$ c_{1} \equal{} \frac {3}{2}$
$ (c_{n \plus{} 1} \minus{} 1)c_{n \plus{} 1} \equal{} c_{n}^{2}$   $ n \equal{} 1,2,3,...$
find $ c_{n}$        $ n \equal{} 1,2,3,...$
 ($ c_{n} > \equal{} \frac {1}{2}$)

additional:
Is there any material(doc,pdf,etc.,)or website which introduce
advanced and particular method that can solve the sequence problem (find $ c_{n}$) like this? Thanks!! :lol:

There is no general theory of recursive equations -- like differential equations, there are a lot of specialized methods and then a lot of equations that no one knows how to solve.  One good place to start is with the theory of generating functions.  Wilf's [url=http://www.math.upenn.edu/~wilf/DownldGF.html]generatingfunctionology[/url] is a good (free!) introduction into the area.

By the way, this does not look like a very nice sequence you've got here.

the origional problem is:
$ a_{1}\equal{}1$
$ a_{n\plus{}1}\equal{}a_{n}\plus{}\sqrt{a_{n}}\plus{}\frac{1}{2}$
$ n\equal{}0,1,2...$
find
$ a_{n}$

I handled this with
$ {b_{n}}^2\equal{}a_{n}$ ......
$ c_{n}\equal{}b_{n}\plus{}1$
...
and then,be the problem above.
If you are confused with the problem above,try the origional problem  :P
Because I am not sure I was in the right way to handle this. :roll:


[quote]Find all $k \in \{1,2,...,89\}$ such as $sin k^o \in Q$[/quote]
$\sin 30^o=\frac 1 2$
So $k=30$


calculus
integration
function
conics
parabola
geometry
calculus computations

Let $ a,\ b$ be the real numbers such that $ 0\leq a\leq b\leq 1$. Find the minimum value of $ \int_0^1 |(x\minus{}a)(x\minus{}b)|\ dx$.

$ \int_{0}^{1}\left|(x\minus{}a)(x\minus{}b)\right|dx$
$ \equal{}\int_{0}^{a}(x\minus{}a)(x\minus{}b)dx\plus{}\int_{b}^{1}(x\minus{}a)(x\minus{}b)dx\minus{}\int_{a}^{b}(x\minus{}a)(x\minus{}b)dx$
$ \equal{}\frac{1}{3}(b^3\minus{}a^3)\minus{}ab(b\minus{}a)\plus{}ab\minus{}\frac{1}{2}(a\plus{}b)\plus{}\frac{1}{3}$. Let's denote this function by $ f(a,b)$.

$ f$ is defined in the unit $ 1: 1: \sqrt{2}$ triangle $ R\equal{}\left\{(a,b): 0\leq a,b\leq 1, a\leq b\right\}$.
Minimum value of $ f$ when $ b\equal{}1$ or $ a\equal{}1$ is $ \frac{1}{2}$, and when $ a\equal{}b$ is $ \frac{1}{12}$ which includes the case $ \nabla f\equal{}0$. $ f$ has it's minimum value when $ (a,b)$ is on the boundary of $ R$ or $ \nabla f\equal{}0$, so the minimum value we want to find is $ \frac{1}{12}$(when $ a\equal{}b\equal{}\frac{1}{2})$.

I am afraid, this can't be right. Just draw a picture of a parabola and think about
what happens to the total area under it between 0 and 1. When you consider
the change of area when moving the parabola in $ x$-direction and in $ y$-direction,
it is easy to see that the minimum is attained for $ a\equal{}1/4$ and $ b \equal{} 3/4$.
(And it is not too difficult to turn this argument into a rigorous proof.)

My approach was right, but I had a mistake when consideing $ \nabla f \equal{} 0$, which brought the wrong answer.

$ \nabla f \equal{} ((a \minus{} b)^2 \plus{} a \minus{} \frac {1}{2}, \minus{} (a \minus{} b)^2 \plus{} b \minus{} \frac {1}{2})$, so $ \nabla f \equal{} 0$ is equivalent to $ (a,b) \equal{} (\frac {1}{2},\frac {1}{2})$ or $ (\frac {1}{4},\frac {3}{4})$.
The case $ (\frac {1}{4},\frac {3}{4})$ makes minimum value of $ f$, and the right answer is $ \frac {1}{16}$.


conics
parabola
quadratics
trigonometry
analytic geometry
graphing lines
slope

The line $ y\minus{}x\sqrt{3} \plus{} 3\equal{}0$ cuts the parabola $ y^{2}\equal{}x\plus{}2$ at $ A$ and $ B$. If $ P\equal{}(\sqrt{3},0)$ then what is the value of $ PA.PB$?

hello,inserting $ y\equal{}x\sqrt{3}\minus{}3$ in $ y^2\equal{}x\plus{}2$ we get the quadratic equation
$ 3x^2\minus{}6x\sqrt{3}\minus{}x\plus{}7\equal{}0$, with the solutions
$ x_1\equal{}\sqrt{3}\plus{}\frac{1}{6}\plus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}$
$ x_2\equal{}\sqrt{3}\plus{}\frac{1}{6}\minus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}$
Inserting this in the equation above we get
$ y_1\equal{}\frac{\sqrt{3}}{6}\plus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}$
$ y_2\equal{}\frac{\sqrt{3}}{6}\minus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}$
Now we have
$ |PA|\equal{}\sqrt{\left(\sqrt{3}\minus{}\sqrt{3}\minus{}\frac{1}{6}\minus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2\plus{}\left(\frac{\sqrt{3}}{6}\plus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2}$
$ |PB|\equal{}\sqrt{\left(\sqrt{3}\minus{}\sqrt{3}\minus{}\frac{1}{6}\plus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2\plus{}\left(\frac{\sqrt{3}}{6}\minus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2}$
and the product is $ |PA|\cdot|PB|\equal{}\frac{8}{3}\plus{}\frac{4}{3}\sqrt{3}$.
Sonnhard.

You are correct but I wanted another approach!

[hide="EASY"]Write the equation of line in parametric form, x= x1+rcos@ , y = y1 + rsin@ ; where x1,y1 is any point on the parabola, \sqrt{3} ,0 satisfies the line , so put x1 and y1 in the equation of the parabola , you will get quadratic in r. Find product of the roots. Are you in BT?[/hide]

[quote="Conjurer"]Write the equation of line in parametric form, x= x1+rcos@ , y = y1 + rsin@ ; where x1,y1 is any point on the parabola, \sqrt{3} ,0 satisfies the line , so put x1 and y1 in the equation of the parabola , you will get quadratic in r. Find product of the roots.[/quote]

Clever solution!

[hide="Solution."]
=====
The parametric form of the line with respect to $ P$ becomes:
$ x\equal{}\sqrt{3} \plus{} r\cos \theta$
$ y\equal{}r\sin \theta$,

where $ r$ is the distance from $ P$. The slope of the line is $ \tan \theta \equal{} \sqrt{3}$, so $ \theta\equal{}\frac{\pi}{3}$. Thus $ A,B$ satisfy:

$ y^2\equal{}x\plus{}2$
$ \Leftrightarrow r^2\sin^2 \theta \equal{} 2\plus{}\sqrt{3} \plus{} r\cos \theta$
$ \Leftrightarrow r^2\sin^2 \theta \minus{} r\cos \theta \minus{} (2\plus{}\sqrt{3})$.

By Viete, the product of the roots is:

$ \minus{}\frac{2\plus{}\sqrt{3}}{\sin^2 \theta} \equal{} \minus{}\frac{2\plus{}\sqrt{3}}{(\sin \frac{\pi}{3})^2} \equal{} \minus{}\frac{2\plus{}\sqrt{3}}{\frac{3}{4}} \equal{} \minus{}\left(\frac{8}{3}\plus{}\frac{4}{3}\sqrt{3}\right)$

Of course, the distance is positive, so we take the absolute value: $ \boxed{\frac{8}{3}\plus{}\frac{4}{3}\sqrt{3}}$.
=====
[/hide]


Hi.  I'm in 8th grade, and I wanted to take the AMC 10 this year.  I asked my mom about it, and she said no, we don't have enough time.  Do you have any good tips to:

A: Convince her to let me take it;

or

B: Prep intensively for next year's AMC 10

Thanks!

Don't get me wrong, though.  My mom is pretty cool.   :)

You don't have enough time?  It's only a 75-minute test, so I guess she means you don't have enough time to prepare for it?  Even if she won't let you prepare for it, it will be a good experience for you to take it.  Even if you aren't prepared, exposure to hard problems is always good.

Don't answer me if this is too personal a question, but what does your mom want you to be doing instead?  I can't think of too many things more worthy of your time, especially from a mother's perspective, than the AMC.

Yup, and I just remembered I have a science class that day (I think.)  I have a feeling that that's what I'm supposed to be doing with my time that day.   :P And yes, I have a pretty full schedule right now, so I don't have that much prep time.  And one more point that I just realized: I'm supposed to take the AMC 8 this year, and my school gives the AMC 8 and the AMC 10 at the same time.

Thanks for your response!

EDIT:  My mom is awesome.  She just knows my schedule better than I do (which is why I can't take the AMC 10).  Isn't that sad?

You can always take it on your own after it ends.  :)

[quote="westiepaw"]...my school gives the AMC 8 and the AMC 10 at the same time.
[/quote]

 :huh: What? The AMC 8 is in November and the AMC 10 is in February...

[quote="westiepaw"]Hi.  I'm in 8th grade, and I wanted to take the AMC 10 this year.  I asked my mom about it, and she said no, we don't have enough time.  Do you have any good tips to:

A: Convince her to let me take it;

or

B: Prep intensively for next year's AMC 10

Thanks![/quote]

Why does she say it's too much time? Tell her the contest format

Are you fairly confident that you can make AIME? If not, you can take it after the test day (whenever they release the test) for practice, simulating the actual testing situation to benefit from it. If you are fairly confident you can make AIME, explain to your mom the significance of AMC/AIME/USAMO/MOP/TST/IMO.
Take some [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43]previous tests[/url] to see whether or not you have a shot at making AIME.

To study (whether or not your mom says no) take all the AMC 10 practice tests (there arent that many) and see where your weaknesses are. Buy the corresponding AoPS introductory book to study for. (eg. I was weak at counting so I got intro to Counting and Probability. It was really helpful). Also, read through and understand AoPS Volume 1.

[quote="Thunder365"]
To study (whether or not your mom says no) take all the AMC 10 practice tests (there arent that many) and see where your weaknesses are. [/quote]

uh, taking [i]all[/i] of them isn't the best idea since they run out quickly. maybe 2 or 3 at most.

anyway, my parents are somewhat like this. they let me take it (well, they can't really refuse that), but they get mad at me when i prepare a lot, mainly 'cause they think i do too much math and need to be well-rounded.

but just try to like.. explain why this is a good test to take (i'm sure you can find many reasons why [like in this thread, for example], as i'm too lazy to list anything). if you [i]have[/i] to, you can always pull the college card by saying that colleges like MIT and caltech (i'm sure there are more, i just don't know them off the top of my head) look at amc scores and beyond (aime, etc.).

[quote="westiepaw"]OK...  I actually really doubt that I could make AIME.  I took a practice test and scored like a 90...  I'm just gonna prep for another year.  Then take the test.  Thanks for all your responses![/quote]

Some suggestions, work the AoPS Introduction Series. Worked for me by a lot. Raised my AMC 10 average 40 points!!

Good luck:)

90 is pretty good score, but the thing is that you probably will not make AIME anyway, so don't be too sad even if you can't take it.

Although I have to admit, not being able to take AMC exam because of "time restraint" is somewhat ridiculous.

Wouldn't prepping intensively (B) take more time than taking the 75 minute test?

If I were you, I would try to take it, even if I don't study. Being timed and in the official test-takers is more beneficial than taking them as practice tests.


Any tips? I did really well in Math but allright in Verbal. I wanted to know if y'all have any tips. I almost finished doing the top 500 SAT words (if I remember them  :? ) but what else should I do?  :?

Err what I learnt from everyone else to improve your verbal score was:
[size=200][u][b]READ READ READ[/b][/u][/size]

But don't necessarily trust me on this; I got a 440 on my verbal (yes it's low I know) but i'm basically ESL so I'm not too worried about verbal for now.

Oh and I forgot one more thing. READ

by verbal, do u mean reading comprehension/sentence completion or indetifying sentence/gramatical errors/essay?or both?

tip: reading is good
perhaps incorporating SAT words into your vocabulary might help...

i basically just read magazines, the newspaper, online news, books, it all helps.

read. anything, everything. im serious about this- junky magazines, comics, kids books, the backs of cereal boxes, instructions... read ANYTHING you can get your hands on. read in the car (not if you are driving though...). read in line at the grocery store (but half pay attention to what you are doing so people don't get uber mad at you). read at the breakfast table. any time you are alone- READ! carry a book with you. read harry potter. then pick up something a little heavier, like maybe [i]the once and future king[/i] if fantasy is your fortay, or [i]pride and prejudice[/i] or [i]tom sawyer[/i]... and then go look for new stuff, like maybe [i]me talk pretty one day[/i] or [i]the da vinci code[/i]. read short stories. read poems. when you get your literature book at the beginning of the year in school, flip through it. if a picture looks interesting, stop and read the story/poem that goes along with it. reading a lot of things, and a variety of things, will help your speed tremendously, which will GREATLY improve your verbal score, as well as your vocabulary. i hear there are some test-prep company who has published some books and highlighted vocab words in the text. this sounds like its probably a good study skill to me.
another hint: WRITE. get a spiral bound (the pages dont fall out as easily) notebook from your local Walmart/Kmart/Target and start filling it with words. write how your day went. write what you really think about your little brother. write about your hopes and wishes. write stories. just... WRITE!
the verbal section of the SAT is designed to test your mastery of the english language. the best way to improve your skills is to actually use them.

note: the above advice comes from a native english speaker. however, said "native english speaker" recently earned a 710 on said native english speaker's SAT. this person has also been an avid reader since the age of six, and has filled five spiral bound notebooks with writing. while this person does not consider his/herself an expert, (s)he is hoping that his/her advice will be worth something... :rotfl:  :rotfl:  :rotfl:

I found that my verbal score jumped rougly 80 points after I learned one strategy from a test prep company(8th grade CTY SAT to PSAT 10th grade).

Basically, don't read the passages themselves, read the questions first and then look for the answers in the passages.

Especially, look for questions with line numbers and answer those first; then, answer questions which ask you about the general idea of the question.

Now, does anyone know of how to improve in grammar?

[quote="agolsme"]
Now, does anyone know of how to improve in grammar?[/quote]

Insane amounts of practice. Take like 50 tests and you'll be good to go. You'll have seen every grammar trick in the books.

[quote="agolsme"]Now, does anyone know of how to improve in grammar?[/quote]

Learn how to talk English.

If all else fails, try convincing them Yoda's your teacher.

for grammar, practice is the key. sophomore year my teacher gave the class a lot of practice materials and went over them so that i pretty much knew all the possible sentence errors that are tested. after this review, my writing score went up by 200 points (well, actually 20 points because it was the PSAT but same thing)

yep, I'm just saying what other people have said but you HAVE TO READ. it helps if you start early and if you actually enjoy reading books like grapes of wrath (which is a pretty good book) and wuthering heights. for sentence completion sometimes even if I don't know or forgotten a definition, if I've read the word somewhere before, I can sort of remember the general "connotation/impression"? lol, kinda confusing. i guess it comes naturally from reading a lot.

I had to do INSANE amounts of memorization, though, for vocabulary. insane for even aops-ers and chinese families. haha.

well anyways good luck!

Yes, if the person opening the thread is in middle school now, there is plenty of time to improve the score. I'm not aware of anyone in the entire country born the year that I was born who took the SAT at middle school age, but now that talent search testing has been expanding since the early 1980s, a lot of young people wonder how to boost their scores. The answer already given, to read a lot, is the most useful general strategy. Read about what interests you, and you will read with more attention to how the writer expresses ideas, which is the beginning of developing critical reading skills.

[quote="eryaman"]IversonFan is in middle school right?  Well for me...the improvement just came natural as I got older.  In 7th grade I got 550 on verbal portion of SAT.  But now I'm entering 10th grade, I took a practice test, without any preparation and am getting 700+ in every section now.
Same happened with my sister too...[/quote]

No, I'm gonna be a Junior and my Verbal was like less than a 600 which is 3/4 of my Math score, and I'm aiming for a 1550.

the same advice from me too :) READ!...and a lot....I would personally recommend 19th century British literature...it has the most SAT words....but if you find it boring you can always try something else....also you can find a lot of vocab lists on the web( because in my opinion as a non-native English speaker- you will need more than just 500 words)....as for grammar....try to focus on the type of errors that are common on the test...if you learn them you won't have any kind of problems on the test...and I also forgot to tell you....reading also heps you improve your writing style ;)
well...good luck to you!

Shobhit:

For whatever it's worth, here is something I said here  before. If you can, take Latin. My sons found it fun (I wish I  can go back and learn more languages ... and latin would help)

And if a small sample is some guide, almost every  one  i know   (specially those where English is a second language to  the parents)  who got a  800 in Math,   and took Latin,  also  go a t perfect (or near perfect) verbal score too.  But that is besides the point, more important is that they enjoyed   the latine class a lot.

i'm in a similar boat as iversonfan. cept i'm now a senior (i would assume i'm a senior even tho its summer and i haven't started my senior year). anywho, my old sat verbal was 590. i need a 650 to get into the honors college at the school i want to go to. i have borrowed 6 books from my friend to read this summer. also, i have seen the cliffs notes books that use many of the sat words. can anyone here tell me if these are that helpful?

and is there a website that has practice tests, particularly the grammer sections. i guess those would have to be psat tests to include the grammer.

yeah i agree, reading [especially short magazine articles] helps.

for novels, does anyone recommend any books?

er... harry potter? but i'm guess you've probably read that.
a few of the ones i like:
lord of the rings, and, if you liked those, read 'beowulf'... now, i know that sounds like a challenge (it IS in verse) but i just started reading it, and i was hit forcefully by how much they reminded me of lotr
dan brown
tamora pierce
the "sabriel" books (basically anything by garth nix, i've found that guys and girls like his books pretty equally)
er... watership down (if you like talking rabbits... *laughs*)
redwall
the once and future king... its EXCELLENT, i've never actually finished it, to be honest, but that was only because of school work and other book distractions (because if one begins reading harry potter, its almost maniacal).
the ear, the eye, and the arm
artemis fowl (if you like HP, you'll probably like these)
the darwin awards (not a "novel" but this is an great read anyways)
1984 (see my siggie... this is one of my ALL TIME favorites, even though the class i studied it in almost ruined the book)
kurt vonnegut is really good

and if anyone gets REALLY stuck for a book to read (im talking to middle schoolers here, now) go look up the newberry (newbery?) awards. all of them are prize winners for a very good reason- some of them look rather boring at first glance, but i've almost never been let down. you may find that the more recent ones (like holes) are more to your liking than the olders ones.

For the sentence completions, it's vocab period. I had those down cold after a few insane word lists.. Try to expand on that 500 though. Memorizing vocab is painful, but it yields results.

For the passages though, I'd recommend reading (like everyone else) but writing also. I found that I understood reading passages better after this school year, but my non-textbook reading this year was close to nil. I figured that writing helped--long nonfiction essays especially--because you get a better sense of the makeup of an essay (main idea, relationship between paragraphs, etc) by actually writing one. Besides, writing essays has got to help on the essay section of the new SAT you'll probably be taking, right? Long story short, make the most out of your English assignments.

Good luck on your next test!

As everyone else has said: READ.. and for a long time too, and make sure u understand EVERYTHING. Magazines, books.. everything. From like 6-11 i read a lot of books, and got a 630 in V in 7th grade.. not too bad, but i stopped reading so much after 11 and i took it at 13... so...

some books id recommend off hand:

Harry Potter (Again!)
Eragon (Christopher Paolini)
the His Dark Materials series by Phillup Pulman

thats all i can think of.

As for grammar and writing, I believe that just practicing a lot will help you greatly. Write some practice essays, and have someone else read it [i]out loud[/i]. The person can critique you, and after u take the advice you write ANOTHER essay or story, seeing how you improve.

I can give you some links....but they are posted on other specialized forums so I guess you most probably know them.
anyway....here they are:
http://www.freevocabulary.com
http://www.number2.com
http://www.takesat.com
that's about all I can remember right now.....hope these help you...they are for vocab, but also for the other sections....
enjot them  ;)

[quote="paladin8"][quote="Iversonfan2005"][quote="Max300"]umm....considering that this topic was started on April 10th this year and iversonfan is a sophomore wouldn't he be taking the new SAT?
so a 1550 would not be the best of scores lol[/quote]

Well, I know  :rotfl: But I'm just talking about Verbal and Math for now....I'm not too worried about writing. I think colleges still focus primarily on Math and Verbal.[/quote]

I'm not sure about that. I wish it was true, but I get the feeling that writing is just as important which is why the old SAT II: Writing was pretty much as much of a requirement as the SAT I.[/quote]

Yes, but when it became part of the SAT I, the section was criticized heavily.  Many insititutions are not factoring in the writing section this year, putting it the test on a 1600 scale.  Note, however, that most of these institutions are using this year's writing section submissions as data as basis for future judgement (later entering classes).

I agree with you there. This year and probably next year it's probably not going to be used very heavily. But I'm the high school graduating class of 2007, so personally I think I have reason to pay attention to the writing section. As does iversonfan, whom I believe to be in the same class as me. In any case, I would still take care not to completely ignore the writing section and come up with a respectable score.

Sorry if this is off-topic, but since we're discussing the new SAT vs. the old SAT, I'd like to state my opinion.

The writing section definitely matters, even to the class of 2006 (me), as sonar said. What I don't understand is why critics and colleges acclaim the new SAT. Basically, everything is the same thing: writing section is the same as SAT II Writing, math has somewhat more advanced algebra, and verbal has more passages and no analogies. The way I see it, the new SAT is not much different from the old SAT, so I don't see why colleges require the new one when they previously required Writing and Math IC or IIC.

Anyways, the best way to prepare for the new verbal section, as stressed by other members, is to READ. Reading improves your analytical skills as well as vocabulary.


number theory
least common multiple
combinatorics unsolved
combinatorics

Let $p$ and $q$ be natural numbers stuch that their $lcm$ is $r^{2}s^{4}t^{2}$ where $r$,$s$ and $t$ are prime numbers.Then determine the number of ordered pairs $(p,q)$.

[hide]Let $p=r^{x}s^{y}t^{z}, q=r^{x'}s^{y'}t^{z'}$
Since their $lcm =r^{2}s^{4}t^{2}$ we have
$\max \{x,x'\}=2$ - 5 possibilities: 02, 12, 22, 21, 20
$\max \{y,y'\}=4$ - 9 possibilities: 04, ... , 40
$\max \{z,z'\}=2$ - 5 possibilities
Multipliing we have in total 225 required pairs.[/hide]

Exactly...... i also tried and got the same answer, but the answer was given to be 224. How is this possible.

I think they assume p and q be distinct that is why you and me have 1 extra pair.


factorial
limit
floor function
logarithms
real analysis
real analysis unsolved

Let $ a_{n}$ be the number of zeros at the end of $ n!$.

Prove that:

$ \textbf{1.}$ $ \lim\limits_{n\rightarrow \infty}\frac{a_{n}}{n}=\frac{1}{4}$.

$ \textbf{2.}$ $ \lim\limits_{n\rightarrow \infty}{n^{\alpha}\left(\frac{a_{n}}{n}-\frac{1}{4}\right)=0}$, where $ \alpha\in [0,1)$.

Find, if exists, the following limit:

$ \textbf{3.}$ $ \lim\limits_{n\rightarrow \infty}n\left(\frac{a_{n}}{n}-\frac{1}{4}\right)=?$

$ a_n \equal{} \sum_{k \equal{} 1}^{\infty} \left\lfloor \frac {n}{5^k} \right\rfloor$.  Let $ b_n \equal{} \sum_{k \equal{} 1}^{\infty} \frac {n}{5^k} \equal{} \frac {n}{4}$ and note that

$ b_n \minus{} a_n < \log_5 n$

from which 1) and 2) follow immediately.  The above estimate is not tight, so I'm not sure about the last.

Edit:  let $ n \equal{} 5^r \minus{} 1$.  Then $ \frac {n}{5^k} \minus{} \left\lfloor \frac {n}{5^k} \right\rfloor \equal{} \frac {5^k \minus{} 1}{5^k}$, so the above estimate is tight for numbers of this form.  On the other hand, letting $ n \equal{} 5^r$ the corresponding value is $ 0$, so the last limit does not exist.

if we note that $ s_p(n)$ means sum of digits of $ n$ in base p and p is prime, than it's known that the largest power  of a $ p$ dividing $ n!$ is
\[ a_n(p)\equal{}\frac{n\minus{}s_p(n)}{p\minus{}1}\]
It's clear that $ a_n(2)\ge a_n(5)$, so
\[ a_n\equal{}a_n(5)\equal{}\frac{n\minus{}s_5(n)}{4}\]
then 1) and 2) follows immediately, because $ s_5(n)\le 4 \frac{\ln n}{\ln 5}$. For the 3) consider $ n_1\equal{}5^k\minus{}1$, and $ n_2\equal{}5^k$. then
\[ n_1\left(\frac{a_{n_1}}{n_1}\minus{}\frac{1}{4}\right)\equal{}(5^k\minus{}1)\left(\frac{\frac{5^k\minus{}1\minus{}4k}{4}}{5^k\minus{}1}\minus{}\frac{1}{4}\right)\equal{}\minus{}k\\
n_2\left(\frac{a_{n_2}}{n_2}\minus{}\frac{1}{4}\right)\equal{}5^k\left(\frac{\frac{5^k\minus{}1}{4}}{5^k}\minus{}\frac{1}{4}\right)\equal{}\minus{}\frac{1}{4}\]
so we have no limit.


inequalities
Euler
inequalities proposed

[quote="shaaam"]Given that, $ a, b, c$ are positive real numbers, and satisfy:

$ abc \equal{} 1$

and

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$ ,

Prove that

$ 3a^4 \plus{} 3b^4 \plus{} 3c^4 \geq a^7 \plus{} b^7 \plus{} c^7 \plus{} 6$[/quote]

By Euler's Identity you have $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc \Leftrightarrow a\equal{}b\equal{}c$
hence $ a\equal{}b\equal{}c\equal{}1$
And your inequality is true as equality ($ 9 \geq 9$) :P

But... $ abc\equal{}1$ and $ a^3\plus{}b^3\plus{}c^3\equal{}3$ doesn't mean $ a\equal{}b\equal{}c$ by AM-GM?

[quote="shaaam"]Given that, $ a, b, c$ are positive real numbers, and satisfy:

$ abc \equal{} 1$

and

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$ ,

Prove that

$ 3a^4 \plus{} 3b^4 \plus{} 3c^4 \geq a^7 \plus{} b^7 \plus{} c^7 \plus{} 6$[/quote]
Another proof :)
$ 3a^4 \plus{} 3b^4 \plus{} 3c^4 \equal{} 3(a^4\plus{}b^4\plus{}c^4) \equal{} (a^3\plus{}b^3\plus{}c^3)(a^4\plus{}b^4\plus{}c^4) \geq a^7 \plus{} b^7 \plus{} c^7 \plus{} 6$
$ \iff$
$ \sum_{cyc} a^4(b^3\plus{}c^3) \ge 6 \equal{} \sum_{cyc} 2abc \equal{} \sum_{cyc} 2a^{7/3}b^{7/3}c^{7/3}$. Which is easy to prove.(fx by AM-GM, muirhead or factoring)

[quote="FelixD"]The condition $ abc \equal{} 1$ implies that $ 3abc \equal{} 3$. With $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$ we get $ 3 \equal{} 3 \iff 3abc \equal{} a^3 \plus{} b^3 \plus{} c^3$ but with AM-GM we have $ \frac {a^3 \plus{} b^3 \plus{} c^3}{3} \geq abc$ hence $ a \equal{} b \equal{} c \equal{} 1$ (cause equality holds if and only if all the three variables are equal).[/quote]

FelixD , thank you very much for the explanation!  :D

[b]Question: is it true over all reals?[/b]

Note \[ 0\equal{}a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}\frac12(a\plus{}b\plus{}c)( (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 )\]
Thus either $ a\plus{}b\plus{}c\equal{}0$ or the second expression is zero, which is only when $ a\equal{}b\equal{}c$ and treated above.

So one of $ a,b,c$ is negative. Then by $ abc\equal{}1$ exactly two are negative. Relabel so $ c>b$ are negative.
Take $ x,y,z \equal{} a,\minus{}b,\minus{}c$.
Then $ xyz\equal{}1,x\equal{}y\plus{}z$.
We must prove \[ 3(x^4\plus{}y^4\plus{}z^4)\geq x^7\minus{}y^7\minus{}z^7\plus{}6\]
take $ y\equal{}z$, so that $ x\equal{}4^{1/3}$, $ y\equal{}z\equal{}2^{\minus{}1/3}$.
Then \[ 1/4>y^7>z^7\text{ and }x^7\minus{}3x^4\equal{}x^4(x^3\minus{}3)>4\], so that
\[ x^7\minus{}y^7\minus{}z^7\plus{}6>x^7\plus{}5.5>3\plus{}3x^4>3(x^4\plus{}y^4\plus{}z^4)\]

[quote="shaaam"][quote="NickNafplio"][quote="shaaam"]Given that, $ a, b, c$ are positive real numbers, and satisfy:

$ abc \equal{} 1$

and

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$ ,

Prove that

$ 3a^4 \plus{} 3b^4 \plus{} 3c^4 \geq a^7 \plus{} b^7 \plus{} c^7 \plus{} 6$[/quote]

By Euler's Identity you have $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc \Leftrightarrow a \equal{} b \equal{} c$
hence $ a \equal{} b \equal{} c \equal{} 1$
And your inequality is true as equality ($ 9 \geq 9$) :P[/quote]

Can you please provide some more information on Euler's Identity. Thanks...[/quote]

This is Eulers Identity: for real numbers $ a,b,c$ we have

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} \frac {1}{2}(a \plus{} b \plus{} c)((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)$

Here $ a \plus{} b \plus{} c > 0$, and since $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} a^3 \plus{} b^3 \plus{} c^3 \minus{} 3 \equal{} 0$, we must have $ a \equal{} b \equal{} c$

But this follow from AM-GM Inequality too


Yes , that's inequalitie proved very easy for method of geometry.


As the figure shown,

$ ABCD$ is a trapezium .$ AB \parallel CD$ .
$ M$ is the mid-point of $ BC$, and
$ MN \perp AD$

If $ MN\equal{}3 ,AD\equal{}4$ , then the area of $ ABCD$=?

Alternatively, extend $ DC$ to $ E$ so that $ EM$ is parallel to $ DA$ and extend $ EM$ to $ F$ on $ AB$.

Then triangles $ CEM$ and $ BFM$ are congruent and so the area of the trapezium $ ABCD$ is equal to the area of the parallelogram $ AFED$ which is $ 4\times3\equal{}12$, as kunny has.

Take another congruent trapezoid, turn it upside down, and connect the two sides of length BC together. It makes a parallelogram with base 4 and height 6, so the area of the trapezoid is half the area of this paralellogram. So the answer is 12.

[quote="kunny"]$ 4*(3 \plus{} 3)*\frac {1}{2} \equal{} 12.$[/quote]

Would you please post the details?

My solution is exactly same as dgreenb8017s one.

i agree your result ,it 's math exercise of grade 7  :rotfl: and it isn't too hard  :rotfl:


A manufacturer built a machine which will address $500$ envelopes in $8$ minutes. He wishes to build another machine so that when both are operating together they will address $500$ envelopes in $2$ minutes. The equation used to find how many minutes $x$ it would require the second machine to address $500$ envelopes alone is:

$\textbf{(A)}\ 8-x=2 \qquad
\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2} \qquad
\textbf{(C)}\ \dfrac{500}{8}+\dfrac{500}{x}=500 \qquad
\textbf{(D)}\ \dfrac{x}{2}+\dfrac{x}{8}=1 \qquad\\
\textbf{(E)}\ \text{None of these answers}$

er...
this seems too easy for a 29
[hide="29"]
so the rate of the first machine is $ 500/8$ envelopes per minute
the second machine is $ 500/x$ (x unknown) envelopes per minute
combined, they go $ 500/2$ envelopes per minute
the rate of the first two machines add up to get combined so
$ 500/8\plus{}500/x\equal{}500/2$
you can cancel the 500's on the top so it's
$ 1/8\plus{}1/x\equal{}1/2$   (B)
[/hide]

sorry about the ugly fraction thingys  again :(

There's 50 problems, so maybe 29 isn't too hard. And also, consider that it's the first one ever written...


complex analysis
function
complex numbers
complex analysis unsolved

1, Let $ f_1,f_2,f_3$ be holomorphic functions on $ \mathbb{C}$ such that $ e^{f_1}\plus{}e^{f_2}\equal{}e^{f_3}$. Prove that there are complex numbers $ C_1$ and $ C_2$ such that $ f_3\equal{}f_1\plus{}C_1\equal{}f_2\plus{}C_2$.

2, Let $ f,g$ be non-constant meromorphic functions on $ \mathbb{C}$ such that $ f^5\plus{}f^3\equal{}g^5\plus{}g^3$. Prove that $ f\equal{}g$.

Ad 1.: Wlog we can assume $ f_3\equal{}0$. If $ f_1$ is not constant, then we have $ f_1(z_0)\equal{}2 ki\pi$ for some $ z_0\in\mathbb C$ and $ k\in\mathbb Z$, by Little Picard. This yields the contradiction $ e^{f_2(z_0)}\equal{}0$.


number theory
greatest common divisor
least common multiple

Two positive integers have their sum equal to the sum of their greatest common divisor and least common multiple. Prove that one of the two numbers divides the other.

[hide="Hint"]Write the two numbers in terms of the GCD.[/hide]
[hide="Solution"]Let $d$ be the GCD of the two numbers, and let $dk$ and $dm$ be the two numbers where $k$ and $m$ are relatively prime. Then the LCM is $dkm$. The problem says that $dk+dm = d+dkm$. So $k+m = 1+km$. $k-1 = km-m$. $k = 1$ is clearly a solution. If $k \neq 1$, we divide $k-1$ out to get $m = 1$. The result is that either $k = 1$ or $m = 1$. WLOG let the solution be $k = 1$. Then our two numbers are $d$ and $dm$. Since $d \mid dm$, we're done.[/hide]

You could finish more quickly by saying that $mk+1=k+m\implies 0=km-m-k+1=(k-1)(m-1)$

[hide="possible idea"]
Isn't the product of 2 numbers equal to the product of the GCF and LCM?
[/hide]

yes, and my solution is based on it. :D
[hide="source of the problem"]Russia 1995[/hide]
[hide="my solution"]let the two numbers be $m$ and $n$ and let $(m,n)=g$,$[m,n]=l$. we have \[m+n=g+l\quad (1)\]and \[mn=gl\quad(2)\]from $(1)\times n$, we get \[mn+n^{2}=gn+ln\]or\[gl+n^{2}=gn+ln\]this is equivalent to \[(n-g)(n-l)=0\]so $n=l$ or $n=g$.
WLOG $n=g$. then $(m,n)=n$ thus $n|m$. [/hide]

[hide="One line"] Let the numbers be $a, b$, and WLOG $a \le b$.  Then $lcm(a, b) < a+b \le 2b \implies lcm(a, b) = b$.  [/hide]


Argh, I'm saying 22-23 for blue and 9-10 for red. You aren't making it easy :(.

6.

Nah, it'd never be 6, I wish it were though  :P

I think 9-10.

Questions 1 and 4 would be the main score-collectors for 9th graders...(including me!  :D )

Well, I'm pretty confident about Red....So I have achieved my goal! Yay~ :P

[quote="ProtestanT"]I voted for 10-11. There are more freshmen this year, and people said this year's is easier than last year's.

[quote="diophantient"]I'm thinking 9-10 for red, because of the difficulties in rigorizing 4, and because partial credit outside of 1 and 4 was really hard for most freshmen to scrounge.[/quote] That makes me think I will end up 1 point short of red. I think I might be in the situation you are describing (difficulty rigorizing 4, low chance of receiving partial credit outside of 1 and 4).[/quote]
No, there are 69 freshman this year taking the USAMO versus 72 last year.
Also, people say this year is easier referring to 2/5 which were more approachable than last year's 2/5, and last year's 5 was practically 2 free points for people aiming for red mop, and freshmen are unlikely to get number 5 this year.

The thing is, when it comes to red MOP cutoffs #1 and #4 are basically all that matters.  Even though #2 and #5 were easier than normal (I disagree that #2 was easier but maybe that's just b/c I suck at geometry), they weren't easy to get points on unless you completely solved the problem, and I doubt very many freshman completely solved either of those.

I would say this year's #1 is harder than last year's, but this year's #4 is easier than last year's.  So the cutoff will probably be about the same.

I'm hoping I have a 12.....would that be guarenteed? :P

[quote="diophantient"]Also, people say this year is easier referring to 2/5 which were more approachable than last year's 2/5, and last year's 5 was practically 2 free points for people aiming for red mop, and freshmen are unlikely to get number 5 this year.[/quote]
I am a freshman, and I believe that I got a pretty good score on number 5... At least better than numbers 1 and 4.   :D


geometry
3D geometry
number theory unsolved
number theory

We have a rectangular box with integers sides. If one places as many cubes as possible, each with volume 2, in the box, with their edges parallel to the edges of the box, one can fill exactly 40% of the box. Determine the possible dimensions of the box.

it is [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=367424#p367424]problem#3[/url] at [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1976]IMO 1976[/url]
a solution can be found [url=http://www.kalva.demon.co.uk/imo/isoln/isoln763.html]here[/url]


Let $ a$ and $ b$ be real numbers satisfying
\[ a^4 \plus{} 8b \equal{} 4(a^3 \minus{} 1) \minus{} 16\sqrt{3}\]
and
\[ b^4 \plus{} 8a \equal{} 4(b^3 \minus{} 1) \plus{} 16\sqrt{3}.\]
Find $ a^4 \plus{} b^4$.

[hide="kind of a giveaway"]
The "ugly" numbers in the problem don't play as large a role as you think at first glance
Think perfect squares and the trivial inequality!
[/hide]


function
superior algebra
superior algebra unsolved

Hi,

I would appreciate it if someone could explain to me how to do group multiplications of permutations.

e.g.

$ \alpha\equal{}(123)(456)$

$ \beta\equal{}(12)(346)$

$ \alpha \circ \beta \equal{} (123)(456) \circ (12)(346)\equal{} (1356)(2)(4)$

I can see that we go from $ \beta$ to $ \alpha$.

So, 1 from $ \beta$ goes to 2 in $ \alpha$ which in turns goes to 3 in $ \alpha$. So, we put $ (13...)$.

Why does 3 in solution go to 5?

If someone could help me with the principle, I would be most grateful.

thanx a lot.

The same observation you made for $ 1\minus{}>3$ works for $ 3$.  $ \beta$ takes $ 3\minus{}>4$, and $ \alpha$ takes $ 4\minus{}>5$.  Therefore, $ \alpha \circ \beta$ takes $ 3\minus{}>5$.

I'm sure that you can work out the rest from here, and hopefully it will make sense.

[quote="Antonio_26"]Hi,

I would appreciate it if someone could explain to me how to do group multiplications of permutations.

e.g.

$ \alpha \equal{} (123)(456)$

$ \beta \equal{} (12)(346)$

$ \alpha \circ \beta \equal{} (123)(456) \circ (12)(346) \equal{} (1356)(2)(4)$

I can see that we go from $ \beta$ to $ \alpha$.

So, 1 from $ \beta$ goes to 2 in $ \alpha$ which in turns goes to 3 in $ \alpha$. So, we put $ (13...)$.

Why does 3 in solution go to 5?

If someone could help me with the principle, I would be most grateful.

thanx a lot.[/quote]

$ \mathfrak{S}_n$ is the group of the bijective function from a set to itself ($ n$ is the cardinality of the set). The group multiplication is the function composition.

Now...
$ \alpha \circ \beta \equal{} (123)(456) \circ (12)(346) \equal{} (123)(456)(12)(346) \equal{} (1356)$
$ \circ$ can be omitted.

$ 1 \mapsto 2 \mapsto 3$
partial $ (13...)$
$ 3 \mapsto 4 \mapsto 5$
partial $ (135...)$
$ 5 \mapsto 6$
partial $ (1356...)$
$ 6 \mapsto 3 \mapsto 1$
so $ (1356)$

$ 2 \mapsto 1 \mapsto 2$
$ 2$ doesn't change.
$ 4 \mapsto 6 \mapsto 4$
$ 4$ doesn't change.


topology
limit
function
calculus
calculus computations

Techniques vary depending on the description or construction of your set. Do you have a more specific question?

What I mean is the  set $X = (-\infty, +\infty) = \mathbb{R}$ is closed although unbounded. How do I prove this?

What exactly do you mean by "closed set"?

What I found on the net confuses me...

Lets go with the following definition - a set $X\subset\mathbb{R}$ is closed if its complement is open.  A set $Y\subset\mathbb{R}$ is open if every point $y\in Y$ has an $\epsilon-$neighborhood $B(y,\epsilon)$ such that $B(y,\epsilon)\subset Y$.

The complement of $\mathbb{R}$ is empty, hence there are no points in the complement and the result is vacuously true.  So the empty set is open and the real line is closed.

More generally, if you endow $\mathbb{R}$ with some topology and make $\mathbb{R}$ into a topological space, then $\mathbb{R}$, being the 'whole space' will always be closed by definition.

Another characterization of closed sets is that they contain all of their limit points.

We could prove, as a theorem, that this is equivalent to what blahblahblah said.

$a$ is a limit point of $X$ if there exist $x_n\in X\setminus \{a\}$ such that $\lim x_n=a.$

There are bounded sets that are closed and bounded sets that are not closed; similarly, there are unbounded sets that are closed and unbounded sets that are not closed.

The interval $[3,\infty)$ is closed.

The interval $(3,\infty)$ (or $]3,\infty[$ if you prefer that notation) is not closed.

The interval $[3,6)$ is not closed. (It's also not open.)

The set $\mathbb{Z}$ of all integers has no limit points. Thus it (vacuously) contains all of its limit points and is closed.

The set $\left\{\frac1n: n\in\mathbb{N}\right\}$ has the limit point $0$ and is thus not closed.

But the set $\left\{\frac1n: n\in\mathbb{N}\right\}\cup\{0\}$ is closed.

It seems like you are saying $x_n \in \mathbb{R} \,\backslash\, \{a\}$ and that $\lim x_n = a$. But $a \in \mathbb{R}$, so $\mathbb{R}$ is closed. Why would $a \in \mathbb{R}$? Is it because we assume that $a$ must be finite (therefore it must be in $\mathbb{R}$), otherwise $\lim x_n = a = \infty$ would mean $x_n$ is not converging to a point but diverges instead?

[quote="blahblahblah"]The complement of $\mathbb{R}$ is empty, hence there are no points in the complement and the result is vacuously true.  So the empty set is open and the real line is closed.[/quote]
Correct if I am wrong. You are saying since complement of $\mathbb{R}$ is empty, so definitely there cannot exist any ball in this complement. Empty (ball) set is in the empty (complement) set, so the complement is therefore open.

[quote="Kent Merryfield"]The set $\mathbb{Z}$ of all integers has no limit points. Thus it (vacuously) contains all of its limit points and is closed.
[/quote]
I don't quite get this one. Could you elaborate more?

[quote="Hazel"]Why would $a \in \mathbb{R}?$ Is it because we assume that a must be finite (therefore it must be in $\mathbb{R}$), otherwise $\lim x_n = a = \infty$ would mean $x_n$ is not converging to a point but diverges instead? [/quote]
That's right. In the standard metric topology for $\mathbb{R},$ a sequence that tends to $\infty$ is a divergent sequence. If you introduce a different topology (the one or two point compactification) in which such a sequence converges, then you've changed the whole premise of the problem; we must assume that you meant the standard topology.

As for $\mathbb{Z}$: the only way a sequence of integers could converge would be for it to be eventually constant. But then it's converging to a point that is already a member of the sequence. Such a sequence cannot converge to a number that isn't already a member of the seqence; therefore the set of integers has no limit points.

Here's a theorem:

$a$ is a limit point of $X$ iff for every $\delta>0,$ there are infinitely many members of $X$ in the punctured ball $B(a;\delta)\setminus\{a\}.$

You can see that that can't ever happen for $X=\mathbb{Z}.$

[quote="Kent Merryfield"][quote="Hazel"]Why would $a \in \mathbb{R}?$ Is it because we assume that a must be finite (therefore it must be in $\mathbb{R}$), otherwise $\lim x_n = a = \infty$ would mean $x_n$ is not converging to a point but diverges instead? [/quote]
That's right. In the standard metric topology for $\mathbb{R},$ a sequence that tends to $\infty$ is a divergent sequence. If you introduce a different topology (the one or two point compactification) in which such a sequence converges, then you've changed the whole premise of the problem; we must assume that you meant the standard topology.
[/quote]

I am not familiar with topology. Do I need to state that I am dealing with "[b]standard metric topology[/b] for $\mathbb{R}$"? Or it is understandable and "$\mathbb{R}$" is adequate?

[quote="Kent Merryfield"]
As for $\mathbb{Z}$: the only way a sequence of integers could converge would be for it to be eventually constant. But then it's converging to a point that is already a member of the sequence. Such a sequence cannot converge to a number that isn't already a member of the seqence; therefore the set of integers has no limit points.

Here's a theorem:

$a$ is a limit point of $X$ iff for every $\delta>0,$ there are infinitely many members of $X$ in the punctured ball $B(a;\delta)\setminus\{a\}.$

You can see that that can't ever happen for $X=\mathbb{Z}.$[/quote]

Thank you for this explanation, I understand this part now. But how is it related to the following?

[quote="Kent Merryfield"]
...Thus it (vacuously) contains all of its limit points and is closed.
[/quote]

First you claim it contains no limit points, then you said it contains all of its limit points. Sound contradicting to me. By the way, what do you mean by [i]vacuous[/i].

Every set contains the empty set. The statement "$a$ implies $b$" is vacuously true when $a$ is always false; we can recast "$S$ contains its limit points" as "If $x$ is a limit point of $S$, $x\in S$".

[quote="Hazel"]I am not familiar with topology. Do I need to state that I am dealing with "standard metric topology for $\mathbb{R}$"? Or it is understandable and "$\mathbb{R}$" is adequate? [/quote]
I was getting a little too fancy there. At heart, "topology" is all that stuff about open sets, closed sets, convergent sequences, continuous functions, and so on. To say "the standard metric topology in $\mathbb{R}$" is to say that everything depends on the notion that the distance from $x$ to $y$ is $|x-y|.$ Yes, just saying "$\mathbb{R}$" implies that, if no one specifies anything different.

[quote="Hazel"]First you claim it contains no limit points, then you said it contains all of its limit points. Sound contradicting to me. By the way, what do you mean by vacuous.[/quote]
There's no contradiction here.

Consider the logical statement "If P then Q." The only way this statement could ever be false would be for P to be true and Q false. So, if P is always false, then this statement is true.

Consider this statement: "Every integer that is both even and odd is a perfect seventh power." That statement, weirdly enough, is true. That's because there are no integers that are both even and odd; since there aren't any, I can say anything at all about them and it will be true. This is precisely what I meant (and blahblahblah meant, in a slightly different setting) by [i]vacuously[/i] true. The root of that word is the same as the word "vacuum": there's nothing there.

Now consider the logical statement "If $a$ is a limit point of $\mathbb{Z},$ then $a\in\mathbb{Z}.$"

The front end of this - "If $a$ is a limit point of $\mathbb{Z}$" - is never true, since there are no such limit points. Hence, I can then say anything I want to and the logical statement is still true.

[quote="Hazel"]My naive conclusion would be $\mathbb{Z}$ is open.[/quote]

Why? Does it have a limit point which is not contained in $\mathbb{Z}$?

(supposing that I understood what Kent and others said)

[quote="Hazel"]I still find it hard to grasp the concept although I believe I understand your explanation on the meaning of the word [i]vacuous[/i] used.

Looking at the statement again,
[quote="Kent Merryfield"]The set $\mathbb{Z}$ of all integers has no limit points. Thus it (vacuously) contains all of its limit points and is closed.[/quote]
The first half of the statement says [b]The set $\mathbb{Z}$ of all integers has no limit points[/b]. Which is true (as has been verified). In this case, why talk about [i]vacuous[/i] while there is some integer, in fact all integers, are not limit points. My naive conclusion would be $\mathbb{Z}$ is open.[/quote]

Of course it isn't.  There are no open intervals $(a-\epsilon,a+\epsilon)$ contained in $\mathbb{Z}$.  One way of looking at that is as follows.  $A\subset B$ implies that the cardinality of $A$ is smaller than that of $B$, but $(a-\epsilon,a+\epsilon)$ has continuum cardinality whereas $\mathbb{Z}$ is countable.

On second thought,

[b]If $a$ is a limit point of $\mathbb{Z}$, then $a \in \mathbb{Z}$[/b]. But you have just proven that $\mathbb{Z}$ has no limit points (at all). So the boldfaced statement is vacuously true since $\mathbb{Z}$ vacuously contains all its limit points, hence $\mathbb{Z}$ is closed.

I think I get what you mean now.

One final warning: "open" is not the negation of "closed." If you prove that a set is not closed, that's all you can say: not closed. There are many, many sets that are not closed and also not open.

The relationship between open and closed is this: a set is closed if and only if its complement is closed.

You'll notice that blahblahblah and I worked opposite sides of this. blahblahblah defined "open" and then worked to show that the complement of the set being asked about was open; by invoking limit points, I sought to define "closed" directly. One can prove that these approaches are equivalent.

The only subsets of $\mathbb{R}$ that are simultaneously open and closed are the empty set and $\mathbb{R}$ itself. There's a name for this property: $\mathbb{R}$ is [i]connected[/i].

Sorry for keeping this thread longer than it should. I am already asking questions other than what I have originally initiated.

[b]The set $\left\{\frac1n: n\in\mathbb{N}\right\}\cup\{0\}$ is closed.[/b]

It is taken from the example given by Kent. It is one of the most popular examples, I can say. I have come across this in many books and web tutorials. I thought I fully understood the example, but actually I don't really.

From  the set we can extract a sequence $\left\{\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots \right\}$. It looks quite obvious that it is converging to zero.  Since zero belongs to the set, we conclude that the set is closed. By definition, a set is said to be closed if it contains all its limit points. But we have only shown one.

Here is my reason and hope that it is correct: Let's construct another sequence from the set, say a sequence converging to 1/50. But this element is not a limit point (by the definition of limit point). In fact, none of the elements are limit points, except zero. So it suffices to show only the sequence that converging to zero. In other words the set only has one limit point, namely, the zero.

Kent: Are you saying $\mathbb{R}$ is also open? Or I just misunderstood what you wrote.

[quote="blahblahblah"]Yes, $\mathbb{R}$ is open.[/quote]

How to prove this?

For every $x\in\mathbb{R}$, we have $(x-1,x+1)\subset\mathbb{R}$.  Thus $\mathbb{R}$ is open.

:o That was what I first thought about $\mathbb{R}$ before I was asked to prove it was closed.


I have a price of an object in dollars, with 2 decimal digits, stored in a variable a.
I have a:=15.01.
When I use trunc(a*100), the result is 1500. Why is it? Shouldn't it be 1501?

It could be beacause the floating point variables are not 100% reliable. Instead of 15.01, your variable a could be holding 15.009999999999 and in that case, trunc (100 * a) returns 1500. The best way to aviod this is to use integers instead of reals and store in them the value * 100. (var a: integer; a := 1500;) Or use round, not trunc.


Find all the sequences $(u_n)_{n=1}^{\infty}: u_1=0;u_2=1$
                                                                   $u_n=2(n-1)(u_{n-1}+u_{n-2})\forall n>2$
 :) If you can solve it,can you post the answer or the complete solution or the links has the solution?

[quote="kunterbunt"]What makes you think that this sequence is nice?[/quote]
I don't know but I think it's nice! :)

A difficult problem! who has the solution?


analytic geometry
graphing lines
slope
vector
trigonometry
logarithms

A small disc A is placed on an inclined plane forming an angle $\alpha $ with the horizontal and is imparted an initial velocity $ v_0 $ . Find how the velocity of the disc depends on the angle $ \phi $ between the +x - direction (along incline downwards along line of greatest slope) and velocity vector , if the friction coefficient $ k =\ \tan{ \alpha } $ and at the initial moment $ \phi _0 =\ \frac{\pi}{2} $

Put the positive x-axis in the direction of greatest slope downwards as specified and the positive y-axis in the horizontal direction of the initial disk velocity. Denote the x- and y-velocity components $v_x = \dot x,\ v_y = \dot y$. As long as $v_y \ge 0$, the direction of the velocity vector $\vec v = (v_x, v_y)$ is given by

$\vec n = (\frac{v_x}{v}, \frac{v_y}{v}) = (\cos \phi, \sin \phi),\ \ \tan \phi = \frac{v_y}{v_x}$

and the friction force $\vec F$ opposes the motion in this direction:

$\vec F = -k mg \cos \alpha\ \vec n = -mg \sin \alpha\ \vec n$

The equations of motion of the disk sliding on the inclined plane are then

$m\ddot x = mg \sin \alpha - k \cos \phi\ mg \cos \alpha = mg \sin \alpha (1 - \cos \phi)$

$m\ddot y = -k \sin \phi\ mg \cos \alpha = - mg \sin \alpha \sin \phi$

or

$m\dot v_x = mg \sin \alpha (1 - \cos \phi)$

$m\dot v_y = -mg \sin \alpha \sin \phi$

Consequently,

$\frac{\text dv_y}{\text dv_x} = \frac{\dot v_y}{\dot v_x} = \frac{\sin \phi}{\cos \phi - 1} = \frac{\tan \phi}{1 - \sqrt{\tan^2 \phi + 1}} = \frac{\frac{v_y}{v_x}}{1 - \sqrt{(\frac{v_y}{v_x})^2 + 1}}$

This is a 1st order homogenous differential equation, separable by the standard substitution $z = \frac{v_y}{v_x}$. Differentiating $v_y = v_x z$ with respect to $v_x$, we get $v_y' = z + v_x z'$ and substituting this into the above differential equation,

$z + v_x z' = \frac{z}{1 - \sqrt{z^2 + 1}}$

$\frac{\text dz}{z\sqrt{z^2 + 1}} - \frac{\text dz}{z} = \frac{\text dv_x}{v_x}$

$-\ln \frac{1 + \sqrt{z^2 + 1}}{z} - \ln z = \ln v_x - \ln C$

$-\ln (1 + \sqrt{z^2 + 1}) = \ln \frac{v_x}{C}$

$\frac{1}{1 + \sqrt{z^2 + 1}} = \frac{v_x}{C}$

$v_x(1 + \sqrt{z^2 + 1}) = v_x + \sqrt{v_y^2 + v_x^2} = C$

where C is an arbitrary constant. At the time t = 0, we have $v_x = 0, v_y = v_0$, hence $C = v_0$.

$v_x + \sqrt{v_y^2 + v_x^2} = v_0$

Substituting $v_y = v_x \tan \phi$, we get

${v_x(1 + \sqrt{\tan^2 \phi + 1}) = v_x(1 + \frac{1}{\cos \phi}}) = v_0$

$v_x = \frac{v_0 \cos \phi}{1 + \cos \phi}$

$v_y = v_x \tan \phi = \frac{v_0 \sin \phi}{1 + \cos \phi}$

$v = \sqrt{v_x^2 + v_y^2} = \frac{v_0}{1 + \cos \phi}$

for $0 \le \phi \le \frac \pi 2$, which leads to $v = v_x = \frac{v_0}{2},\ v_y = 0$ for $\phi = 0$.

You've got a greta solution from the basics which basically involves solving some differential eqns. But I have a two line solution.It comes only afetr extensive thought. Think you can come up with it??!!!! :lol:

Just observe your own solution. I think yo'll be able to make out!!!

I can set up a simpler (no square roots) differential equation, if you do not like the one I used. Not that it makes any substantial difference.

$\frac{\text dv_y}{\text dv_x} = -\frac{\sin \phi}{1 - \cos \phi}$

was established from the equations of motion. Differentiating $v_y = v_x \tan \phi$ with respect to $v_x$,

$\frac{\text dv_y}{\text dv_x} = \tan \phi + \frac{v_x}{\cos^2 \phi}\ \frac{\text d\phi}{\text dv_x}$

Consequently,

$-\frac{\sin \phi}{1 - \cos \phi} = \tan \phi + \frac{v_x}{\cos^2 \phi}\ \frac{\text d\phi}{\text dv_x}$

which is a separable 1st order differential equation for $v_x(\phi)$

$-\frac{\sin \phi}{1 - \cos \phi} - \tan \phi = -\frac{\sin \phi}{\cos \phi (1 - \cos \phi)} = \frac{v_x}{\cos^2 \phi}\ \frac{\text d\phi}{\text dv_x}$

$\frac{\text dv_x}{v_x} = -\frac{1 - \cos \phi}{\cos \phi \sin \phi}\ \text d\phi = -\frac{(1 - \cos \phi)\sin \phi}{\cos \phi (1 - \cos^2 \phi)}\ \text d\phi = -\frac{\sin \phi\ \text d\phi}{\cos \phi (1 + \cos \phi)}$

Subsituting $z = \cos \phi,\ \text dz = -\sin \phi\ \text d\phi$,

$\frac{\text dv_x}{v_x} = \frac{\text dz}{z(1 + z)} = \frac{\text dz}{z} - \frac{\text dz}{1 + z}$

$\ln v_x = \ln z - \ln (1 + z) + \ln C = \ln \frac{Cz}{1 + z}$

$v_x = \frac{Cz}{1 + z } = \frac{C \cos \phi}{1 + \cos \phi}$

$v_y = v_x \tan \phi = \frac{C \sin \phi}{1 + \cos \phi}$

By the initial conditions, $v_y = v_0$ for $\phi = \frac \pi 2,\ \sin \phi = 1,\ \cos \phi = 0$, hence $C = v_0$, etc.

However, I cannot read your mind to find the solution you like among all possible ones, hence, I pass.

Yetti

Consider Newton's law along the tangent to the trajectory.

Also consider the acceleration along X direction.

You'll see omething beautiful from which the answer will follow in one line!!!!

Rushil... Yetti has spend time to post two quite impressive solutions and clearly indicated that he has exhausted his effort... If I were him, I wouldn't go randomly searching my solution either.

Why don't you just post it already... Because I'm sure there are many people who would like to see it.

Fine I will.I'm in a bit of hurry now and I'll post them later!!!

I double posted because the site froze for a long time right in the middle of my post being done. I actually had to submit it about 7 times before I thought it went through. Meanwhile no other pages would load so I couldn't check. Trust me, we'll all be fine...

hey , whya re you double posting spoon.
I told you I will give the solution!!!!

[quote="Rushil"]Hey, yetti !!!!

You are participating in the physics contest, right???[/quote]
Unfortunately, I am too old for this, but I did mathematical olympiads, when I was a teenager.

This problem was confusing to me :blush:, until I tilted my dining table and flicked a quarter (25 cent coin) across it in the horizontal direction :idea:.

Yetti

That's great!!!

Nothings better than Experiment.
I'll post the 2-line solution I have in some time!!!

The acceleration of the disc along the plane is dtermined by the projection of the force of gravity on this plane $F_x = mg \sin{\alpha} $ and the friction force $ F_{fr} = kmg \cos{\alpha} $. In our case, k $ = \tan{\alpha} $ and thus

$ F_{fr} = F_{x} = mg \sin{\alpha} $

the projections of the acceleration on the direction of the tangent to the trajectory and on the x axis are -
$ m w_{\tau} = F_x \cos{\phi} - F_{fr} = mg \sin{\alpha} ( \cos{\phi} -1 ) \\
\\ m w_{x} = F_x  - F_{fr}\cos{\phi} = - mg \sin{\alpha} ( \cos{\phi} -1 )  $

Thus $ w_{\tau} = - w_x $ which means that the velocity $v$ and its projection $ v_x $ differ only by a constant . The constant can be found using the initial condition that $ v= v_0 $
$v_x = v \cos{\phi} $

Thus $ v = -v_{x} + v_0 $
thus, $ v = \frac{v_0}{1 + \cos{\phi}}  $


[quote="now a ranger"]HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:[/quote]

Stop spamming up this place. :mad:

You have posted this same message 50+ times.

Same words but in different communities.

And in a time span of 5 minutes.

And, by the way, if you didn't spam up this place, you might want to know that India has a 'lot' of new years as each religion and each state in India celebrates it at different times.

 :wink:

Example: Tamil New Year is like April 14th or something around there

[quote="junggi"][quote="anirudh"][quote="now a ranger"]HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:[/quote]

Stop spamming up this place. :mad:
[/quote]
yea, way to ruin his new year
and you're NOT the one to be lecturing about spam.
In many people's eyes, you're one of the most annoying/worst spammers aops had/has.(thank god it died down a bit, but it seems to be starting again)[/quote]

I have sent you a private message. Open and respond.

[quote="anirudh"][/quote][quote="junggi"][quote="anirudh"][quote="now a ranger"]HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:[/quote]

Stop spamming up this place. :mad:
[/quote]
yea, way to ruin his new year
and you're NOT the one to be lecturing about spam.
In many people's eyes, you're one of the most annoying/worst spammers aops had/has.(thank god it died down a bit, but it seems to be starting again)[/quote][quote="anirudh"]

I have sent you a private message. Open and respond.[/quote]

ANIRUDH CHILLL!!! You have a horrible temper and bad behaviour. You act like a childish 7 year old!!!  :mad:

It is a good thing that I already celebrated last night with a party for Happy NEw Year, while you were probably on AoPS writing that annoying post. We sang HAppy New Year songs and had a blast. What is wrong with saying Happy New year? It IS the new year!

Dude improve your adamant attitude , and your utterly paranoid posts are very representable of your "behaviour".

And just admit it. You spam a lot more than anyone else.

And junggi was right. You DO spam a lot. just check lots of the threads you have posted in. How many of them are really good posts? And as to your AMCs and stuff, that is just copying and pasting problems and altering them a bit. Not very original.

There.  :lol: I shall now end my post

I bet your private message was harsh and cruel. and you've spammed up this thread with two useless posts.

ANYWAY HAPPY NEW YEAR!!!

Happy New Year  :cool:  :rotfl:  :laugh:


limit
function
symmetry
trigonometry
calculus
calculus computations

Given $ f: (a,b)\rightarrow\mathbb{R}$ and $ x\in (a,b)$ consider the statements:

(i) $ \lim_{h\rightarrow 0}|f(x\plus{}h)\minus{}f(x)|\equal{}0$ and (ii) $ \lim_{h\rightarrow 0}|f(x\plus{}h)\minus{}f(x\minus{}h)|\equal{}0$

Show that (i) always implies (ii).

Thank you.

$ 0\equal{}\lim_{h\to 0}|f(x\plus{}h)\minus{}f(x)|\equal{}\lim_{\minus{}h\to 0}|f(x\plus{}h)\minus{}f(x)|\equal{}\lim_{h\to 0}|f(x\minus{}h)\minus{}f(x)|\equal{}0$

$ \lim_{h \to 0} |f(x\plus{}h)\minus{}f(x\minus{}h)| \equal{}\lim_{h \to 0} |f(x\plus{}h)\minus{}f(x)\plus{}f(x)\minus{}f(x\minus{}h)| \le$
$ \lim_{h \to 0} |f(x\plus{}h)\minus{}f(x)|\plus{}|f(x)\minus{}f(x\minus{}h)| \equal{} 0\plus{}0\equal{}0$

Thank you.  Much simpler than I thought.

Find a counterexample to show that (ii) does not imply (i).

Apparently Kent knows the question.  I don't get your "spoiler", though.

Clearly the counterexample has to be discontinuous at x. You can take (ir)rational sequences tending to x to verify nonexistence of limit.

What you really want is that the function $ f(x \plus{} h) \minus{} f(x \minus{} h)$ is continuous in $ h$ for fixed $ x$ but that the function $ f(x \plus{} h) \minus{} f(x)$ is not.  In other words, a symmetry about $ x$.

spoiler is the answer to kents question =(

$ \lim_{h\rightarrow 0}|f(a \plus{} h) \minus{} f(a \minus{} h)| \equal{} b$
$ \lim_{h\rightarrow 0}|f(a \plus{} h) \minus{} f(a)| \equal{} c$

Expample:
$ b \equal{} 0, c \equal{} \pi$: $ f(x) \equal{} 0$ if $ x\ne a$ and $ f(a) \equal{} \pi$.
$ b \equal{} 0, c \equal{} \infty$: $ f(x) \equal{} \frac {1}{|x \minus{} a|}$ if $ x\ne a$ and $ f(a) \equal{} \pi$.
$ b \equal{} 0, c$ does not exist: $ f(x) \equal{} \frac {\sin(\frac {1}{|x \minus{} a|})}{|x \minus{} a|}$ if $ x\ne a$ and $ f(a) \equal{} \pi$.
$ b \equal{} \pi, c$ does not exist: $ f(x) \equal{} \pi$ if $ x\ge a$ else $ f(x) \equal{} 0$.


geometry
analytic geometry
geometry proposed

Find the minimum of the area that a convex pentagon with the vertices lattice points (i.e. their coordinates in an orthonormal system are integers) can have.

The trick used to solve this seems "fallen from heaven" to say the least.

Is there more than one solution to this problem?

the leading idea should be that you cannot have a pentagon whit less than one lattice-point  inside. Then the minimum area is 5/2.


Let  $ F\equal{}{2}^{{2}^{n}}$ , posotive integer n , satisfying  F|($ 3^\frac{F\minus{}1}{2}\plus{}1$)
Prove that : F is a prime

Let F=$ 2^{{2}^{n}}\ + 1$, posivite integer n, satisfying F|(${ 3^\frac{F-1}{2}}+1$)
Prove that : F is a prime


For example:
k00lperson was having a good day.
Please bold, and if you want you can post consecutively.
Just nothing bad except for if you are playing around

I will start:
[b]Spring is coming![/b]

Spring is coming! [b]Mr. Williamson decided to go outside to enjoy the air.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. [b]He wanted to go to the park, but he had work to do.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. [b]First, he decided to go to the mall, because he needed some new pants.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. [b]When he got in his car, the engine wouldn't start.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. [b]He decided to go borrow a jumper from his neighbor, since he did not have one.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. [b]He managed to trip over poor k00lperson[/b]

:D :D :D

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. [b]k00lperson burst out into ABC's.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. k00lperson burst out into ABC's. Spaceguy started :rotfl: ing at k00lperson.

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. k00lperson burst out into ABC's. Spaceguy started Rotfl ing at k00lperson.[b]Then, Spaceguy died thanks to Mr.Williamson who took a gun and shot him  :P [/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. k00lperson burst out into ABC's. Spaceguy started  :Rotfl:  ing at k00lperson.Then, Spaceguy died thanks to Mr. Williamson who took a gun and shot him  : Razz :  . [b]Someone saw this and decided to call  :police: .[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. k00lperson burst out into ABC's. Spaceguy started :Rotfl: ing at k00lperson.Then, Spaceguy died thanks to Mr. Williamson who took a gun and shot him : Razz : . Someone saw this and decided to call Police . [b]Soon, the police came and caught Mr. Williamson, and of course Mr. Williamson started to protest.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. k00lperson burst out into ABC's. Spaceguy started :Rotfl: ing at k00lperson.Then, Spaceguy died thanks to Mr. Williamson who took a gun and shot him : Razz : . Someone saw this and decided to call Police . Soon, the police came and caught Mr. Williamson, and of course Mr. Williamson started to protest. [b]Spaceguy's ghost started to haunt Mr. Williamson, and the latter went insane.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. k00lperson burst out into ABC's. Spaceguy started :Rotfl: ing at k00lperson.Then, Spaceguy died thanks to Mr. Williamson who took a gun and shot him : Razz : . Someone saw this and decided to call Police . Soon, the police came and caught Mr. Williamson, and of course Mr. Williamson started to protest. Spaceguy's ghost started to haunt Mr. Williamson, and the latter went insane.[b] Meanwhile, K00lperson finished ABC's and started counting digits of pi.[/b]

Spring is coming! Mr. Williamson decided to go outside to enjoy the air. He wanted to go to the park, but he had work to do. First, he decided to go to the mall, because he needed some new pants. When he got in his car, the engine wouldn't start. He decided to go borrow a jumper from his neighbor, since he did not have one. He managed to trip over poor k00lperson. k00lperson burst out into ABC's. Spaceguy started :Rotfl: ing at k00lperson.Then, Spaceguy died thanks to Mr. Williamson who took a gun and shot him : Razz : . Someone saw this and decided to call Police . Soon, the police came and caught Mr. Williamson, and of course Mr. Williamson started to protest. Spaceguy's ghost started to haunt Mr. Williamson, and the latter went insane. Meanwhile, K00lperson finished ABC's and started counting digits of pi. [b]Then at Mr. Williamson's office Bill his cubicle buddy, wondered where he was and started to investigate[/b].

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized.[b] However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious.[/b]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious.  [b]Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5.[/b]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". [b]Meewhee then danced, to the beat of k00lperson's cry.[/b]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". Meewhee then danced, to the beat of k00lperson's cry.[b]However, her clumsy dancing knocked over a mafia member, who called his boss and goons with machine guns started to come out of the back alleys and shoot[/b]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". Meewhee then danced, to the beat of k00lperson's cry.However, her clumsy dancing knocked over a mafia member, who called his boss and goons with machine guns started to come out of the back alleys and shoot.  [b]Sabien5 leaps from inside the lead mafia members shoe and throws a primed mortar at the goons.[/b]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". Meewhee then danced, to the beat of k00lperson's cry.However, her clumsy dancing knocked over a mafia member, who called his boss and goons with machine guns started to come out of the back alleys and shoot. Sabien5 leaps from inside the lead mafia members shoe and throws a primed mortar at the goons.[b] However, the freak wind comes and blows on it so hard that the primed mortar goes back into her face, 2 milliseconds after it leaves her hands, killing her.[/b]

[hide="FYI for some people who dont understand"]ok
1.)im a boy
2.)SUPER BOB just transformed into Sabien5 so obviously Sabien5 is invincible from such things[/hide]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". Meewhee then danced, to the beat of k00lperson's cry.However, her clumsy dancing knocked over a mafia member, who called his boss and goons with machine guns started to come out of the back alleys and shoot. Sabien5 leaps from inside the lead mafia members shoe and throws a primed mortar at the goons. However, the freak wind comes and blows on it so hard that the primed mortar goes back into [b]his[/b] face, 2 milliseconds after it leaves [b]his[/b] hands, killing [b]him[/b]. [b]But obviously he didn't die and just went with plan B and kicked the goons with giant bowling shoes.[/b]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". Meewhee then danced, to the beat of k00lperson's cry.However, her clumsy dancing knocked over a mafia member, who called his boss and goons with machine guns started to come out of the back alleys and shoot. Sabien5 leaps from inside the lead mafia members shoe and throws a primed mortar at the goons. However, the freak wind comes and blows on it so hard that the primed mortar goes back into his face, 2 milliseconds after it leaves his hands, killing him. But obviously he didn't die and just went with plan B and kicked the goons with giant bowling shoes. [b]Meanwhile, the flying primed mortar flies back into the mafia group and explodes, waking up Yoshi, who then decides to load Krypton Eggs into a launcher aimed at the mafia.[/b]

On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". Meewhee then danced, to the beat of k00lperson's cry.However, her clumsy dancing knocked over a mafia member, who called his boss and goons with machine guns started to come out of the back alleys and shoot. Sabien5 leaps from inside the lead mafia members shoe and throws a primed mortar at the goons. However, the freak wind comes and blows on it so hard that the primed mortar goes back into his face, 2 milliseconds after it leaves his hands, killing him. But obviously he didn't die and just went with plan B and kicked the goons with giant bowling shoes. Meanwhile, the flying primed mortar flies back into the mafia group and explodes, waking up Yoshi, who then decides to load Krypton Eggs into a launcher aimed at the mafia.[b] However, he discovered that the rubber bands that had allowed the launcher to launch had been burned, and he ran off 10,000,000,000 light years away to find more.[/b]

[quote="Richard_Min"]On a dark and rainy day, Bob decided to go to the mall. Suddenly, a heat seeking missle hits the ground 1 kilometer away. Bob hears this with his super hearing and becomes SUPER BOB!!! However Super Bob's nemisis Yoshi came by, and had Krypton, which forced Super Bob to be immobile within it's presence without a Diamond shield in the way.Then Yoshi decides to egg Richard with Krypton Eggs, causing him to be immobilized. However, at the time Richard happened to be 1 million miles away, and a freak tornado sweeps by and hits Yoshi on the head, knocking him unconscious. Super Bob, recovering from the Krypton decides to use his Super Awesome Dangerous Crazy Super Awesome Transformation Sequence and turn into Sabien5. K00LPERSON CRIED"ZOMG...U SHOULDN'T HAVE STARTED ANOTHER ONE WITHOUT TELLING ME!". Meewhee then danced, to the beat of k00lperson's cry.However, her clumsy dancing knocked over a mafia member, who called his boss and goons with machine guns started to come out of the back alleys and shoot. Sabien5 leaps from inside the lead mafia members shoe and throws a primed mortar at the goons. However, the freak wind comes and blows on it so hard that the primed mortar goes back into his face, 2 milliseconds after it leaves his hands, killing him. But obviously he didn't die and just went with plan B and kicked the goons with giant bowling shoes. Meanwhile, the flying primed mortar flies back into the mafia group and explodes, waking up Yoshi, who then decides to load Krypton Eggs into a launcher aimed at the mafia. However, he discovered that the rubber bands that had allowed the launcher to launch had been burned, and he ran off 10,000,000,000 light years away to find more. Then, Sabien seriously died and by the way, he is not immortal!!!!!!!![b]Then aliens came to have revenge on Yoshi's planet, and the planet united to defeat the invading aliens after killing Yoshi for calling them to this planet.[/quote]
[b]You guys.this is getting too long, and it's getting seriously boring, so let's start over.......
Srry..for people who want to continue this, I want to start a new one, so ya..
[b]Joseph was standing in the yard, watching over his dog when he saw a black car drive buy, and the guy inside the car looked very suspiscious.[/b][/b]

Joseph was standing in the yard, watching over his dog when he saw a black car drive buy, and the guy inside the car looked very suspiscious.[b] Suddenly, the car exploded after a eerie humming noise.[/b]

Joseph was standing in the yard, watching over his dog when he saw a black car drive buy, and the guy inside the car looked very suspiscious. Suddenly, the car exploded after a eerie humming noise. Joseph then saw a tiny leprechaun, dancing and cackling, pointing to the burning wreckage.

(idk)


inequalities
trigonometry
geometry proposed
geometry

prove that in any triangle
$ p<8R^3\sin\alpha\sin\beta\sin\gamma$    :roll:


What is the least integer divisible by 11 that is greater than
2003?

First of all, $ 2\minus{}0\plus{}0\minus{}2\equal{}0$, which is divisible by $ 11$, so $ 2002$ is divisible by $ 11$.

Next integer greater than that is just $ 2002\plus{}11\equal{}\boxed{2013}$.


trigonometry
geometry
area of a triangle
Heron\u0027s formula

In triangle $ ABC$, let $ s$ be the semiperimeter, and $ r$ the inradius. Prove the relationship

$ (s\minus{}a)^2(1\minus{}\cos A)^2\plus{}r^2\cos^2 A \equal{} (s\minus{}b)^2(1\minus{}\cos B)^2\plus{}r^2\cos^2 B$

[hide="Proof"]\begin{eqnarray*}1 - \cos A & = & 1 - {b^2 + c^2 - a^2\over 2bc} \\
& = & {a^2 - (b - c)^2\over 2bc} \\
& = & {(a - b + c)(a + b - c)\over 2bc} \\
& = & {2(s - b)\cdot 2(s - c)\over 2bc} \\
& = & {2(s - b)(s - c)\over bc}\end{eqnarray*}

If $ P$ is the area of the triangle, the LHS (with the help of Heron's formula and $ P = rs = {bc\over 2}\sin A$) now becomes

\begin{eqnarray*}(s - a)^2{4(s - b)^2(s - c)^2\over b^2c^2} + r^2\cos^2A & = & {4P^4\over s^2b^2c^2} + r^2\cos^2A \\
& = & \frac {4P^2r^2}{\left({2P\over\sin A}\right)^2} + r^2\cos^2A \\
& = & r^2\sin^2A + r^2\cos^2A \\
& = & r^2\end{eqnarray*}

Hence LHS doesn't depend on the choice of the angle/side. QED.[/hide]


If none of the numbers $ a,a\plus{}d,a\plus{}2d,...,a\plus{}(n\minus{}1)d$ is divisible by $ n$, then $ d$ and $ n$ are coprime.

What if $ a\equal{}1$ and $ d\equal{}2$ and $ n\equiv{0}\pmod{2}$?  Then $ a$, $ a\plus{}d$, $ a\plus{}2d$, $ \ldots$ are all odd and therefore not divisible by $ n$, but $ n$ and $ d$ are not coprime...

The conclusion should be that $ d$ and $ n$ are not coprime...

Proof by contradiction
[hide]If $ d$ and $ n$ are coprime, then $ x \equiv y \pmod n \iff a \plus{} xd \equiv a \plus{} yd \pmod n$. Hence, these number are all distinct modulo $ n$. But there are $ n$ numbers, so one must be divisible by $ n$. Contradiction.
[/hide]


limit
function
logarithms
algebra
polynomial
calculus
calculus computations

Consider equation
$ x^{n}\equal{}x\plus{}n$
a)Prove that equation has unique solution on $ (0,\plus{}\infty)$
b)$ \lim_{n\to\infty}x_{n}\equal{}1$
c)$ \lim{n\to\infty}\frac{n(x_{n}\minus{}1)}{ln n}$=?

a) The function $ f_{n}(x) \equal{}\frac{x\plus{}n}{x^{n}}$ is decreasing, continuous on $ (0,\infty)$, $ \lim_{x\to\infty}f_{n}\equal{} 0 < 1, f(1) \equal{} n\plus{}1 > 1$, therefore we're done.

[hide="b)hint "]$ 1 < x_{n}< u_{n}\to 1$ when $ n\to\infty$.[/hide]

[hide="c)hint"]$ n \equal{}\frac{\ln (x_{n}\plus{}n)}{\ln x_{n}}$[/hide]

Let $ f(n)$ is a polynomial of n
$ x^{n}\equal{} x\plus{}f(n)$
Prove that
1)There exist  unique solution on $ (0,\plus{}\infty)$ .Call $ x_{n}$
2)Find $ \lim_{n\to\infty}x_{n}$
3)Find all k such that
$ lim_{n\to\infty}\frac{n^{k}(x_{n}\minus{}1)}{ln n}$ exist and finite.

See the first exercise from http://www.mathlinks.ro/Forum/viewtopic.php?t=70659

Hint : $ \lim_{n\to\infty}f(n)^{\frac{1}{n}}\equal{} 1$(least cofience >0)


linear algebra
matrix
graph theory
linear algebra unsolved

Let $ M (a, b, c)$ denote a matrix with order $ n \equal{} a \plus{} b \plus{} c \minus{} 2$ composed of
three intersecting block matrices $ J_a \minus{} I_a$, $ J_b \minus{} I_b$, $ J_c \minus{} I_c$,
where $ J_n$ denote the matrix with all ones with dimension $ n$ and $ I_n$ is the identity matrix.
\[ M (a, b, c) \equal{} \left[
\begin{array}{ccccc}
J_{a\minus{}1} \minus{} I_{a\minus{}1} & 1 & 0 & 0 & 0 \\
1 & 0 & 1 & 1 & 0 \\
0 & 1 & J_{b\minus{}2} \minus{} I_{b\minus{}2} & 1 & 0 \\
0 & 1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & J_{c\minus{}1} \minus{} I_{c\minus{}1} \\
\end{array}
\right ],\]
where $ 1$ denote the matrix of ones with appropriate size.

Prove that among all matrices $ A \equal{} M (a, b, c)$ of fixed size,
the trace of $ A^{2010}$ is maximized for $ a \equal{} 2$, $ b \equal{} n \minus{} 2$, $ c \equal{} 2$.

A loose and unconvincing piece of intuition:

This is a graph. The trace of a high power of the matrix is some kind of measurement of how well-connected it is.

The graph consists of the following elements: sets $ A,B,$ and $ C$ and extra points $ x$ and $ y.$ $ A,B,$ and $ C$ are complete graphs with $ a\minus{}1,b\minus{}2,$ and $ c\minus{}1$ vertices, respectively. Point $ x$ is stuck between sets $ A$ and $ B$ and has edges connecting it to every point in $ A$ and every point in $ B.$ Point $ y$ is similarly stuck between sets $ B$ and $ C.$ There's also a direct edge connecting $ x$ to $ y.$

The extreme case that's supposed to maximize that trace: just the complete graph on $ n\minus{}2$ vertices with two outlier points, each connected to a solitary (and different) vertex of that complete graph. That sounds about as connected as you can make it.

Not a proof.

May I suggest alternate notation? Let $ U$ be the $ (a\minus{}1)\times b$ matrix with ones in the first column and zeros elsewhere, and let $ V$ be the $ b\times (c\minus{}1)$ matrix with ones in the bottom row and zeros elsewhere. Then we can write
\[ M(a,b,c)\equal{}\begin{bmatrix}J_{a\minus{}1}\minus{}I_{a\minus{}1}&U&0\\U^T&J_b\minus{}I_b&V\\0&V^T&J_{c\minus{}1}\minus{}I_{c\minus{}1}\end{bmatrix}\]

You are right, that is exactly what I had in mind :)

One need to compare the number of closed walks of length 2010 in graph $ M(a, b, c)$.
Probably the best and easiest way to do this is to establish some injections.
If a > b, from the set of closed walks of the graph $ M (a, b, c)$ to the set of closed walks of the graph $ M (b, a, c)$ of length 2010,
and from $ M (a, b, c)$ to $ M (a \minus{} 1, b \plus{} 1, c)$.
Finally, we get $ M (2, n \minus{} 2, 2)$ is the extremal graph with the maximum number of closed walks of length 2010.


Quisiera que dedicasemos este topic para que vayan colocando cuales son los participantes que conforman el equipo Ibero de cada pais.

Nuestro selectivo se hace el 18 y 19 de agosto, es decir en tres semanas, asi que hasta el 25 no vamos a saber. De todas formas, uno supone que Severius (plata en esta IMO) y Gaby (Bronce) iran y los demas saldran del resto del equipo de IMO (contando a los suplentes) mas algunos mas viejos  que no pudieron participar en IMO.
De todas formas, espero verlo en Cartagena!!

Bueno, ya los selectivos fueron hechos para el equipo Venezolano. El equipo Ibero es
Leonardo Urbina (yo!, IMO 2005)
Roland Hablutzel (Mencion IMO 2005)
Victor Villamizar
Rafael Guedez

Espero oir de otros equipos pronto

Ya está la selección de México:
Guevara Guevara
Joshua Hernández
Pablo Soberón
David Torres  :)
Los tres primeros sacaron bronce en la IMO y yo saqué mención.
Nos vemos en Cartagena

El quipo de El Salvador ya esta
Gabriel Chicas (Mencion IMO 2005)
Mario Castro (IMO 2005)
Otto Tang
Jose Hernandez (IMO 2005)
 :ninja:

Hola chicos ...  soy nueva en el forum

mi nombre es Fabiola Bravo ...de Peru  :lol:

y quisiera saber cuando seran los examenes de la IBERO este año  :?
...

Gracias...

 :rotfl:

Bueno, ya esta el equipo de Argentina, va a ser:

-German Gieczewski (Severius, Plata IMO 2005) (perdon Severius por haber escrito el apellido mal)
-Marcos Cossarini
-Maximiliano Camporino (Yo :), IMO 2005)
-Gabriel Carvajal (Bronce IMO 2005, Mencion IMO 2004)

Bueno eso es todo, nos vemos en Cartagena.

hola esta es la lista parcial de la delegacion de Peru..

Claudio Espinoza (BRONCE IMO 2003, MENCION HONROSA IMO 2002)
Sergio Vera Patiño (BRONCE IMO 2004, MENCION HONROSA IMO 2003)
Franco Vargas Pallete (BRONCE IMO 2005)
OTROS

son chicos del cole en el q estudio : BERTOLT BRECHT:
[url=http://www.aduni.com.pe]www.aduni.com.pe[/url]

espero les vaya bien en colombia...

bye

Bueno aqui el equipo anfitrion:

Pascual Restrepo Mesa (imo)
Betty Angelica Rincon (primera participacion)
Alejandro Samper (primera participacion) (esta en octavo!)
Federico Castillo (Primera Partcipacion) (tambien esta en octavo creo)

Como ven nuestra politica es llevar solo uno de la Imo y foguear el resto de los participantes mas jovenes.

Dada estas circunstancias no creo que colombia gane la ibero, pero si sera interesante ver como les va a estos jovenes.

Por cierto todos los participantes de los ultimos 5 años iran de guias y muchos participantes de la decada de los 80, ahora importantes matematicos iran de coordinadores

Yo soy el mas viejo del equipo y esta es mi ultima olimpiada...

Nos vemos en Cartagena

Pascual, conoces el proceso de elaboracion de los examenes de la IBERO?..
como sabemos.... para la IMO, existe un shortlist ...de donde se sacan las 6 preguntan que conforman los dos examenes...
mi pregunta es ... si hay algo parecido para la elaboracion de la IBERO...
y si fuese asi, como conseguir estos "IBERO's shortlist"...

Garcias por la informacion que puedas brindarme..

Carlos
Lima - Peru

yo tengo como 3 shorlists de la ibero.

otra cosa: nos mantienen avisado sobre lo que pasa en la ibero, los problemas y los resultados. que no voy este año  :( ,  gracias y mucha suerte a todos

si, hay shortlist de ibero, se consigue con los lideres de sus paises en los respectivos años.

las pruebas son esta semana y ya hay un pdf con los equipos e informacion en la pagina de la olimpiada. mirenla si tienen tiempo.

Suerte

Bueno, tengo una idea ..... colgar los shortlist
 de las IBERO y otras competencias regionales de por aqui ... asi como en el caso de las IMO's... (q ya se hace)
espero compartan mi idea...

bueno.... la delegacion de peru esta saliendo el miercoles..

creo que ya alguien puso una lista previa, aqui la lista completa:  :lol:

 Claudio Espinoza.
 SERGIO Vera
 David Uribe
 Franco Vargas

Suplentes:

 Jossy Alva
 Oscar Garay

Suerte a todos y ... a colgar los examenes ...
para su solucion..  ;)

Carlos Bravo
Lima - peru  :lol:


1)A bucket of water weghing 20k.g is slowly raised from a depth of 20m to ground level by a rope.The mass of the rope is 1.2kg per metre.Find the work done.

2)When a ball is thrown up,the magnitude of its momentum decreases and then increases.Does this violate the conservation of momentumprinciple?

3)Twoparticles of mass 1k.g and 3k.g move towards each other under their mutual force of attraction.No other force acts on them.When their relative velocity of approach is 2m/s,their centeres of mass have a velocity of 0.5m/s.When the relative velocity of approach becomes 3m/s,the velocity of centre of mass is 0.75m/s.True or False?

4)A test tube of mass 10gm closed with a cork of mass 1gm contains some ether.When the test tube is heated,the cork flies out under the pressure of ether gas.The test tube is suspended by a weightless rigid bar of lenght 5cm.What is the minimum velocity with which the cork is to fly out of the test tube so that the test tube describes a full vertical circle about the pivot?Neglect mass of ether.$g=9.8m/s^2$.

5)A ball moving with a speed of 9m/s strikes an identical ball such that after the collision the direction of each ball makes an angle 30degrees with the original line of motiton.Find speeds of the two balls.

[quote="bus"]Oh, c'mon [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=10016]shadysaysurspammed[/url], you can do these. Or is it just an exercise for the rest of us? :P[/quote]

 :oops: i am not that good art physics.....okay i'll tell you the problems in which my answer isn't mathching,the 3rd and 4th one.

[quote="bus"]To question 3: At the beginning the two particles are both at rest? Or in other words, what do you mean by "the velocity of center of mass is 0.75 m/s" ? What is the velocity relative to?

To question 4: Is the test tube at the beginning oriented horizontally?[/quote]

The test tube is horizontal and the third one,that's how the question is given,i don't know it myself :lol:

The problem #3 is really odd. The velocity of the center of mass of the whole system shouldn't change and the velocities of the two particles should be in 3:1 ration relative to the center of mass of the system. The nubers in the problem statement therefore could be velocities of the heavier particle realtive to the center of mass of the system.

[quote="bus"]Then why don't you use energy? The tube needs to have enough kinetic energy to get to the highest point of the circle:

$\frac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

If the velocity of the cork is $v_c$ and its mass $m_c$ then we get:

$m_cv_c=mv=m\sqrt{2gh}$

$v_c=\frac{m}{m_c}\sqrt{2gh}=10\sqrt{9.8\times0.1\text{ m}^2\text{/s}^2}\approx9.9\text{ m/s}$[/quote]

Dude,that is what i did,though I don't remember what answer i had got,the answer given is 15.65m/s...maybe my calculkations are screwed up.Are you saying [b]false[/b] for the second one?

If the problem asked about the velocity of center of mass [b]of the second paricle[/b] then the answer would be true, otherwise false. I think :).

In the other one, I made a mistake, I only calculate $\sqrt{gh}$ instead of $\sqrt{2gh}$. The right answer is:

$v_c=\frac{m}{m_c}\sqrt{2gh}=10\sqrt{2\times9.8\times0.1\text{ m}^2/\text{s}^2}=14\text{ m/s}$

That's weird...


$ f(x)$ is a given polynomial whose degree at least 2. Define the following polynomial-sequence: $ g_1(x)\equal{}f(x), g_{n\plus{}1}(x)\equal{}f(g_n(x))$, for all $ n \in N$. Let $ r_n$ be the average of $ g_n(x)$'s roots. If $ r_{19}\equal{}99$, find $ r_{99}$.

The sequence $ r_n$ is constant, so $ r_{99}= 99$ as well. To see this we will use a simple formula for $ r_n$: If $ p(x) = \alpha_d x^d + \alpha_{d-1}x^{d-1} + \ldots$ is a polynomial of degree $ d$, then the sum of its roots is $ -\alpha_{d-1}/\alpha_d$ and the average is

$ r = -\frac{\alpha_{d-1}}{d \alpha_d}$.

Let $ f(x) = a_m x^m + a_{m-1}x^{m-1} + \ldots +a_1 x + a_0$. For a fixed but arbitrary $ n$, let us write

$ g_n(x) = c_N x^N + c_{N-1} x^{N-1} + \ldots$

$ g_{n+1}(x) = d_M x^M + d_{M-1} x^{M-1} + \ldots$.

We have $ g_{n+1}(x) = f(g_n(x))$ so

\begin{align*}
  d_M x^M + d_{M-1}x^{M-1} + \ldots & = & \\
  & = & a_m(c_N x^N + c_{N-1}x^{N-1} + \ldots)^m + a_{m-1}(c_N x^N + c_{N-1}x^{N-1}+\ldots)^{m-1} + \ldots
\end{align*}

The top coefficient $ d_M$ is easy to identify: the largest power of $ x$ on the right is $ x^{Nm}$ and it is uniquely obtained by raising $ c_N x^N$ to the $ m$th power, so we have

$ M = Nm$

$ d_M = a_mc_N^m$.

How can we get $ x^{M-1} = x^{Nm-1}$? One way is to raise $ c_Nx^N$ to the $ (m-1)$st power and multiply by $ c_{N-1}x^{N-1}$, which appears in the multinomial with coefficient $ m$. We claim that this is the only contribution to $ d_{M-1}$. Indeed, raising $ x^{N-1}$ to the $ m$th power yields $ x^{Nm-m}$ and since $ m>1$ this is not $ x^{M-1}$. This is where the assumption "degree at least $ 2$" of $ f$ comes in. It's not hard to see that other combinations in the multinomial aren't going to yield $ x^{Nm-1}$ either.

Also, the second term (the one following $ a_{m-1}$) has degree $ Nm-N < Nm-1$ since $ N>1$ follows from $ m>1$. Therefore,

$ d_{M-1} = ma_mc_N^{m-1}c_{N-1}$.

We can now compare the average $ r_n$ of $ g_n$'s roots with that of $ g_{n+1}$. We have:

$ r_n = - \frac{c_{N-1}}{N c_N}$

$ r_{n+1} = -\frac{d_{M-1}}{M d_M} = -\frac{m a_m c_N^{m-1}c_{N-1}}{Nm a_m c_N^m}$

and these two are obviously equal. Since this holds for an arbitrary $ n$, the average $ r_n$ remains constant. QED