.  And if you can't build anything off your beliefs, then you'll never figure anything new out.  How could Einstein or Euler or anyone else have figured out all the things they figured out if they had decided that 1 + 1 doesn't equal two?  I'd rather figure out interesting stuff based on something I have to assume is true rather than sit on my butt and wonder all day what if we used the numeral "3" in place of the "2" and vice versa...

Let \sum_{k=0}^{\infty} x^k = 1/(1-x). For x = 2 we have -1 = 2 + 4 + 8 + 16 + ..., for x = -1 gives 2 = 1 - 1 + 1 - 1 + ..., x = 1/10 gives 10/9 = 1 + 1/10 + 1/100 + ... Cool huh? If you can't figure it out then look below.

[hide]

The given sum only converges for -1 < x < 1. Consider (x^n - 1)/(x-1) = x^(n-1) + ... + 1 which for -1 < x < 1 we can write x = 1/a where |a| >= 2 for integers. Thus (1/a^n - 1)/(1/a - 1) which clearly has a limit of 1/(x-1) when n goes to infinity.

[/hide][/hide]

Have a look under the Paradox thread, theres been heaps posted there.

Another interesting thing that I saw today:
Suppose you have a box with 30 red balls, and 60 others that are either yellow or black.
I pick out a ball at random. You have to choose between one of these options:

1) You win $100 if the ball is red
2) You win $100 if the ball is yellow

Which of the two would you pick? (no trick here, this isn't the weird bit)

Now what if you had to choose between

3) You win $100 if the ball is yellow or black
4) You win $100 if the ball is red or black

Which would you pick?

Most people would pick 1 out of the first two, and 3 out of the second two. Which means are willing to bet there are more red then yellow, and more yellow then red..

Yes, that's probably because of our personalities.  If we are "resourceful", then we'll take it the safe way, which is not necessarily the way with the most profit.

-interesting_move

[color=cyan]Three guys walk into a hotel.  They're sharing a room, and the manager charges them $30 dollars for the night.  They each give him ten dollars.  After they've gone up to their room, the manager, who is an honest guy, realizes that he was only supposed to charge them $25.  So he sends the bellboy up with the extra $5.  Now, the bellboy is walking up the stairs, and he thinks, \"This is silly.  They don't know they were overcharged, and they can't split the $5 between the three of them anyhow.  I'll just pocket 2 and give them the other 3, in case the manager asks them about it."  So he goes and gives $3 to the guys in the room and keeps $2.

At this point, out of the original $30, each of the guys has paid $9, which makes $27, and the bellboy has $2.  But that only makes $29.  So what happened to the other dollar?[/color]

Hehe, just a few days ago, I heard one just like it:

Three people goes to a restaurant for lunch.  They each pay 5 dollars to the waitress.  The waitress gives the money to the manager, who says that they can have a discount today.  The manager takes only 10 dollars and gives the extra 5 back to the waitress. She then keeps two for herself and returns the rest to the three people (1 person gets $1 back each).  But that means everybody payed only 4 dollars and the waitress kept 2 making a total of 4*3+2=14.  What happened to the other dollar that they all payed?

-interesting_move

That probably the most famous puzzle ever. Basically (using JBLs puzzle) the thing is that.. [hide] each guy paid 27, the bellboy has 2, when is 2 we don't add it.[/hide]

This thread got a little silly, particularly considering the section it's in (Challenge Others), so I cleaned it up...

Sorry, Mr. Rusczyk.

Here's a pretty famous false proof:

P. Halsey published this false proof in 1959, as an example of a convincing false proof.

False proof: Here angle A = angle C, and AB = DC. We draw BE perpendicular to AD at E, and DF perpendicular to BC at F. Draw BD. Triangles ABE and CDF are congruent, so BE = DF and AE = CF. Then triangle BDE is congruent to triangle DBF, so ED = FB. So AE+ED = BF+FC or AD = BC. Since AB = DC and AD = BC, the quadrilateral is a parallelogram. Can you find the flaw in the proof?

Personally, I like this one better:

Every natural number can be unambiguously described in fourteen words or less.
The proof of this statement is as follows:
Step 1: Suppose there is some natural number which cannot be unambiguously described in fourteen words or less.
Step 2: Then there must be a smallest such number. Let's call it n.
Step 3: But now n is "the smallest natural number that cannot be unambiguously described in fourteen words or less".
Step 4: This is a complete and unambiguous description of n in fourteen words, contradicting the fact that n was supposed not to have such a description.
Step 5: Since the assumption (step 1) of the existence of a natural number that cannot be unambiguously described in fourteen words or less led to a contradiction, it must be an incorrect assumption.
Step 6: Therefore, all natural numbers can be unambiguously described in fourteen words or less.

http://www.everything2.com/index.pl?node_id=961354
This site gives some interesting methods of mathematical 'proof'.

Enjoy.

[quote="Rep123max"]if by false you mean one slight mistake i have a good one:
you probably already know it, its 2=1[/quote]

http://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html This has a more subtle version of the proof that uses complex numbers.


Let $ f: [0,\infty) \to \mathbb{R}$ a continuous function.
Define $ g: [0,\infty) \to \mathbb{R},g(x) \equal{} \sup \{f(t)\mid 0 \le t \le x\}$
Prove that $ g$ is a continuous function.

That's a classical problem.. Any way..
Let $ P$ be a positive real number. $ f$ is continuous on $ [0,P]$, so it is uniformly continuous : $ \forall _{\epsilon >0}, \exists _{\alpha},\  \forall_{u,v\in[0,P]}, |u\minus{}v|<\alpha \Longrightarrow |f(u)\minus{}f(v)| <\epsilon$
Take for ex $ v > u$ , so $ g(v)\equal{}max(g(u),m)$ , where $ m\equal{}\sup_{t\in [u,v]}(f(t))$ .
Because of $ f$ is continuous, there is $ x_m\in [u,v]$ such that $ m\equal{}f(x_m)$ . Manifestly $ |v\minus{}u|<\alpha \Longrightarrow |x_m\minus{}u| < \alpha\Longrightarrow |f(x_m)\minus{}f(u)| < \epsilon$
Two cases are possible :
(1) $ g(v)\equal{}g(u)$ so $ |g(v)\minus{}g(u)|<\epsilon$
(2) $ g(v)\equal{}m\equal{}f(x_m)$, and from $ f(u) \leq g(u) \leq g(v)\equal{}f(x_m)$  I deduce $ |g(v)\minus{}g(u)|\leq |f(x_m\minus{}f(u)| < \epsilon$

Manifestly $ g$ is uniformly continuous on $ [0,P]$, so it's continuous .. That's all..
:cool:


A boy, a girl and a dog go for a 38 mile journey.  The girl takes 16 minutes to walk a mile. The boy takes 15 minutes to walk a mile and the dog can trot a mile in 10 minutes.  They also have a bicycle which only one of them (including the dog!) can use at a time.  When riding, the girl can travel a mile in 6 minutes, the boy can travel a mile in 5 minutes and the dog can pedal a mile in 4 minutes (what a marvelous dog!).  What is the shortest time in which all three can complete the trip?

If the boy rides the bike for 6 miles, or 30 minutes and then gives it to the girl, the girl cannot use it right away; she must first catch up to the 6 mile mark, where the bike was left, right?

That's right.  The bike can be left on the road.

I think I might have a solution...

[hide="Solution"]
Answer = 399

The last person to finish determines the time the journey took, so we want the three people to finish as close as possible (preferably at the same time). I will go under the assumption that they do finish at the same time.

Let:
b = The number of miles that the boy spent on the bike
g = The number of miles that the girl spent on the bike
d = The number of miles that the dog spent on the bike

Time for the girl to finish: 16(38-g) + 6g
Time for the boy to finish: 15(38-b) + 5b
Time for the dog to finish: 10(38-d) + 4d

So we have the equations:
16(38-g) + 6g = 15(38-b) + 5b
16(38-g) + 6g = 10(38-d) + 4d
g + b + d = 38

Solving, we get a negative mileage for the dog. That is impossible so the dog's mileage on the bike must have been 0 and the dog must have finished first.

So, now we have these equations:
16(38-g) + 6g = 15(38-b) + 5b
g + b = 38

Solving, we see that g = 20.9 and b = 17.1.

So the girl and the boy finish in 399 minutes while the dog finishes in 380 minutes. This is achieved when:

Dog walks the entire way. The Girl starts on the bike and rides for 20.9 miles and walks the rest of the way. The boy walks for the first 20.9 miles and then picks up the bike and rides the rest of the way.
[/hide]

I tried to solve it the same way, but didn't see that the dog's time on the bike was negative  :oops:

Make the dog bike the entire way, with the boy and girl sitting on top.  If he gets tired, let the boy bike :D

Anything related to math can appear at Math Zoom.

[quote="PowerOfPi"]Is this a kind of question that would appear at MathZoom?[/quote]

one of the types

Let me give a hint:

[hide]The optimal time is 394 minutes.  Now you need to figure out how this is done.[/hide]

[hide="Here is the solution"]
The answer is 394 minutes.
Assume the girl rides the bike for X miles, then runs for 38-X miles.
Assume the boy runs first for Y miles, then rides for 38-Y miles.
The dog runs first for X miles, picks up the bike and rides BACK X-Y mile. Leaves the bike there (the boy hasn't arrived yet), and runs to the end for 38-Y miles.
The shortest time is to assure everyone get to the end at the same time.

girl: T=6X+16(38-X)
boy: T=15Y+5(38-Y)
dog: T=10X+4(X-Y)+10(38-Y)

X=21.4miles, Y=20.4miles, and T=394min
So the dog needs to ride back 1 mile and leaves the bike there for boy to ride later. The dog runs 39 miles (390min) plus rides 1 mile (4min). Good dog!  :)
When you got the negative distance or time for the dog riding the bicycle, that should have told you something  :D [/hide]


geometry
geometric transformation
reflection
calculus
Stanford
college

I just have two questions regarding AP results..

1. Out of numerous AP exams (say more than 10 tests), how much
would one bad (less than 4 or 3 and it's World History for me) score affect u in getting into colleges? Is it just matter of getting the credits depending on your score?

2. Once i got my score and unsatisfied with it, can I retake it and replace the bad one with better one?

Thanks,

I think the standard response to question 1 is that AP scores don't generally count against you.  Getting a 2 or 3 on a history test doesn't seem like it would be worse than not taking the test at all.  There's not much you can do after the fact anyway, so don't worry too much.

About question 2, I don't know if it's possible to redo a bad score, but it doesn't seem like it would be a good idea.  You'd have to study a lot to maintain your history knowledge from this year to next year, and you probably would have trouble getting a high score without actually taking a class next year.  (At least I'd have trouble.)  It would be more worthwhile to take a class in something you haven't studied already instead of something you have.

A bad AP Score usually isn't TOO bad, but it's always nice to look consistent. Also, getting an A in the class and a 3 on the AP Test could reflect grade inflation as well (as could be the case at my school). I'm basically in the same situation with my AP English and US History tests :( .

Thanks for the reply.

Yes, I know that retaking the same AP is not a good idea (i can't even imagine myself studying WH again), but I'm just wondering whether it's possible or not...

but u can't replace your bad score, right?

Thanks,

[quote="bleumoose"]Yes, it's possible. A guy at our school was unsatisfied with his AP Chemistry score, so he retook the test the following year. It can be done. I would think it would make you look incredibly anal and overly worried about numbers, but that's your call.[/quote]

I don't think so.

Retaking AP tests just look like the kid really wants some college credit, in my opinion. Nothing to do with wanting high numbers like the SATs.

I think it depends. If you're going to end up taking the same class over just as a refresher...what's the point? You could just as easily be learning new material and taking a different AP test. Now if you think you can wing it, possibly with minimal self study, I don't see how that can hurt you. Then again, history isn't exactly a subject that's very memory friendly to me, so I know this would be impossible for me. I would much rather learn something new instead of re-studying through the same material.

either way, good luck

If it is one score out of ten, I would just keep on pursuing OTHER subjects. You could take an AP test in those subjects or not as you please.

Good luck in your applications.

[quote="jzzsxm"]The impression I've received is that colleges don't really care what you get on AP tests unless they're giving you credit for them, but even then it doesn't really count in admissions.  High scores are nice because they look good, but from what I've read/heard, colleges just want you to TAKE the test because if you don't it makes them think you were unprepared.  Other than that, I think scores are mostly about credit requirements and less about admission.[/quote]
Thanks! this certainly relieves me .. :) (not sarcastic)

Also, most colleges don't really give credit for World History anyways, so don't worry too much.

[quote="PTynan89"]I heard from a friend that we can call in already for scores, and he did so about 3 or 4 days ago.  Not that I'm going to, I'll wait until they come in the mail.  Has anyone else heard back yet?[/quote]
Neither did I...(i didn't want to waste money for faster score report, since i didn't do well in the test..)

Yeah, you can call in now and get your scores for eight bucks.

I called in since my whole family will be out of town when the scores come in paper, so I decided to get it now

kinda feel guilty about the eight bucks, but hey, I guess I won't be eating dinner tonight, eh? heheh

[quote="mysmartmouth"][quote="aznphatso"]
+ statistics-5 from last year and that makes me an AP national scholar :)[/quote]

What are the levels of AP scholarship, and what do you get as a...prize?[/quote]

See the collegeboard website [url]http://www.collegeboard.com/student/testing/ap/scholarawards.html[/url]

[quote="juicybooty911"]ap statistic 5
ap physics b mech 4 (errrr) :|[/quote]

No wonder you got a 4. Physics B contains basic Mech, basic fluid/thermo, basic waves/optics, and basic atomic.

 :D

[quote="chess64"]5 on calc bc and 5 on stat

happy about calc since i learned the whole thing in about 15 hours :D[/quote]

You learned calculus or stat in 15 hours?

[quote="joml88"][quote="chess64"]
happy about $\boxed{\text{calc}}$ since i learned the whole thing in about 15 hours :D[/quote]

I'm pretty sure he was talking about calculus ;)  Though I'm fairly certain that he's grossly underestimating how much time he's spent learning calculus in his life.[/quote]

ok well i spent about 15 hours reading the calculus textbook :D

[quote="kimby_102"][quote="PTynan89"]AP Bio - 5
AP Chem - 5
AP Physics C: Mechanics - 5
AP Physics C: E&M - 5
AP Physics B - 5[/quote]
PTynan89, you took both Physics B and C??[/quote]

Yeah, cuz I'm from the streets, B****!

But seriously, I just took the alternate version for Physics B.

[quote="MysticTerminator"]maybe the calculus book was travelling at some significant fraction of the speed of light or something[/quote]

It'd be an awfully long book to read, though.

[quote="PTynan89"]
Yeah, cuz I'm from the streets, B****!
[/quote]

Delaware...


limit
real analysis
real analysis unsolved

Let $ (x_n)_{n \in \mathbb{N}^*}$ be a sequence such that: $ x_1\equal{}1, x_2\equal{}1, x_{n\plus{}1}\equal{}\sqrt{n \cdot x_n\plus{}x_{n\minus{}1}}$, $ \forall n \in \mathbb{N}$. Prove that the sequence $ \frac{x_n}{n}$ has the limit $ 1$.

1. Let's prove that $ x_{n} > n\minus{}2$ by induction.
It's easy to see, that $ x_i > i\minus{}2$ for $ i\equal{}1,2,3,4$ by checking, then if $ x_{n\minus{}1} > n\minus{}3$ and $ x_{n} > n\minus{}2$ we have: $ x_{n \plus{} 1} \equal{} \sqrt {n \cdot x_n \plus{} x_{n \minus{} 1}} > \sqrt {n (n\minus{}2) \plus{} n \minus{} 3} \equal{} \sqrt {n^2  \minus{} n \minus{} 3} > n\minus{}1$ for $ n>4$ , q.e.d
2. Let's prove that $ x_{n} < n\plus{}1$ by induction.
It's easy to see, that $ x_i < i\plus{}1$ for $ i\equal{}1,2$ by checking, then: if $ x_{n\minus{}1} < n$ and $ x_{n} < n\plus{}1$ we have:
 $ x_{n \plus{} 1} \equal{} \sqrt {n \cdot x_n \plus{} x_{n \minus{} 1}} < \sqrt {n ( n\plus{}1) \plus{} n} \equal{} \sqrt {n^2 \plus{}2n} <(n\plus{}1) \plus{} 1$ for $ n>2$ , q.e.d
3. From $ (1)$ and $ (2)$:
$ n\minus{}2 <x_{n} < n\plus{}1$
I.e.:
$ 1\minus{} \frac {2}{n} <\frac {x_{n}} {n} < 1\plus{}\frac {1} {n}$
I.e.
$ \lim (\frac {x_n}{n}) \equal{} 1$
Q.e.d.


It is the end of the year and students are signing eachother's yearbooks.  If there are 100 students in the 6th grade and it takes 1 minute to sign a book, what is the least possible time it takes for everyone to sign eacthoer's yearbooks.  (say it doesn't take any time to transition from one person to another).

[quote="amirhtlusa"]lol, today was the day that the yearbooks came out in our school--
the answer would simply be $99$, ?[/quote]

yes...this was too easy...yearbooks are being handed out tommarow


Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers with $a_{1}\leq a_{2}\leq \cdots \leq a_{n}$, such that the arithmetic mean of $a_{1}^{2}, a_{2}^{2}, \cdots, a_{n}^{2}$ is $1$. If the arithmetic mean of $a_{1}, a_{2}, \cdots, a_{n}$ is $m$, prove that if $a_{i}\leq m$ for some $1 \leq i \leq n$, then $n-i \geq n(m-a_{i})^{2}$.

see here http://www.mathlinks.ro/Forum/viewtopic.php?p=212353#212353  :wink:


Hi, I am trying to create a Ferrers diagram in LaTeX to represent the partitioning of 6 into 3+2+1, and I am using \bullet to represent the dots in the diagram. However, I can't seem to get them alligned properly (for the other partitions, I generally use \begin{eqnarray*}, but I don't know how to align the dots. Can someone help?

Thanks in advance!

In my paper, which you can find [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]here[/url] (it's the one on partitions), I used \begin{tabular} and \bullet.  Here's what I would do for 6=3+2+1:

[code]
\begin{tabular}{cccc}
3 & \bullet & \bullet & \bullet\\
2 & \bullet & \bullet \\
1 & \bullet \\
\end{tabluar}
[/code]

\[\begin{tabular}{cccc}
3 & \bullet & \bullet & \bullet\\
2 & \bullet & \bullet \\
1 & \bullet \\
\end{tabluar}\]


I think students can benefit more from math question if replies are posted with diagrams.

Agree?

Do you meen that the person who asks the question includes diagrams, or that the person that answers it does/ either way , I think it would make it easier to answer/figer out the answer to problems.

If one is included I agree. It is often hard to draw one yourself.

and most of the time people interpret the question wrong, so they draw the incorrect diagram and gets an incorrect answer

If think students can somehow draw the SAME geometric shapes given in their textbooks if they had a special program.  I think it would be a good idea for this site to include a special program (like LaTex) to make any geometric shape there is in geometry.  Think about all the geometry, trigonometry, calculus, physics and astronomy math questions that can be easily solved using shapes.

What do you think?


Don't solve for $x$, find the sum $\tanh+%Error. "secx" is a bad command.
$ if
\[2004\tanh=2004%Error. "secx" is a bad command.
+2005\].

latex error :P

it should be:
find $\tan{x}+\sec{x}$

if $2004\tan{x}=2004\sec{x}+2005$

i dont know if this counts.

you can say
$\tan{x}-\sec{x}=\frac{2005}{2004}$ (i)
$\tan{x}+\sec{x}$=$k$.  (ii)  [  we want to find $k$.]

So adding (i) and (ii) gives $\tan{x}=\frac{\frac{2005}{2004}+k}{2}$ (iii)

squaring (i) and (ii) and adding them, then using $1+\tan{x}^2=\sec{x}^2$, we  get $2(1+2\tan{x}^2)=\frac{2005^2}{2004^2}+k^2$.  Substituting for $\tan{x}$ using eqn. (iii) on the LHS we get a quadratic equation in $k$, solving it gives $k=-\frac{2004}{2005}$.

Well cats..., your method is nearly similar to mine.Now let see how I get the solution:
from the relation\[2004tanx=2004secx +2005\]
we have $tanx-secx=\frac{2005}{2004}$   $ (i)$
suppose that the sum we are finding is $s$, then we have the relation
$[tanx+secx=s$     $(ii)$
multiply $(i)$ and $(ii)$ ,we obtain:
$tan^2x-sec^2x=\frac{2005}{2004}s$
but $tan^2x-sec^2x=tan^2x-\frac{1}{cos^2x}=-1$
then $s=-\frac{2004}{2005}$
 :lol:

limkimsrun: use \tan x, \sec x.  You need the \ to make it look nice, but you need to put a space after the end of it (so it understands that the command is just the tangent).

thank you JBL for your advice. I have just learn about latex for only one month, so I am still not familar with it.
Excuse me if my English and my latex make you all crazy.