Is it ok if my solutions are like 5 page per problem(I want to be throuogh[is that how u spell it??])...
Because even I get tired reading my solutions...
btw, what is the "bad box" thing when you compile, and how u get rid of them?????

That's a bit on the long side, but it won't cost you points.

As for the bad boxes, as long as you're happy with how it looks, then don't worry about it.  LaTeX has a lot of ticky-tacky mystery warnings that aren't terribly important.


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Let R be a quotient ring of the polynomial ring C[X]. Prove that
for any maximal ideal M of R, the ring R/M is isomorphic with C

This is simply a consequence of the fundamental theorem of algebra. The only irreducible elements in $ \mathbb{C}[x]$ are polynomials of degree one, the quotients by which are isomorphic to $ \mathbb{C}$.

And the so-called Third Isomorphism Theorem.


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274251]Puerto Rico PRMO Team Selection Test 2009[/url]

Find all integers $ b$ and $ c$ such that the equation $ x^2 \minus{} bx \plus{} c \equal{} 0$ has two real roots $ x_1, x_2$ satisfying $ x_1^2 \plus{} x_2^2 \equal{} 5$.

sorry, dont know latex.

using quadratic you get the two solutions to be x=(b+(sqrt(b^2-4c))/2,
x=(b-(sqrt(b^2-4c))/2. squaring both yields b^2-2c=5. There are infinitely many solutions for b>1 or b<-1 and b is odd, however; b^2-4c>0. since we know b^2-2c=5, c<5/2. Thus, the only solutions are b=3, c=2 or b=-3, c=2.

EDIT:  :blush: sorry, i forgot the other two cases (-1, -2) and (1, -2).

You can skip a bit of work in that first part using Newton's sums:
$ S_1\equal{}b$ from Vieta's.
$ S_2\equal{}bS_1\minus{}2c\equal{}b^2\minus{}2c\equal{}5$.

Since the sum of the squares of the roots is 5 and the roots are real, the roots are between + and - sqrt(5) inclusive and so -5 <= c <= 5. So we have -5 <= c <= 2. Checking, I get (b, c) = (-3, 2) and (3, -2) as in lukezli's post and also (-1,  -2) and (1, -2).


geometry
parallelogram
geometry proposed

In the parallelogram $ ABCD$ consider $ M$ on $ CD$ such that $ M \not\equal{}C$ and $ M \not\equal{}D$. Denote by $ N$ the intersection between $ AM$ and $ BC$. Prove that the triangles $ DMN$ and $ BCM$ have the same area.

$ \frac {[\triangle DMN]}{[\triangle ADM]} \equal{} \frac {MN}{AM} \ \ , \ \ \angle ADM \plus{} \angle BCM \equal{} 180^{\circ}$

$ \frac {[\triangle ADM]}{[\triangle BCM]} \equal{} \frac {AD \cdot DM}{BC \cdot MC} \equal{} \frac {DM}{MC}$

$ \frac {MC}{DM} \equal{} \frac {MN}{AM}\Longrightarrow \ \frac {[\triangle ADM]}{[\triangle BCM]} \equal{} \frac{[\triangle ADM]}{[\triangle DMN]} \Longrightarrow \ [\triangle BCM] \equal{} [\triangle DMN]$

[quote="moldovan"][color=darkred]Let $ ABCD$ be a parallelogram. Consider the point $ M\in CD \minus{} \{C,D\}$

and the intersection $ N\in AM\cap BC$ . Prove that $ [DMN] \equal{} [BCM]$ .[/color] [/quote]
[color=darkblue][b][u]Proof[/u].[/b] $ [DNC]\stackrel{(AD\parallel NC)}{\ \ \equal{} \ \ }[ANC]\implies$ $ [DNC]\pm [MNC] \equal{} [ANC]\pm [MNC]$ $ \implies$

 $ [DMN] \equal{} [ACM]\stackrel{(AB\parallel CM)}{\ \ \equal{} \ \ }[BCM]\implies$ $ [DMN] \equal{} [BCM]$ .[/color]


superior algebra
superior algebra unsolved

If you know some results about conjugacy classes
 or size of conjugacy classes in finite group, please write here.

On a less snarky note, what you probably want is the [url=http://en.wikipedia.org/wiki/Conjugacy_class#Conjugacy_class_equation]class equation[/url].

They also have an intimate connection with irreducible representations.


geometry
rectangle
analytic geometry
geometry unsolved

"Given N distinct points in the unit square [0,1]x[0,1], including the origin (0,0) as one of the N points. Can you construct N rectangles, contained in the unit square, with sides parallel to the coordinate axes, pairwise non-intersecting, such that each of our N given points is the lower-left-hand corner of one of the rectangles, and such that the total area of the rectangles is at least 1/2?

For example, N=3 and the points are (0,0), (0.2,0.4) and (0.8,0.6). The three rectangles could be [0,1]x[0,0.39], [0.2,0.79]x[0.4,1], and [0.8,1]x[0.6,1]. Their areas are 0.39, 0.354 and 0.08, summing to 0.824, which is larger than 1/2.
(We allow degenerate rectangles -- a point or a line segment -- but still do not allow them to intersect with other rectangles, even in a single point.)

Either PROVE that such a construction is always POSSIBLE, or give a set of N distinct points (including the origin) for which it is IMPOSSIBLE."

It seems to me that "worst case" is when our N points are equally spaced along the main diagonal: (i/N, i/N) for i=0,1,...,N-1, then any solution will have area less than  W = (N+1)/(2*N), which is slightly larger than 1/2. In any other situations we have a greater W. But a general solution is not known?!? (I cannot prove it entirely).
I already check this for N = 2,3,4,5.

Does anyone has an ideea?

[quote="amir2"] ░َ7,ô|7╧○○8○[12☺-||5Dطé[/quote]

Can u be more specific, plz?  :?


i am thinking about going to University of Illinois because i hear it has a good engineering program.  I live only three hours away from it, and the tuition is not terribly high, either.  However, I know there are better schools for engineering that i could probably get into.

My question is: Would U of I be a good school to attend, or would it make a terrific difference if i went to a Ivy League-type school.  Am i sacrificing anything by wanting to go there?  and does it have a good math department too.

I think i want to study mechanical engineering, but that isnt set in stone.  my big things are math and physics.

I think it's probably not worth it to go to an Ivy League school, although you should probably apply anyway.  The University of Illinois, assuming you like Champaign-Urbana, the environment, etc., is not much worse than the Ivy Leaguers and whatnot.  Anyway, it's only undergrad (I'm assuming).

what about purdue and university of michigan?  According to US News Report, michigan has the second best mechanical engineering school in the nation, while illinois has the sixth best.  That difference is not that big of a deal, right?  What makes that difference?

Hi, my name is Geehoon and I am currently a high school senior.

Yes, I did apply to and got accepted to University of Illinois at Urbana Champaign (along with many other schools), and had a chance to research the school a little bit.

Rankingwise, the engineering program is really awesome, I would say definitely top 20 in nation, if not top 10.
In terms of Environment, I have a friend there (he won IOI gold once) and it seems that he likes being there.

It will be a good choice for you especially since you live around there.
It is not that hard to get in, so it will be safer for you also.

However, I will say these two things.

First, you don't have to decide where youre going right now. What I mean by that is, apply to all the schools that you like, and see what happens. as JBL once mentioned, college admission is somewhat random and you cannot expect what is going to happen. For example, I did not get into Georgia Tech, but got into MIT and Caltech. If I were afraid of getting rejected and didnt apply to MIT/Caltech, I might have gotten into serious problem after getting rejected from Georgia Tech.

Second, don't worry about money so much. I'm closer to the poor than to the rich, but never hesitated to apply Ivy League schools. I see many people who do not apply to Ivy League schools simply because they are expensive. However, if you get into more than one school, you still have a chance to compare the financial aid packet from all schools that you got in, and obviously you have more options if you got into more schools.

To make the long story short, my opinion is:
1. UIUC is wonderful school, especially their engineering.
2. Apply Ivy leagues. It only widens your option.
3. Worry about money after you get financial aid packet, not before.

thank you for your response.  I am going to be a senior next fall, but i need to start thinking about these things now.

i kind of like the fact that U of I is only 3 hours away.  i have to consider travel costs as well.

what is IOI?
what school are you going to?

IOI = [url=http://olympiads.win.tue.nl/ioi/]International Olympiad in Informatics[/url]

Geehoon is going to MIT next year.

UIUC is significantly better than most of the Ivies for Engineering, anyways.

"most," probably.  Columbia and Cornell are excellent, but the others are not.

But is it better than Purdue and University of Michigan?

I have heard U of I has a pretty good math department too.  If i can handle engineering i want to get a degree in mathematics.

My husband was a math major at U of I more than 20 years ago so this info is a little old but he thinks the quality of the science departments hasn't changed much.  The math and physics departments were excellent.  The intro classes were large but the upper level classes were small.  Upon graduation he was accepted by U Chicago for math graduate studies so the U of I math dept is well-respected by other schools.

What Geehoon said (post #4) is excellent advice for all the readers of this forum. My one comment about your specific question of whether or not you should study in your home state is that there is an advantage to seeing the rest of the world. I went to the same state university that most people from my high school went to. I lived close enough to it that I could take a bus to the campus and live at home. Later I went overseas and lived there for three years, after I had completed my undergraduate degree. Knowing what I know now as a forty-something-year-old, my advice is apply to all the schools you desire, including your state university, but if you can go somewhere far away at affordable cost, go to some place you have never lived in before to go to college, and prefer colleges, if you are accepted to more than one, that have students from all over the country and all over the world. You learn by meeting new kinds of people who didn't have the same kinds of life experiences you had when growing up. Changing your environment builds more flexibility of thinking and helps increase your creativity.

Good luck in your applications. It sounds to me like your in-state choice is a very good choice indeed. Apply to various schools, and see what application results you end up with.


find all integers like a ,b and c so that:
b+c|bc , c+a|ac , b+a|ab .

no idea?
this is making me crazy.I hate this kind of problem in number theory.

Strange - maybe I misunderstood something but:

let $a = \prod p_{i}^{\alpha_{i}}$ ,  $b = \prod p_{i}^{\beta_{i}}$,  $c = \prod p_{i}^{\gamma_{i}}$. Then our condition rewrites as:

$\alpha_{i}\leq \beta_{i}+\gamma_{i}$
$\beta_{i}\leq \alpha_{i}+\gamma_{i}$
$\gamma_{i}\leq \alpha_{i}+\beta_{i}$.

so $\alpha_{i}, \beta_{i}, \gamma_{i}$ have to be sides of triangle (maybe degenerated) and I suppose it's sufficient and necessary condition...  :maybe:

I made a big mistake in saying the problem.so I edited it.excuse me.


algebra
polynomial
superior algebra
superior algebra unsolved

let $ \varphi$ be  a ring epimorphism  $ \varphi : R \to S$ how to show that $ J(R)\subseteq J(S)$ where $ J(R)$ is Jacobson radical

First, it should be $ \varphi\left(J\left(R\right)\right)\subseteq J\left(S\right)$.
Then, what definition of Jacobson radical are you using? The $ J\left(R\right) \equal{} \left\{x\in R\mid \text{for every }r\in R\text{ there exists some }u\in R\text{ such that }u\left(1 \minus{} rx\right) \equal{} 1\right\}$ definition makes this completely obvious.

And: problems of this kind belong into College Playground >> Superior Algebra, [b]not[/b] into the Olympiad Section >> Algebra board. In the Olympiad Section, algebra means number sequences, functional equations, polynomials, finite sums etc., but definitely not rings and ideals.

  darij


Let $ f(x)=4x-x^{2}$  for $ x_{0}\in\mathbb{R}$ we consider the finite sequence $ x_{0}, x_{1}=f(x_{0}), x_{2}=f(x_{1}),....$ Prove that there exists many $ x_{0}$ such that $ x_{0},x_{1},x_{2},...$ consists of finitely many different values


I think that it was said about two weeks ago that scores will be coming in mid-December, but possibly late December. Are there any further updates on when we are going to get the scores? Is it more likely to be mid-December or late December? Thank you.

Mid-december is a few days after today...

So I think then

I know but I'm a little nervous so I'm sort of impatient to get it over with...

We will make an announcement via this forum, the USAMTS website, and the email mailing list when the scores are available.


Yep, you're right.  :) I will try to progress.

[b]Generalisation[/b]

Compute the sum: $ n^0\plus{}n^1\plus{}...\plus{}n^m$ where $ m$ and $ n$ are positive integers greater than $ 1$.

[b]Solution[/b]

$ S\equal{}n^0\plus{}n^1\plus{}...\plus{}n^m$

$ n \cdot S\equal{}n^1\plus{}n^2\plus{}...\plus{}n^{m\plus{}1}$

$ \Rightarrow (n\minus{}1) \cdot S\equal{}n^{m\plus{}1}\minus{}1$

$ \Rightarrow S\equal{}\frac{n^{m\plus{}1}\minus{}1}{n\minus{}1}$

WELL OBVIOUSLY IT'S NOT AN INTEGER. I MEAN DUH!

I am afraid (3^3000-1)/2 is an integer Ace of Diamonds!

I am (actually not really) afraid that you did not use parentheses in the post that AoD responded to :wink:

The result is $ S\equal{}\frac{3^{2010}\minus{}1}{2}$.


find the development/expansion in (-2) of $ \sqrt[7]{x^2\minus{}2}$;
why does't it work (the first two terms are the same, the third is different) if I substitute $ x^2\minus{}2$ with $ u\plus{}2$ and then make the expansion in zero?

The substitution you want to use is $ u \equal{} x \plus{} 2,$ which means that $ x \equal{} u \minus{} 2$ and $ x^2 \minus{} 2 \equal{} 2 \minus{} 4u \plus{} u^2.$

I get an expansion that starts out with

$ \sqrt[7]{2}\left(1\minus{}\frac27u\minus{}\frac{17}{98}u^2\plus{}O(u^3)\right)$

$ \equal{}\sqrt[7]{2}\left(1\minus{}\frac27(x\plus{}2)\minus{}\frac{17}{98}(x\plus{}2)^2\plus{}O((x\plus{}2)^3)\right).$

thanks, but why is not possible to substitute $ x^2\minus{}2$ with u directily, maybe because you have to use linear powers?

The problem asks for a series expansion in terms of $ x\plus{}2$; we have to get that into the picture.


[quote="Interval"]A piggybank contains 2 quarters, 3 dimes, 4 nickels, and 5 pennies. One coin is removed at random. What is the probability that the coin is a dime or a nickel?[/quote]
[hide]There are 7 coins that are either a dime or a nickel and there are 14 coins over all so then the probability is 7/14 which is 1/2[/hide]

[quote="Quevvy"][quote="ckck"][quote="Interval"]Did you reduce 7/14?[/quote]
yes he did. If you can reduce a fraction,always do that. you always want a fraction in simplest form.[/quote]
Unless it tells you not to. (usually it'll say put it in a common fraction which is in lowest terms)[/quote]

and not in mixed numbers I believe

[hide]
There are $14$ coins in all. There are $4+3$ dimes and nickels. $7/14=1/2$[/hide]


1. Lead storage batteries can be recharged, but why can't dry cells recharge?

2. Why do boiling points increase as you go down the halogens, but decrease as you go down the alkali metals?

3. Name a nonmetallic oxide that reacts with water to form a base and a metallic oxide that reacts with water to form an acid.

No.2:
[hide]
On descending Group I, the boiling points decrease because the metallic bonding in the metals becomes longer and weaker (the metallic radius increases on descending).

On descending Group VII, the boiling points increase because there are more electrons available for intermolecular bonding. Thus, with a larger and more easily polarisable electron cloud, there is stronger and more extensive intermolecular van der Waal's forces of attraction.[/hide]

[quote="g1sk8terz"]for # 3.............

[hide]$Na_{2}O\ +\ H_{2}O\ \to \ 2NaOH$

[/hide][/quote]

I tried that too! I just didn't realize that $Na_{2}O$ was nonmetallic...  :blush:  Well, is there a way you can tell?

A nonmetal oxide (like $\text{SO}_2$) usually (meaning almost always) forms an acid in water, and a metal oxide (like $\text{Na}_2\text{O}$) almost always forms a base in water.  Maybe (huge huge maybe) something like polonium oxide might form an acid ($\text{H}_2\text{PoO}_4$?), but I'm not sure about that.

Lead acid batteries are not fully rechargable.  For example, during rechage, some of the electrons are energic enough to react and sepertate the water in the sulfuric acid into hydrogen and oxygen.  They migrate away from their respective elecrodes and are lost to the battery.  More water needs to be added to a battery to restore the orginial condition.

To address your question, sometimes the products of the redox reaction that releases energy in a battery happen (or are designed) to be reactants to consume far more energy than given off to restore (mostly) the material to the orginal condition.  (Entropy will always get its share.)

In a lead acid battery, lead reacts to lead oxide, which can be readily converted back to lead.

For a metallic oxide that forms an acid, I think some sort of chromium oxide (VI?) would do it.  Maybe something like:

$2CrO_3 + H_2O \rightarrow H_2Cr_2O_7$

No idea about a nonmetallic base.


(1) If n is a positive integer and d a divisor of 2n^2, prove that n^2+d cannot be a perfect square.

(2) For positive integers m,n, prove that 3^m+3^n+1 can never be a perfect square.

(3) If n is a positive integer > 11, prove that n^2-19n+89 is never a perfect square.

a) Let d*m= 2n 2 .
Then m 2 *(n 2 +d) = n 2 ((m+1) 2 2 - 1)
Thus, if n 2 +d is a square so is (m+1) 2 2 - 1, from which it follows that (m+1) 2 2  = 0 (since 0 and 1 are twe only two squares differing by 1). A contradiction.

b)  An odd square is known to be equal to 1 mod[8].
But, according to the parities of m and n, it is easy to see that 3 m + 3 n is never equals to 0 mod[8].

c) For n > 11, it is easy to verify that
(n-9) 2  < n 2 -19n + 89 < (n-10) 2 , then it cannot be a square....

Pierre.


$ ABCD$ is a parallelogram,$ O$ is the circumcenter of of $ ABC$ and $ BOB^{\prime}$ is a diameter of this circle. Prove that $ DB^{\prime}$ is perpendicular to $ AC$. :)

[b]lemma:$ XY \bot MN \iff XM^{2}-XN^{2}=YM^{2}-YN^{2}$[/b]
now according to the lemma its sufficient to show that:
$ DA^{2}-DC^{2}=B'A^{2}-B'C^{2}$
now we have:
$ BB'$ is a diameter so $ \angle B'AB=90$ so $ BB'^{2}=B'A^{2}+BA^{2}\Rightarrow B'A^{2}=BB'^{2}-AB^{2}$
in a similar way we get that $ B'C^{2}=BB'^{2}-BC^{2}$
hence we have:
$ B'A^{2}-B'C^{2}=(BB'^{2}-AB^{2})-(BB'^{2}-BC^{2})=BC^{2}-AB^{2}=DA^{2}-DC^{2}$ so we have:
$ B'A^{2}-B'C^{2}=DA^{2}-DC^{2}$ as wanted...

Wow really nice solution Babak :)
my solution was this:
as $ BOB^{\prime}$ is diameter we have $ CB^{\prime}$ perpendicular to $ BC$ (or prpendicular to $ AD$) and $ AB^{\prime}$ is perpendicular to $ AB$(or perpendicular to $ DC$)! as a consequence $ B^{\prime}$ is the orthocenter of triangle $ ADC$ and $ DB^{\prime}$ is perpendicular to $ AC$. :)

Let $E$ be midpoint of $AC$; since $O$ was the circumcenter and $AC$ a chord of the circle, $OE\bot AC$, but $OE$ is midline of $\Delta BB'D$, done!

Best regards,
sunken rock


dayereye delkhahe C1 az ro oose  B ,C mosallase ABC migozarad
dayereye C2 ra tori rasm mikonim ke bar C1 dar X va bar azlaae AB ,AC
az mosallas momaas shavad.

sabet konid nimsaaz e BXC  az mahalle barkhorde nimsaaz haye mosallase ABC migozarad

ghablan [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59059]inja[/url]hal shode

albate man sheklhasho pak kardam :D


What is the pressure in atmospheres on the body of a diver if he is 36 feet below the surface of the water when atmospheric pressure at the surface is 0.95 atm?

Are you referring to the pressure exerted by the water on the body, or the gas pressure inside the lungs of the diver?


limit
trigonometry
logarithms
calculus
derivative
calculus computations

Evaluate of the folowing limit [b]without using l'Hopital's theorem[/b].

\[\lim_{x\rightarrow 0} \frac{e^{x\sin 2005x}-1}{x\ln (1+x)}\]

It is obviously $2005$ (replace each expression by its equivalent), but replacement by equivalents here is equivalent to knowledge of certain derivatives, which is equivalent to using L'Hopital... What I mean is that I doubt it is possible to do it without knowing that $\lim_{x\to 0}\frac{\log(1+x)}x=1$, but this is equivalent to saying that $\log(1+x)'=1$ at $0$ (just by the definition of the derivative) and once we have a derivative, we are using L'Hopital. So I am really confused about what the problem asks to do   :?

Thank you for your reply, fedja.

This problem is a practice for using the following basic definition of the limit.

$\lim_{h\rightarrow 0}\frac{e^h-1}{h}=1,\lim_{x\rightarrow 0}\frac{\ln (1+x)}{x}=1,\lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta}=1$

kunny


inequalities
real analysis
abstract algebra
projective geometry
geometry
function
trigonometry

What your favorite theorem?
My favorite is Sylow's First Theorem ... duh!

Well this thread is in the [b]Polls[/b] section so either make a poll or a mod is going to move this to G&FF or something.

My favorite theroem? I'm too tired to think so i'll just say Pythagorean Theorem (i probably didn't even spell it right)

Is a theorem just something that can be proved?

Hack, you have to be kidding me.
This is way too hard and involves what you know, and what type of math you like.

I think my favorite theorem is Fermat's Little Theorem. However, when I learn a new ,cooler one this will probably change.

Karamata's Weighted Inequality and Schur's Convexity.

Why isn't anyone saying Pappus's Theorum?

[quote="spix"]what I liked most until now is Wedderburn's theorem  :love:[/quote]
Yes, Wedderburn is nice.
Some other nice ones are the residue theorem, local-to-global principles (especially Hasse-Minkowski), the main theorem of class field theory, Chebotarev's density theorem, the theorem of Kronecker-Weber and a lot of other quasi-cyclotomic stuff :D

[quote="davidyko"]Stewart's for the win.[/quote]

I like Stewart's too.

How about everyone actually posts the theorem when they declare them? ("Declare"? I've been programming too much)
And the proof, if it's within reason.

The Chicken McNugget theorem states that for two relatively prime numbers m and n, the largest positive number you cannot make using an infinite amount of m and n is mn-m-n.

I don't really know many of them ... but for now my favourite is Binomial theorem  :lol:

Here are my favorites:
Ptolemy
Pick's Theorem
Pythagorean Theorem
Fermat's Last Theorem

My criteria was to pick theorems that are very generalized and natural and in order to prove them you have to somehow break the natural symmetry of the problem.

[quote="dynamo729"]I thought the Ptolemy Theorem was that the earth was the center of the galaxy or whatever.....  :ninja:  :huh:  :|[/quote]
....

I really, really, really, hope that this wasn't a serious post.

Ptolemy did indeed support the false geocentric model, instead of the heliocentric one, but...

[quote="7h3.D3m0n.117"]The Squeezing Theorem is a fun one to use  :)[/quote]

*squeeze*  :what?:

that isn't ptolemy theorem :o .....but indeed...ptolemy said that the earth is in the center of tha galaxy and that is a sphere...but this is not math:))

The main part of galoin theory(like some one who gave the same choice i think)

You maybe mean "Galois".
Most are lacking the more advanced really interesting stuff here.

The non-existence of the quintic formula using standard notation (addition, subtracton, multiplication, division, radicals)! (The Abel-Ruffini Theorem)

yes i meant 'galois".it's a superb theorem i think.

My favourite one is the Pythagorean Theorem :)

I like DeMoivre's theorem and the Cubic Formula.

$ x^3\plus{}ax^2\plus{}bx\plus{}c\equal{}0$

$ r_1  \equal{}  \minus{}\frac{a}{3} \plus{} \sqrt[3]{\frac{\minus{}2a^3 \plus{} 9ab \minus{} 27c \plus{} \sqrt{(2a^3\minus{}9ab\plus{}27c)^2 \plus{} 4(\minus{}a^2\plus{}3b)^3}}{54}} \plus{} \sqrt[3]{\frac{\minus{}2a^3 \plus{} 9ab \minus{} 27c \minus{} \sqrt{(2a^3\minus{}9ab\plus{}27c)^2 \plus{} 4(\minus{}a^2\plus{}3b)^3}}{54}}$

$ r_2 \equal{} \minus{}\frac{a}{3} \minus{} \frac{1\plus{}i\sqrt{3}}{2} \sqrt[3]{\frac{\minus{}2a^3 \plus{} 9ab \minus{} 27c \plus{} \sqrt{(2a^3\minus{}9ab\plus{}27c)^2 \plus{} 4(\minus{}a^2\plus{}3b)^3}}{54}} \plus{} \frac{\minus{}1\plus{}i\sqrt{3}}{2} \sqrt[3]{\frac{\minus{}2a^3 \plus{} 9ab \minus{} 27c \minus{} \sqrt{(2a^3\minus{}9ab\plus{}27c)^2 \plus{} 4(\minus{}a^2\plus{}3b)^3}}{54}}$

$ r_3 \equal{} \minus{}\frac{a}{3} \plus{} \frac{\minus{}1\plus{}i\sqrt{3}}{2} \sqrt[3]{\frac{\minus{}2a^3 \plus{} 9ab \minus{} 27c \plus{} \sqrt{(2a^3\minus{}9ab\plus{}27c)^2 \plus{} 4(\minus{}a^2\plus{}3b)^3}}{54}} \minus{} \frac{1\plus{}i\sqrt{3}}{2} \sqrt[3]{\frac{\minus{}2a^3 \plus{} 9ab \minus{} 27c \minus{} \sqrt{(2a^3\minus{}9ab\plus{}27c)^2 \plus{} 4(\minus{}a^2\plus{}3b)^3}}{54}}$


Let a,b,c be the positive number satisfying ab+bc+ca=1.
Prove that:
$ \frac{ab}{a^2\plus{}1}\plus{}\frac{bc}{b^2\plus{}1}\plus{}\frac{ca}{c^2\plus{}1}$≤$ \frac{1}{16}(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}2(\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c})\plus{}3)$

[quote="SaYaT"][quote="SUPERMAN2"]Let a,b,c be the positive number satisfying ab+bc+ca=1.
Prove that:
$ \frac {ab}{a^2 + 1} + \frac {bc}{b^2 + 1} + \frac {ca}{c^2 + 1}$≤$ \frac {1}{16}(\frac {a}{b} + \frac {b}{c} + \frac {c}{a} + 2(\frac {b}{a} + \frac {c}{b} + \frac {a}{c}) + 3)$[/quote]

This inequality equivolent to:
$ \frac {ab}{(a + b)(a + c)} + \frac {bc}{(b + c)(a + b)} + \frac {ac}{(a + c)(b + c)} \le \frac {1}{16}(\frac { 2\sum_{cyclic}{a^2b} + \sum_{cyclic}{ab^2} + 3abc}{abc})$
or
${ 16(3abc + \sum_{cyclic}{ab^2})abc} \le ( 2\sum_{cyclic}{a^2b} + \sum_{cyclic}{ab^2} + 3abc)(\sum_{cyclic}{a^2b} + \sum_{cyclic}{ab^2} + 2abc)$

Let's $ x = \sum_{cyclic}{a^2b}$, $ y = \sum_{cyclic}{ab^2}$, $ z = abc$ then our inequality become:

$ (2x + y + 3z)(x + y + 2z) \ge 48z^2 + 16yz$
or
$ 2x^2 + y^2 + 3xy + 7xz \ge 42z^2 + 11yz$
So it's left to prove
$ 2x^2 + y^2 + 3xy + 7xz \ge 42z^2 + 11yz$
we have:
$ 2x^2 \ge 18z^2$
$ 7xz \ge 21z^2$
$ \frac {2}{3}y^2 \ge 2yz$
$ \frac {y^2}{3} \ge 3z^2$
$ 3xy \ge 9yz$
So by summing this 5 inequalities we will have desired result.[/quote]
Your solution is right but it is not very nice.
Here is my sol.
Note that:
$ \frac{ab}{a^2+1}=\frac{ab}{(a+b)(a+c)}=\frac{ab}{a+b} \frac{1}{a+c}$≤$ \frac{a+b}{4} \frac{1}{4}(\frac{1}{a}+\frac{1}{c})$
=> $ \frac{ab}{a^2+1}$≤$ \frac{1}{16}(a+b)(\frac{1}{a}+\frac{1}{c})$
Similarly,$ \frac{bc}{b^2+1}$≤$ \frac{1}{16}(b+c)(\frac{1}{a}+\frac{1}{b})$ and $ \frac{ca}{c^2+1}$≤$ \frac{1}{16}(c+a)(\frac{1}{b}+\frac{1}{c})$
Then sum,we get LHS≤RHS.

[quote="Jaswinder"]just use a=cotA and so on where A,B,C are angles of triangle[/quote]
I have thought the same way to you,but when I do that,I can not solve the problem.


[quote="hell_ever"]The inradius of $ \Delta ABC$ is 1 and the sides are all integers. Prove that $ \Delta ABC$ is right angled.[/quote]

$ S\equal{}pr$  --> $ S\equal{}p$  $ a\equal{}x\plus{}y,b\equal{}x\plus{}z,c\equal{}y\plus{}z$   ---->  $ S\equal{}\sqrt{xyz(x\plus{}y\plus{}z)}\equal{}x\plus{}y\plus{}z$ --->   $ xyz\equal{}x\plus{}y\plus{}z$ --->
$ 2(x\plus{}y\plus{}z)\equal{}a\plus{}b\plus{}c$  and $ c\equal{}y\plus{}z$ integer ----> $ 2x$ -> integer ---> $ (2x)yz\equal{}2(x\plus{}y\plus{}z)$  ---> $ yz$ integer ----> $ x,y$ integer ---> $ x,y,z$   integer --->  $ xyz\equal{}x\plus{}y\plus{}z$< --->$ (x;y;z)\equal{}(1;2;3)$ ---> $ a\equal{}3,b\equal{}4,c\equal{}5$  right angled


x^4 - 8y^4 = z^2
Assuming x,y,z is the minimal solution,
I understand that x and z are both odd, but how do I explain :
- the remainder of x modulo 4 is 1, and
- the remainder of z modulo 4 is 3?

Is that true?
3^4-8x0^4=9^2
Did you solve it in positive int or in int.

first of all, what do you mean by minimal solution? there are a number of ways to define minimal.

$ x,y,z$ are positive integers and
we assume there's a solution where $ x$ is minimal.

Then you can show there are no solutions by the method of infinte descent.

$ x \equiv 1 \bmod 4$
$ z \equiv 3 \bmod 4$

This is the part that escapes me.

Edit:
Okay, this is the context :
'We assume that there's a positive solution with minimal $ x$.
Then $ (x,y) \equal{} 1$
if $ x, z$ are both even, $ y$ would be odd and $ z^2 \equiv 8 \bmod 16$, which is impossible.
Hence $ x, z$ are both odd.
$ x^2 > z$ --> $ 8y^4 \equal{} x^4 \minus{} z^2 \equal{} (x^2 \minus{} z^)(x^2 \plus{} z^)$
Since $ x \equiv 1 \bmod 4$ and $ z \equiv 3 \bmod 4$,  -> [b]What goes wrong if you negate this part?[/b]
$ x^2 \minus{} z \equiv 2 \bmod 4$ and $ x^2 \plus{} z \equiv 0 \bmod 4$
so $ m \equal{} (x^2 \minus{} z)/2$ is odd, $ n \equal{} (x^2 \plus{} z)/4$ is an integer......'


[quote="phucvy"]Given a,b,c >0 and a+b+c=3. Prove :
\[ \sqrt {\frac {a}{b^2 \plus{} 1}} \plus{} \sqrt {\frac {b}{c^2 \plus{} 1}} \plus{} \sqrt {\frac {c}{a^2 \plus{} 1}} \geq 6\sqrt {\frac {1}{(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)}}\]
[/quote]
IT's wrong! :P
Let $ a\equal{}b\equal{}0,c\equal{}3\equal{}>RHS \leq LHS$ :P

[quote="2424"][quote="phucvy"]Given a,b,c >0 and a+b+c=3. Prove :
\[ \sqrt {\frac {a}{b^2 \plus{} 1}} \plus{} \sqrt {\frac {b}{c^2 \plus{} 1}} \plus{} \sqrt {\frac {c}{a^2 \plus{} 1}} \geq 6\sqrt {\frac {1}{(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)}}\]
[/quote]
IT's wrong! :P
Let $ a \equal{} b \equal{} 0,c \equal{} 3 \equal{} > RHS \leq LHS$ :P[/quote]

Take wrong value of $ (a,b,c)$. Please look at the statement of the problem carefully, [b]the condition is[/b] $ a,b,c>0$

[quote="mathVNpro"][quote="2424"][quote="phucvy"]Given a,b,c >0 and a+b+c=3. Prove :
\[ \sqrt {\frac {a}{b^2 \plus{} 1}} \plus{} \sqrt {\frac {b}{c^2 \plus{} 1}} \plus{} \sqrt {\frac {c}{a^2 \plus{} 1}} \geq 6\sqrt {\frac {1}{(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)}}\]
[/quote]
IT's wrong! :P
Let $ a \equal{} b \equal{} 0,c \equal{} 3 \equal{} > RHS \leq LHS$ :P[/quote]

Take wrong value of $ (a,b,c)$. Please look at the statement of the problem carefully, [b]the condition is[/b] $ a,b,c > 0$[/quote]
@:Chao ban,voi bdt nay thi gt $ a,b,c>0$ hay $ a,b,c \geq 0$ khong khac nhau! :P
Co the cho $ a,b\to 0,c\to3$


calculus
integration
geometry
real analysis
real analysis unsolved

Following Evans' book "PDE":
$ \Phi (x,t)\equal{}\frac{1}{(4\pi t)^{n/2} } e^{\minus{}\frac{|x|^2}{4t}}$,when $ t>0$. $ \Phi \equal{}0$ when $ t<0$. is fundamention solution to heat equation $ u_t\equal{}\Delta u$.

For $ r>0$, define "heat ball" around point (0,0) as: $ E(r)\equal{} \{ (y,s) \in R^{n\plus{}1}| s<\equal{}0, \Phi(y,\minus{}s)>\equal{} \frac{1}{r^n} \}$

How to show:

$ \iint_{E(1)} \frac{|y|^2}{s^2} dyds\equal{}4$

(since $ \Phi$ is $ \infty$ at (0,0), so (0,0) is also in heat ball. the integrand would be singular?Maybe heat ball defination should be $ s<0$ instead of $ s<\equal{}0$?)

I think that reading Evans with full understanding requires more knowledge of integration. For instance, you should know that $ s\le 0$ is no different from $ s<0$ as far as integration is concerned, and that a singularity is not a death sentence for an integral. As for the integral, the substitution $ t=-s$, $ y=z\sqrt{t}$ turns it into
\[{ \int_{e^{-|z|^2/4}\ge (4\pi t)^{n/2}}|z|^2t^{n/2-1}\,dz\,dt =\int_{\mathbb R^n}|z|^2\,dz \int_0^{\frac{1}{4\pi}e^{-|z|^2/(2n)}}t^{n/2-1}dt
=\frac{2}{n(4\pi)^{n/2}}\int_{\mathbb R^n}|z|^2e^{-|z|^2/4}}dz\]
The last integral is the area of $ S^{n-1}$ times $ \int_0^{\infty} r^{n+1}e^{-r^2/4}dr$. It's quite possible that I made a mistake or a few here; it's getting late.  :sleeping:

[quote="mlok"]I think that reading Evans with full understanding requires more knowledge of integration. For instance, you should know that $ s\le 0$ is no different from $ s < 0$ as far as integration is concerned, and that a singularity is not a death sentence for an integral. As for the integral, the substitution $ t = - s$, $ y = z\sqrt {t}$ turns it into
\[{ \int_{e^{ - |z|^2/4}\ge (4\pi t)^{n/2}}|z|^2t^{n/2 - 1}\,dz\,dt = \int_{\mathbb R^n}|z|^2\,dz \int_0^{\frac {1}{4\pi}e^{ - |z|^2/(2n)}}t^{n/2 - 1}dt = \frac {2}{n(4\pi)^{n/2}}\int_{\mathbb R^n}|z|^2e^{ - |z|^2/4}}dz
\]
The last integral is the area of $ S^{n - 1}$ times $ \int_0^{\infty} r^{n + 1}e^{ - r^2/4}dr$. It's quite possible that I made a mistake or a few here; it's getting late.  :sleeping:[/quote]

thanks. let me finish the last integral: by change variable: $ r^2/4=u$ it became:
$ 2^{n+1} \int_0^{\infty} u^{n/2} e^{-u} du=2^{n+1} \Gamma (n/2+1)$ by defination of gamma function. But $ S_{n-1}=\frac{n \pi^{n/2}}{\Gamma (n/2+1)}$, ...


linear algebra
matrix
linear algebra unsolved

Let $f$ be a continious morphism of the additive groupes $f: \mathbb{R} \to M_n(\mathbb{K})$. Prove that there exists $A \in M_n(\mathbb{K})$ such that $f(t)=e^{tA}$ for all $t \in \mathbb{R}$.


a rope of length $l$ and linear density function $\lambda(x)$ where $x$ is the vertical position on the rope; holds up a block of mass $m$ What is the function $T(x)$ where $T$ is the tension of the rope at a vertical position $x$

At any point the tension is only due to the mass of the rope Below it , the upper part does not contribute to the tension.

Hence the tension due to the weight of the rope at a point at a distance x from the bottom is given by

$T(x) =$  $\int^{x}_{0}{\lambda(x)g\,dx}$

Hence ,the net tension at any point = tension due to rope+tension due to the overhanging mass which is equal to

$T(x) =$ $mg$ $+$ $\int^{x}_{0}{\lambda(x)g\,dx}$


i was givin this in class (the picture is the work sheet) and my teacher wants us to find the most efficient way of filling the truck bed without going over 3375 in^3
i got 1800 rectangular prisms with the dimensions of 9.6x6x10
p.s dont mind the diagram i did at the top.

WOrk sh33t:
[url=http://imageshack.us][img]http://img462.imageshack.us/img462/843/s3pg.jpg[/img][/url]

any got any bit of an answer lower than mine?

Well if you're not worry about surface area, it's easy to do a 12" x 12" x 12" box. So are you trying yo conserve surface area AND volume? Which is the most important? Also, rule out cylinder and triangular prism.

yea got any answers lower than 1800 boxes?


Let a,b,c be nonnegative real numbers.  Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {4018}{2000}\sqrt [2009]{(abc)^{2000}} \plus{} \frac {1982}{2000}\geq 2(ab \plus{} bc \plus{} ca)$

these numbers really threaten people in a scary way  :D
However, we have

$ \frac{4018(abc)^{\frac{2000}{2009}}}{2000}\plus{} \frac{1982}{2000} \ge 3 \sqrt[3]{(abc)^2}$

Now we turn to the our well known inequality!

$ a^2\plus{}b^2\plus{}c^2 \plus{}3\sqrt[3]{(abc)^2} \ge 2(ab\plus{}bc\plus{}ca)$


inequalities
real analysis
real analysis unsolved

Does there exsist positive constant $ C$ and $ \alpha \in (0,1]$ such that the following inequality
holds for all $ x_i\in [ \minus{} \pi,\pi]$ and $ y_i\in ( \minus{} \infty, \plus{} \infty), i\equal{}1,2,3$:
$ |1 \minus{} cos(x_1y_1 \plus{} x_2y_2 \plus{} x_3y_3)|\leq C |x_1 \plus{} x_2 \plus{} x_3|^\alpha |y_1 \plus{} y_2 \plus{} y_3|^\alpha$.

If $ x_1\plus{}x_2\plus{}x_3\equal{}0$, but not all $ x_j\equal{}0$, the right hand side is $ 0$ but the left hand side can be anything between $ 0$ and $ 2$. :roll:

thank you fedja,
What if we add following condition: the product $ (x_1\plus{}x_2\plus{}x_3)(y_1\plus{}y_2\plus{}y_3)$ is not equal to zero.


Find all functions $ f: \mathbb R^2 \to \mathbb R$ such that it qualifies both conditions below:
i)  $ f(x,y)+f(y,z)+f(z,x)=x^2+y^2+z^2$, for all reals $ x,y,z$, and
ii) $ f(x^2+y-f(0,x),0)=x+f^2(y,0)-f(0,y^2-f(0,y))$, for all reals $ x,y$.

[quote="mathVNpro"]Find all fuction $ f(x,y)$ such that it qualifies both condition below:
i)  $ f(x,y) \plus{} f(y,z) \plus{} f(z,x) \equal{} x^2 \plus{} y^2 \plus{} z^2$, for all $ x,y,z$ real numbers
ii) $ f(x^2 \plus{} y \minus{} f(0,x),0) \equal{} x \plus{} f^2(y,0) \minus{} f(0,y^2 \minus{} f(0,y))$, for all $ x,y$ real numbers[/quote]

Let $ g(x,y)\equal{}f(x,y)\minus{}y^2$

The two conditions may be writen :
A1(x,y,z) : $ g(x,y)\plus{}g(y,z)\plus{}g(z,x)\equal{}0$
A2(x,y) : $ g(y\minus{}g(0,x),0)\equal{}x\plus{}g^2(y,0)\minus{}g(0,\minus{}g(0,y))\minus{}g^2(0,y)$

A1(0,0,0) $ \implies$ $ g(0,0)\equal{}0$
A1(x,0,0) $ \implies$ $ g(x,0)\equal{}\minus{}g(0,x)$

Using this last property in A2, we get :
A3(x,y) : $ g(y\plus{}g(x,0),0)\equal{}x\plus{}g(g(y,0),0)$

Let then $ h(x)\equal{}g(x,0)$. A3 becomes :
A4(x,y) : $ h(y\plus{}h(x))\equal{}x\plus{}h(h(y))$ with $ h(0)\equal{}0$

A4(x,0) $ \implies$ $ h(h(x))\equal{}x$
A4(0,x) $ \implies$ $ h(x)\equal{}h(h(x))$

So $ h(x)\equal{}x$ and $ g(x,0)\equal{}x$

A1(x,y,0) $ \implies$ $ g(x,y)\plus{}g(y,0)\plus{}g(0,x)\equal{}0$, so $ g(x,y)\plus{}g(y,0)\minus{}g(x,0)\equal{}0$, and so $ g(x,y)\equal{}x\minus{}y$

And so $ f(x,y)\equal{}x\minus{}y\plus{}y^2$

And it is easy to verify that this necessary condition works.


vector
function
superior algebra
superior algebra unsolved

Hello,
I just "finished" a proof only to find a hole.  The problem runs thus:

Prove that if $V$ is a finite dimensional vector space and $W$ is a subspace of $V$, then there exists an idempotent endomorphism of $V$ with image $W.$

I (incorrectly) considered the "endomorphism" $\phi$ defined a restriction on the
identity function $\phi$ on $V$ where if $w\in W,$ then $\phi(w)=w$ and if $w\in V\backslash W,$ then $\phi(w)=0.$  This, of course, falls apart when one notes that if you have two elements from $V,$ one of which is in $W$ and the other in $V \backslash W$ (say $w\in V$ and $v\in V\backslash W.$)

$\phi(w+v)\neq \phi(w)+\phi(v)$ for all $w,v\in V$  is not well defined for the operation.

Any one have any other considerations?

-JNG

Make $\phi(x)=0$ not in all $V\setminus W$, but on some (sum)complement of $W$.


algebra
polynomial
linear algebra
matrix
linear algebra theorems

Suppose $ F \equal{} C$, $ T\in L(V)$, $ p \in P(C)$, and $ a \in C$. prove $ a$ is an eigenvalue of $ p(T)$ iff $ a \equal{} p(S)$ for some eigenvalue $ S$ of $ T$.

starting with an eigenvalue of $ T$ and proving the implication is an easy 1 step solution. Going the other way should be easy but im a bit stuck. We can bring the eqn $ p(T)v \equal{} p(S)v$, $ v \in V$ to the form where $ (T \minus{} SI)(very\: messy\: polynomial\: in\: T,S)v \equal{} 0$ (LHS-RHS and factor out) so either one of those two linear transforms takes $ v$ to zero. If it is $ (T \minus{} SI)$ then we are done, but I am not sure how to show this?

That argument won't work, because the inverse of $ p$ is multiple-valued. If $ p(S)$ is an eigenvalue, $ S$ might not be.

Here's what I would do: Triangularize $ T$. The diagonal entries $ t_{ii}$ are the eigenvalues. The diagonal entries of $ p(T)$ are $ p(t_{ii})$, and are the eigenvalues of $ p(T)$.

[quote="jmerry"] That argument won't work, because the inverse of p is multiple-valued. [/quote] I know very little about polynomials, so can you explain a bit more what you are saying here?

oh ok, the next question is to show why this is not necessarily true for the field of reals; will any 'un-diagonal' matrix do? how do we find an un-diagonal matrix? and how do we show p(T) does not have some $ S$ as an eigenvalue? or do we find a matrix with no eigenvalues and show $ p(T)$ has one?

An extremely simple counterexample in $ \mathbb{R}$: $ T\equal{}\begin{bmatrix}0&1\\\minus{}1&0\end{bmatrix}$ has no eigenvalues (because they're $ \pm i$). $ T^2\equal{}\minus{}I$ does have eigenvalues.


vector
geometry
circumcircle
analytic geometry
search
articles
absolute value

Well I had a simple question on my Geometry test yesterday that asked to find the circumcenter of a triangle given the coordinates of the vertices. I know that it can be solved using some algebra, but I wondered if there was a more efficient method to reach the same answer. After a quick search, I found [url=http://en.wikipedia.org/wiki/Circumcenter]this Wikipedia article.[/url] I was wondering how one would:

1. Go about subtracting $ P_2\minus{}P_3$. My dad says that to subtract two vectors, you have to "line them up" and then find the distance between the two "ends." (Not too specific, I know).

2. Why do the have the absolute value signs when that part of the expression is already squared?

To add two vectors, stick the tail of one onto the head of another.  It turns out that the components of the new vector are obtained by summing the corresponding components of the two vectors you added.  I could try to go into more detail, but it'd probably be good to find an "introduction to vectors" website, maybe the Wikipedia article on vectors, and dig in.

The vector $ P_2 \minus{} P_3$ is the vector that goes from the tip of $ P_3$ to the tip of $ P_2$ when you give $ P_2$ and $ P_3$ the same initial point.  (This makes complete sense: it's what you have to add to $ P_3$ to get $ P_2$.)  Algebraically, it's just the coordinatewise difference.

For the second question, the absolute values are lengths of vectors.  If $ v$ is a vector, $ v^2$ doesn't mean anything, but $ |v|^2$ means the length of $ v$, squared.


induction
inequalities
inequalities unsolved

Let $ x_{1},x_{2},\cdots,x_{m},y_{1},y_{2},\cdots,y_{n}$ be positive real numbers. Denote by $ X \equal{} \sum_{i \equal{} 1}^{m}x,Y \equal{} \sum_{j \equal{} 1}^{n}y.$ Prove that $ 2XY\sum_{i \equal{} 1}^{m}\sum_{j \equal{} 1}^{n}|x_{i} \minus{} y_{j}|\ge X^2\sum_{j \equal{} 1}^{n}\sum_{l \equal{} 1}^{n}|y_{i} \minus{} y_{l}| \plus{} Y^2\sum_{i \equal{} 1}^{m}\sum_{k \equal{} 1}^{m}|x_{i} \minus{} x_{k}|$

Oh of course  :blush:
Well assume $ x_1\ge x_2\ge \dots \ge x_n$ and $ X\ge Y$, then make an induction :
if the inequality is true for $ m \minus{} 1$, then we can prove for $ x \equal{} x_1 \plus{} x_2$ the following statement, which will solve our problem in matter of fact:
$ X\sum_{j \equal{} 1}^{n} |x \minus{} y_j| \minus{} Y\sum_{i \equal{} 3}^{m} |x \minus{} x_i|\le X\sum_{j \equal{} 1}^{n} (|x_1 \minus{} y_j| \plus{} |x_2 \minus{} y_j|) \minus{}$ $ Y\left(\sum_{i \equal{} 3}^{m} (|x_1 \minus{} x_i| \plus{} |x_2 \minus{} x_i|) \plus{} |x_1 \minus{} x_2|\right)$ (*).

First $ LHS \equal{} X\sum_{j \equal{} 1}^{n} |x \minus{} y_j| \minus{} Y\sum_{i \equal{} 3}^{m} |x \minus{} x_i| \equal{} \sum_{j \equal{} 1}^{n}\sum_{i \equal{} 1}^{m} x_i|x \minus{} y_j| \minus{} \sum_{j \equal{} 1}^{n}\sum_{i \equal{} 3}^{m} y_j|x \minus{} x_j|$

$ RHS \equal{} X\sum_{j \equal{} 1}^{n} (|x_1 \minus{} y_j| \plus{} |x_2 \minus{} y_j|) \minus{}$ $ Y\left(\sum_{i \equal{} 3}^{m} (|x_1 \minus{} x_i| \plus{} |x_2 \minus{} x_i|) \plus{} |x_1 \minus{} x_2|\right) \equal{} \sum_{j \equal{} 1}^{n}\sum_{i \equal{} 1}^{m} (x_i|x_1 \minus{} y_j| \plus{} x_i|x_2 \minus{} y_j|)$ $ \minus{} \sum_{j \equal{} 1}^{n}\sum_{i \equal{} 3}^{m}\left(y_j|x_1 \minus{} x_i| \plus{} y_j|x_2 \minus{} x_i|) \plus{} y_j|x_1 \minus{} x_2|\right)$

and now using the inequality $ |xx_i \minus{} y_jx_i| \minus{} |y_jx \minus{} y_jx_i|\le$ $ x_i|x_1 \minus{} y_j| \plus{} x_i|x_2 \minus{} y_j| \minus{} y_j|x_1 \minus{} x_i| \minus{} y_j|x_2 \minus{} x_i|$ which follows by verifying all cases - $ x_1\ge x_2\ge y_j \cup x\ge y_j$  etc. , we will take what we need.

$ \Rightarrow 2XY \sum_{i \equal{} 1}^m\sum_{j \equal{} 1}^n|x_i \minus{} y_j|$ $ \minus{} X^2 \sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n|y_i \minus{} y_j| \minus{} Y^2 \sum_{i \equal{} 1}^m\sum_{j \equal{} 1}^m|x_i \minus{} x_j|\ge$ $ 2XY \sum_{i \equal{} 2}^m\sum_{j \equal{} 1}^n|x'_i \minus{} y_j| \minus{} X^2 \sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n|y_i \minus{} y_j| \minus{} Y^2 \sum_{i \equal{} 2}^m\sum_{j \equal{} 1}^m|x'_i \minus{} x'_j|\ge 0$ by (*), where $ x'_2 \equal{} x_1 \plus{} x_2$ and $ x'_i \equal{} x_i$ for $ i\ge 3$...

Could you please show us where we can find $ \sum y_j |x_1\minus{}x_2|$? :)

[quote="hxy09"]This is the shortest proof(not mine) for the hard problem
I have to say this brilliant approach is owed to the official solution for a TST problem from Poland :)[/quote]
So this problem is from a Poland TST? If so, can you tell me which year? By the way, the solution you posted is very nice :)!

Well not all of us can be a genius who comes up with above brilliant approach...

Relax the condition in the problem statement such that the $x_{i}$ and $y_{j}$ are nonnegative real numbers.

Let $P(x_1, x_2, \cdots, x_m, y_1, y_2, \cdots, y_n)$ denote the desired inequality. Let $A = \sum_{i = 1}^{m}\sum_{j = 1}^{n}|x_{i} - y_{j}|$, $B = \sum_{j = 1}^{n}\sum_{l = 1}^{n}|y_{i} - y_{l}|$, and $C = \sum_{i = 1}^{m}\sum_{k = 1}^{m}|x_{i} - x_{k}|$.

We will write $(x)$ and $(y)$ to denote the numbers $x_1, x_2, \cdots, x_m$ and $y_1, y_2, \cdots, y_n$ respectively.

Notice that if $m = 0$ or $n=0$, the LHS and the RHS of the desired inequality are both equal to zero. In what follows, we induct on $m+n$.

[b]Lemma 1.[/b] If $x_m = 0$, we can delete $x_m$ to yield an equivalent desired inequality. In other words, $P(x_1, x_2, \cdots, x_m, y_1, y_2, \cdots, y_n)$ is implied by $P(x_1, x_2, \cdots, x_{m-1}, y_1, y_2, \cdots, y_n)$ if $x_m = 0$.

[i]Proof.[/i] Note that $P(x_1, x_2, \cdots, x_{m-1},y_1, y_2, \cdots, y_n)$ is the following: \[2XY(A-Y) \ge X^2B+Y^2(C-2X).\] This is easily seen to be equivalent to $2XYA \ge X^2B + Y^2C$. $\blacksquare$

Time for the key idea of the proof.

[b]Lemma 2.[/b] Suppose there are $d \ge 3$ distinct values present among the numbers $(x),(y)$. Then we can find nonnegative reals $x_1', x_2', \cdots, x_m', y_1', y_2', \cdots, y_n'$ (henceforth abbreviated $(x'), (y')$) such that either there are at most $d-1$ distinct values present among them, or at least one of them is equal to zero, and such that $P((x'),(y'))$ implies $P((x),(y))$.

[i]Proof.[/i] Let $a<b<c$ be three distinct values present among the numbers $(x),(y)$. Let $r_1, s_1$ denote the number of times $a$ appears among $(x)$, respectively $(y)$. Define $r_2, s_2, r_3, s_3$ similarly using $b$ and $c$ respectively.

Partition $(x)$ into four disjoint multisets $(x)^a, (x)^b, (x)^c, (x)^0$ such that $(x)^a, (x)^b, (x)^c$ contains those $x_i$ which are equal to $a$, $b$, $c$ respectively. The set $(x)^0$ contains those $x_i$ which are not equal to any of $a,b,c$. Partition $(y)$ into four sets $(y)^a, (y)^b, (y)^c, (y)^0$ similarly.

It is possible to find real numbers $u,v,w$, not all zero, such that $r_1u+r_2v+r_3w = s_1u+s_2v+s_3w = 0$.

Now we fix a real number $\epsilon$ and define the set $(x')^a$ to be equal to $(x)^a + u\epsilon$; i.e. the multiset consisting of every element of $(x)^a$ increased by $u\epsilon$. Define $(x')^b, (x')^c$ similarly using $v$ and $w$. Finally take $(x')$ as the multiset consisting of $(x')^a, (x')^b, (x')^c, (x)^0$. We can let $(x')$ consist of numbers $x_1', x_2', \cdots, x_m'$, with the same labeling as the $x_i$'s. (For each $x_i$, its index $i$ follows it through the partitioning process and then through the recombining process.)

Define $(y')$ and the $y_j'$'s similarly.

The punch line is that the inequality $P((x'),(y'))$ is linear in $\epsilon$, as long as the relative ordering of all the $x_i$ and $y_j$ is the same as the relative ordering of all the $x_i'$ and $y_j'$. Indeed, $X,Y$ are constant with respect to $\epsilon$ because we chose $r_1u + r_2v + r_3w = s_1u + s_2v + s_3w = 0$. Furthermore, $A,B,C$ behave like linear combinations of the $x_i'$ and $y_j'$, so the LHS and RHS of $P((x'),(y'))$ are both linear functions in $\epsilon$. Hence we can smooth $\epsilon$ to one of its extremes in order to obtain a tighter inequality.

$\epsilon$ reaches an extremum precisely when one of the $x_i'$ or $y_j'$ is equal to zero, or when $(xy')^a, (xy')^b, (xy')^c, (xy)^0$ are no longer disjoint. (Here $(xy')^a$ means the multiset consisting of the elements of $(x')^a$ and also the elements of $(y')^a$.) $\blacksquare$

By repeatedly applying Lemma 2, and by repeatedly deleting any $x_i$ or $y_j$ that is equal to zero, it suffices to prove $P((x),(y))$ where there are at most two distinct values present among the numbers $(x), (y)$. (We already took care of the case $m = 0$ or $n = 0$.)

If there is only one value present among the numbers $(x), (y)$, then all the $x_i$ and $y_j$ are equal. Hence $A = B = C = 0$ and $P((x),(y))$ is trivially true.

Assume that $a<b$ are the values present among the numbers $(x), (y)$, and let $r_1, r_2$ be the number of indices $i$ such that $x_i = a$, respectively $x_i = b$. Define $s_1, s_2$ similarly.

We calculate $X = r_1a + r_2b$, $Y = s_1a + s_2b$, and $A = (r_1s_2 + r_2s_1)(b-a)$, $B = 2s_1s_2(b-a)$, $C = 2r_1r_2(b-a)$. Our inequality $P$ becomes $XY(r_1s_2 + r_2s_1) \ge X^2s_1s_2 + Y^2r_1r_2$, which factors as \[0 \ge (Xs_1 - Yr_1)(Xs_2 - Yr_2).\] This rewrites as \[0 \ge (b(r_2s_1-r_1s_2))(a(r_1s_2 - r_2s_1)),\] which is obviously true. $\blacksquare$


Subtract $ 1.5$ from each number to get $ 0,4,8,\ldots,44$.  There are $ \frac{44\minus{}0}{4}\plus{}1\equal{}11\plus{}1\equal{}\boxed{12}$ numbers in this list.


How many single operations (multiplications, subtractions) you should do to solve a $ n\times n$ system by the elimination process? Do the same thing interpreting the "operations" as row operations (multiplications and subtractions).

i believe $ n^2$ is approximately the number of operations required.
Start from top left of the matrix.
Multiply by its reciprocal, then subtract from the rest of the column.
Do the same for the next column.

From this you can see the pattern.

Counting row operations seems a little weird to me since the row operations are of a range of sizes. At the level of individual operations, it should take (asymptotically) $ \frac13\,n^3$ mulitiplications/divisions and $ \frac12\,n^3$ additions/subtractions.


calculus
integration
trigonometry
function
search
symmetry
calculus computations

Compute the following integral:

$ \int^\infty_{\minus{}\infty} \frac{\sin^2{x}}{x^2}\,dx$

I was thinking maybe to exploit the fact that the function was even, but other than that, I had no other ideas.

Thank you for your help in advance.

I know I've done this at least once and probably several times on this forum, but I'm not sure of what phrase to search for in my own posts.

Step 1: even symmetry and trig identity:

$ \int_{\minus{}\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx\equal{} 2 \int_{0}^{\infty}\frac{\sin^2x}{x^2}\,dx\equal{}\int_0^{\infty}\frac{1\minus{}\cos 2x}{x^2}\,dx$

Step 2: auxiliary integral:

$ \frac1{x^2}\equal{}\int_0^{\infty}te^{\minus{}tx}\,dt.$ (Straightforward to show.)

Step 3: changing the order of integration (Fubini/Tonelli):

$ \int_0^{\infty}\frac{1\minus{}\cos 2x}{x^2}\,dx\equal{}\int_0^{\infty}\int_0^{\infty}te^{\minus{}tx}(1\minus{}\cos 2x)\,dt\,dx$

$ \equal{}\int_0^{\infty}\int_0^{\infty}te^{\minus{}tx}(1\minus{}\cos 2x)\,dx\,dt$

Step 4: final computational details:

--- I'll let you finish it. Suffice it to say that you should get an arctangent on the last integral with respect to $ t$ and that the answer will contain $ \pi.$


[hide]Simplify the second equation and you'll see that x cancels, and y=-1.
So x = 3. [/hide]


Let a,b,c be non negative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$ . Prove that

1.
$ \frac {1}{{1 \plus{} a^2 (b \plus{} c)}} \plus{} \frac {1}{{1 \plus{} b^2 \left( {a \plus{} c} \right)}} \plus{} \frac {1}{{1 \plus{} c^2 \left( {a \plus{} b} \right)}} \le \frac {3}{{1 \plus{} 2abc}}$

[code]\frac{1}{{1 + a^2 (b + c)}} + \frac{1}{{1 + b^2 \left( {a + c} \right)}} + \frac{1}{{1 + c^2 \left( {a + b} \right)}} \le \frac{3}{{1 + 2/27(a+b+c)^3}}[/code]
 by Jensen, but [code]27abc\le(a+b+c)^3 [/code] by Mur for example, i think its true :maybe:

[quote="thanhak5"]Let a,b,c be non negative real numbers. Prove that

1.
$ \frac {1}{{1 \plus{} a^2 (b \plus{} c)}} \plus{} \frac {1}{{1 \plus{} b^2 \left( {a \plus{} c} \right)}} \plus{} \frac {1}{{1 \plus{} c^2 \left( {a \plus{} b} \right)}} \le \frac {3}{{1 \plus{} 2abc}}$[/quote]
It's wrong. Try $ a\equal{}2$ and $ b\equal{}c\equal{}1.$ :wink:

Sorry . It is not ringht . I repaired it

[quote="thanhak5"]Let a,b,c be non negative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$ . Prove that

1.
$ \frac {1}{{1 \plus{} a^2 (b \plus{} c)}} \plus{} \frac {1}{{1 \plus{} b^2 \left( {a \plus{} c} \right)}} \plus{} \frac {1}{{1 \plus{} c^2 \left( {a \plus{} b} \right)}} \le \frac {3}{{1 \plus{} 2abc}}$[/quote]
$ \sum_{cyc}\frac {1}{1 \plus{} a^2 (b \plus{} c)}\le \frac {3}{1 \plus{} 2abc}\Leftrightarrow\sum_{cyc}\left( \frac {1}{1 \plus{} 2abc}\minus{}\frac {1}{1 \plus{} a^2 (b \plus{} c)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{a^2(b\plus{}c)\minus{}2abc}{1 \plus{} a^2 (b \plus{} c)}\geq0\Leftrightarrow\sum_{cyc}\frac{ac(a\minus{}b)\minus{}ab(c\minus{}a)}{1 \plus{} a^2 (b \plus{} c)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)\left(\frac{ac}{1 \plus{} a^2 (b \plus{} c)}\minus{}\frac{bc}{1\plus{}b^2(a\plus{}c)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{c(1\minus{}abc)(a\minus{}b)^2}{(1\plus{}a^2(b\plus{}c))(1\plus{}b^2(a\plus{}c))}\geq0,$ which is obvious.

[quote="thanhak5"]Let $ a,b,c$ be nonnegative real numbers such that $ bc + ca + ab = 3$ . Prove that

$ \frac {1}{1 + a^2 (b + c)} + \frac {1}{1 + b^2 (c + a)} + \frac {1}{1 + c^2 (a + b)} \le \frac {3}{1 + 2abc}.$[/quote]$ \frac {3}{{1 + 2abc}} - \sum{\frac {1}{{1 + a^2 (b + c)}}}$

$ \frac {(1 - abc)\left[2\sum{b^3c^3} - \sum{a^3(b^2c + bc^2)} + \sum{a^2(b + c) - 6abc\right]}}{(1 + 2abc)\left[1 + a^2 (b + c)\right]\left[1 + b^2 (c + a)\right]\left[1 + c^2 (a + b)\right]}\geq0,$

which is clearly true by Muirhead theorem.

[hide="Remark."]If $ a,b,c$ be real numbers such that $ a + b + c\geq0$, $ bc + ca + ab\geq0$ and $ - \frac {1}{2} < abc\leq 1$, then the original inequality holds.[/hide]


let S = {x1,....,xn} be a linearly independent subset of a vector space over the field Z2, how many elements are there in span(S)? justify your answer..
my idea was that you have linear combinations of x1,...,xn  axi where a is always 0 or 1, then you have one 0 vector so you only really need to look at combinations of xi. you'll have the {x1,.....,xn} set, combinations of 2 vectors, 3 vectors... up to n vectors but I don't know how to put this into something concrete.. please help, thanks

elements in the span of $ S$ have the form $ a_1x_1\plus{}\cdots\plus{}a_nx_n$, where each $ a_1$ is 0 or 1. thus every point of $ \mbox{span}(S)$ has $ n$ coordinates $ (a_1, ..., a_n)$. there are clearly $ 2^n$ possible choices for this $ n$-tuple, and moreover, different $ n$-tuples give you different elements of $ \mbox{span}(S)$, so the answer is $ 2^n$. (if $ a_1x_1\plus{}\cdots \plus{}a_nx_n\equal{}b_1x_1\plus{}\cdots \plus{}b_nx_n$, then $ (a_1\minus{}b_1)x_1\plus{}\cdots\plus{}(a_n\minus{}b_n)x_n\equal{}0$, so by linear independence, we must have $ a_i\minus{}b_i\equal{}0$ for every $ i$, i.e. $ a_i\equal{}b_i$ for every $ i$.)


How to tie the [b]Overhand Knot[/b] without lifting the rope from your hands ? I hope u get what I need .

Usual way :

http://www.iwillknot.com/overhand_knot/

you can see he lift his hand when he  pulled  the rope .

Thx  :)

Just make the knot with your hands (i.e., fold them the usual way used when expressing defiance; I'm too lazy to draw a picture now) and take the straight piece of rope afterwards. When you are unfolding the hands, the knot will pass to the rope.

I don't get it in fact. Could u express more please ? :)

That really nice.

O.k.

What about if someone give us the rope and ask to do it directly , that means I can't fold my hands before taking the rope .

I appreciate your help  :)

The knot should be somewhere from the very beginning unless you cheat on what it means "to never lift your hands from the rope" (which I understand as "the ends of the rope are glued to your palms by the strongest glue ever made and tearing them off will result in an immediate death penalty") or on what it means "to tie an overhead knot on the rope" (which I understand as "there should be a neat fully tightened knot on the rope in the end without an arm or anything else caught in it") and such cheating stuff does not belong to the realm of mathematical questions. If it is not on the rope, it should be on your arms. So, there is no other way.

I DON"T CHEAT DUDE  :wink: ( No goal to cheat  :) )

That what I was looking for :

[quote="fedja"] If it is not on the rope, it should be on your arms. So, there is no other way.[/quote]

Thx  :)


linear algebra
matrix
combinatorics proposed
combinatorics

Assume that in each small square of a big $m\times n$ square, one put down a real number such that the sum of all numbers in each row or column is integer. Prove that, one can replace each number in each small square by its lower integer part or upper integer part such that sum of all numbers in each row or column doesnot change.
Where the lower integer part of the number $x$ is the largest integer number not exceed x, similar definition to the upper one.

We call the sum of the $k^{th}$ column as $a_k$ and sum of $k^{th}$ row as $b_k$. It is easy to verify that $\sum a_k=\sum b_k$ .

Then , we take the lower part of all the numbers in the square and again denote the sum of $k^{th}$ column as $a'_k$ and sum of $k^{th}$ row as $b'_k$ . Again we know that $\sum a'_k=\sum b'_k$

We know that $a_k\geq a_'k$ , $b_k\geq b'_k$ and $\sum a_k-a'_k=\sum b_k-b'_k$
This means the the total " error " ( error in the sense that their sum are different )of the column is same as the total " error " of the row .  Now to " fix " the " error " , we start from first column . If $a_1-a'_1=5$ , that means the first column has $5$ error and we have to add five to that some of the square in the first column ( this is the same as taking the upper part of the number ) . Then , we then trace row from the first row and see which row also have error . For the particular square which is the intersection of "error " column and " error " row , we add one to it .

Proceed it until all error is fixed  ( this can be done since when each "error " in column is fixed ,  the "error" in row also being fixed at the same time . And we know that the total "error" of column and row are the same means all error can be fixed .)

[quote="kevinatcausa"]I think we need to be a little more careful here.  Suppose our matrix was

$[ 0.4 \, \, 0.0 \, \, 0.6]$
$[ 0.9 \, \, 0.4 \, \, 0.7]$
$[ 0.7 \, \, 0.6\, \, 0.7]$

Our row errors are 1, 2, 2
Our column errors are 2,1,2.

Now I try and fix errors column by column.  If I fix from the top down, I may be in this situation after 2 columns:

$[1 0 0]$
$[1 1 0]$
$[0 0 0]$

where the third row and third column both have an error of 2, which can't be fixed.

Therefore we need to make sure we pick the right rows to use to fix an error in column $i$.  If the error in column $i$ is $e_i$, I think picking the $e_i$ rows with the largest error left after the first $i-1$ columns have been fixed would work.[/quote]
maybe you got some mistakes, after fixed,
100
101
011

That would fix the matrix, but the question is how you knew to pick that particular matrix to fix your example.

If you're doing it column by column (as the original solution suggests), how do you know the second column is 001 and not 010 without just looking ahead to see that one works and the other doesn't.


Prove that:
[tex]\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{9}{10}[/tex]
with x,y,z are reals and x+y+z=1.

I hope your short solves.

it was posted here
http://www.mathlinks.ro/Forum/topic-27450.html
are you have the solution?


I have been looking for any tutorials that can guide me about the arithmetic of polynomial with coefficients in finite field. Could somebody please help me to locate where can I find it on the internet? I have refer to books etc... but no detailed explanation on it. TQ

Faisal :(

The arithmetic is the same as over any other field, like $\mathbb{Q}$ or $\mathbb{R}$ or $\mathbb{C}$.
What do you need in particular?
Do you know all finite fields?

The arithmethic isn't the same... The behaviour of roots is completely different.
But I recommend any algebra book, they often contain some special chapters about finite fields.


There's an Oklahoma forum!

Fingers crossed that even one other person from OK visits this board. (Maybe Syntax will come again.)

Well, I grew up in Oklahoma. (Bartlesville, to be specific.) But it's been over 30 years since I lived there.

I've been there, really nice performing arts center.

Have you been back since?

That was built after I moved away, but my younger sister had her wedding reception there. But everyone in my family now lives on the West Coast.

me had middle school in oklahoma!  :)  now we need greater than one person in the oregon forum.....

yes i am....
EDIT: that was also me that is

lol physics_man, you just said go arizona in the arizona forum.

[size=200][color=blue]GO INDIANA!!!!!!!!!!!!!!!!!!!!!!!!!![/color][/size]

physics_man = arizona?
[size=200] GO WHATEVER STATE I AM IN[/size]

Hi....I've lived in Oklahoma for 10 years, and will probably be here for another 5 years.

hi i live in indiana.

please copy my avatar. (which was originally magixsters)

I have visited Oklahoma before in fact my cousins live there.

Quite nice except being in the midst of Tornado Valley.

haha indiana, of all boards, why oklahoma?

And I hear some oregon guy  :wink:  is coming back to Oklahoma? Sets up pretty interesting year next year.

Hey numberdance, it's good to see another Oklahoman on here. even though I'm pretty much not active lol

pieterminate (YCHWYO)  Is that true?  If so, your MC nat coach says welcome back!

[size=150][color=indigo]GO OKLAHOMA!!!!!!!!!!!!!!!!!!!![/color][/size]

I beat umimpossible  :lol:


its way too bad

sorry that was a spam :rotfl:  :rotfl:  :rotfl:


please prove
chebyshev"s inequality
Holder"s inequality
Minkowski"s inequality
Jensen"s inequlaity

You should have asked for the reordering inequality (which gives Chebyshev's equality), for Young's inequality (which yields Holder's, and Minkowski's), and Jensen's is given by the concavity of $ln$.
All of that you can google and probably find proofs in a twinkle ...


[hide]you would need to post 86 consecutive tags. 951 must equal 75% of the new number of tags, which will be 1268. 25% of 1268 is 317, so, 86 more tags are necessary.[/hide]

[hide]
So we have a total of 951.  since 231 are not "Click to reaveal hidden content" ones that means 720 are.  So 720 needs to be 25% so 720/.75=960.  25% of 960 is 240 so 240-231= $\boxed{\boxed{\boxed{9}}}$[/hide]

[quote="236factorial"]If there are a total of 951 hide tags on the basic forum, and 231 do not say "click to reveal hidden content". How many hide tags without the labelling "click to reveal hidden content" must be posted consecutively for the percentage of these hide tags to rise to 25%?

Please hide your answers  :D[/quote]
[hide]

951-231=720
720 has to be 75% of all of the hide tags.
720 needs to be 75%, so 720/.25=288
288-231=$\boxed{57}$[/hide]
I don't think that's correct :fool:

You sort of interpreted the problem wrong... there has been several different answers  :lol:

[quote="math92"]has anyone got it yet?[/quote]

You did.

really???
when i wrote my answer i thought it was wrong.
 :D


Let's suppose that we are in a country where there are no cents, that is, every price and every amount of money is given as an exact integer number. They only use coins, no bills. And suppose that by law the currency denominations have to be of the form $2^n5^m$ where $-1\leq m-n\leq2$ (obviusly, both non-negative). If all the possible denominations are taken up to 1,000 (included), then:

(a) How many different coin denominations are there in this country?
(b) What is the minimum number of coins that you need to have (of course, they may include repited denominations and different denominations) to be sure that you can choose a subset of the coins that add up to $n$ for $1\leq n \leq 1,000$.

Enjoy!

[quote="rcv"]Part (a) is an interesting MATHCOUNTS-level challenge.

Part (b) is harder, but still in the range of MATHCOUNTS.

Bumped from "Basics" to "MATHCOUNTS".[/quote]

OK, May be I just misjudge the difficulty of the problem  :P . Thank you for moving it!

Part a doesn't seem horribly difficult, I haven't tried part b yet.

[hide="Click to reveal hidden content...if you dare!"]Okay, so we can have four different relationships between m and n: m-n=-1, m-n=0, m-n=1, and m-n=2. Let's take them one at a time. Suppose m-n=-1, or n-m=1. Since the exponent of 2 is one greater than the exponent of 5, it's just 2 times powers of ten. So that gives us 2, 20, and 200. Not 200, because 2000>1000. Okay, now suppose m-n=0. Note that if five and two have equal exponents, it ends up just being a power of ten, so we have 1, 10, 100, and 1000. Nothing higher, obviously. Now suppose m-n=1. Just as m-n=-1 gave us 2 times powers of ten, this gives us 5 times powers of ten. So we have 5, 50, and 500. Lastly assume m=n=2, giving us 25 times powers of ten. This yields 25, and 250. Put 'em all together and what do you get? 1, 2, 5, 10, 20, 25, 50, 100, 200, 250, 500, 1000. That's 12 denominations.

Despite all that writing, it shouldn't take more than 20 seconds mentally.[/hide]

[hide] a) Do some casework:

When m-n=2:
Then possibilities for (m,n) are (2,0) and (3,1), which equals 25 and 250 dollars, respectively.

When m-n=1:
Then possibilities for (m,n) are (1,0),(2,1),(3,2), which equal, 5, 50, and 500.

When m-n=0:
Possibilities: (0,0),(1,1),(2,2),(3,3) are 1,10,100, and 1000 dollars.

When m-n=-1:
(0,1),(1,2),(2,3) equal 2, 20, and 200 dollars.

Thus, there are twelve different denominations possible.

b) We need one 1-dollar bill so we can cover the even and odd
#s between 1 and 1000 (as well as 1).

We then need two 2-dollar bills.

We also need two 5's, and one 10, and one 25, and two 50's.

Then we need a 100, and a 250, and finally, a $500.

This gives us 1+2+2+1+1+2+1+1+1=12 coins (the money is not represented by bills).

Thus, we need a minimum of 12 coins to be able to make any value 1-1000 dollars. By the way, there are other sets of 12 coins which work, but I just gave an example.

Answer (to both parts): 12
[/hide]

[hide="Part b"]Ugh, I just did the solution for b but after looking at my teammate's, there's no point in repeating it. I got 12 too.[/hide]


One dozen eggs cost $ \$1.08$. It would cost 24 cents more to buy the eggs individually. What is the number of cents it would cost to buy one egg?

1.08/12=0.09 so each egg costs 0.12 dollars or $ \boxed{12}$ cents.

Not only does your solution make no sense, but it is incorrect!

Because it costs 24 cents more it will cost 108+24=132 cents for 12 eggs, or 11 cents per egg.

I am sorry.  I thought that a dozen was 8 for a second.

Explaining my solution:

The unit price per egg buying the dozen is 9 cents.  Then 24/12=2, so we add 2 cents to get 11.


Let  $ A, B, C, D, E, F$  be six consecutive points around the circumference of a circle.
Prove that the chords $ AD, BE, CF$ concur if and only if: $ |AB|.|CD|.|EF| \equal{} |BC|.|DE|.|FA|$.

for trigonometric Ceva on the triangle ACE we have that AD, CF, EB concur iff

$ 1\equal{}\frac{\sin \angle DAC}{\sin \angle DAE} \ \frac{\sin \angle FCE}{\sin \angle FCA} \ \frac{\sin \angle BEC}{\sin \angle BEA} \equal{} \frac{\sin \angle DEC}{\sin \angle DCE} \ \frac{\sin \angle FAE}{\sin \angle FEA} \ \frac{\sin \angle BAC}{\sin \angle BCA} \equal{}$ $ \frac{DC}{DE} \ \frac{FE}{FA} \ \frac{BA}{BC}$

:) very simple and nice. I'm sorry for the stupid question.
The proven statement may be used for proving 7-circles theorem.

It is the 7-circles theorem: http://mathworld.wolfram.com/SevenCirclesTheorem.html

[b]borislav_mirchev wrote:[/b][quote]The proven statement may be used for proving 7-circles theorem.[/quote]

How? Can you give a complete proof? :?:

At the moment I haven't enough time. I have hundred of interesting problems to solve and I have no even started.
But I found that proof on the Web: http://www.mathpropress.com/stan/bibliography/sevenCircles.pdf  :wink: enjoy...


calculus
integration
function
trigonometry
calculus computations

How would you integrate sin(x)ln(x)dx? All help greatly appreciated!

This is not a beginning integration  :ninja:

See [url=http://integrals.wolfram.com/index.jsp]The Integrator[/url]

The primitive of that function doesn't seem to expressable as a sum of elementary functions.

Not only "doesn't seem to be", but "is not"; integration by parts shows that if we were able to find this integral, we would also be able to find $\int\frac{\cos x}{x}$, which is a classical example of a non-elementary antiderivative.


Let $ a,b,c$ be positive distinct numbers. Prove that

(a) $ \frac 1{(a\minus{}b)^2}\plus{}\frac 1{(b\minus{}c)^2}\plus{}\frac 1{(c\minus{}a)^2} > \frac 9{a^2\plus{}b^2\plus{}c^2}$;

(b) $ \frac 1{(a\minus{}b)^2}\plus{}\frac 1{(b\minus{}c)^2}\plus{}\frac 1{(c\minus{}a)^2} > \frac {5\plus{}4\sqrt 2}{a^2\plus{}b^2\plus{}c^2}$.

If I'm not mistaken anything, using mixing variable absolutely works in this case!
Since I see if decreasing a,b,c by the same amount 0<c=min(a,b,c)(the equality can not be happened, then we see that the LHS is the same as it is, but the RHS is increasing(since (a-c)^2+(b-c)^2+0<a^2+b^2+c^2)
Therefore, we just need to prove in the case when c goes to 0! And we just have to prove:
1/(a-b)^2+1/b^2+1/c^2>9/(a^2+b^2), which is easy proven by mutiplying everything and substracts :lol:
On the problem 2, which is stricter than previous, we also have to prove:
1/(a-b)^2+1/b^2+1/c^2>5+4square2/{a^2+b^2}
W.L.O.G, assume that b=1(we can do that since c goes to 0, means c is very small).Now we will have inequality for one-variable, which will be true if this inequality is true and vice versa!

[quote="Vasc"]Let $ a,b,c$ be positive distinct numbers. Prove that

(a) $ \frac 1{(a \minus{} b)^2} \plus{} \frac 1{(b \minus{} c)^2} \plus{} \frac 1{(c \minus{} a)^2} > \frac 9{a^2 \plus{} b^2 \plus{} c^2}$;

(b) $ \frac 1{(a \minus{} b)^2} \plus{} \frac 1{(b \minus{} c)^2} \plus{} \frac 1{(c \minus{} a)^2} > \frac {5 \plus{} 4\sqrt 2}{a^2 \plus{} b^2 \plus{} c^2}$.[/quote]
The best is
\[ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2} \ge \frac {11 \plus{} 5\sqrt {5}}{2(a^2 \plus{} b^2 \plus{} c^2)}
\]
See here http://www.mathlinks.ro/viewtopic.php?t=168974
It just is a simple application of the entirely mixing variables method.

[quote="ehku"]
The best is
\[ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2} \ge \frac {11 \plus{} 5\sqrt {5}}{2(a^2 \plus{} b^2 \plus{} c^2)}
\][/quote]
You are right, but for
 $ \frac 1{(a \minus{} b)^2} \plus{} \frac 1{(b \minus{} c)^2} \plus{} \frac 1{(c \minus{} a)^2} > \frac {5 \plus{} 4\sqrt 2}{a^2 \plus{} b^2 \plus{} c^2}$,

there is a nice solution with AM-GM inequality (after setting $ a\equal{}0$).

[hide]If $ c\equal{}\min \{a,b,c\}$, the inequality
\[ (a^2\plus{}b^2\plus{}c^2)\left( \frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{(a\minus{}c)^2}\plus{}\frac{1}{(b\minus{}c)^2}\right) >5\plus{}\alpha\]
reduces to
\[ (a^2\plus{}b^2)\left( \frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{a^2}\plus{}\frac{1}{b^2}\right) \geq 5\plus{}\alpha\]
since $ c\neq 0$ and $ (a\minus{}c)^2\leq a^2,(b\minus{}c)^2\leq b^2$.

Now writing $ x\equal{}\frac{(a\minus{}b)^2}{ab}$ (use $ a^2\plus{}b^2\equal{}(a\minus{}b)^2\plus{}2ab$ and expand) reduces it to
\[ x^2\plus{}4x\plus{}\frac{2}{x}\geq \alpha\]
Now for any positive $ x_0$ (ideally the $ x$ where the left hand side is minimized), letting $ r\equal{}\frac{x}{x_0}$ gives
\[ x_0^2r^2\plus{}4x_0r\plus{}\frac{2}{x_0r}\geq \left( x_0^2\plus{}4x_0\plus{}\frac{2}{x_0}\right) r^\frac{2x_0^2\plus{}4x_0\minus{}\frac{2}{x_0}}{x_0^2\plus{}4x_0\plus{}\frac{2}{x_0}}\]
We want the bound to be independent of $ r$, so choose $ x_0$ such that
\[ 0\equal{}x_0^3\plus{}2x_0^2\minus{}1\equal{}(x_0\plus{}1)(x_0^2\plus{}x_0\minus{}1)\Rightarrow x_0\equal{}\frac{\sqrt{5}\minus{}1}{2}\]
and in this particular case,
\[ x_0^2\plus{}4x_0\plus{}\frac{2}{x_0}\equal{}3x_0^2\plus{}8x_0\equal{}3\plus{}5x_0\equal{}\frac{1\plus{}5\sqrt{5}}{2}\]
so we can choose this for $ \alpha$ and it gives ehku's result.

There is equality when $ c\equal{}0,b\equal{}1,$ and
\[ a\equal{}\frac{3\plus{}\sqrt{5}\plus{}\sqrt{6\sqrt{5}\minus{}2}}{4}\][/hide]

[quote="scorpius119"][hide]If $ c \equal{} \min \{a,b,c\}$, the inequality
\[ (a^2 \plus{} b^2 \plus{} c^2)\left( \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(a \minus{} c)^2} \plus{} \frac {1}{(b \minus{} c)^2}\right) > 5 \plus{} \alpha
\]
reduces to
\[ (a^2 \plus{} b^2)\left( \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{a^2} \plus{} \frac {1}{b^2}\right) \geq 5 \plus{} \alpha
\]
since $ c\neq 0$ and $ (a \minus{} c)^2\leq a^2,(b \minus{} c)^2\leq b^2$.

Now writing $ x \equal{} \frac {(a \minus{} b)^2}{ab}$ (use $ a^2 \plus{} b^2 \equal{} (a \minus{} b)^2 \plus{} 2ab$ and expand) reduces it to
\[ x^2 \plus{} 4x \plus{} \frac {2}{x}\geq \alpha
\]
Now for any positive $ x_0$ (ideally the $ x$ where the left hand side is minimized), letting $ r \equal{} \frac {x}{x_0}$ gives
\[ x_0^2r^2 \plus{} 4x_0r \plus{} \frac {2}{x_0r}\geq \left( x_0^2 \plus{} 4x_0 \plus{} \frac {2}{x_0}\right) r^\frac {2x_0^2 \plus{} 4x_0 \minus{} \frac {2}{x_0}}{x_0^2 \plus{} 4x_0 \plus{} \frac {2}{x_0}}
\]
We want the bound to be independent of $ r$, so choose $ x_0$ such that
\[ 0 \equal{} x_0^3 \plus{} 2x_0^2 \minus{} 1 \equal{} (x_0 \plus{} 1)(x_0^2 \plus{} x_0 \minus{} 1)\Rightarrow x_0 \equal{} \frac {\sqrt {5} \minus{} 1}{2}
\]
and in this particular case,
\[ x_0^2 \plus{} 4x_0 \plus{} \frac {2}{x_0} \equal{} 3x_0^2 \plus{} 8x_0 \equal{} 3 \plus{} 5x_0 \equal{} \frac {1 \plus{} 5\sqrt {5}}{2}
\]
so we can choose this for $ \alpha$ and it gives ehku's result.

There is equality when $ c \equal{} 0,b \equal{} 1,$ and
\[ a \equal{} \frac {3 \plus{} \sqrt {5} \plus{} \sqrt {6\sqrt {5} \minus{} 2}}{4}
\]
[/hide][/quote]
It's the same as my solution posted above :wink: . Since c>0, you can let c goes to 0(which means c is ver small).It's still the same as c=0!

[quote="Vasc"][quote="ehku"]
The best is
\[ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2} \ge \frac {11 \plus{} 5\sqrt {5}}{2(a^2 \plus{} b^2 \plus{} c^2)}
\]
[/quote]
You are right, but for
 $ \frac 1{(a \minus{} b)^2} \plus{} \frac 1{(b \minus{} c)^2} \plus{} \frac 1{(c \minus{} a)^2} > \frac {5 \plus{} 4\sqrt 2}{a^2 \plus{} b^2 \plus{} c^2}$,

there is a nice solution with AM-GM inequality (after setting $ a \equal{} 0$).[/quote]

Assume $ c\equal{}\min(a,b,c)$, then $ (a\minus{} c)^2 \le a^2,(b \minus{} c)^2 \le b^2, a^2\plus{}b^2\plus{}c^2 \ge a^2\plus{}b^2$, hence
$ (a^2\plus{}b^2\plus{}c^2)(\sum \frac {1}{(a \minus{} b)^2}) \\ \ge (a^2\plus{}b^2)(\frac {1}{(a \minus{} b)^2}\plus{}\frac {1}{a^2}\plus{}\frac {1}{ b^2}) \equal{} \frac{1}{1\minus{}\frac{2ab}{a^2\plus{}b^2}}\plus{}\frac{(a^2\plus{}b^2)^2}{a^2b^2} \equal{} \frac{1}{1\minus{}x}\plus{}\frac{4}{x^2}\equal{}f(x)$
where $ x\equal{}\frac{2ab}{a^2\plus{}b^2}$, $ x \in [0,1)$.
We can check easily that $ \min f(x)\equal{}f(\frac{3\minus{}\sqrt{5}}{2})$.
But if you want a solution by AM-GM:
By AM-GM we have
$ f(x)\equal{}\frac{1}{1\minus{}x}\plus{}a\frac{4}{ax^2} \ge (a\plus{}1)(\frac{4^a}{(1\minus{}x)x^{2a}})^{\frac{1}{a\plus{}1}}$
and
$ (1\minus{}x)x^{2a} \le \frac{(2a)^{2a}}{(2a\plus{}1)^{2a\plus{}1}}$
Hence
$ f(x) \ge (4^a\frac{(2a\plus{}1)^{2a\plus{}1}}{(2a)^{2a}})^{\frac{1}{a\plus{}1}}$.
The inequality holds if and only if $ \frac{1}{1\minus{}x}\plus{}\frac{4}{ax^2}$ and $ 1\minus{}x\equal{}\frac{x}{2a}$, that means $ x\equal{}\frac{3\minus{}\sqrt{5}}{2}$ and $ a\equal{}\frac{1\plus{}\sqrt{5}}{2}$.


It is given that $ a^2\plus{}b^2\plus{}(a\plus{}b)^2\equal{}c^2\plus{}d^2\plus{}(c\plus{}d)^2$. Prove that $ a^4\plus{}b^4\plus{}(a\plus{}b)^4\equal{}c^4\plus{}d^4\plus{}(c\plus{}d)^4$.

Square both sides of the equation $ a^2\plus{}b^2\plus{}(a\plus{}b)^2\equal{}c^2\plus{}d^2\plus{}(c\plus{}d)^2$ and divide both sides by $ 2$ ,then we get the result.

[quote="stephencheng"]Square both sides of the equation $ a^2 \plus{} b^2 \plus{} (a \plus{} b)^2 \equal{} c^2 \plus{} d^2 \plus{} (c \plus{} d)^2$ and divide both sides by $ 2$ ,then we get the result.[/quote]
Umm.. What do we get then? :maybe:  :maybe:  :(

ith_power wrote:
[quote]Umm.. What do we get then?[/quote]

$ \frac {[a^2 \plus{} b^2 \plus{} (a \plus{} b)^2]^2}{2} \equal{} 2(a^2 \plus{} ab \plus{} b^2)^2 \equal{} 2a^4 \plus{} 4a^3 b \plus{} 6a^2 b^2 \plus{} 4a b^3 \plus{} 2b^4 \equal{} a^4 \plus{} b^4 \plus{} (a \plus{} b)^4$
Similarly for $ R.H.S.$


Something occurred to me the other day... supposedly you can qualify for AIME through USAMTS.  But how does the AMC committee determine your USAMO qual index with your USAMTS score?  Is there some special formula?  I know that most people who take USAMTS (specifically, those who do well on USAMTS) also take AMC10/12, but consider those few who don't...

You get 100 on AMC12 if you take the USAMTS and not the AMC test. That's why it's not a very good idea to try to qualify like that...

And rightly so too, noting that USAMTS is a take-home exercise based on the honor system.

Fierytycoon

Indeed, but do you think it would make much of a difference? Suppose it counted as 130, say. You'd still need a 10 on AIME in 2003 to make USAMO, which isn't trivial. I suppose most people who score 10s on AIME make it to USAMO. Meanwhile, with a 100 on the AMC12, a 13 on AIME is necessary, which is very difficult. (I never got that high or even close, even if it was because careless errors kept me back.)

[quote="Fierytycoon"]And rightly so too, noting that USAMTS is a take-home exercise based on the honor system.[/quote]

Honor system or not, 100 is, in my opinion, far too low.  After all, even though you get as much time as you want (within the space of however many weeks, I know...but that's lots of time compared to AMC, AIME, and USAMO), you ought to get credit for actually working on the problems.  I mean, if it takes you 100 hours (this is an exaggeration) to do a round of USAMTS, then you should get credit for working hard...

But then, if you get 100 JUST for participating in one round of USAMTS, then maybe that is justified.

So my advice to anyone who wants to qualify for USAMO via USAMTS... take AMC10/12.  It's a lot easier, a lot faster, and you can get a lot higher score.

[quote]Honor system or not, 100 is, in my opinion, far too low. After all, even though you get as much time as you want (within the space of however many weeks, I know...but that's lots of time compared to AMC, AIME, and USAMO), you ought to get credit for actually working on the problems. I mean, if it takes you 100 hours (this is an exaggeration) to do a round of USAMTS, then you should get credit for working hard... [/quote]

I understand, but the USAMTS and AMC are two different types of math activities. The USAMTS is more of a long-term investigation of interesting math phenomenons while the AMC is a 75-min test on problem-solving. Yes, one should get credit for doing the USAMTS, as they do by means of USAMTS awards given by the NSA people. The credit should not be in the form of AMC points though.

Fierytycoon

[quote="Fierytycoon"]I understand, but the USAMTS and AMC are two different types of math activities. The USAMTS is more of a long-term investigation of interesting math phenomenons while the AMC is a 75-min test on problem-solving. Yes, one should get credit for doing the USAMTS, as they do by means of USAMTS awards given by the NSA people. The credit should not be in the form of AMC points though.[/quote]

But that's exactly my point: if you are working hard on the USAMTS, then that's a signal that you [i]like math and math competitions[/i] and [i]deserve[/i] to be [i]invited[/i] to take the AIME.

The fact is this: the AMC and AIME are two separate exams.  The AMC is a test where, if you distinguish yourself as good at math, you are invited to the AIME.  Similarly, the USAMTS is a test where, if you distinguish yourself, you can get invited to AIME.  Don't think of AMC score as AMC points, because then you get the idea that anything else, like USAMTS, is just a way to "get" AMC points.  There are no such things.  There are just two ways of getting to AIME.

I imagine that in the future, the AIME may live up to its name of [i]Invitational[/i] when some of the AMC people send [i]invitations[/i] to other kids who didn't take the AMC but in some other way distinguished themselves as "good at math."

Also, if you keep in mind that the AMC's intent is to figure out which students should represent the USA at IMO, then all of the minor exams, like AMC, don't matter so much - they're just ways of figuring out who the best math competitors are.  Besides, USAMTS is more like an olympiad than AMC/AIME, so USAMTS participants ought to be able to get into USAMO and IMO.


Suppose that $A$ is an algebra over a non denombrable field that is finitely generated and which is a field. Then it is a finite dimensional vectorial space over that field. What happens when the field is denombrable?

I know this is one useless post, but... you mean countable, right?

Yes, of course : dénombrable (french) = countable. ;)

Sorry, I made two mistakes: I forgot to put accent on e (french variant) and I forgot the adjective countable in english. Mea culpa.  :blush:


Given a triangle with sides a,b,c a nd angles A,B,C, in the usual notation.

Drop the perpindicular from A to a. Draw in the bisector of B. Draw in a median from C to c. Find conditions on a,b,c, so that these three lines intersect at a common point.

It's the old problem about determining when an altitude, an angle bisector and a median of a triangle concur. Look at http://www.mathlinks.ro/Forum/viewtopic.php?t=33634 .

  Darij


We know that if $ A$ and $ B$ are matrices $ n \times n$ over $ F$ then $ (I\minus{}AB)$ is invertible iff $ (I\minus{}BA)$ is invertible, i.e. $ (I\minus{}BA)^{\minus{}1} \equal{} I \plus{} B(I\minus{}AB)^{\minus{}1}A$, this is easy to see doing the computations. The problem is the following: Using that fact show that if $ A$ and $ B$ are matrices $ n \times n$ over $ F$ then $ AB$ and $ BA$ have the same eigenvalues in $ F$.

Follow [url=http://www.mathlinks.ro/viewtopic.php?t=240806]this topic[/url].Using the same arguments from that topic you can easily prove that  they have the same characteristic polynomial.


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

2(y^3)+6(x^2)y-12(x^2)+6y=1

(a) Write an equation of each horizontal tangent line to the curve. Justify your work analytically.

(b) The line through the origin with slope of -1 is tangent to the curve at point P. Find the x- and y-coordinates of point P.

Well as you suggest, use implicit differentiation.  $ 6y^{2}(y')\plus{}12xy\plus{}6x^{2}(y')\minus{}24x\plus{}6y'\equal{}0 \rightarrow y'\equal{}\frac{24x\minus{}12xy}{6y^{2}\plus{}6x^{2}\plus{}6}$.  To find the horizontal tangents, this derivative must equal 0 so work from there.


calculus
integration
complex analysis
complex analysis unsolved

find a closed form for $ \sum_{n\equal{}\minus{}\infty}^{\infty}\frac{1}{(z^3\minus{}n^3)}$

use a partial fraction for the general term

and probably the expansion in series of simple fractions from

the Table integral serie product (fith edition)

Gradshteyn; Ryzhik; Jeffrey

formulaes

1.421
1.422
1.423

will help to get a closed form


symmetry
geometry
geometric transformation
reflection

Source: ACoPS
Find the length of the shortest path from the point $(3,5)$ to the point $(8,2)$ that touches the x-axis and also touches the y-axis.

[/hide]

[hide]
You have to reflect across both the x- and y-axes.
[/hide]

I did reflect across the x-axis.

But do you reflect the two points across the y-axis?

If u reflect across x-axis, then u get a point on the x-axis, right?

Reflect $(8,2)$ over both axises (how do you spell the plural of axis?), leaving the problem to finding the distance between $(3,5)$ and $(-8,-2)$, which by Pythagorean's is $\sqrt{170}$.


Apples are priced at 5 apples for $ \$1.49$ or 8 apples for $ \$2.41$. How many dollars, to the nearest cent, does it cost to buy 40 apples at the better price?

If you buy 5 apples for 1.49, each one is 0.298. 8 apples for 2.41 is about 0.3, so clearly the first deal is better. The cost of 40 apples is therefore $ 8 \times 1.49 \equal{} 11.92$.

Or you could just compare $ 8 \times 1.49$ and $ 5 \times 2.41$.


limit
real analysis
real analysis unsolved

$1\blacktriangleright$ $\{\begin{array}{c}n^{2}\cdot(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2)\rightarrow-\frac{1}{4}\\\\ n^{2}\cdot(\sqrt [p]{1+\frac{1}{n}}+\sqrt [p]{1-\frac{1}{n}}-2)\rightarrow-\frac{p-1}{p^{2}}\mathrm{\ ,\ where\ }p\in N\ ,\ p\ge 2\end{array}$

$2\blacktriangleright \underline{\mathrm{Generally}}\mathrm{\ (without\ the\ l'Hospital's\ rules)\ : }$

$\{p,n\}\subset N\ ,\ p\ge 2\ ,\ n\ge 2\ ,\ \sum_{k=1}^{n}a_{k}=0$ $\Longrightarrow$ $\lim_{x\to 0}\frac{1}{x^{2}}\cdot(\sum_{k=1}^{n}\sqrt [p]{1+a_{k}x+b_{k}x^{2}+\ldots }-n)=\frac{1}{p}\cdot\sum_{k=1}^{n}b_{k}-\frac{p-1}{2p^{2}}\cdot\sum_{k=1}^{n}a^{2}_{k}\ .$

I would do this

$(1+x)^{1/p}\approx 1+\frac{x}{p}+\frac{1}{p}\left(\frac{1}{p}-1\right)\frac{1}{2}x^{2}$
thus

$\sqrt[p]{1+a_{k}x+b_{k}x^{2}}\approx 1+(a_{k}x+b_{k}x^{2})\frac{1}{p}+\left(a_{k}x+b_{k}x^{2}\right)^{2}\frac{p(1-p)}{2p^{2}}$

Adding these n roots and substractind n we get that

$S\approx\sum_{k=1}^{n}b_{k}x^{2}\frac{1}{p}+\frac{p(1-p)}{2p^{2}}x^{2}\sum_{k=1}^{n}(a_{k}+b_{k}x)^{2}$, where $S$ is the sum of those n roots.

Dividing by $x^{2}$ and taking the limit as $x$ approaches 0 the desired result follows.
The other two limits can be proven by using a similar technique.

Without l'Hospital's rules ! Taylor's means at least l'Hospital's !

I'll never understand what Taylor series and L'Hospital ever did to you, that you so fiercely oppose them.... :?:

Probably it's more exciting to write pages of horrible fractions and rationalisations than to use some things which are really useful, nice and not at all complicated. Anyway, we have a tradition in Romania to do that...

[quote="harazi"][color=darkblue]Probably it's more exciting to write pages of horrible fractions and rationalisations than to use some things which are really useful, nice and not at all complicated. Anyway, [b][u]we have a tradition in Romania to do that...[/u][/b][/color][/quote]

In the my opinion, using the l'Hospital's rules and the Taylor's series, 80% from the exercises (limits) of this section are very simply, noninteresting and ... unsightly.

In my humble opinion, as long as they reduce to the simple computation of limits, they remain uninteresting and simple. Just imagine the huge "literature" written in our magazines by some persons and which says absolutely nothing, is completely sterile and with no interest. Not to mention that it is absolutely trivial using asymptotic estimates. People should compute limits in order to apply the results to some more enlightening problems, not just to develop some useless automathisms.

[quote="harazi"]In my humble opinion, ... not just to develop some useless automathisms.[/quote]

I didn't understand why "humble opinion". Yoy are right : the l'Hospital's and Taylor's are also automathisms, but ... [u]usefully[/u] !


Can someone give an example of inequalities that can be proved using Holder's Inequality, Schur's Inequality, and Muirhead Inequality?

Thanks.

Muirhead can be done with AM-GM instead [b]ALWAYS[/b], and as [url=http://web.mit.edu/~tmildorf/www/Inequalities.pdf]Thomas Mildorf[/url] suggests, it's preferable to use AM-GM.
Holder, it's just somekind generalized Cauchy-Schwarz...

Here are some very easy examples:

[hide="Muirhead"]
$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\ge8abc$[/hide]
[hide="Schur"]
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc\ge a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}ca^2\plus{}ac^2$[/hide]

[quote="Bugi"]Muirhead can be done with AM-GM instead [b]ALWAYS[/b], and as [url=http://web.mit.edu/~tmildorf/www/Inequalities.pdf]Thomas Mildorf[/url] suggests, it's preferable to use AM-GM.
[/quote]
[hide="muirhead"]$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \ge abc(a\plus{}b\plus{}c)$[/hide]
[hide="holder"]$ (a\plus{}1)(b\plus{}1)(c\plus{}1) \ge (1\plus{}\sqrt[3]{abc})^3$[/hide]


limit
inequalities
induction
real analysis
real analysis unsolved

Squaring, qsumming and using Cauchy-Schwarz,, we get :

with $S_n = \sum_0^n u_k$
$S_n + ln(n) + O(1) \geq S_n^2 / n^2$
or $n + O(1) \leq S_n \leq n + ln(n)/n + O(1)$

This proves that $u_n$ must be close to $1$ for some $n$ we can take as large as we want.

Be if $n$ is large enough and $u_n$ close to $1$ enough $1 \leq u_{n+1} \leq u_n$

$u_n \geq 1$ becomes decreasing for $n$ enough so it converges and its limit must be $1$

------------

Is that enough :P  ?

[quote="tµtµ"]

with S_n = \sum_0^n u_k
S_n + ln(n) + O(1) \geq S_n^2 / n^2
or n + O(1) \leq S_n \leq n + ln(n)/n + O(1)

This proves that u_n must be close to 1 for some n we can take as large as we want.

Be if n is large enough and u_n close to 1 enough 1 \leq u_{n+1} \leq u_n

u_n \geq 1 becomes decreasing for n enough so it converges and its limit must be 1

[/quote]

I'm not so sure that a such proof is accepted for the Oral test  :D
Anyway ...

Let $M$ be such that $M= max(2,u_1)$ , so $\sqrt{M+\frac 1n} < M$ , so $1< u_n < M, \forall n > 1$.
Let $A$ be the adhesion point set of $u_n$ (adhesion point = limit of a convergent sub-sequence). Manifestly $A$ is a non empty bounded set ..
I denote , $I=\inf(A) , S =\sup(A)$ , so for all $a\in A$ I can write : $1 \leq \sqrt{I} \leq a \leq \sqrt{S}$.
$a = S \Longrightarrow S \leq \sqrt{S} \Longrightarrow S \leq 1$ .Finally $I=S=1$ , so $u_n$ converges to $1$

 :cool:  Do you follow me ?  :cool:

Since $n\geq 1$ we can show that $u_n\geq 1$ by induction.Thus using the convexity of $y=\sqrt{x+1}$,

we have $u_{n+1}=\sqrt{u_n+\frac{1}{n}}\leq \sqrt{u_n+1}\leq \frac{1}{2}(u_n+1)\Longleftrightarrow u_{n+1}-1\leq \frac{1}{2}(u_n-1)$, therefore

$0<u_{n}-1\leq \frac{1}{2}(u_{n-1}-1)\leq \left(\frac{1}{2}\right)^2 (u_{n-2}-1)\leq \cdots\leq \left(\frac{1}{2}\right)^n (u_0-1)$,yielding $\lim_{n\to\infty}(u_n-1)=0$ $\therefore \lim_{n\to\infty}u_n=1$

[quote="kunny"]$\sqrt{u_n+1}\leq \frac{1}{2}(u_n+1)$[/quote]
How do you prove this ?

If $u_i\in(-1,3)$, that inequality is reversed.

[quote="Diogene"]
I'm not so sure that a such proof is accepted for the Oral test  :D
[/quote]
with some $\epsilon$ arounds ....

[quote="Diogene"]
Do you follow me ?
[/quote]
I'm not sure you would have a better rank  ;)  :P

[quote="wisemaniac"]Study the sequence defined by $u_0$ and the following relation :
\[ u_{n + 1} = \sqrt{u_n + \frac{1}{n}}  \][/quote]

Let see for $u(0)>0$

By induction $u(n)\geq 1$ for any $n \geq 1$

$u(n+1)\leq \sqrt{u(n)}+\frac{1}{\sqrt{n}}$

for $x\geq 0$ we have $\sqrt{x}\leq \frac{1}{2}(x-1)+1$

$0\leq u(n+1)-1\leq \frac{1}{2}(u(n)-1)+\frac{1}{\sqrt{n}}$

We deduce that

$0\leq u(n+1)-1\leq \frac{u(1)-1}{2^n}+a(n)$  where $a(n)=\sum_{k=0}^{n-1}\frac{1}{2^k\sqrt{n-k}}$

Since $a(n)$ tends to zero see Kent proof http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57157

We have $u(n)$ tends to $1$

Here an extra question

Compute $\lim_{n\rightarrow + \infty} u(n)^{n}$