inequalities
linear algebra
matrix
logarithms
inequalities theorems

[quote="sinajackson"]In the 38'th Annual Iranian Mathematics Confrence that was $ 3\minus{}6$ September in Zanjan, J.Rooin and M.Tahghighi (Speaker):

For nonnegative real numbers $ x_{1},x_{2},\cdots,x_{n}$ let $ A_{n}\equal{}\frac{1}{n}\sum^{n}_{i \equal{} 1}x_{i}$ and $ G_{n}\equal{}\prod^{n}_{i \equal{} 1}x^{1/n}_{i}$

AGM inequality:  $ A_{n}\geq G_{n}$ with equality holds if and only if $ x_{1}\equal{}\cdots \equal{} x_{n}$.

$ \textbf{|SOME REFINEMENTS OF AGM INEQUALITY|}$

$ \text{Theorem 1.}$ If $ A_{n\minus{}1}\leq x_{n}$, then:

$ A_{n}\minus{}G_{n}\geq\frac{1}{n}\sum^{n}_{k \equal{} 2}\begin{pmatrix}n\\ k\end{pmatrix}A^{\frac{n\minus{}k}{k}}_{n\minus{}1}\left(x^{1/n}_{n}\minus{}A^{1/n}_{n\minus{}1}\right)^{k}\plus{}\frac{n\minus{}1}{n}\left(A_{n\minus{}1}\minus{}G_{n\minus{}1}\right)$

$ \text{Corollary 2.}$ If $ 0\leq x_{1}\leq x_{2}\leq\cdots\leq x_{n}$, then:

$ \left(A_{n}\minus{}G_{n}\right)\geq\frac{1}{n}\sum^{n}_{m \equal{} 2}\sum^{m}_{k \equal{} 2}\begin{pmatrix}m\\ k\end{pmatrix}A^{\frac{m\minus{}k}{m}}_{m\minus{}1}\left(x^{1/m}_{m}\minus{}A^{1/m}_{m\minus{}1}\right)^{k}$

$ \text{Theorem 3.}$ If $ x_{i}> 0\;(1\leq i\leq n)$, then:

$ \frac{A^{n}_{n}}{G^{n}_{n}}\geq 1\plus{}\frac{1}{4}\left(\sqrt{\frac{x_{1}}{x_{2}}}\minus{}\sqrt{\frac{x_{2}}{x_{1}}}\right)^{2}$

$ \text{Theorem 4.}$ If $ 0\leq x_{1}\leq x_{2}\leq\cdots\leq x_{n}$, then

$ A^{n}_{n}\minus{}G^{n}_{n}\geq\frac{x^{n\minus{}2}_{1}}{2}\sum^{n\minus{}1}_{k \equal{} 1}\frac{k}{k\plus{}1}\left(x_{k\plus{}1}\minus{}A_{k}\right)^{2}$

$ \geq\frac{x^{n\minus{}2}_{1}}{2}L^{2}(n\minus{}C_{n}\minus{}\ln n)$

$ \geq\frac{x^{n\minus{}2}_{1}}{2}L^{2}(n\minus{}1\minus{}\ln n)$

and

$ A^{n}_{n}\minus{}G^{n}_{n}\leq\frac{x^{n\minus{}2}_{n}}{2}\sum^{n\minus{}1}_{k \equal{} 1}\frac{k}{k\plus{}1}\left(x_{k\plus{}1}\minus{}A_{k}\right)^{2}$

$ \leq\frac{x^{n\minus{}2}_{n}}{2}M^{2}(n\minus{}C_{n}\minus{}\ln n)$

$ \leq\frac{x^{n\minus{}2}_{n}}{2}M^{2}(n\minus{}1\minus{}\ln n)$

where

$ L \equal{}\min\left\{x_{k\plus{}1}\minus{}A_{k}\;:\;1\leq k\leq n\minus{}1\right\},$

$ M \equal{}\max\left\{x_{k\plus{}1}\minus{}A_{k}\;:\;1\leq k\leq n\minus{}1\right\},$

and

$ C_{n}\equal{}\sum^{n}_{k \equal{} 1}\frac{1}{k}\minus{}\ln n.$

$ \textbf{|APPLICATION|}$

Let $ n$ be a positive integer and $ A$ be an $ n\times n$ matrix with all entries $ a_{ij}$ positive. Let $ P$ be the permanent of $ A$. Then

$ P \equal{}\sum_{\sigma\in S_{n}}a_{1\sigma(1)}\cdots a_{n\sigma(n)}\geq n!\left(\prod_{\sigma\in S_{n}}a_{1\sigma(1)}\cdots a_{n\sigma(n)}\right)^{1/n!}\equal{} n!\left(\prod^{n}_{i \equal{} 1}\prod_{\sigma\in S_{n}}a_{i\sigma(i)}\right)^{1/n!}$

$ \equal{} n!\left(\prod^{n}_{i \equal{} 1}\sum^{n}_{j \equal{} 1}\prod_{\sigma(i) \equal{} j}a_{i\sigma(i)}\right)^{1/n!}\equal{} n!\left(\prod^{n}_{i \equal{} 1}\prod^{n}_{j \equal{} 1}\left(a_{ij}\right)^{(n\minus{}1)!}\right)^{1/n!}\equal{} n!\left(\prod_{1\leq i,j\leq n}a_{ij}\right)^{1/n},$

and so $ P\geq n!\left(\prod_{1\leq i,j\leq n}a_{ij}\right)^{1/n}.$[/quote]

Very nice works ,sinajackson  :D . But can you post all solution for your inequalities  :wink: .
I have seen some inequalities .

1, $ A_{n}\minus{}G_{n}\ge\frac{1}{n}(\sqrt{a_{1}}\minus{}\sqrt{a_{n}})^{2}$.

2,$ A_{n}\minus{}G_{n}\ge\frac{n\minus{}1}{n}(A_{n\minus{}1}\minus{}G_{n\minus{}1})$

But it is easier than all your inequalities  :blush:


AMC
AIME
calculus
articles
Science Olympiad

We are all familiar with the popular Olympiad competitions:

$ \text{IMO, IChO, IPhO, IBO, IOI}$ but what about the newer ones that seem to be less 'popular' (in my view anyway):

-$ \text{International Philosophy Olympiad}$
-$ \text{International Linguistic Olympiad}$
-$ \text{International Astronomy Olympiad}$
-$ \text{Internation Olympiad on Astronomy and Astrophysics}$
-$ \text{International Geography Olympiad}$
-$ \text{International Junior Science Olympiad}$
-$ \text{International Earth Science Olympiad}$

Do you think they will ever become as popular or widely acknowledged as the Mathematics, Chemistry, Physics, Biology and Informatics Olympiad?

Personally I don't think so because some of them (e.g. Philosophy, Astronomy) aren't really part of the curriculum of most schools (unlike Mathematics and the Sciences) and Social Science (Geography) seems to be less popular with bright students.

Thinking about the Linguist one, is that like "How many languages" someone knows, or how well they know them?

I agree I can't see them being as popular, especially things like Astro-phsyics and Philosophy.

Linguistics is actually like puzzles. You can figure them out without any knowledge of the language beforehand.

I would like the Astronomy and the Astrophysics Olympiads to become popular. :)
 :oops: Though, believe it or not, my school did not know about even the popular competitions or olympiads, such as the AMC, AIME, IMO, and USAMTS.

[quote="Nerd_of_the_Ages"]Just curious, how would the philosophy one work?[/quote]

Well:

'Students competing in the Olympiad are given four hours to write a philosophical essay on one of four topics in a language other than their own'

According to Wikipedia. So if you are English, you can't write in English.

There's your answer right there!

Personally, I would love to compete in the IPO, but I don't write or speak any of the accepted languages other than English.  I disagree with your idea about the popularity of the topic - most smart people I know are only too willing to share their opinion on any philosophical topic.  The real issue, to me, is that language constraint.

True, not everyone can write in a different language (that is you exclude Mathematics  :lol: ). However, in Australia, we actually have a distinction course in Philosophy that students in Year 12 (who have finished certain subjects) can take.

When I was browsing the IPO site, I found that it's held in pretty much the same places each year. I think in order to make it more popular, they have to invite more countries to participate.

What about the fact that we constantly hear about 'Mathematical Olympiad Genius' (e.g. Reid Barton, Ciprian Manolescu) but never really things like 'Physics Olympiad Genius' or 'Chemistry Olympiad Genius'. Now why is that?

Is it because the Sciences seem to lack the creative element? Like a great number of the questions can be answered with extensive study of a certain subject?

I have certain comments to make. These are largely based in Australia:

1. Calculus is quite essential in Physics and Chemistry (it is in Australia anyway). Most Junior students who manage to get into the Camp have little/no clue of Calculus while Calculus is a big part of the Senior Mathematics curriculum.

2. Building on fact 1) the students who eventually make it into the Olympiad Team are almost always Senior Students and hence it is usually their last year to sit the Olympiad. Hence we don't get people who sit the Olympiads 4/5 times like the IMO.

3. Mathematics is one of the most widely promoted subject in the country. Every good high school has a strong Mathematics Department. Sciences, despite being popular, are less promoted (by a large degree). A huge problem in Australia is that the Science curriculum is absolutely appalling (every subject is English really, the science element is missing). The Mathematics courses are engaging and challenges good students.

In the USA there are "Physics Olympiad geniuses". But they are not as 'famous' as the math ones. I agree it is probably because people don't start on the sciences until much later than maths, generally.

By the way, one of my friends here went to the linguistics olympiad. So those things do actually happen...

In Australia we only have Mathematics, Informatics, Biology, Chemistry and Physics Olympiad (as far as I know).

In New Zealand they have Mathematics, Informatics, Biology and Chemistry Olympiad (apparently Physics is under funded).

So some of the Olympiads (which I never actually heard about until last year) aren't offered in certain countries.

On the Philosophy Olympiad you have to write in a language other than your own?

...

That is actually dumber than the idea of having a "philosphy competition" in the first place.

* * *

On the topic of why the IMO is better known than the others: from looking at sample problems, I get the feeling that IBO/ICO/IPhO problems are to some extent plug-and-chug—they just require a really, really thorough understanding of their subject matter. (Unless the biology sample problems are seriously misrepresenting the actual contest, this seems to be especially true for the IBO; the other two I'm not so sure about, because I can't really understand what the problems are talking about.) If that's true, the IMO (and IOI) is fundamentally different from the other olympiads, which is at least a reason why the IMO might be held in such high regard. What IMOers do meshes better with the popular conception of genius than what other IO participants do. (If this offends anybody, keep in mind that I'm mostly talking out of my ass.)

I think the fact that you could take certain IMO problems and explain it (and the solution) in a way that an average person would understand and appreciate is a large part of the appeal of it.

I remember reading a newspaper article about Reid Barton (I think I was in middle school then?) and it had one of his sample solutions from the IMO. Absolutely beautiful.

[quote="SuperThread"]

On the topic of why the IMO is better known than the others: from looking at sample problems, I get the feeling that IBO/ICO/IPhO problems are to some extent plug-and-chug—they just require a really, really thorough understanding of their subject matter.[/quote]

Indeed, the Olympiads in Chemistry and Biology (not so much Physics I think, which might explain its popularity) are really a matter of how well you know the subject and not so much creativity. I personally find the Biology Olympiad quite a 'bad' (not really sure what the best word is) competition.

IBO really lacks what the IChO and IPhO possess and less popular. The structure of the IBO is quite poor and the questions aren't too good (plus they are hard to find).

I actually participated in the IJSO (International Junior Science Olympiad)... 1 gold, 1 silver, 1 bronze medal, so I'm Germany's best participant in history :D (There were only 4 IJSOs yet though)

sylow_theory you have quite a valid point. I am currently doing the IChO 2006 Preparatory Problem Set, I am zooming through the thing like it's nothing (actually this is probably the easiest year).

Yes like Zuming Feng said, 'The heart of the IMO is about creating new Mathematics' (or somethig like that). IMO is popular among 'smart' students because memorizing formula is limited, it is more about thinking about problems differently.

[quote="sylow_theory"]Not to offend other olympiads but the math olympiad the is only that that truly makes sense to me. Because it is more about problem solving than any other olympiad.

Is there anybody who won gold in different olympiads?[/quote]

The linguistics olympiad is all problem-solving too. Although, IMHO, from looking at problems, the ILO is a lot easier than the IMO for a mathematical mind. But the ILO also requires a smaller set of knowledge. Sadly, a lot of people who would be good at it brush it off as "ugh, linguistics, who cares."

Meanwhile, I gave my math team some linguistics puzzles for homework  :)

The Astronomy Olympiads (IAO and IOAA) are quite popular in my country. In my opinion, the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170589]problems[/url] require quite a bit of creativity and only minimal astronomy (but more of Physics).
I know a person who has won two silvers and a bronze at IMO and two golds at IAO.

[quote="motal"]I think that these olympiads will not become as popular as IMO or IPho because subjects like philosophy and linguistic aren't problem-solving subjects. Philosophy is not for competitions! How can a judge say his own opinion without being influenced by his ideas and his previous studies?[/quote]

I imagine it works similar to how debate competitions work.  You can judge who gives a more solid argument...it doesn't matter you believe.  I don't see how it's that hard for a judge to place his own views aside and judge the competitors solely on the merit of the their argument.  I assume these judges will themselves be philosophers and I can guarantee you that any philosopher worth his salt out there cares about the argument being placed forth moreso than whether or not the argument agrees with his/her views.

With respect to bio/chem and math/physics, I think the way the olympiads are generally falls in line with how those disciplines are in real life.  Math and physics are traditionally more theoretical involving more creative thinking while chem and bio are traditionally more experiment-based.  Those are the general trends of those fields at least.  Of course, there are rich fields of applied math and physics and bio/chem can be a little theoretical but those fields are usually on the boundary between two disciplines such as math and biology, e.g.

[quote]Is there anybody who won gold in different olympiads?[/quote]

Well, I have, apart from success in national math olympiad, participated in International Philosophy Olympiad and got honorable mention - not quite gold medal, but still nice :).  In fact, that olympiad is hardly a competitive contest in any sense (how could you judge one's philosophical competence, even if it were possible, by a single essay?), it's rather a purely social-cultural event.

I know this is irrelevant but I really can't bring myself to start another topic for something so small.

Is it really true that Olympians can only be trained for a maximum of two weeks? As in officially trained (not self-study).


A club provides 30 snacks to 18 members, and each member orders 3 different snacks. It is known that every snack is ordered by at least one member, and that any two members order at most one same snack. Is it possible to find 12 snacks, such that the snacks ordered by any member is not completely in these 12 snacks?

this should have beeen solved long ago.
use a probabilistic argument; consider a random set of $S\subset X$(the set of snacks) of size $12$ and let$1_A(S)=1$ if $A\subset S$, and $0$ otherwise.then
$N=\sum_{A\in\mathcal{A}} 1_A$ counts the number of subsets of $\mathcal{A}$ contained in $S$, so that $E(N)=\sum_{A\in\mathcal{A}} E(1_A)$. now $E(1_A)=P(A\subset S)=\frac{\binom{27}{9}}{\binom{30}{12}}$ and so $E(N)=18\cdot \frac{10\cdot 11\cdot 12}{30\cdot 29\cdot 28}=\frac{198}{203}$ and so there always exists a set of size $12$ with none of the elements of $\mathcal{A}$ contained in it.

We take a snack from each of the 18 ordering lists of the members. There must exist 12 snacks not containing any of the above 18 (possibly repeated) snacks, and those 12 snacks are what we are looking for.


Ok, I haven't officially started this yet until I finish telling you guys what to do!  So, I won't start speaking backwards..just yet
I start. But, w/e I say I will say it backwards, and the next person has to translate it and then that same person gets to say another sentence backwards..and the next person does the same thing
k. I will start now
[b]ih, woh era uoy yadot?[/b]

[quote="spaceguy524"]I ma yugecaps.[/quote]

Spaceguy am I.

You know, that sentence makes sense if the words are reversed, too.

Olleh, I ekil syelims.

Hello, I like smileys.

I ma yltnerruc derob.

I am currently bored.

Hosgimho, I ekil sllorgge.

Ykao... eht gnitalsnart gniht si yrev diputs.  Re'we ton tsrif sredarg.

[quote="royalflower"]Ohmigosh, I like eggrolls.

I yllatot ekil sllorgge oot esuaceb I ma esenihc.[/quote]

Yours: I totally like eggrolls too because I am chineses

And ernie's is:

Okay... the translating thing is very stupid. Were not first graders

Mines is:
[u]
[i][b]" Eraepsekahs I ma ton" [/b][/i][/u]

BLEH!.well
w/e
jampa's is: shakespeare  I am not.
lol
mine is
:hem em si derob, yerv derob.

[quote="1=2"]meh: me is bored, very bored.

!seireS rolyaT a otni mrofsnart lliw enoyreve ro noom lluf txen eht erofeb eniL noissergeR serauqS tsaeL eht dnif tsum I[/quote]
ok
Sooooo
I must find the Least Squares Regression Line before the next full moon or everyone will transform into a Taylor Series!
lol
and ernie's was

that'S too mathy for me!
hm
mine isL
WOW.....HUH..LOL...HEETEE..LLA SEMORDNILAP!!!! RACECAR...HEH...

kool's
Hi, How are you doing today.
Mines

LOL ti htiw yako saw rehcaet eht dna ssalc ni deklat sdarwkcab enco sdrawkcab gniklat evol I. srouh rof siht od dlouc I.

mg.. u were supposed to do my post above that..like not the first one.
 mgs:
I could do this for hours. I love talking backwards once backwards talked in class and the teacher was ok with it LOL
mine:
HELB


AMC
AMC 12
AIME
AMC 10
AMC 12 A
USA(J)MO
USAMO

Here is my question: is there anyone here who decided that they will compete in the AMC12?

I have an ambitious goal this year around: compete in the AMC12, defend the championship for my college and, in doing this, access the AIME (2nd round).

Yeah, I have been lucky to find a school (finally) near me that is offering it, so I'm going to take it there... Good Luck!!!  :D

My school has been offering it, so I will definitely be taking it :)
I took it last year too... and bombed it..  :(
 i should've taken the amc 10 last year

My school offers it, and I'm definitely taking it.

I will be taking it (unless the math department head hasn't sent in registration like he apparently did last year).

I'm taking it. I got 140 on AMC10 last year, and I hope to do well on this year's as well and get to AIME and do better than I did last year, and maybe get to USAMO.  :)

I was told that taking part in the AIME in itself was a significant achievement due to the extreme difficulties to qualify...

100+ is ok..you could just answer 14 questions and leave the rest blank, that would still get you 100.5...and I'm sure you can do better than that.

I'd say that the concern is more about how well you do in the AIME, not whether or not you make it,
albeit, making the AIME is a good place to start.  :)

And yes, I'll be taking both the AMC12A and B.

Yes, yes, AIME is important.

Why take both A and B?

Oh, mainly because my school is offering both.. and I don't have to pay for either of them.

Never heard of "The most important thing is not to win but to participate"? This is true for the AIME (from a Canadian viewpoint) but not for the COMC. Because AIME and USAMO lead nowhere for us, though the CMO does.

well usamo's part of the canadian imo team selection, so it might be worthwhile to have a shot :)
not that i have any chance of making it to imo team though :(

Yeah, if you're gonna do a contest, do well.

USAMO seems hard...

Good luck to you all who are writing the AMCs.

Although I am Grade 12, I retired from contests since I recieved a perfect score in AMC 12 last year  :P

This year I'm coaching other students in my school to help them win

Any tips you have for what i should study before I take it. Anything I should go over?


I thought this was a nice problem. :)

25 boys and 25 girls are sitting around a circular table. Prove that two girls must be sitting next to each other.

Um, I don't think so...
Consider the following circular table(x a boy, z agirl)
xzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxzxz

yea check the problem again cuz its definitely possible (try smaller cases like 5 boys 5 girls etc)

Wow, I didn't remember the problem well enough. The problem is "prove that you can always find someone that has two girls as neighbors".

hm i think there are lots of ways to approach this but here's one

[hide]

we see that given 5 spots, _ _ G _ _, where G denote girl, then the two ends must be guys in order to assure that there are no two girl neighbors (so B_G_B and we can fill in the empty spots)
so, if we were to order these, we would have:
_ _ G _ _ G ...

giving numerical values to the 50 spots in the circle, starting with 1 and going to 50, we see that this arrangement is equivalent to adding 3 to each previous spot occupied by a girl

hence the spots with girls are (for the maximal case):
$1, 4, 7, \ldots, 46, 49$

but, since we have $49$ and $1$ occupied by a girl, and since this is a circle, we have two girls with exactly one space in between, meaning that regardless of what gendered person goes in that spot, he/she will have two neighbors that are girls

hence we are done

[/hide]


calculus
integration
inequalities
function
trigonometry
real analysis
real analysis unsolved

Let $f: [0,1]\rightarrow\mathbb{R}$ be a continuous function satisfying $xf(y)+yf(x)\le 1$ for every $x,y\in[0,1]$.
(a) Show that $\int^1_0 f(x)dx \le \frac{\pi}4$.
(b) Find such a funtion for which equality occurs.

Whoops, I thought of trig but didn't quite see how it would work.  Here's a ... funny solution.

Let $y = \sqrt{1-x^2}$ to obtain
\[
x f(\sqrt{1-x^2}) + \sqrt{1-x^2}f(x) \leq 1\text{ for all }x\in[0,1].
\]
Multiplying the given equality by $\sqrt{1-x^2}$ and integrating over the interval $[0,1]$ gives
\[
\int_0^1 x\sqrt{1-x^2}f(\sqrt{1-x^2})\,dx  + \int_0^1 (1-x^2)f(x)\,dx \leq \int_0^1 \sqrt{1-x^2}\,dx = \frac\pi 4.\quad(*)
\]
For the first integral, write $u = \sqrt{1-x^2}$.  Then $u\,du = x\,dx$, so
\[
\int_0^1 x\sqrt{1-x^2}f(\sqrt{1-x^2})\,dx = \int_0^1 u\cdot uf(u)\,du = \int_0^1 u^2 f(u)\,du.
\]
Hence the left hand side of $(*)$ simplifies to $\textstyle\int_0^1 u^2f(u)\,du + \textstyle\int _0^1 (1-x^2)f(x)\,dx = \textstyle\int_0^1 f(x)\,dx$, proving the desired inequality.  Equality occurs, of course, when $f(x) = \sqrt{1-x^2}$.

[b][color=blue]Remark.[/color][/b] Trying to prove a pointwise bound on $f$ does not work, because the function $f(x) = \chi_{[0,\frac12]}(x)$ also satisfies the given conditions.  This means one needs to work with the integral directly.


I was wondering what books and articles (online or newspaper, etc) are interesting to read in economics if I just completed introductory courses in micro and macro? Thanks

Well, can anyone help me out here? :)  :(  :?  :D

Paul Krugman is interesting (Pop Internationalism , Peddling Prosperity...) and has a good website.
You'd probably enjoy Levitt and Dubner "Freakonomics"

I requested freakeconomics from the library, I think I can read that Krugman book online, what is it about? U an econ major ?

Krugman has a lot of stuff online, lots of the stuff in his books
is a collection of articles for The Slate or New York Times.
Peddling prosperity (1994, I think) is about the U.S. economy and
policy since the 1970's inlcuding a blinding attack on supply-sided economics.

Accidental Theorist is very enjoyable, a collection of articles.

His style is very good as he points outs fallacies, common mistakes that people make, leaving you feeling he (maybe even you) would be a better economic advisor than the current bunch.

Pop Internationalism is all about trade and international economics. He is the (co)author of the book we used for these courses. Its good provoking stuff, without the textbook maths.
He will also explain better than I your question regarding savings = investment and trade deficit stuff.

I bought another of his (cant remember the name, pic of Bush with his mouth zipped on the front). This is more political than the others, also a collection of articles.

Freakonomics is accessible for the untrained in economics. The most technical it gets is in using regressions (using excel or... to find correlations between events.) Quickest book I ever read, featuring chapters like "Why do drug-dealers live with their moms?" and "Would a Roshanda by any other name smell as sweet?" about the names people give their children.

Yes, I single-honoured in econ

If you are intrested in micro you can try to atack H R Varian's book -Intermediate Microeconomics . Have a nice day.
Marius :)


ok, i have a questions but ill first post the problem that inspired this question and see what u guys think.

[b]If $ \frac{x}{y}\equal{}\frac{4y}{x}\equal{}z$, then find all possible values of $ z$[/b]

[hide="my incorrect solution"]ok, u know how there are those properties of proportions. well for this problem i used the particular one that goes like this
if $ \frac{a}{b}\equal{}\frac{c}{d}$, then $ \frac{a}{b}\equal{}\frac{c}{d}\equal{}\frac{a\plus{}c}{b\plus{}d}$. thus i got $ z\equal{}\frac{x\plus{}4y}{x\plus{}y}$. therefore, i thought there was an infinite number of solutions[/hide]

[hide="solution"]$ \frac{x}{y}\equal{}\frac{4y}{x}$

$ x^{2}\equal{}4y^{2}$

$ x\equal{}\pm2y$

Thus, we have $ z\equal{}\frac{\pm2y}{y}\equal{}\pm2$.[/hide]

@your stuff:  what you have is correct, but not strong enough.

[quote="b-flat"][hide="solution"]$ \frac{x}{y}\equal{}\frac{4y}{x}$

$ x^{2}\equal{} 4y^{2}$

$ x \equal{}\pm2y$    <==Divide by $ y^{2}$ here, then take $ \sqrt{4}$=ans.

Thus, we have $ z \equal{}\frac{\pm2y}{y}\equal{}\pm2$.[/hide]

@your stuff:  what you have is correct, but not strong enough.[/quote]

Just a minor deviation, just saves one step  :lol:

[quote="mihail911"]ur answer is right b-flat. why do u believe that mine was wrong? that's what im struggling to find out.[/quote]

Like I said before, it is because your answer is simply not strong enough.  You would have to add more to get a full solution.

Example of not strong enough stuff:

Find all perfect squares.

Solution:
The quadratic residues $ \pmod4$ are 0,1.  Thus, all squares are either 0 or 1 mod 4.  However, when we take all numbers 0 & 1 mod 4, not all of them are perfect squares, so the argument isn't strong enough.

I can give you another (better; probably with AM-GM or something) example if you need...


[quote="SnowEverywhere"]Hello. I was just wondering how you guys see the tricks involved in AIME and USAMO type problems. For instance, how do you know what lines to draw in geometry problems (ones which involve creating your own points not just connecting ones unconnected in the problem statement), how do you determine what will be a suitable method of approach (i.e. crossover tactics) or how do you determine what algebraic manipulation or substitution to use (to simplify the problem).

Sorry if this is an abnormally large question to be asking in this forum. Although it tackles the basis of problem solving, I was just wondering what your views and approaches are?

I was just wondering because on many occasions, I have failed to draw the right line or see the problem in the right way such as to use some tactic which is not immediately obvious. Does this come from experience, is it acquired, or is it something which you are either good at or not?[/quote]

As everyone has noted, tricks probably should not be memmed, because you should learn the concepts and besides, no matter how good you think your memory is, you [b]will[/b] make a mistake when using the "trick" some time or another.

Okay, the big competitions, like AMC and stuff like that are used to determine the best math geeks in the country. If everyone could just mem tricks, it would be easy to ace the AMC and AIME and all that. Basically, there's no magic bullet to acing all the contests.

It's not beneficial to do math competition if all you do is use all these tricks there.

EDIT: Sorry for an abnormally large, nerdy answer

There's no easy answer. Personally, I have a big library of tricks, and I'm willing to briefly try a technique, then discard it if it doesn't look like it will work.
On a contest, that decision cycle needs to be very fast. You don't have time to spend a while fully working through the technique that doesn't work.
Then, there's the intuition gained from lots of experience on which tricks to even look at. With that many techniques, you can't try all of them even minimally, so you immediately filter down to those that might work for this kind of problem.

[quote="SnowEverywhere"]...because on many occasions, I have failed to draw the right line or see the problem in the right way...[/quote]
Don't be afraid to just abandon whatever isn't working. Move on to another problem and then come back later, or look at a completely different approach to the same problem. Once in a while, you'll miss something that just took a lot of work. More often, you'll spend your time more productively on something else. Once in a while, you'll spend that time more productively, and then spot something new when you come back.

The only way to actually "learn" a "trick" is to use it over and over.  Now, practice helps a tremendous amount while learning these tricks.  You cannot just know the trick, you have to know when to apply it.  For example, in the time that you are finding the trick, someone else is already thinking about what he/she is going to do with the trick.  Ideally, you should be able to look at a problem and know exactly what to do when you see it.  (This is ideally, and no one is perfect.)

For example, if I gave you this:
$ x \plus{} 3 \equal{} 6$.  Solve for $ x$.

You would know exactly what to do, just subtract the whole equation by 3.

However, if I gave you this:
Find any rational root to the equation:
$ 2009x^{2009} \plus{} 2008x^{2008} \plus{} 2007x^{2007} \plus{} 2006x^{2006} \cdots 2x^2 \plus{} x \equal{} 0$ it wouldn't be as obvious, but still a simple problem.  (Just pull out $ x$, $ x \equal{} 0$)

That's the same way the AMC problems work.  They give you a seemingly difficult problem, but by using the techniques learned from easier problems, the difficult problem simplifies dramatically.

Another thing that helped me a lot: You need to truly believe that each and every one of those problems are solvable.  Not only solvable, but simple, given the tools necessary.

If you go into the competition thinking that the problems are difficult, people who know the simplicity will have an advantage over you.

Basically, just don't think they are impossible.  Every one of those problems has a solution.


define a binary operation $ *$ on all real numbers such that $ (a*b)*c\equal{}a\plus{}b\plus{}c$, show that $ a*b\equal{}a\plus{}b$.

[hide]
Let $ k\equal{}0*0$. Then $ k*0\equal{}(0*0)*0\equal{}0$. Thus $ 2k\equal{}k*(k*0)\equal{}k*0\equal{}0$, implying $ k\equal{}0$. Therefore $ 0*0\equal{}0.\quad (1)$

Using (1), $ a*0\equal{}a*(0*0)\equal{}a$. Thus $ a*0\equal{}a$ for all $ a.\quad (2)$

Now $ a\plus{}b\equal{}(a*b)*0\equal{}a*b$ (using (2)) for all $ a,b$. Hence we are done.[/hide]

[hide="Tentative solution"]

Set $ a \equal{} b \equal{} 0$. Then we have that $ \boxed{(0 * 0) * c \equal{} c}$.

Now let's determine the value of $ 0 * 0$. If we plug $ a \equal{} b \equal{} c \equal{} 0 * 0$ in the original equation, we have that

$ [(0 * 0) * (0 * 0)] * (0 * 0) \equal{} (0 * 0) \plus{} (0 * 0) \plus{} (0 * 0)$
$ (0 * 0) * (0 * 0) \equal{} 3(0 * 0)$
$ (0 * 0) \equal{} 3(0 * 0)$
$ \boxed{0 * 0 \equal{} 0}$ (1)

Set $ a \equal{} 0 * 0$. Then we have that
$ [(0 * 0) * b] * c \equal{} (0 * 0) \plus{} b \plus{} c$.
$ b * c \equal{} (0 * 0) \plus{} b \plus{} c$
$ \boxed{b * c \equal{} b \plus{} c}$
[/hide]

EDIT: Beaten by 2 minutes! Funny how we used the same identities...


Sue has \$3.08 in pennies, nickels, and quarters. She has four more penniesthan quarters and one more nickel han pennies. How many nickels does she have?

Hi;

[hide]9 quarters, 14 nickels, 13 pennies[/hide]

First turn all the words into algebra.
$ 0.01p \plus{} 0.05n \plus{} 0.25q \equal{} 3.08$
$ q \equal{} p \minus{} 4 \equal{} n \minus{} 5$
$ p \equal{} n \minus{} 1$

Now plug in the $ q$ and the $ p$ into the first equation and solve for $ n$.

$ 0.01(n \minus{} 1) \plus{} 0.05 \plus{} 0.25(n \minus{} 5) \equal{} 3.08$
$ 0.01n \minus{} 0.01 \plus{} 0.05n \plus{} 0.25n \minus{} 1.25 \equal{} 3.08$
$ 0.31n \equal{} 4.32$
$ n \equal{} 14$

So, we find that Sue has 14 nickels.


Men's in the Final Four, Women's will obviously pwn.

Nah, they're gonna lose to villanova in the semis.

Women's just pwned, again.  What are they, 36-0?

Oh sure, UNC can go all the way, until they meet a team known as the Huskies! :P

huskies might not even be able to get past mich. state

even though my bracket has uconn and unc going for the championship which btw is UNC's

OH, btw, ernie, opposite of progress is congress....lol,

Twins had a nice win yesterday... post your brackets here!

Red Sox
----------- Red Sox in 4
Angels
----------------------- Red Sox in 6
Twins
----------- Yankees in 4
Yankees
----------------------------------- [b]Red Sox in 6 [/b]
Cardinals
----------- Cardinals in 5
Dodgers
------------------------ Phillies in 7
Rockies
----------- Phillies in 4
Phillies

If the Red Sox win over the Yankees, and that is a HUGE if, it's not going to be in 6 games. The Yankees look unbelievably good.


What is the median of all values defined by the expression $ 2^{x} \minus{}1$, where $ x$ is a prime number between 0 and 20?

Of all 21 values, we know the median is the 11th one (11th from top, 11th from bottom), which is

$ 2^{10} \minus{} 1 \equal{} 1024 \minus{} 1 \equal{} \boxed{1023}$

Actually, since x is PRIME, it should be $ \frac{2095\plus{}127}{2}\equal{}\boxed{1111}$

Oh wait isn't $ 2^{11}$ equal to $ 2048$?

Then we would have $ \frac{2047 \plus{} 127}{2} \equal{} 1087$

Oh, oops you're right. Wow I got $ 2^{12}$ and $ 2^{11}$ mixed up......

$ \text{\LaTeX}$ tip: For exponents $ >10$, use:

[code]$x^{y}$[/code]

Thanks ernie. I knew that, but I forgot. Maybe it would be a better habit for me if I just always used {} for stuff, even if its one digit


A Leap Year has 366 days and the year 2000 will be a Leap Year.  Rick's 49th birthday on  April 1, 1999 will be on a Thursday.   Rick's 50th birthday will be on a

A.Tuesday    B.     Wednesday	C. Friday	    D.    Saturday	      E.   Sunday

[hide] Since $ 366\equiv2\pmod7$, the day will move $ 2$ forward. Thus, the day will be Saturday, choice D.[/hide]


linear algebra
matrix
algebra solved
algebra

Compute the following determinant of order 3!
a11=sha;a12=cha;a13=sh(2a);
a21=shb;a22=chb;a23=sh(2b);
a31=shc;a32=chc;a33=sh(2c);

cheers! :D

this is also from  a romanian gazette! :D

[quote="Lagrangia"]Compute the following determinant
a11=sha;a12=cha;a13=sh(2a);
a21=shb;a22=chb;a23=sh(2b);
a31=shc;a32=chc;a33=sh(2c);
[/quote]

For matrix of size three we have Sarrus trick to expand determinant 3x3

|a b c |
|d e f |
|g h i|

= aei+dhc+bfg-ceg-afh-ibd

oubi

i think that what he wants is not Sarrus' rule, but a closed expression for the determinat, isn't that right Lagrangia?

And what is the result? There should be a very smooth result! I have also applied Sarrus, but I can't find a cute result!
:D :D

cheers!

[quote="Lagrangia"]Compute the following determinant of order 3!
a11=sha;a12=cha;a13=sh(2a);
a21=shb;a22=chb;a23=sh(2b);
a31=shc;a32=chc;a33=sh(2c);
[/quote]

sh(2t) = 2sh(t)ch(t)

|sha cha 2sha.cha|
|shb chb 2shb.chb|
|shc chc 2shc.chc|

=2cha.chb.chc.D

D=
|tha  1  sha|
|thb  1  shb|
|thc  1  shc|
= -
|1  tha sha|
|1  thb shb|
|1  thc shc|

elementary operations on line
L2 -> L2-L1 , L3->L3-L1
D= -
|1 tha sha|
|0 (thb-tha) (shb-sha)|
|0 (thc-tha) (shc-sha)|
= (thc-tha)(shb-sha) - (thb-tha)(shc-sha)

the initial determinant is
2cha.chb.chc[(thc-tha)(shb-sha) - (thb-tha)(shc-sha)]


[quote="epsilonist"]Let $ x$ be real number such that $ 3^x \plus{} 5^x \equal{} 2\cdot 4^x$.
Since, $ 4^{x} > 0$, $ \left(\frac {3}{4}\right)^x \plus{} \left(\frac {5}{4}\right)^x \equal{} 2$.
By the Mean Inequality, x=0,1 are the only values that state the equation.  :)[/quote]
 :wink:  :mad:


This is above MathCounts level; I suggest a moderator moves this to HSB.

$ \left|\sqrt {x - 1} - 1\right| + \left|\sqrt {x - 1} + 1\right| = \frac {1}{x - 1}$

[hide="Solution"]We know $ x\ge 1$ from the LHS, and $ x\ne 1$ from the RHS, so $ x > 1$. Since $ \sqrt {x - 1}$ and $ 1$ are both positive, $ \left|\sqrt {x - 1} + 1\right| = \sqrt {x - 1} + 1$.

Let $ y = \sqrt {x - 1}$. The equation now becomes $ \left|y - 1\right| + y + 1 = \frac {1}{y^2}$.

[b][color=darkred]Case 1:[/color][/b] $ y\ge1$. In this case, $ \left|y - 1\right|\ge0$, so $ \left|y - 1\right| = y - 1$. We can now solve for $ y$ directly:
\begin{align*}y - 1 + y + 1 & = \frac {1}{y^2} \\
 \\
2y^3 & = 1 \\
 \\
y & = \frac {1}{\sqrt [3]{2}}\end{align*}Now, solve for $ x$:
\begin{align*}\sqrt {x - 1} & = \frac {1}{\sqrt [3]{2}} \\
 \\
x - 1 & = \frac {1}{\sqrt [3]{4}} \\
 \\
x & = 1 + \frac {1}{\sqrt [3]{4}} \\
 \\
x & = 1 + \frac {\sqrt [3]{2}}{2} \\
 \\
x & = \frac {2 + \sqrt [3]{2}}{2}\end{align*}
[b][color=darkred]Case 2:[/color][/b] $ y < 1$. In this case, $ \left|y - 1\right| < 0$, so $ \left|y - 1\right| = - (y - 1) = 1 - y$. We now solve for $ y$:
\begin{align*}1 - y + y + 1 & = \frac {1}{y^2} \\
 \\
2y^2 & = 1 \\
 \\
y & = \pm\frac {1}{\sqrt {2}}\end{align*}Note that $ y\ge0$ since $ y$ is the square root of a number, so $ y = \frac {1}{\sqrt {2}}$. Now, we solve for $ x$:
\begin{align*}\sqrt {x - 1} & = \frac {1}{\sqrt {2}} \\
 \\
x - 1 & = \frac {1}{2} \\
 \\
x & = \frac {3}{2}\end{align*}
In conclusion, $ x = \left\lbrace\frac {3}{2},\ \frac {2 + \sqrt [3]{2}}{2}\right\rbrace$.[/hide]

There's someting wrong in case 2.
In case 2 :-1<y<1
so:
case 1.    y>1
case 2.    -1<y<1
case 3.    y<-1
Is that right?
But the answer is right.

[quote="ericlou"]| \sqrt {x - 1} - 1 |+| \sqrt {x - 1} + 1 |= 1 / {x - 1}
Solve for x[/quote]

It looks like you're trying to use $ \text{\LaTeX}$. To do that here, use the dollar signs ($ \$$) around the $ \text{\LaTeX}$ code.

And also, since we are taking the positive square root, $ y \geq 0$.

[quote="distracted523"]Since we are taking the positive square root, $ y \geq 0$.[/quote]
Yes, I messed up a little by saying $ y\equal{}\pm\frac{1}{\sqrt{2}}$, but the error doesn't matter because I square $ y$ later anyway. :)

[quote="i_like_pie"]This is above MathCounts level; I suggest a moderator moves this to HSB.

$ \left|\sqrt {x - 1} - 1\right| + \left|\sqrt {x - 1} + 1\right| = \frac {1}{x - 1}$

[hide="Solution"]We know $ x\ge 1$ from the LHS, and $ x\ne 1$ from the RHS, so $ x > 1$. Since $ \sqrt {x - 1}$ and $ 1$ are both positive, $ \left|\sqrt {x - 1} + 1\right| = \sqrt {x - 1} + 1$.

Let $ y = \sqrt {x - 1}$. The equation now becomes $ \left|y - 1\right| + y + 1 = \frac {1}{y^2}$.

[b][color=darkred]Case 1:[/color][/b] $ y\ge1$. In this case, $ \left|y - 1\right|\ge0$, so $ \left|y - 1\right| = y - 1$. We can now solve for $ y$ directly:
\begin{align*}y - 1 + y + 1 & = \frac {1}{y^2} \\
 \\
2y^3 & = 1 \\
 \\
y & = \frac {1}{\sqrt [3]{2}}\end{align*}
Now, solve for $ x$:
\begin{align*}\sqrt {x - 1} & = \frac {1}{\sqrt [3]{2}} \\
 \\
x - 1 & = \frac {1}{\sqrt [3]{4}} \\
 \\
x & = 1 + \frac {1}{\sqrt [3]{4}} \\
 \\
x & = 1 + \frac {\sqrt [3]{2}}{2} \\
 \\
x & = \frac {2 + \sqrt [3]{2}}{2}\end{align*}
[b][color=darkred]Case 2:[/color][/b] $ y < 1$. In this case, $ \left|y - 1\right| < 0$, so $ \left|y - 1\right| = - (y - 1) = 1 - y$. We now solve for $ y$:
\begin{align*}1 - y + y + 1 & = \frac {1}{y^2} \\
 \\
2y^2 & = 1 \\
 \\
y & = \pm\frac {1}{\sqrt {2}}\end{align*}
Note that $ y\ge0$ since $ y$ is the square root of a number, so $ y = \frac {1}{\sqrt {2}}$. Now, we solve for $ x$:
\begin{align*}\sqrt {x - 1} & = \frac {1}{\sqrt {2}} \\
 \\
x - 1 & = \frac {1}{2} \\
 \\
x & = \frac {3}{2}\end{align*}
In conclusion, $ x = \left\lbrace\frac {3}{2},\ \frac {2 + \sqrt [3]{2}}{2}\right\rbrace$.[/hide][/quote]
I'm pretty sure he didn't say only solve for reals.

Let's notice something nice...

Square both sides... =P. See what you get.

[quote="Ihatepie"][quote="ZhangPeijin"]
I'm pretty sure he didn't say only solve for reals.[/quote]

while unreals are not mc level.[/quote]
Yeah but they still need to be a part of the solution.

Just move the freaking thread to high school basics and stop arguing.  This isn't supposed to be in mathcounts anyways.


Solve for $x$: $\{3 + [2 + (1 + x^2)^2]^2\}^2 = 144$.

[color=darkred]...where x is a real number.[/color]

There are more solutions than just that. Here's one  .

[hide]

 3+[2+(1+x^2)^2]^2 = 12

[2 + (1 + x^2)^2]^2 = 9

2 + (1 + x^2)^2 = 3

(1 + x^2)^2 = 1

1+x^2 = -1

x^2 = -2

x = +/- i :sqrt: 2

[/hide]

There are other uglier solutions than this, but this is just to get your mind working.

Of course, at MATHCOUNTS it is assumed that you are only looking for real solutions.

[hide]144={3+[2+(1+x^2)^2]^2}^2
12=3+[2+(1+x^2)^2]^2
9=[2+(1+x^2)^2]^2
3=2+(1+x^2)^2
1=(1+x^2)^2
1=x^2+1
0=x^2
x=0[/hide]
I bet there's more solutions but you only asked for real x solutions which removes the chances of having solutions with imaginaries in them.


[quote="Ngông"]
2/Let $a,b,c> 0$ .Prove that:
$\frac{ \sqrt{a^{2}+bc}}{b+c}+\frac{\sqrt{b^{2}+ca}}{c+a}+\frac{\sqrt{c^{2}+ab}}{a+b}\leq\frac{3 \sqrt{2}}{2}$

I don't know if 1/ is right. But I think 2/ is correct.  :D[/quote]
Try $a=b=1$ and $c\rightarrow0^{+}.$ :wink:

[quote=Ngông]1/Let $a,b,c> 0$ .Prove that:
$\frac{ \sqrt{a^{2}+bc}}{c+a}+\frac{\sqrt{b^{2}+ca}}{a+b}+\frac{\sqrt{c^{2}+ab}}{b+c}\geq\frac{3 \sqrt{2}}{2}$

2/Let $a,b,c> 0$ .Prove that:
$\frac{ \sqrt{a^{2}+bc}}{b+c}+\frac{\sqrt{b^{2}+ca}}{c+a}+\frac{\sqrt{c^{2}+ab}}{a+b}\leq\frac{3 \sqrt{2}}{2}$

I don't know if 1/ is right. But I think 2/ is correct.  :D[/quote]

$\frac{ \sqrt{a^{2}+bc}}{b+c}+\frac{\sqrt{b^{2}+ca}}{c+a}+\frac{\sqrt{c^{2}+ab}}{a+b}\geq\frac{3 \sqrt{2}}{2}$

So nice!
why no one do?

It was here:
http://www.artofproblemsolving.com/community/c6h97690


limit
logarithms
real analysis
real analysis unsolved

find limes
$ \lim\limits_{n\to\infty}\left(\frac{\left(1\plus{}\frac{1}{n}\right)^{n}}{e}\right)^{n}$

The English is "limit".

Ah, power series. We want
$ \lim_{n\to\infty}\exp\left(n\ln\left(\frac{(1\plus{}\frac1n)^n}{e}\right)\right)$
$ \equal{}\lim_{n\to\infty}\exp\left(n\left(n\ln\left(1\plus{}\frac1n\right)\minus{}1\right)\right)$
$ \equal{}\exp\left(\lim_{n\to\infty}n\left(n\left(\frac1n\minus{}\frac1{2n^2}\plus{}O(n^{\minus{}3})\right)\minus{}1\right)\right)$
$ \equal{}\exp\left(\lim_{n\to\infty}n\left(1\minus{}\frac1{2n}\plus{}O(n^{\minus{}2})\minus{}1\right)\right)$
$ \equal{}\exp\left(\lim_{n\to\infty}\minus{}\frac12\plus{}O(n^{\minus{}1})\right)$
$ \equal{}e^{\minus{}\frac12}$


Circle path Inc makes bicycles, tricycles and unicycles. Last week they made 88 more bicycles then unicycles and 5 times as many tricycles as unicycles. If they made 40 more bicycles than tricycles, how many unicycles did they make?
Note: this is a mom trying to get help for her student and I need to be able to explain how the answer is acquired. help

oops sorry appearently the 1st was wrong, i totally read the problem wrong. so using systems of equation the answer is  12 unicycles :] and i guess you can see how in middle school, classmates just want to point  out your mistakes for everyone to see.

[hide]
ok, to solve the problem, you need to 1st list out all the givens with equations

88 more tricycles than unicycles, T+88=U OR T=U-88
and 5T=U    B+40=T

so to find U all you do is combine 2 equations that equal the  same thing. T+88=U=5T so 4T=88 and T is 22. so plug it back into T+88=U U=110
[/hide]
HOPE IT HELPS
: ][color=blue][size=75]kyyuanmathcount: Added hide tags :)[/size][/color]

[quote="Walk Around The River"]*JINZI, YOURE ANSWER IS INCORRECT. PLEASE DONT POST INCORRECT SOLUTIONS.  IT CAN BE MISLEADING, ESPECIALLY IN THIS CASE WHERE A MOTHER WANTS HELP FOR HER SON*

this is simple system of equations.

b=u+88
t=5u
t=b-40

[hide]so 5u=b-40.  or b=5u+40 AND b=u+88.
so 5u+40=u+88. subtract u and subtract 40.
4u=48.
u=[b]12[/b][/hide][/quote]
No need to freak out. Everyone makes mistakes.
Also to cmckeehan in general when you have x variables you need x equations to solve for the variable(s) (other than some special cases). Isolate one variable in terms of the others and substitute that into the other equations if you are stuck. Not always the best way but it works.

[quote="Walk Around The River"]*JINZI, YOURE ANSWER IS INCORRECT. PLEASE DONT POST INCORRECT SOLUTIONS.  IT CAN BE MISLEADING, ESPECIALLY IN THIS CASE WHERE A MOTHER WANTS HELP FOR HER SON*[/quote]
Uhh.. WATR, you should point out where he went wrong.

[hide="solution"]We have this system of equations:

$b=88+u$
$t=5u$
$b=40+t$

We will solve:

$40+t=88+u$
$t=48+u$
$48=4u$
$\boxed{u=12}$[/hide]

[quote="Walk Around The River"]*JINZI, YOURE ANSWER IS INCORRECT. PLEASE DONT POST INCORRECT SOLUTIONS.  IT CAN BE MISLEADING, ESPECIALLY IN THIS CASE WHERE A MOTHER WANTS HELP FOR HER SON*

this is simple system of equations.

b=u+88
t=5u
t=b-40

[hide]so 5u=b-40.  or b=5u+40 AND b=u+88.
so 5u+40=u+88. subtract u and subtract 40.
4u=48.
u=[b]12[/b][/hide][/quote]
Don't mind him. He's just teasing bfrock since they are in the same class.  :wink:


How can we find sin(18) and cos(18) without using a calculator?

One nice way is to find sin(72) and cos(72) first. To do this, we solve x4+x3+x2+x+1=0 in two ways. First we divide by x2 and solve the quadratic equation in x+1/x. The other way is to multiply by x-1 to get x5-1, which has roots at 72, 144, 216, 288, and of course 0 degrees in the complex plane.

Another way to find them is to look at a regular pentagon inscribed in a unit circle. The length of the edges of the pentagon will be x = 2*sin(36). Can you see why?

Then use Ptolomy's Theorem on the pentagon and it's diagonals to find a quadratic expression for x.

Also, see problems #33-35 in AoPS2.

Thanks for the responses. I think I get it now.


I am trying to solve the following problem but cannot get anywhere...

The task is to find two indepedent non-normal rvs X and Y such that
(X+Y)/|X-Y| has t distribution.

Any help would be much appreciated.

one thing i forgot: X and Y are identically distributed.


geometry
geometric transformation
reflection
real analysis
real analysis solved

What exactly do people mean when they say taht they want to bell curve the data because it is too high or because it is too low.

fr eg.
A recent test had a class average which was normally distributed with mean 83% and standard deviation 10%. The teacher wants to "bell curve" the marks so that the new scores should reflect a mean of 67% and standard deviation of 5%. What happens to a students mark if it was originally 93%?

How would you do this type of question?

It means to transform data so that it exhibits some more desirable statistical properties, just like in your example.

The easiest way to do this is via a linear transformation, i.e. $Y = aX + b$, where $X$ is the old grade, $Y$ is the new grade, and $a$ and $b$ are constants.  There are formula for $E[Y]$ and $Var(Y)$ in terms of $E[X]$ and $Var(X)$.  Use these to find $a$ and $b$.


function
real analysis
real analysis unsolved

Prove that the function $ f:\mathbb{R}\to\mathbb{R}$, \[ f(x)\equal{}(C_{n}^{0})^{2}x^{n}\plus{}(C_{n}^{1})^{2}x^{n\minus{}1}\plus{}\ldots\plus{}(C_{n}^{n})^{2},\]
  has $ n$ distinct roots on the interval $ (\minus{}\infty, 0)$, where $ C_{n}^{k}\equal{}\frac{n!}{k!(n\minus{}k)!}$.

Somebody was faster then you. :P  But no answer.[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=903197#903197[/url] If no solution appears I'll ask him to give it.


The set $M$ of $4$ positive integers is called good set if $\forall x \in M$ we have only one of the numbers $x-1, x+1$ in $M$. Let $A_{n}$ be the number of good subsets of $\left\{ 1,2, ... , n \right\}$.
(i) Evaluate $A_{7}$.
(ii) Find the minimal n so that $A_{n}\geq 2007$

Assume that $M$ is a good subset of $\{1,2,...,n+1\}$. If $n+1\not\in M$ then number of those sets is equal to $A_{n}$. If $n+1\in M$ then $n\in M$ and $n-1\not\in M$ therefore $M=\{n+1,n,b+1,b\}$, here $b\in\{1,2,...,n-3\}$.

By above $A_{n+1}=A_{n}+n-3(n=4,5,...)$.


inequalities
trigonometry
limit
AMC
AIME
triangle inequality
trig identities

For a and b positive reals, compute the minimum value of:

sqrt[a^2 + (7sqrt2)a + 49] + sqrt[a^2 + b^2 - (sqrt2)ab] + sqrt[b^2-10b+50]

[quote="pda"]For $ a$ and $ b$ positive reals, compute the minimum value of:

$ \sqrt{a^2 \plus{} 7\sqrt2a \plus{} 49} \plus{} \sqrt{a^2 \plus{} b^2 \minus{} \sqrt2ab} \plus{} \sqrt{b^2\minus{}10b\plus{}50}$[/quote]

hard is right.
How in the world do you start? square? (yeah right)

or Fermat Point or minimumization of a sum of 3 segments.

[hide]Let $ O,P,Q,R,S$ be points in a plane such that $ OP\equal{}7$, $ OQ\equal{}a$, $ OR\equal{}b$, $ OS\equal{}5\sqrt{2}$, $ \angle POQ\equal{}135^\circ$, and $ \angle QOR \equal{} \angle ROS \equal{} 45^\circ$.

[asy]pair P,Q,R,S,O,A,B;
P=(-7,0);
Q=(3,3);
R=(5,0);
S=(5,-5);
O=(0,0);
A=(10,10);
B=(10,0);
dot(P);
dot(Q);
dot(R);
dot(S);
dot(O);
draw(P--Q--R--S);
draw(O--P);
draw(O--A,EndArrow);
draw(O--B,EndArrow);
draw(O--S);
label("P",P,W);
label("Q",Q,NE);
label("R",R,E);
label("S",S,SE);
label("O",O,SW);[/asy]

Using the Law of Cosines:
$ PQ\equal{}\sqrt{7^2\plus{}a^2\minus{}2\cdot7\cdot a \cdot \cos 135^\circ}\equal{}\sqrt{a^2 \plus{} 7\sqrt2a \plus{} 49}$
$ QR\equal{}\sqrt{a^2\plus{}b^2\minus{}2\cdot a\cdot b\cdot \cos45^\circ}\equal{}\sqrt{a^2 \plus{} b^2 \minus{} ab\sqrt2}$
$ RS\equal{}\sqrt{b^2\plus{}(5\sqrt{2})^2\minus{}2\cdot b\cdot5\sqrt{2}\cdot\cos 45^\circ}\equal{}\sqrt{b^2\minus{}10b\plus{}50}$

So, we must minimize $ PQ\plus{}QR\plus{}RS$.

Since $ \angle POQ > 90^\circ$, $ PQ \ge PO$ with equality when $ Q\equal{}O$.
By the triangle inequality, $ QR\plus{}RS \ge QS$ with equality when $ R$ is on $ QS$.
Since $ \angle QOS \equal{} 90^\circ$, $ QS \ge OS$ with equality when $ Q\equal{}O$.

Thus, $ PQ\plus{}QR\plus{}RS \ge PO\plus{}OS \equal{} 7\plus{}5\sqrt{2}$ with equality when $ Q\equal{}O$ and $ R$ is on $ QS$.
However, this yields $ a\equal{}b\equal{}0$, which aren't positive reals.
But it is easy to see that $ \lim_{a \to 0^\plus{}}\lim_{b \to 0^\plus{}}(PQ\plus{}QR\plus{}RS) \equal{} 7\plus{}5\sqrt{2}$, so any value slightly greater that $ 7\plus{}5\sqrt{2}$ can be achieved.
[/hide]

But the last 5 problems on the Australian MC are like AIME problems, which means that the answer must be an integer from 0 - 999 :)

A friend of mine who sat the test claims that this question in fact asked for the minimum integer value of the expression.

Actually, the problem just says "minimum value", not minimum integer value  :) I'm reading the problem booklet right now

why cant you just substitute 0? since a,b are all positive intergers. we could assume that the minimum value of that is when a,b is at 0. so that thing is greater than 7+sqrt(50)? am i correct?

In this case, it just happens to work, but for problems of this type, you have to justify your answer as well. Why is your reasoning correct?

In particular, the third term of the original expression is minimized when $ b\equal{}5$.

It's not just enough to minimize one of the terms of a sum -- you have to work with all of them together.

The minimum value of the expression is 13. Hint: Consider cosine rule. sqrt of 2, what can you link it to?

I did it the same way as Xantos, what is wrong? $ 7\plus{}5\sqrt{2} \ge 14$, so why should it be 13?

Consider a figure OPQRS, with OP=7, OQ=a, OR=b, OS=5sqrt2 and angle POQ=angle QOR=angle ROS= 45 degree.

As a result, one can easily see that PQ=sqrt(49+a^2-7sqrt2), QR=sqrt(a^2+b^2-sqrt2 ab) and RS=sqrt(50+b^2-10ab). Then, the question is actually asking what is the minimum value of PQ, QR and RS and it only occurs when the sum is equal to the length PS. Using cosine rule again, we can find that PS is equal to 13.

Hmmm.. Mathematica seems to say its limiting value is 14...
I may be missing something...


For the second step of hydroboration where $ \ce{H2O2}$ is used to produce the alcohol. Does anyone know the mechanism for this process? Can someone propose one or cite a reference?

Thanks.

That step is a beautifull rearrangement reaction: three consecutive migrations from boron to oxygen occur. Here's the detailed mechanism:

The reaction of trialkylboranes with carbon monoxide is even more beautifull, since all three migrations are [i]different[/i].


videos
Alcumus
algorithm
probability
expected value
Support

I was watching the first expected value video and finished it, but when i returned to the videos page, the whole section (along with another topic) was gone.

Any reason?

P.S.
I didn't do any problems after the video

Have you done any problems since you noticed this issue? Sometimes you might answer a couple of questions wrong and your completion percentage in that topic might go down as a result. Also, do you remember what topics you were working on?

In Alcumus, the system determines what topics you are ready for based on what topics you have passed, and what topics the algorithm thinks you are safe to skip. Sometimes, when you miss a question in a topic that the system previously determined that you were safe to skip, it might make other topics unavailable until you get a higher completion percentage in the topic that is no longer safe to skip. I suspect this is what happened in your case.

Thanks for using Alcumus!

Thanks for answering the disappearing video question for me. I had the problem too. Also, one day I passed this subject after a while of work and a few days later on that subject instead of having it passed and 100% it was back to 8%. Why?


geometry
circumcircle
geometric transformation
reflection
geometry proposed

AA', BB' and CC' are diameters of the circumcircle of $\triangle{ABC}$.
Prove: $\triangle{A'BC}+\triangle{B'AC}+\triangle{C'AB}=\triangle{ABC}$.

Let the parallel through $A'$ to $BC$ cut the circle again in $A''$. Since $\angle AA''A'=\frac\pi 2$ and $A'A''\|BC$, it means that $AA''\perp BC$. This means that $A''$ is the reflection of the orthocenter $H$ of $ABC$ in $BC$, so $S_{[A'BC]}=S_{[A''BC]}=S_{[HBC]}$. We add up the analogous identities obtained for the other sides, and we get the desired result.


$px^4+qx^3+r=0$

$rx^4+qx^3+p=0$

These two polynomials have a common real root.

Find all possible values of q, if p and r are distinct numbers and $p+r=100$

[hide]
i think my solution is wrong, so help me out please. thanks

let the common real root be $\alpha$

$\implies P(\alpha) = 0$

since $p+r=100, \implies r=100-p$

$\implies p\alpha^4+q\alpha^3+100-p=0 --- (1)$

$\implies (100-p)\alpha^4+q\alpha^3+p=0$

$\implies 100\alpha^4-p\alpha^4+q\alpha^3+p=0 --- (2)$

We add equations $(1) \& (2)$

We get $100\alpha^4+2q\alpha^3+100=0$

$\implies q=\frac{-50(\alpha^4+1)}{\alpha^3}$

$\implies \alpha^3|50$

Only possibilities are when $\alpha=1 or \alpha=-1$

Subbing into $100\alpha^4+2q\alpha^3+100=0$, yields, $q=-100$ and $q=100$

[/hide]

[quote="sen"]$\implies q=\frac{-50(\alpha^4+1)}{\alpha^3}$

$\implies \alpha^3|50$
[/quote]

That doesn't seem like a safe step to me.  I think you're assuming that $\alpha$ is an integer, but at this point, you don't even know if $\alpha$ is rational.

Here's a hint:
[hide]Instead of adding the equations, try subtracting the equations.[/hide]

if we assume say  alpha to be the common root and then eliminate q  we get alpha=+-1
n substituting gives value of q = +- 100


calculus
integration
trigonometry
logarithms
calculus computations

For $ a\in\mathbb{R}$, find the minimum value of $ \int_0^{\frac {\pi}{2}} \left|\frac {\sin 2x}{1 \plus{} \sin ^ 2 x} \minus{} a\cos x\right| dx$.

Let $ f(x) = \frac {\sin 2x}{1 + \sin^{2}x} - a\cos x,\forall x\in \left[0,\frac {\pi}{2}\right]$

Case1
$ a\in ( - \infty,0]\rightarrow f(x)\geq 0$
${ I = \int_{0}^{\frac {\pi}{2}} \left|f(x)\right| dx = \int_{0}^{\frac {\pi}{2}} f(x) dx = \left[\ln\left(1 + \sin^{2}x\right) - a\sin x\right]_{0}^{\frac {\pi}{2}} = \ln 2 - a}$

Case2
$ a\in (0,1]\rightarrow \left[f(x) = 0\leftrightarrow x = \arcsin\left(\frac {1 - \sqrt {1 - a^{2}}}{a}\right)\right]$
$ I = \int_{0}^{\frac {\pi}{2}} \left|f(x)\right| dx = \int_{\arcsin\left(\frac {1 - \sqrt {1 - a^{2}}}{a}\right)}^{\frac {\pi}{2}} f(x) dx - \int_{0}^{\arcsin\left(\frac {1 - \sqrt {1 - a^{2}}}{a}\right)} f(x) dx$
$ I = 2 + 4\ln a - 2\ln\left(1 - \sqrt {1 - a^{2}\right) - 2\sqrt {1 - a^{2}} - a - \ln 2}$

Case3
$ a\in (1,\infty)\rightarrow f(x) < 0$
$ I = \int_{0}^{\frac {\pi}{2}} \left|f(x)\right| dx = - \int_{0}^{\frac {\pi}{2}} f(x) dx = - \left[\ln\left(1 + \sin^{2}x\right) - a\sin x\right]_{0}^{\frac {\pi}{2}} = a - \ln 2$

The minimum value of $ \int_{0}^{\frac {\pi}{2}} \left|\frac {\sin 2x}{1 + \sin^{2}x} - a\cos x\right| dx=\ln\frac{32}{25}$ when $ a=\frac{4}{5}$

That's right. :) , but Case 2, $ \frac {dI}{da} \equal{} .....$. :wink:

The minimum value of Case3 does not exist right?
But all value in Case3 is greater than $ 1\minus{}\ln 2$ :lol:

Here is my solution.

Let $ I(a) \equal{} \int_0^{\frac {\pi}{2}}\left|\frac {\sin 2x}{1 \plus{} \sin ^ 2 x} \minus{} a\cos x\right|dx \equal{} \int_0^1 \left|\frac {2t}{1 \plus{} t^2} \minus{} a\right|dt\ [t \equal{} \sin x]$

For $ c$ such that $ \frac {2c}{1 \plus{} c^2} \minus{} a \equal{} 0\ (0 < c < 1)$,

$ I(a) \equal{} \int_0^c \left|\frac {2t}{1 \plus{} t^2} \minus{} a\right|dt$

$ \geq \minus{} \int_0^c \left(\frac {2t}{1 \plus{} t^2} \minus{} a\right)\ dt \plus{} \int_c^1 \left(\frac {2t}{1 \plus{} t^2} \minus{} a\right)\ dt$

$ \equal{} \minus{} [\ln (1 \plus{} t^2) \minus{} at]_0^c \plus{} [\ln (1 \plus{} t^2) \minus{} at]_c^1$

$ \equal{} (2c \minus{} 1)a \minus{} 2\ln (1 \plus{} c^2) \plus{} \ln 2$.

Thus determinimg the value of $ c$ such that this equation takes the constant number independent of the value of $ a$, we obtain $ c \equal{} \frac {1}{2}\Longleftrightarrow a \equal{} \frac {2c}{1 \plus{} c^2} \equal{} \frac {4}{5}.$

If $ a$ takes this value, $ I_{min} \equal{} 5\ln 2 \minus{} 2\ln 5$.

[quote="kunny"]
$ I(a) \equal{} \int_0^c \left|\frac {2t}{1 \plus{} t^2} \minus{} a\right|dt$

$ \geq \minus{} \int_0^c \left(\frac {2t}{1 \plus{} t^2} \minus{} a\right)\ dt \plus{} \int_c^1 \left(\frac {2t}{1 \plus{} t^2} \minus{} a\right)\ dt$
[/quote]
How can like this :?:
Sorry if I ask a silly question. :maybe:
And also typo on the first line of your solution too. :lol:


Can someone help me with this? I don't understand how to solve a system of linear congruences:

$N \equiv 1 (mod 2)$

$N \equiv 3 (mod 5)$

For some integers $a,b$, we have $N=2a+1=5b+3.$
Rearranging this gives $2a-5b=2\Longleftrightarrow 2(a-1)=5b=10m$ for some integer $m$, yielding $a=5m+1, b=2m.$ Thus $N=5\cdot 2m+3=10m+3.$

[quote="alkjash"]How did you get from $2a-5b$ to $22(a-1)$ ????[/quote]
Sorry I made a typo. I have just edited.


Well considering that many of us here shud be luving hogging(or so i presume) wat all food do u hog :?:
I can eat anything veg.
particularly miss the madras pongal sambar....  :( Its been YEARS (atleast 2) since i came there.
But ow i can hog ne kinda food.
Kerala food s Awesome.
my list will go on....
 :rotfl: but u can first start off urs

I am a vegeterian.... and don't have any special preference except that it should be cooked properly and must taste good...

and thought that I could eat anything veggie on earth until I stumbled upon real Chinese food in Beijing and Thai food in Chiang Mai.... :rotfl:

And came to the conclusion I am much more comfortable with [b]Indian[/b] veggie food and cooked food with spices and salt and not semi-boiled near-raw mushroom, tofu, brocolli... :huh:  :rotfl:

So Mr. Raghavan I think you are not too late to take back your claim that
You can eat anything veg... :wink:

I am extremely picky about food, but not very specific only about Indian food. I can manage with a sandwich and fries just fine :D :D

Shreyas on the other hand, cannot go even one day without having pork chops, lamb fry and whale stew..

:D

[quote="shruthibaskaran"]
Shreyas on the other hand, cannot go even one day without having pork chops, lamb fry and whale stew..

:D[/quote]

You forgot Beef Kabab, Parrot Brains, Monkey Kidneys and Roast Ostrich.

On a serious note, I'm pure veggie as well. :) I can eat almost anything except Plain idli (a decent sidedish is a must)..

Have you heard of a Brahmin who eats anything from Thayir sadham oorugai  to Mexican crispy chicken , Shrimp cocktail, Mutton pizza.. :D
[hide]Thats me :rotfl:  :rotfl: [/hide]

I like food. PERDIOD!

but, i'm a veggie too (so far  :wink:) though, i absolutely love eggs   :D

no preferences as long as i can eat!!!!!!!!!

[quote="Chronoz"]Rags, next time engha varumbodhu sollu. Rendu perum serndhu poi pongal sambar sappidalam....  :lol:[/quote]

I too shall join you, but only if it's at Murugan Idli Kadai  :rotfl:

Lol. Idli with or without side dish sucks. Period.

I dont mind dosas though. But easiest to make, and best to eat would be a sandwich. I live on them mostly :D

easier to make--noodles :lol:
[hide="more easy...."]spaghetti :P [/hide]

well...
here are a few of my favorites:
favorite starter:
[hide] [b] veg[/b] balls schezwan
dont be a perverted bum
[/hide]

and ofc ne kinda manchoorian...

then i love hogging chats.
and among desserts, i like gudbud and triple sundae.

I am strictly vegetarian
I like all types of food but the favourites are south indian(dosa &idli with other things) and paneer dishes.
how many of you know to cook? i am unique one
knows to cook nothing practically, but knows to cook every dish in theory.

I was thinking what was my favorite snack.

[hide]
MOMOS! (veg)
[/hide]

All time fav-

nobody can deny that they like maggi   :P

maggi sux. i only have it whenever i stay up at nite, if dere's nothing else available. :blush:

rather we suck up maggi :rotfl:  :rotfl:  :rotfl:

Actually Maggi isn't very nutritious nor filling. Its tasty but the "Taste Bhi, Health Bhi" is bull.

I just love dosas and I can live with indian food to some extent but please note that I am an omnivore and eat meat occasionally.

so are you from south india originally???  :D  :P

the sigma proportional isnt good and the question on amc is not upto the standards american national standards so better get a question on quantum physics ya anything else

indian!

the ques was wat  u like :rotfl:  :rotfl:  :rotfl:

how did quantum physics come up when the discussion was of food???? i think rt_08 is a gifted minded person to derive physics out of food comments.  but u could have mathematical statements, thus: if i have samosa, then i have my favorite snack.----a contitional statement.
ALSO: with my favorite dessert, rassogulas i am greater than (>) happy.
see math applies to evrything IT ROX!!!!! yay math! continue with quantum phiysical food if u must :rotfl:  :lol:  :D

rt_08 thank you very much for making my head hurt and me feel stupid........ :D

[quote="rt_08"]QUANTUM PHYSICS X^0214+.0012)^56*117
THATS THE VALUE OF THE SIGMA PROPORTIONAL![/quote]

is that wat u like to eat????? :rotfl:  :P

well the topic just reminded me of my summer camp at bangalore
the food at Audi's was great :D

but that at mess--well i didn't feel like eating breakfast any day :rotfl:
@da--do u really like that rasam thing??? :oops:

PS -the best part was the Ice CReam--well i had 8 flavours in 5 days
and most of them i had eaten 4 1st tym :)


BC has moved left by 4 units and up by 4 units.

The midpoint is therefore left 2 units and up 2 units.

$ M\equal{}(5\minus{}2,1\plus{}2)\equal{}(3,3)$

[quote="moldovan"]Consider $ B(5,1)$ and $ C(1,5)$. The middle of $ BC$ has the coordinates:
$ (A)$ $ (3,0)$
$ (B)$ $ (2,3)$
$ (C)$ $ (0,0)$
$ (D)$ $ (1,3)$
$ (E)$ $ (3,3)$[/quote]

Use the midpoint formula.
For coordinates (x1,y1) and (x2,y2), the midpoint is
((x1+x2)/2,(y1+y2)/2).
In this case the midpoint is ((5+1)/2,(1+5)/2)=(3,3).
So the answer is [b]E[/b]

[quote="moldovan"]Consider $ B(5,1)$ and $ C(1,5)$. The middle of $ BC$ has the coordinates:
$ (A)$ $ (3,0)$
$ (B)$ $ (2,3)$
$ (C)$ $ (0,0)$
$ (D)$ $ (1,3)$
$ (E)$ $ (3,3)$[/quote]

You find the midpoint of a line segment by finding the average of the endpoints' coordinates, in this case the midpoint would be;

$ (\frac{5 \plus{} 1}{2}, \frac{1 \plus{} 5}{2})$

$ (\frac{6}{2}, \frac{6}{2})$

$ (3, 3)$

E is the answer.


topology
geometric series
advanced fields
advanced fields unsolved

Let $ \ell_1(\mathbb{Z}) \equal{} \{(x_n)_{n\in\mathbb{Z}}: \sum_{n\in\mathbb{Z}}|x_n| < \infty\}$. Addition is defined in the usual way ( by components ), and we define $ z \equal{} xy\Leftrightarrow z_n \equal{} \sum_{k\in\mathbb{Z}}x_ky_{n \minus{} k}, (x_n),_{n\in\mathbb{Z}}(y_n)_{n\in\mathbb{Z}}$. Prove that the spectrum $ \sigma(\ell_1(\mathbb{Z}))$ is homeomorphic  to $ S \equal{} \{\lambda\in\mathbb{C}: |\lambda | \equal{} 1\} \equal{} \{e^{i\alpha}: \alpha\in\mathbb{R}\}$, so by that, the Gefland transformation is $ x \equal{} (x_n)_{n\in\mathbb{Z}}\in\ell_1(\mathbb{Z})\Rightarrow \check{x}(e^{i\alpha}) \equal{} \sum_{n\in\mathbb{Z}}x_ne^{in\alpha}$ ("Fourier series")

So far, I have showed that $ \ell_1(\mathbb{Z})$ is generated by $ x \equal{} (x_n)_{n\in\mathbb{Z}}, x_n \equal{} \delta_{n,1},n\in\mathbb{Z}$ and it's inverse $ x^{ \minus{} 1} \equal{} (x^{ \minus{} 1}_n)_{n\in\mathbb{Z}}, x^{ \minus{} 1}_n \equal{} \delta_{n, \minus{} 1},n\in\mathbb{Z}$, and one can easily see that $ \ell_1(\mathbb{Z})$ is a unital commutative algebra, so $ \sigma(\ell_1(\mathbb{Z}))$ is isomorphic to $ \sigma(x)$.
Now, I can't prove that in fact $ \sigma(x) \equal{} S$ ( and by this I would finish of the first part of the problem ). $ x$ reminds me of a shift operator, but $ \sigma(x)$ is way to abstract for me to make some kind of analogy between $ x$ and shift operator ( in $ \mathcal{B}(X)$. )
Any hints are welcome!

elements of $ l^1(\mathbb{Z})$ may be written as laurent series $ \sum_{n} a_n T^n$ where $ a_n$ is an absolutely convergent series of comples numbers. this motivates the multiplication. the adjoint operation is defined by extending $ T^* \equal{} T^{ \minus{} 1}$. then it's clear that $ l^1(\mathbb{Z})$ is generated, as a C*-algebra, by $ T$. thus the spectrum of this algebra is isomorphic to the spectrum of $ T$. since $ T$ has norm $ 1$, the spectrum lies in the closed unit disk. but if $ \lambda \in \mathbb{C}$ satisfies $ |\lambda| < 1$, the geometric series explicitely gives an inverse of $ T \minus{} \lambda$ in $ l^1(\mathbb{Z})$, and the same idea (working first in $ \mathbb{C}[ [T] ]$ and then comparing coefficients) shows that $ T \minus{} \lambda$ is not invertible if $ |\lambda| \equal{} 1$. thus the spectrum is the unit circle. the gelfand transform is clear.


My guess is that you are actually looking for

$\sum^{\infty}_{i=2}(\zeta(i)-1)$

We have

\begin{eqnarray*} \sum^{\infty}_{i=2}(\zeta(i)-1)&=&\sum^{\infty}_{i=2}\sum^{\infty}_{j=1} \frac{1}{j^i} \\ &=&\sum^{\infty}_{j=1}\sum^{\infty}_{i=2}\frac{1}{j^i} \\ &=&\sum^{\infty}_{j=1}\frac{1}{j(j-1)} \\ &=&\sum^{\infty}_{j=1}\left(\frac{1}{j}-\frac{1}{j+1}\right) \\ &=&1 \end{eqnarray*}

(you can interchange order of summation since the summands are positive (or because the sums converge absolutely))


I walked at 5 km per hour down Main Street and left town. A car driving down Main Street at 35 km per hour left town at 10:15 am the same day and passed me at 10:40 am. At what time that morning did I leave that town?

Source: Math Team team round question (don't worry, that was the last one!)

Feel free to correct

[hide]

Answer: $ \boxed{6: 35}$

Solution:

Ok because we know that the distance that both you and the car have traveled is the same we can set both D=RT equations equal to eachother.  The first equation (you) is $ D\equal{}R(5) \times {T(X)}$ and the car $ D\equal{}R(35) \times{T(\frac{7}{12})}$  Because both of these equations equal $ D$ we know that $ 5r \equal{} \frac{35\times{7}}{12}$ solving for $ x$ yeilds $ 4 \frac{5}{60}$ which means that you have been walking for 4 hours and five minutes before passing the car so we subtract 4hours 5mins from 10:40 and get that you left town at $ \boxed{6: 35}$ [/hide]


induction
combinatorics proposed
combinatorics

There are some blue points and some red points in the plane, each red point is connect with at least one blue point, each blue point is connect with one or two red points. Prove that: we can delete at most half of number points so that with all remain points, each blue point is connect with exactly one red point.

What exactly do you mean :?:
Suppose we have 1 red point connected with 4 blue points.
Then deleting 3 blue points gives 1 blue point connected with a red point and then 3>(1+4)/2.
Deleting the red point gives 0 blue points connected with a red point and then 1<(1+4)/2.
kweenie

[quote="kweenie"]
Suppose we have 1 red point connected with 4 blue points.
[/quote]

In this case you only can delete at most $ 2$ points!

I still don't understand what you mean.
Do you mean:
Deleting at most half the number of points it's possible to have each blue point being connected with exactly 1 red point? (In the example I gave with 5 points this is not possible as you have to delete 3 points to have this).

Or do you mean something else?

kweenie

It mean: "Exist a way" in which we delete at most half of number points so that with all remain points, each blue point is connect with exactly one red point.

[quote="kweenie"](In the example I gave with 5 points this is not possible as you have to delete 3 points to have this).[/quote]  No: in your example, we can delete 0 points to have each blue point connected with exactly one red point.

OK, I realised this already with your previous mail.
Now I get the picture and see what you mean. I'm thinking about a proof (preferable an elegant one). Please don't post your solution too quickly (let me and others have two days).
kweenie

PS: if the little boy in your avatar is really wearing a gun, please say him not to shoot "by accident" his friendly dog.

Working the reverse way:
Suppose we have a good colored graph ( it must have at least 1 red point) and we add now as much red points as possible. Then the maximum of added red points is equal to the number of blue points and thus smaller than half the total number of vertices.

But this does not solve the problem as there exist original graphs where we have also to delete blue points (eg: a 6 circuit with alternating red an blue points).
So I give up. Please come up with your proof tdl (except if other want you to wait).
kweenie

note: my previous mail was directed to tdl, I confused with JBL.

We are gonna solve it by induction on the sum of the numbers of blue and red points. The small cases are true, so assume that we have k red and m blue points. (K is the set of red points, M is the set of blue points)
Take the largest subset of red points with the largest number of neigbours such that any two elements of this subset haven't got any common neighbours. Say this set A, and the blue points that is a neighbour of some element of A are set B. Now look at K-A, some of these vertices have at least one neighbour from M-B. Say this set of blue vertices C, the remaining blue vertices D. Finally let me say E to the set M-B. The sets A, B, C, D, E have respectively a, b, c, d, e elements.
The sets E and C satisfies the conditions of the induction. So we can take at least (e+c)/2 elements from these 2 sets so that all of the blue vertices have degree 1. Assume that there are l points from the set B which are neighbour with one of the red points we took.
1) If we take these l red points. We can take also all the elements of D and their blue neighbours. One can easily check that these (e+c)/2+l+2d points satisfy the condition.
2) If we don't take any of these l red points. We can take the remaining vertices of set B and all the vertices of set A. It's again clear that these (e+c)/2+b-k+a points satisfy the condition. (due to A's selection, the points of E are not connected to A)
Since (e+c)/2+l+2d+(e+c)/2+b-l+a is more than or equal to the total number of vertices, one of them must be at least half of it. So we are done.


What is the sum of the first 100 square numbers?


[quote="indybar"]If $a_1,a_2,...,a_n$ are positive, and $\frac{1}{a_1}+...+\frac{1}{a_n} = n$, prove that
$\frac{1}{1+a_1}+...+\frac{1}{1+a_n}\leq \frac{n}{2}$[/quote]

For every positive a, Cauchy-Schwarz yields

$\left(1+\frac{1}{a}\right)\left(1+a\right)\geq\left(\sqrt{1\cdot 1}+\sqrt{\frac{1}{a}\cdot a}\right)^2=\left(1+1\right)^2=4$,

so that $\frac{1}{1+a}\leq\frac{1+\frac{1}{a}}{4}$. Thus,

$\sum_{i=1}^n\frac{1}{1+a_i}\leq\sum_{i=1}^n\frac{1+\frac{1}{a_i}}{4}=\frac14\left(n+\sum_{i=1}^n\frac{1}{a_i}\right)=\frac14\left(n+n\right)=\frac{n}{2}$,

qed.

  Darij

Darij . You actually prove AM-HM Inequality ;)


trigonometry
inequalities proposed
inequalities

Let x,y,z be the real number such that 0<x,y,z<1 and $ (x\plus{}y\plus{}z\minus{}2)^2\equal{}2xyz$.Prove that $ \sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}$≤$ \frac{3\sqrt2}{2}$

[quote="Flakky"]Lagrange multipliers[/quote]
Excuse me,what is Lagrange multipliers?
My idea is to assume the opposite assertion.

[quote="Flakky"]Lagrange multipliers[/quote]
Oh I see. Lagrange multipliers is a nice choice :lol:

[quote="Marian Dinca"]$ \mbox{Let: } x \equal{} 1 \minus{} a, y \equal{} 1 \minus{} b, z \equal{} 1 \minus{} c \mbox{ and } a,b,c\in(0;1); \mbox{ then: } \\
(x \plus{} y \plus{} z \minus{} 2)^2 \equal{} 2xyz \Leftrightarrow (1 \minus{} a \minus{} b \minus{} c)^2 \equal{} 2.(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\Leftrightarrow \\
\Leftrightarrow 1 \plus{} a^2 \plus{} b^2 \plus{} c^2 \minus{} 2a \minus{} 2b \minus{} 2c \plus{} 2ab \plus{} 2ac \plus{} 2bc \equal{} 2.(1 \minus{} a \minus{} b \minus{} c \plus{} ab \plus{} ac \plus{} bc \minus{} abc)\Leftrightarrow$
$ \Leftrightarrow 1 \minus{} a^2 \minus{} b^2 \minus{} c^2 \minus{} 2abc \equal{} 0\mbox{ rezults: } a \equal{} \cos A, b \equal{} \cos B, c \equal{} \cos C. \\
A, B, C \mbox{ angles of triangle and: } \sum{\sqrt {x}} \equal{} \sum{\sqrt {1 \minus{} \cos A}} \equal{} \\
\equal{} \sum{\sqrt {2}.sin\frac {A}{2}}\le3.\sqrt {2}.\sin\frac {A \plus{} B \plus{} C}{6} \equal{} 3.\sqrt {2}.\frac {1}{2} \equal{} \frac {3\sqrt {2}}{2}.$[/quote]
Nice :lol: Thank you!


Show that for any integer n > 10000, there are integers a, b such that n < a2 + b2 < n + 3 n1/4.

posted in a wrong section and already posted before few times - search for it


Imagine yourself participating in the following BIZARRE contest.
You have a choice of three doors.Behind one door there is a costly car,behind the other two doors are goats!! You will be given the object behind the door you open.
You choose the first door and the quiz master opens the third door revealing a goat.He then offers you the opportunity to change your choice of door.

Prove that you can not calculate the probability that the car lies behind the second door.

The author states that many people think the problem  has a solution ,popular answers being 1/2 and 2/3

This is the famous Monty Hall Problem, in case your book didn't say. Here's a link [url]http://en.wikipedia.org/wiki/Monty_hall_problem[/url] I'm not a probability expert, so I can't really help on this, but external reading might.

The key lies in whether you think the quiz master opened a random door and it just happened to be a goat, or you think he opened a door that he knew was a goat-door.  That's where the two different answers come from.


quadratics
number theory unsolved
number theory

Assume that $a|5+1998^{1998}$.Prove that a can be written as $2x^2+3xy+2y^2$ when and only when $a=20k+3$ or $20k+7$

the solution uses quadratic residue.It's not very hard


Let $ a$ and $ b$ be arbitrary integers. Prove that if $ k$ is an integer not divisible by $ 3$, then $ (a\plus{}b)^{2k}\plus{}a^{2k}\plus{}b^{2k}$ is divisible by $ a^{2}\plus{}ab\plus{}b^{2}$

Equivalently, $ (x\plus{}1)^{2k}\plus{}x^{2k}\plus{}1$ is divisible by $ x^{2}\plus{}x\plus{}1$ over $ \mathbb{Z}[x]$.

Let $ \omega$ be a root of $ x^{2}\plus{}x\plus{}1$.  Then

$ (\omega\plus{}1)^{2k}\plus{}\omega^{2k}\plus{}1 \equal{} (\minus{}\omega^{2})^{2k}\plus{}\omega^{2k}\plus{}1 \equal{}\omega^{2k}\plus{}\omega^{k}\plus{}1 \equal{} 0$

a easy proof is use
$ a\minus{}1|a^{n}\minus{}1$
But I think complex is a good and popular solution.


Me again... I am totally confused by this problem and any help would be greatly appreciated:

Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid $(H_2C_2O_4)$ in a solid given that a $0.7984g$ sample of that solid requires $37.98 mL$ of $0.2283 M NaOH$ for neutralization.

There are two real world problems linked in this question.  The first is the acid used is not pure, the second is that you can't measure what you want to know.

You can't work directly on the first question with the information given, so you have to work on a second question: how much oxalic acid is present in the 0.7984 g impure sample.

To figure out how much oxalic acid is present in the given sample, you have to figure out how much hydroxyl ions (OH-) are provided in the sodium hydroxide.  If you multiply the volume (in L) by the Molarity you will calculate the number of gram-moles of hydroxyl ions provided by the sodium hydroxide.

If you take the number of gram-moles of hydroxyl ions and divide by the number of protons each oxalic acid molecule provides, you will have the number of gram-moles of oxalic acid that was in the impure sample.

Multiplying the number of gram-moles of oxalic acid by the molecular weight ow oxalic acid will yield the number of grams of oxalic acid in the impure sample.

I will leave you to calculate the percentage from the amount present and total mass of the sample.

I'm glad to see that they are giving you a real world problem.  This type of problem is solved every day my folks with real jobs.  The only questionable real world aspect of this problem is the fact that you have a weight and molarity measured to 4 significant digits.  Normally, you might have the molarity to 3 significant digits and the weight to 2 significant digits.

Oh well, you have a good program that is using real world type problems.  You can thank your teacher now or after your graduate.

Wow! Thank you so much for that easy-to-understand explanation! My teacher didn't even go over this, but it's a bonus on the test tomorrow. Thank you _so_ much!

And I'm glad to know that what we're doing actually _does_ come in handy lol.


inequalities
limit
number theory proposed
number theory

Find all pairs of integers $ (x,y)$ satisfying $ x^2\plus{}y^2\minus{}8xy\plus{}6\equal{}0$

Let $ u\equal{}x\minus{}y$. We have
$ u^2\equal{}6(y^2\plus{}uy\minus{}1)$.
We see $ 6|u$. Let $ u\equal{}6v$.
$ 6v^2\equal{}y^2\plus{}6vy\minus{}1$, or
$ (y\plus{}3v)^2\minus{}15v^2\equal{}1$.
You get pell equation.

... except that it doesn't.  For example, $ \lim_{x \to \infty} f(x, x) \equal{} \minus{}\infty$.  The Pell's equation approach reveals that there are infinitely many solutions, and those solutions are otherwise somewhat difficult to isolate.

[quote="RED-and-black"]Don't even need that. Obviosly for large $ x,y$, $ x^2 \plus{} y^2 \minus{} 8xy$ becomes positive. So inequalities would be a better way, expecially if you are in a contest.[/quote]
Wrong! Note that you can not choose two variants x,y too large in f(x,y), you can unless fixed x or y.
[quote="xxp2000"]Let $ u \equal{} x \minus{} y$. We have
$ u^2 \equal{} 6(y^2 \plus{} uy \minus{} 1)$.
We see $ 6|u$. Let $ u \equal{} 6v$.
$ 6v^2 \equal{} y^2 \plus{} 6vy \minus{} 1$, or
$ (y \plus{} 3v)^2 \minus{} 15v^2 \equal{} 1$.
You get pell equation.[/quote]
Can you give the final solution of (x,y) in form of sequences?


linear algebra
matrix
Ring Theory
absolute value
number theory proposed
number theory

I've been running around in circles all day, and still no solution..

The nicest equivalent form I could find is this: let $u=\frac{1+i\sqrt 7}2$, and let $\bar u$ be its conjugate. Find $n$ s.t. $\left |\frac{u^n-\bar u^n}{u-\bar u}\right|=1$.

The $n$'s we're looking for are the $n$'s we find for the problem above $+2$. It's pointless to get into details about how to reach that since I have no idea whether it yields a solution or not. What's nasty about this is that the solutions seem to be (considering only positive $x$'s) $x\in\{1,3,5,11,181\}$. How on Earth does one reach $181$? :) This means that the answer to the question I asked should be $n\in\{1,2,3,5,13\}$ (I'm not sure, it just seems that way).

Another equivalent question: find those $n$ for which the element in position $(1,2)$ of the matrix $M_n=\left(\begin{array}{cc}1&1\\-2&0\end{array}\right)^n$ has absolute value $1$.

Anyway, I hope this helps someone.

Ramanujan never solved this one only conjectured (correctly) it's solutions. The one who did solve it was Nagell, the standard proof: http://www.math.sc.edu/~filaseta/gradcourses/TheMath784Notes.pdf on page 50.

Since $-7=M_4+1$, you want to work with $\mathbb Z[\frac{1+\sqrt {-7}}2]$ (the ring of integers in $\mathbb Q[\sqrt{-7}]$), which is a UFD.

thanks, i read the solution and i saw it uses this idea...


A triangle and a quadrilateral intersect at finitely many points. What is the most number of intersection points they can have?

[quote="236factorial"][hide]Is it 6 or more than 6? [/hide][/quote]

[hide]i guess it is also 6 . Unless you can point out the seventh . But there could be a case where the side of triangle is the same as the side of the quadrilateral . Who knows ...  :P [/hide]

:rotfl:  :rotfl:
I didn't say the quadrilateral needs to be convex  :D

No, think of a concave quad. How can you draw it that makes it intersect the triangle the most number of times?

After 10 triangles, I get
[hide]6  :rotfl: [/hide]

Hint
[hide]each side of the quadrilateral must have intersect the triangle twice (this is possible because the quad is concave) [/hide]

Yes!
(something like a V-shaped quadrilateral)

[quote="236factorial"]I found 8 points of intersection!!!!! Finally!!![/quote]

 :D  :)

Here's a diagram:

Will disappear in about 2-3 weeks  :D


[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=242]Diagram[/url]

A uniform board of mass m and length L is placed on the back of a flatbed truck as shown in the diagram above. There is no friction between the top of the board and the vertical surface of the board. The coefficient of static friction between the bottom of the board and the horizontal surface is u=0.50. The truck always moves in the forward direction. Express your answers to the following as multiples of g.

a. What is the maximum starting acceleration the truck can have if the board is not to slip or fall over?
b. What is the maximum stopping accleration the truck can have if the board is not to slip or fall over?
c. For what value of stopping acceleration is the static frictional force equal to zero?

Any help would be appreciated, thanks! :)

[quote="bus"]It's nothing but mechanics. You need to write down all the forces parametrized by acceleration of the truck and find out when does the board slip/fall and when it doesn't. More exactly:

- gravitational force acting on the board
- force created by the acceleration of the truck
- reaction force of the truck in places where the board touches the truck
- friction force

If all these forces add up to zero, board is not slipping/falling, otherwise it is.[/quote]

I know how to do a and b, but I don't really understand c.

The friction force is zero when the board doesn't want to slip left nor right. If this sounds confusing to you: When the truck doesn't accelerate, the board would like to slip left but the friction prevents this from happening. However, when the truck starts to decelerate too much, the board is pulled right and wants to slip right this time, but the friction force again pulls in the other direction and keeps the board standing. You need to find the right deceleration at which the board is in equilibrium - it is not being pulled left nor right, only down. Write down the equations and solve them.


geometry
3D geometry
sphere
Gauss
integration
LaTeX

A conductor sphere of radius a is placed inside a hollow conductor sphere with inner radius 4a and external radius 5a.
The 2 spheres are initially uncharged. On a subtle dielectric surface of radius 2a there's an electric charge Q equally distributed. The conductor spheres are linked to a battery with an electromotive force V.

a) Find the electric charges Q1, Q2, Q3 on the surfaces 1,2,3, at the electrostatic equilibrium

b) Find the electric potential V(II) of the conductor II.

Now the conductor II is linked to the resistor R by switching T.

c)Find the electric current I which passes through R immediately after T has been switched.
d) Find the electric charge Q3 at the equilibrium.

A very difficult problem...can u do it?
enjoy![/img]

Very nice problem. My answers:
[hide]
a) $ Q_{1}\equal{}\frac{1}{3}(16\pi\varepsilon_{0}a V\minus{}Q)$
$ Q_{2}\equal{}\frac{\minus{}2}{3}(8\pi\varepsilon_{0}a V\plus{}Q)$
$ Q_{3}\equal{}Q$
b) $ V(II)\equal{}\frac{Q}{20\pi\varepsilon_{0}a}$
where $ V(\infty)\equal{}0$
c) $ I\equal{}\frac{Q}{20\pi\varepsilon_{0}aR}$
d) $ Q_{3}\equal{}0$
I'm not sure about the last one...
[/hide]

I've similar results..i'll check it better...meanwhile can you comment your solution? i think it's better we give clear explaination, especially when the problem is nice...and without results :lol:
enjoy!

[hide="My solution"]
a) In the beginning there is no additional charge on the conductors, so
\[ Q_{1}\plus{}Q_{2}\plus{}Q_{3}\equal{}0\]
Using Gauss law we can find the electric field between $ 4a$ and $ 5a$:
\[ E_{4\minus{}5}\equal{}\frac{Q_{1}\plus{}Q_{2}\plus{}Q}{4\pi\varepsilon_{0}x^{2}}\]
This is the field inside a conductor, so at equilibrum $ E_{4\minus{}5}\equal{} 0$, so
\[ Q_{1}\plus{}Q_{2}\plus{}Q\equal{}0\]
From the two equations we get $ Q_{3}\equal{} Q$
The electric field between $ a$ and $ 2a$ and between $ 2a$ and $ 4a$ is:
\[ E_{1\minus{}2}\equal{}\frac{Q_{1}}{4\pi\varepsilon_{0}x^{2}}\]

\[ E_{2\minus{}4}\equal{}\frac{Q_{1}\plus{}Q}{4\pi\varepsilon_{0}x^{2}}\]
Potential difference between the two conductors is:
$ V \equal{}\minus{}\int_{4a}^{2a}\frac{(Q_{1}\plus{}Q)dx}{4\pi\varepsilon_{0}x^{2}}\minus{}\int_{2a}^{a}\frac{Q_{1}dx}{4\pi\varepsilon_{0}x^{2}}\equal{}\frac{1}{4\pi\varepsilon_{0}a}(\frac{Q_{1}}{2}\plus{}\frac{Q_{1}\plus{}Q}{4}))$
so: $ Q_{1}\equal{}\frac{1}{3}(16\pi\varepsilon_{0}a V\minus{}Q)$
and $ Q_{2}\equal{}\frac{\minus{}2}{3}(8\pi\varepsilon_{0}a V\plus{}Q)$

b) The electric field outside the outer conductor is:
\[ E_{5}\equal{}\frac{Q_{1}\plus{}Q_{2}\plus{}Q_{3}\plus{}Q}{4\pi\varepsilon_{0}x^{2}}\equal{}\frac{Q}{4\pi\varepsilon_{0}x^{2}}\]
so $ V(II) \equal{}\minus{}\int_{\infty}^{5a}\frac{Qdx}{4\pi\varepsilon_{0}x^{2}}\equal{}\frac{Q}{20\pi\varepsilon_{0}a}$

c) The current passing through R immediately after T has been switched is:
\[ I\equal{}\frac{V(II)}{R}\equal{}\frac{Q}{20\pi\varepsilon_{0}aR}\]
d) At equilibrum we still have got $ E_{4\minus{}5}\equal{} 0$, so
\[ Q_{1}\plus{}Q_{2}\plus{}Q\equal{}0\]
We also have got $ V(II) \equal{} 0$, so:
$ V(II) \equal{}\minus{}\int_{\infty}^{5a}\frac{(Q_{1}\plus{}Q_{4}\plus{}Q_{3}\plus{}Q)dx}{4\pi\varepsilon_{0}x^{2}}\equal{}\frac{Q_{1}\plus{}Q_{4}\plus{}Q_{3}\plus{}Q}{20\pi\varepsilon_{0}a}\equal{} 0$
\[ Q_{1}\plus{}Q_{2}\plus{}Q_{3}\plus{}Q\equal{}0\]
From these two equations we have:
$ Q_{3}\equal{} 0$
[/hide]

golduck, I think I have a different definition of potential difference.
I define the electric potential difference between two points A and B as:

$ V_{B}\minus{}V_{A}\equal{}\minus{}\int^{B}_{A}E(r)dr$

You have the minus sign in your definition, but you inverted the limits of integration.
It should be:

$ V \equal{}\int_{4a}^{2a}\frac{(Q_{1}\plus{}Q)dx}{4\pi\varepsilon_{0}x^{2}}\plus{}\int_{2a}^{a}\frac{Q_{1}dx}{4\pi\varepsilon_{0}x^{2}}$

I got the same answers for the other question.  :) Maybe I am wrong so I will be glad to be corrected if it's the case.

Astronum...i agree! Infact on the sphere (I) there must be a negative charge. To prove this, look a the battery: it has the negative pole oriented towards sphere (I)...i'll post my own results today (without latex :blush: )

Yes, I didn't look on the orientation of the battery.  :oops:  In that case the answers are:
$ Q_{1}\equal{}\frac{\minus{}1}{3}(16\pi\varepsilon_{0}a V\plus{}Q)$
$ Q_{2}\equal{}\frac{2}{3}(8\pi\varepsilon_{0}a V\minus{}Q)$

finally my results:

Q(1) = -1/3 * (16*3.14*8.85*10^-12*a*V - Q)

Q(2) = 4/3 * (4*3.14*8.85*10^-12*a*V -Q)

Q(3) = Q

then for V(II) i have a minus in front of your results....V(II)= -Q/(20*3.14*8.85*10^-12*a)

The other results are ok, even if we should explain why we considered only V(II) for I...


Let a polynomial $f(x)$ be given with real coefficients and has degree greater or equal than 1. Show that for every real number $c > 0$, there exists a positive integer $n_0$ satisfying the  following condition: if polynomial $P(x)$ of degree greater or equal than $n_0$ with real coefficients and has leading coefficient equal to 1 then the number of integers $x$ for which $|f(P(x))| \leq c$ is not greater than degree of $P(x)$.

[b]Lemma:[/b] For every $M \in \mathbb{R}$ there is a positive integer $n_0$ such that for every $P(x)$ satisfying the conditions and for every set $A \subset \mathbb{Z}$ with $|A| > \textrm{deg} \; P$ there is an element of $P[A]$ greater than $M$.

[b]Proof:[/b] Assume the contrary for a fixed $M$, i.e. for every $n_0$ there are $P(x)$, $A = \{ a_1, \ldots, a_k \}$ with $k > \textrm{deg} \; P \ge n_0 $ such that we have $P(a_i) \le M$ for $i = 1, \ldots, k$. For a fixed $P(x)$, take $A$ with the minimal number of elements (just reduce $A$ to any $\textrm{deg} \; P + 1$ elements) and assume $a_1 < \ldots < a_n$.

Because $k = \textrm{deg} \; P + 1 > \textrm{deg} \; P$, by the Lagrange interpolation formula we have:
\[ P(x) =  \sum_{i = 1}^{k} P(a_i) \prod_{j \neq i} \frac{x - a_i}{a_i - a_j} .\]

The polynomial is monic, so $[x^{k-1}]P(x) = 1$. On the other hand, he have:
\[ [x^{k-1}]P(x) = \sum_{i = 1}^{k} \frac{P(a_i)}{ \prod_{j \neq i} (a_i - a_j)}. \]

Now, $|a_i - a_j| \ge |i - j|$ for each $i$, $j$ because $A$ is a sorted set of positive integers. Each term of the sum is bounded from above by its absolute value:
\[ \frac{P(a_i)}{ \prod_{j \neq i} (a_i - a_j)} \le \frac{|P(a_i)|}{ \prod_{j \neq i} |a_i - a_j|} \le \frac{M}{(i-1)!(k-i)!} = \frac{M}{(k-1)!} \binom{k-1}{i} \]
 \[ [x^{k-1}]P(x) \le \frac{M}{(k-1)!} \sum_{i = 1}^{k}  \binom{k-1}{i} = M \frac {2^{k-1}}{(k-1)!} .\]

The last term is obviously much lower than $1$ for large enough values of $k$. But $k > n_0$, and we can set $n_0$ as large as we want, which gives a contradiction. $\square$

Returning to the original problem, $f(x)$ is nonconstant, so also unbounded. That means that for every $c > 0$ there is a real $M$ such that $|f(x)| > c$ for every $x > M$.

By the lemma, we can choose $n_0$ such that the image of every set $A$ of integers under every $P(x)$ satisfying the conditions with $|A| > \textrm{deg} \; P$ contains a value $P(a)$ greater than $M$. But then the image of $A$ under $f \circ P$ contains a number $f(P(a))$ for which $|f(P(a))| > c$. which means that such a pair $(P(x), A)$ doesn't satisfy $f[P[A]] \subseteq [-c, c]$.

If there were a polynomial $P(x)$ satisfying the conditions and a set of more than $\textrm{deg} \; P$ integers $x$ for which $|f(P(x))| \leq c$, a contradiction would arise with the previous statement, which finishes the proof. $\blacksquare$


Hey,

I recently began reading and working on AoPS volume 1, and plan on going on to volume 2 as well. Working on these books has really sparked my interest in math. Can someone suggest any other math books that are fun/interesting to read?

the resources section on AOPS has plenty of them. you can also find many on the AMC site.

[color=red][i][10000th User EDIT]It would be better if you can provide the link to AMC site instead of saying it. He may not be familiar with it.

Here is a link: [url=http://www.unl.edu/amc/]http://www.unl.edu/amc/[/url] and then click Publications if you are looking for books. Browse AMC, it has a lot of good resources too.[/i][/color]

my website, [url=http://4aunlimited.co.nr]click here to access it, has lots of stuff[/url]


Let $ p: \widetilde{M} \rightarrow M$ be the universal $ k \minus{}$sheeted covering of the closed orientable manifold $ M$ of dimension $ m$.   Show that

1. $ \widetilde{M}$ is orientable
2. If $ [\widetilde{M}]$ (resp. $ M$ ) $ \in H_m(\widetilde{M})$ (resp. $ H_m(M)$) is a fundamental class for $ \widetilde{M}$  (resp. $ M$), then $ p_*[\widetilde M] \equal{} \pm k[M]$.

I guess the way to do this is by using cup products but how should I start approaching this problem?

Thanks in advance

$ p$ is a local diffeomorphism, so I feel like 1 should be trivial. Given an orientation form $ \omega\in \Omega^m(M)$, isn't the pullback $ p^*\omega$ an orientation form on $ \tilde {M}$?

For 2, what is $ H_m(\tilde {M})$? The m-th homology?

Thank you Hilbertthm90!  and yes, it is the $ m\minus{}$th homology.

What if $ M$ is not equipped with a differential structure? Topological orientability is defined by the existence of the fundamental class, a generator of $ H_m(M)$ that restricts nicely to local homology groups $ H_m(M,M\setminus{x})$ for $ x \in M$.

As Nukular said, orientability is equivalent to the existence of a continuous global section of the orientation bundle, that is a a choice of generator $ s_x\in H_n(M,M\setminus\{x\})$ for all $ x\in M$ such that there exists a covering $ \{U_i\}_{i\in I}$ of $ M$ together with elements $ s_i\in H_n(M,M\setminus\overline{U_i})$ restricting to the chosen $ s_x$.
If we have such a continuous choice for $ M$, then we can simply lift it to a continuous choice for $ \tilde{M}$ since for each $ \tilde{x}\in\tilde{M}$ the map $ H_n(\tilde{M},\tilde{M}\setminus\{x\})\to H_n(M,M\setminus\{x\})$ induced by $ \tilde{M}\to M$ is an isomorphism. The continuity is clear from the fact that one can shrink the $ U_i$ above arbitrarily and that the covering is a local homeomorphism.
If $ M$ is compact, then one can glue the choice of generators $ s_x$ together to get a generator of $ s\in H_n(M)$, and similarly one can glue the lifts of local orientations to a generator $ \tilde{s}\in H_n(\tilde{M})$. Now the image $ \tilde{s}$ under $ H_n(\tilde{M})\to H_n(M)$ equals $ k\cdot s$; as all maps $ H_n(M)\to H_n(M,M\setminus \{x\})$ are isomorphisms, this can be checked locally. But $ H_n(\pi)(\tilde{s})_x \equal{} \sum\limits_{y\in\pi^{\minus{}1}(x)} H_n(\pi)(\tilde{s}_y) \equal{} \sum\limits_{y\in\pi^{\minus{}1}(x)} s_x \equal{} k\cdot s_x$ because $ H_n(\pi)(\tilde{s}_y) \equal{} s_x$ by definition of $ \tilde{s}$.


calculus
derivative
calculus computations

Find an equation of the tangent line to the curve at the given point.

1) y = sqrt{x}/(x + 1), (4, 0.4)

2) y = x(sqrt{x}), (1,1)

hello, the tangent line has the form $ y=mx+n$ with $ m=f^'(x_0)$. At first we compute the first derivative
$ f^'(x)=\frac{\frac{1}{2}x^{-\frac{1}{2}}(x+1)-\sqrt{x}}{x+1}=
\frac{\frac{1}{2}(x+1)-x}{(x+1)\sqrt{x}}=
\frac{-\frac{1}{2}x+\frac{1}{2}}{(x+1)\sqrt{x}}=
\frac{-x+1}{2\sqrt{x}(x+1)}$ and from here we get
$ f^'(4)=\frac{-3}{20}$. So we have
$ y=\frac{-3}{20}x+n$. By inserting $ x=4$ and $ y=\frac{2}{5}$ we find $ n=1$.
Sonnhard.


let $\theta_{1},...,\theta_{n}$ and $\lambda_{1}, ..., \lambda_{n}$ be postive numbers such that $\theta_{1}+...+\theta_{n}=1$
Show that $(\sum_{i=1}^{n}\theta_{i}\lambda_{i}) (\sum_{i=1}^{n}\theta_{i}\lambda_{i}^{-1}) \geq 1$

hi koola,

Cauchy–Schwarz inequality states that

$\sum_{i=1}^{n}a_{i}^{2}\sum_{i=1}^{n}b_{i}^{2}\geq \left( \sum_{i=1}^{n}a_{i}b_{i}\right)^{2}$.

So, let $a_{i}^{2}= \theta_{i}\lambda_{i}$ and $b_{i}^{2}= \theta_{i}\lambda_{i}^{-1}$, we get $a_{i}= \sqrt{\theta_{i}\lambda_{i}}$ and $b_{i}= \sqrt{\theta_{i}\lambda_{i}^{-1}}$. Substitute these results into the Cauchy–Schwarz inequality, we get

$\sum_{i=1}^{n}\theta_{i}\lambda_{i}\sum_{i=1}^{n}\theta_{i}\lambda_{i}^{-1}\geq \left( \sum_{i=1}^{n}\sqrt{\theta_{i}\lambda_{i}\theta_{i}\lambda_{i}^{-1}}\right)^{2}= \left( \sum_{i=1}^{n}\theta_{i}\right)^{2}= 1$.