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2005 has been a pretty fun year for AoPS, so, in the spirit of my mildly successful AoPS Flash/Art Contest <[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35620[/url]>, I'm starting a new contest: the AoPS Best of 2005!

This contest is composed of five categories:

[b]Topic of the Year[/b]
(Eligible: all topics started in 2005)

[b]Post of the Year[/b]
(Eligible: all [i]interesting, off-topic, or just plain hilarious[/i] posts started in 2005)
[b]
Blog of the Year[/b]
(Eligible: all blogs hosted by AoPS-Mathlinks and started in 2005)

[b]Blog Post of the Year[/b]
(Eligible: all blog posts written in 2005)

[b]AoPSer-Mathlinker of the Year[/b]
(Eligible: all members who joined and were active {100+ posts} during 2005)

In each category smileys will be awarded to the winner (:first:), second-place finisher (:winner_second:), third-place finisher (:winner_third:), and two runner-ups (:w00tb: :w00tb: ).

[b]Here is a projected timeline:[/b]
01/4/06 - 02/1/06 : Nominations
02/2/06 - 02/16/06 : Voting for Topic of the Year
02/17/06 - 03/3/06 : Voting for Post of the Year
03/4/06 - 03/18/06 : Voting for Blog of the Year
03/19/06 - 04/2/06 : Voting for Blog Post of the Year
04/3/06 - 04/17/06 : Voting for AoPSer-Mathlinker of the Year
04/18/06 : "Prizes" (Smileys) awarded to all 25 winners (5 in each category - see above)

I am currently writing a post describing the nomination process. It will be up in a few minutes. After that the contest will begin. Good luck!

[size=150]Rules for Nominations:[/size]

[list]* All nominations must be "seconded" (funny term to use here) by two people to be official. This is done to limit the number of nominees, as otherwise there would be way too many.[/list]

[list]* You may not nominate yourself, or anything written by yourself.[/list]

[list]* You are limited to one nomination in each of the five categories, but you may second up to three nominees in each category.[/list]

[list]* Please put all your nominations and "second-ations" in one post. If you change your mind, edit your post instead of re-posting.
[/list]
[list]* All posts should be of the form:

I nominate "title" by "author" <link> for Topic of the Year.
I nominate "first few words of post" by "author" <link> for Post of the Year.
I nominate "title" by "author" <link> for Blog of the Year.
I nominate "title" by "author" <link> for Blog Post of the Year.
I nominate "member" <link> for AoPSer-Mathlinker of the Year.
I second ...'s nomination of ... by ... for ... .

For example: I nominate AoPS: Best of 2005! by NoSoupForYou <http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68624> for Topic of the Year.[/list]

---
Well, that's about it! The contest has now begun - I am accepting nominations for all five categories!

EDIT: Instead of editing your posts if you change your mind, just make a new post. That way it will be easier for me to see if you change your mind.

[b][size=200]Current Nominations[/size][/b]

[i](A $\triangleright$ indicates an official nomination. A nomination becomes official once it is seconded by two people.)
[/i]
[u][size=150]Topics[/size][/u]

\begin{tabular}{|c|c|c|c|c|} \hline Topic & Author & Nominated by & Seconded by\\ \hline Problem Solving Marathon 2 & Silverfalcon & NoSoupForYou & \\ \hline iTest poorly written? & GuitarGeek143 & beta & \\ \hline Anyone Anxious Yet? & L\_Li & Sly Si & \\ \hline Nothing to do with this crazy math & arinthros & LadyKn1ght & ccy\\ \hline two 2s make 5---for all of u who like challenges & flamesofdestiny & biffanddoc & ccy\\ \hline\end{tabular}

[b]Links:[/b]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33505]Problem Solving Marathon 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52739]Test poorly written?[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31459]Anyone Anxious Yet?[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48232]Nothing to do with this crazy math[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48329]two 2s make 5---for all of u who like challenges[/url]

[u][size=150]Posts[/size][/u]

\begin{tabular}{|c|c|c|c|c|} \hline Post & Topic & Author & Nominated by & Seconded by\\ \hline \triangleright\text{ (Post \#20)} & iTest poorly written? & bubala & beta & Sly Si, joml88\\ \hline (Post \#51) & iTest poorly written? & bubala & joml88 & \\ \hline (Post \#2) & Another chapter level question & robinhe & Treething & \\ \hline \end{tabular}

[b]Links:[/b]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=332154#p332154]Post #20 in [i]iTest poorly written?[/i][/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=335016#p335016]Post #51 in [i]iTest poorly written?[/i][/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=409970#p409970]Post #2 in [i]Another chapter level question[/i][/url]

[u][size=150]Blogs[/size][/u]

\begin{tabular}{|c|c|c|c|c|} \hline Blog & Author & Nominated by & Seconded by\\ \hline \triangleright\text{ the Search for Intelligence Continues. . . }& rrusczyk & NoSoupForYou & joml88, -Terminator-\\ \hline Billy's Blog & solafidefarms & LadyKn1ght & \\ \hline\end{tabular}

[b]Links:[/b]
[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1]the Search for Intelligence Continues[/url]
[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=34]Billy's Blog[/url]

[u][size=150]Blog Posts[/size][/u]

\begin{tabular}{|c|c|c|c|c|} \hline Blog Post & Blog & Author & Nominated by & Seconded by\\ \hline Hypnotism! & NoSoupForYou's blog & NoSoupForYou & -Terminator- & \\ \hline File Sharing & the Search for Intelligence ... & rrusczyk & solafidefarms & beta\\ \hline Amusing Google search & the Search for Intelligence ... & rrusczyk & Sly Si & \\ \hline Interesting AIM Conversation & My Mathematical Experience & beta & joml88 & \\ \hline \end{tabular}

[b]Links:[/b]
[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3648]Hypnotism![/url]
[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3477]File Sharing[/url]
[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=2016]Amusing Google search[/url]
[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3903]Interesting AIM Conversation[/url]

[u][size=150]AoPSers/Mathlinkers[/size][/u]

\begin{tabular}{|c|c|c|c|c|} \hline AoPSer/Mathlinker & Nominated by & Seconded by\\ \hline chess64 & NoSoupForYou & PenguinIntegral\\ \hline \triangleright\text{ shobber} & eminem & Arne, NoSoupForYou\\ \hline \triangleright\text{ 236factorial} & PenguinIntegral & -Terminator-, ccy\\ \hline \triangleright\text{ SamE} & solafidefarms & joml88, probability1.01\\ \hline Virgil Nicula & joml88 & \\ \hline shadysaysurspammed & smellyfart & \\ \hline JesusFreak197 & h_s_potter2002 & \\ \hline djimenez & eryaman & Scrambled\\ \hline shadysaysurspammed & shyong & \\ \hline\end{tabular}

[b]Links:[/b]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=6912]chess64[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=6601]shobber[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=7080]236factorial[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=8426]SamE[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=9882]Virgil Nicula[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=10016]shadysaysurspammed[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=8525]JesusFreak197[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=6641]djimenez[/url]
[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=9928]shyong[/url]

And now to start things off:

I nominate [i]Problem Solving Marathon 2[/i] by Silverfalcon <[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33505[/url]> for Topic of the Year.

I nominate [i]the Search for Intelligence Continues. . .[/i] by rrusczyk <[url]http://www.artofproblemsolving.com/Forum/weblog.php?w=1[/url]> for Blog of the Year.

I nominate chess64 <[url]http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=6912[/url]> for Member of the Year.

PS Sorry for the four posts in a row, but I felt that they are all necessary posts to make. Mods, you may deduct these from my post count if you wish.

Edit #1: I second shobber for AoPSer/Mathlinker of da Year.

[quote="shadysaysurspammed"]I think the award for best AOPSER is already given away([/quote]

so who's that?

[quote="eminem"]i nominate shobber for Aopser of the year

second:- G-UNIT
third:- frt[/quote]

You can only nominate one person for AoPSer of the Year, so I'm assuming that you're nominating shobber.

[quote="shadysaysurspammed"]I think the award for best AOPSER is already given away(and rightly given away)[/quote]

I'm not sure what you're talking about.  :P

[quote="PenguinIntegral"]I second chess64 and nominate 236factorial as spammer of the year. I nominate the best thread to be, collectively, all the problem solving marathons run in 2005. Those are the only thread ever reallyed payed attention too.[/quote]

By spammer of the year I assume you mean AoPSer of the year.  :D

You cannot nominate more than one thread for Topic of the Year.

[quote="math92"]you should make an award for spammer of the year... :D[/quote]

Hmm.... I wonder who would get it...

Why can't all the voting be done at the same time?

Umm, this isn't AoPS anymore...it's AoPS-Mathlinks ;)

[quote="joml88"]Why can't all the voting be done at the same time?[/quote]

I feel it's better if nominations are done before voting, so the voters will have a list of options to choose from.

[quote="joml88"]Umm, this isn't AoPS anymore...it's AoPS-Mathlinks ;)[/quote]

Right... Gotta be politically correct... *changing all my posts*

EDIT: OK, done. Is a "mathlinker" someone who goes on mathlinks? :P
joml88, aren't you going to make any nominations?

No, I mean why do we vote for one thing for a couple weeks, then another for the next couple weeks and so forth?

Yes, I'm gonna make nominations.  I don't have nominations for all of them yet but I'll edit this post with them:

AoPS-Mathlinker of the year: Virgil Nicula
Blog Post of the Year: http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3903
Post of the Year: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=335016#p335016

I second Blog of the year: rrusczyk's
I second SamE as ML/AoPS er of the year.
I second beta's choice for the post of the year (the AHSIMC thing).

[quote="joml88"]No, I mean why do we vote for one thing for a couple weeks, then another for the next couple weeks and so forth?[/quote]

I think it's easier that way - otherwise a bunch of people would all be voting for a bunch of different things at the same time, and it would be really confusing.

I vote for
Post #20 in iTest poorly written

I vote for Post #51 in iTest poorly written

[quote="solafidefarms"]Can I nominate and vote for http://www.artofproblemsolving.com/Forum/viewtopic.php?p=335315 ?[/quote]

The Topic of the Year contest is over. Sorry.

Wow, this topic seems pretty dead...

Voting for Post of the Year ends in two days!

And [b]Post #20 in [i]iTest poorly written?[/i][/b] wins!

[quote="solafidefarms"]Blogs next? I vote for my blog. :D[/quote]

Err... I don't think you can.  :D

Wow, instead of making a new post for blog of the year voting I replaced the old one... Just noticed that...  :(  Oh well.

[quote="easyas3.14159..."]By the time this voting is finished it'll be 2007.[/quote]

 :D  :D  :D

But seriously, no, it won't. (See the first post for a schedule.)

Woah I just searched for something and then I see this thread and i find out i got nominated - and seconded - for aopser of the year... :huh:


Prove that there isn't a function $f:\mathbb{R}\to [a,\infty)$,$a\in\mathbb{R}$ , monotonic(increasing or descreasing) and bounded. ;)

I think this problem must be stated strictly monotone.
we just consider the case strictly increasing.
Here is my solution:
If this function exists, then we have $f(x)<M$ with integer $M$ ( because it assumes that the function is bounded)
Of course, we take $x_{0}$.
we have $f(x_{0})=m$.  and denote $p=M-m$
Because the function is strictly increasing,
so we can find such $x_{1}>x_{0}$, with $f(x_{1})>m+p=M$
But this contradicts the assumption.

Rmk: This is not complete proof. Just my idea.

This is trivially false as stated; consider $a+1\pm\tanh x$.

Is there something else to the question?


real analysis
calculus
integration
LaTeX
logarithms
function
real analysis solved

Sorry guys , I dont know whow to use Latex Code

If $a >1 $
Prove that Intergral \[\int_a^\infty x^p \ln^q x\,dx\] exist , both as an improper Riemman Integral and as Lebesgue Integrel for all $q$ if $p<1$, or $q< -1$ if $p =1$.
Thank u again

[i]Edited by Myth[/i]

You can easily to learns $\LaTeX$ basic commands from examples on this site.

BTW, I think you meant the integral exists for all $q$ with $p<-1$, and for all $q<-1$ with $p=-1$.
For the first case you are to use Wierstrass, and for second case you can easily find primitive.

One more thing, you posted the problem to the Solved Section, though the proper section for it is Unsolved Problems.

Myth Thank u very much , Can you show me how to use Wierstrass for the first case .Indeed u are right  for the first case for all q and p<-1.
Also how to show that integral is  Lebesgue Integrel

We know that $\ln x$ grows slower than any positive power of $x$, it means that we can find $C>0$ s.t. $\ln x\leq Cx^{\frac{-p-1}{2q}}$, then $x^p\ln^q x\leq C^qx^p\cdot x^{\frac{-p-1}{2}}=C^qx^{\frac{p-1}{2}}$. Note that $\frac{p-1}{2}<-1$, hence $\int_2^\infty x^{\frac{p-1}{2}}dx$ is finite.
That's all.

Your function is continuous, so it is measurable and it is bounded by summable function. It follows that the function is summable.


how can we prove that the set R is uncountanle?


If you need help with latex check this out: [url]http://kwiktex.com[/url]
This is especially helpful to practice or paste in your word or powerpoint documents

Also try the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]AoPS TeXer[/url]


linear algebra
matrix
vector
quadratics
complex numbers
linear algebra unsolved

Let $ A \in M_{n \times n}(\mathbb{R})$. Prove that, if A is positive definite then

$ \left| \begin{array}{cc} A & x \\
x^T & 0 \end{array}\right| < 0$

for every non zero vector $ x \in \mathbb{R} ^n$.

Suppose $ \det(B)\ge 0$. Then $ B$ is positive semi-definite and there is a decomposition $ B \equal{} U^TU$ (*) with $ U \equal{} \left(\begin{array}{cc}C & y \\
0 & \alpha\end{array}\right)$ where $ C\in M_{n \minus{} 1}(\mathbb R)$, $ y\in\mathbb R^{n \minus{} 1}$, $ \alpha\in\mathbb R$. The relation (*) implies $ y^Ty \plus{} \alpha^2 \equal{} 0$, hence $ y \equal{} 0$ and $ \alpha \equal{} 0$. Contradiction to $ 0\neq x\in\mathbb R^{n \minus{} 1}$.

BTW: The statement holds over the complex numbers as well if the hypothesis is amended to $ \left|\begin{array}{cc}A & x \\
\overline{x}^{T} & 0\end{array}\right| < 0$.

I took this to my local matrix guru and came up with the [i]Cauchy interlacing theorem[/i]: If $ A$ is a principal submatrix of a Hermitian matrix $ B,$ then the eigenvalues of $ A$ interlace the eigenvalues of $ B.$

So in our case, we let $ B\equal{}\begin{pmatrix}A&x\\x^*&0\end{pmatrix}.$ Let $ \mu_i$ denote the eigenvalues of $ B$ and $ \lambda_i$ denote the eigenvalues of $ A.$

Then $ \mu_1\ge\lambda_1\ge\mu_2\ge\lambda_2\ge\cdots\ge\mu_n\ge\lambda_n\ge\mu_{n\plus{}1}.$

Since $ A$ and $ B$ have the same trace, $ \sum_{i\equal{}1}^{n\plus{}1}\mu_i\equal{}\sum_{i\equal{}1}^n\lambda_i.$

Rearrange that to $ \mu_{n\plus{}1}\equal{}\sum_{i\equal{}1}^n(\lambda_i\minus{}\mu_i)\le 0.$

But $ \mu_{n\plus{}1}$ cannot be zero, since $ B$ is invertible. Hence $ \mu_{n\plus{}1}<0.$ Since the other $ \mu_i$ are all positive, being at least as great as the corresponding $ \lambda_i,$ we have that $ \det(B)\equal{}\prod_{i\equal{}1}^{n\plus{}1}\mu_i<0.$

Why can't $ B$ be singular? Let's examine its null space:

If $ \begin{pmatrix}A&x\\x^*&0\end{pmatrix}\begin{pmatrix}u\\t\end{pmatrix}\equal{}\begin{pmatrix}0\\0\end{pmatrix}$ then

$ x^*u\equal{}0$ and $ Au\plus{}tx\equal{}0.$ Then $ 0\equal{}u^*(Au\plus{}tx)\equal{}u^*Au\plus{}tu^*x\equal{}u^*Au$ since $ u^*x\equal{}\overline{x^*u}\equal{}0.$

Since $ A$ is positive definite, $ u^*Au\equal{}0$ implies $ u\equal{}0.$ That in turn implies that $ tx\equal{}0,$ so $ t\equal{}0$ since we were assuming $ x\ne 0.$ So the null space of $ B$ is trivial and $ B$ cannot have zero as an eigenvalue.

One can actually directly expand it out.
the required determinant = $ 0\, det A \minus{} x^T (adjA) x \equal{} \minus{}x^T(adj A) x < 0$
where $ adj A$ is the classical adjoint is also positive definite.

More generally,
$ det\left(\begin{array}{cc}A & v \\ u & a\end{array}\right) \equal{} a \,detA \minus{} u (adj A) v$
(Reference: Prasolov's linear algebra p.29 theorem 3.1.3)

That's right. And from that expansion one gets what we might call [i]the determinantal representation of a quadratic form[/i] ($ x \in \mathbb{R} ^n$):

(1) if A is non-singular, then $ x^TAx \equal{} \minus{} \det(A) \left|\begin{array}{cc}A^{\minus{}1} & x\\ x^{T}& 0\end{array}\right|$;

(2) if A is singular, then $ x^TAx \equal{} \minus{} \left|\begin{array}{cc} \widehat{A} & x\\ x^{T}& 0\end{array}\right|$, where $ \widehat{A}$ is the matrix such that $ adj(\widehat{A}) \equal{} A$.

The answer to the problem I posted above follows immediatly from (1).

[quote="Third Edition"]
BTW: The statement holds over the complex numbers as well if the hypothesis is amended to $ \left|\begin{array}{cc}A & x \\
\overline{x}^{T} & 0\end{array}\right| < 0$.[/quote]
  General with $ A \in M_n{\mathbb{(C)}}$ are Hermitian matrix and positive define and let for every non zero matrix $ X \in M_{nm}\mathbb{(C)}$ Then :
1) If $ m$ is odd then $ \left|\begin{array}{cc}A & X \\
X^{*} & 0\end{array}\right| < 0$.
2) If $ m$ is even then $ \left|\begin{array}{cc}A & X \\
X^{*} & 0\end{array}\right| > 0$
 Here $ X^{*}\equal{}\overline{X}^T$ Agree ?


Columbia
University of Chicago
Harvard
college
MIT
Putnam
Princeton

I'll be starting my freshman year as an undergraduate this fall. My goal is to get into a good graduate school in four years. Seeing as how different graduate school and undergraduate school is, I would guess that the admissions requirements would be much different. For example, I doubt a graduate school admissions committee would care much if I were the president of the chess club at my college. Besides maintaining good grades, especially in my field of interest, what else should I be doing to make sure that in four years, I'm not saying, "Shoot, I should have done that, but now it's too late" ?

good references are really important, and it's better if you can get them from people who are well-known in the field you're interested in for grad school.  summer research/summer programs (as early as your first or second summer if you can swing it) are pretty much essential these days.  GRE scores give a small boost if they are good, and can disqualify you if they are really bad.

Basically, you should take at least a couple graduate courses (but don't rush it too much - you want to do well in all of your courses), do some research, and develop relationships with professors so that you can get good references.

Not every school even offers the chance at a senior thesis.  I would doubt it unless you do the REU at the school you're already attending and the professor would continue working on it with you.

[quote="calc rulz"]Does work in an REU ever turn into one's Senior Thesis?[/quote]  No reason why it shouldn't, but obviously this depends far more on individual circumstances than on general principles.

Two issues that deserve mention:
1) Most people (even those seriously planning on going to grad school) don't actually end up going to grad school.  You should keep an open mind about other sorts of things to do with yourself.
2) Many people are prone to constantly planning for the "next step," whether that's college, grad school, or whatever.  You're not in college yet -- you don't have to be planning for what you're going to do [i]after[/i] college.  In the particular case of grad school versus undergrad applications: one major difference between the two is that grad school is a far more focused endeavor than a liberal arts education.  You're almost certainly not in a position yet to have any idea what you want out of a graduate school.  So, be patient.

As a continuation of this topic i was wondering how the undergrad school affects grad school admissions. Currently, I have been accepted to Cornell and UChicago, but I will probably end up at Cornell because of money. I plan on studying math at Cornell, and I am also curious about physics. If I majored in physics rather than math how would this affect grad school admissions for math at places like Harvard, Pton, Caltech, MIT?
      Also, how good is the math program at cornell? I can't seem to get an accurate, non-biased opinion on this. The rankings don't really help because i'm not quite sure how the rankings are determined--personally, I find rankings dubious and unreliable once the schools hit the top 20 or so.
Thanks ahead of time

Don't you find it somewhat doubtful that you would be so interested in physics that you would prefer to major in physics than math, but not prefer to go to graduate school in physics than math?

Cornell has an excellent math department.  Given the size of the school, you may have to take initiative to take full advantage of it (e.g., to get to know professors well), though.

I agree with pretty much everything JBL said, but I will add that math department graduate admissions is pretty much all about recommendations, grades and coursework, and GRE scores.  Research experience mostly comes into play in the recommendations category.  Putnam rankings might give you a little boost, but your mileage will vary as to how much of a boost, if any, you'll get.

But I'll offer a stupid-sounding piece of advice, which is that if your goal is to be a mathematician, you should just focus on learning math, and the rest will fall into place.  I mean this in the most honest way possible.  The whole point of wanting to go to a top math grad school is to further your mathematical career, but the best way to further your mathematical career (at any stage of life), is to actually DO math.

To prove that I'm being honest rather than just idealistic/cheesy, I will point out that this advice is NOT relevant for applying to college.  Wanting to go to a top college is a much more complicated desire than wanting to go to a top grad school.  In other words, there is a wide spectrum of respectable reasons for wanting to attend Harvard College, and in order to achieve that goal you may have to work the system a bit.  But there is only one reason for wanting to get a PhD from the Princeton math department--because you think it will be good for your mathematical education/career.  Therefore what the admissions committee wants out of you should be nearly the same thing that you want out of yourself.

(Reading that over again, I may or may not be making sense.   Others will have to let me know.)

Another bit:  There's nothing weird about getting an undergraduate degree in physics and then a graduate degree in math.  What would be slightly weird would be *planning* to do this from the beginning.

Finally, JBL didn't really address LightEntropy's very reasonable question about how your undergraduate institution affects graduate school admissions.  I feel confident that no one can really know the answer to this question.  The more answerable question is how your undergraduate institution can affect your mathematical education, but the answer is still complicated.

[quote="JBL"][quote="blahblahblah"]summer research/summer programs (as early as your first or second summer if you can swing it) are pretty much essential these days.[/quote]  Well, there are certainly counter-examples -- I never attended an REU or similar program, for example.  But it is certainly true that they are a useful credential, and (more importantly) a more realistic introduction to what the actual work of a graduate student/professor is like.  (Well, modulo teaching duties ;).)[/quote]

you don't have to formally do an REU but you pretty much have to do *something* during the summers, since it reflects very negatively on you if it appears that you have done nothing during the three or four summers of your undergraduate education.  even working with a professor informally is very helpful for recommendations, etc.

[quote="JRav"]My own adviser told me that he spoke to a friend at Columbia who says that to him, REU's are meaningless (of course, one school).[/quote]

well, they aren't meaningless.  exactly how much value is associated with them will vary but they certainly have some value, if (as mentioned earlier) for no other reason than the references you will probably get through completing them.  a reference that can comment on your ability to do original research (even if it is conceptually shallow as most REUs are) is much more valuable than a reference that says that 'student x performed at y level in my introductory real analysis course'.

[quote="yenlee"]Finally, JBL didn't really address LightEntropy's very reasonable question about how your undergraduate institution affects graduate school admissions. I feel confident that no one can really know the answer to this question. The more answerable question is how your undergraduate institution can affect your mathematical education, but the answer is still complicated.[/quote]
Cornell and Chicago are both known for their rigorous course loads.  No grad school is going to look at a Chicago/Cornell transcript and say, "It's good, but I'm not convinced student X had a hard enough work load."

@yenlee
Your response does make sense and i thank you, JBL, and Jrav for your contributions. I understand how weird a question i posed: physics then math. I think i am confused as to what I want to do currently. Ever since sophomore year in HS i have been contemplating a career in mathematics. However recently, as my question indicates, i have been thinking of physics as well. Perhaps the first year of college can help me narrow the choice down, but as of right now i'm not as keen on math as i used to be.

More to the point, i seem to be disillusioned by math lately; i was hoping calculus would be awesome this year but so far it's been pretty easy and boring. Don't get me wrong, some of the topics have been interesting and challenging, but on the whole it's kind of dull. I'm hoping that math will revert back to the fun-loving AOPS kind of math which i loved. However, even AOPS math hasn't been able to attract me in the same way to pull myself to study it on my own; perhaps a result of a busy schedule coupled with senioritis. Additionally, I fear that a pure math degree will place severe limitations on job options as it isn't-in my perception- as valuable as the other degrees. I know many of you might have some qualms with the previous statement, but i would argue that a financially sound future supersedes passion. Perhaps someone here can put my fears to rest?
Thanks in advance

[quote="LightEntropy"]Additionally, I fear that a pure math degree will place severe limitations on job options as it isn't-in my perception- as valuable as the other degrees. I know many of you might have some qualms with the previous statement, but i would argue that a financially sound future supersedes passion. Perhaps someone here can put my fears to rest? [/quote]

see e.g. [url=http://online.wsj.com/article/SB123119236117055127.html]this article[/url]; i don't necessarily put too much faith in the methodology used but the point is that a math degree is probably more valuable than just about any other degree, especially if you pair it with a fair bit of computer science (or even double major).  i think the distinction between pure math and applied math is a bit artificial; courses like analysis or PDE or probability have many applications and would be very useful for non-academic jobs but are really 'pure math'.

Note that the top [b]three[/b] jobs in blahblahblah's article are all math or direct offsprings of math, and several of the other top 15 jobs (such as economist, physicist).

@LightEntropy:  I was a math and physics major who didn't decide about math vs physics grad school until junior year at least.  The fact is that right now, you don't know enough to make the decision.  Real proof-based college math is absolutely nothing like high school calculus, and it's also very different from competition/AoPS-type stuff, so you'll have to wait and see how you like it.  On the other hand, college physics is not as qualitatively different from AP Physics.  (However, imho nothing quite has the same beauty as mechanics.  It's too bad one only gets to learn it once or twice at most.)  But to make matters worse, college courses are still very different from research, with physics courses even more divorced from research-style thinking than math courses (especially experimental physics).  So again, wait and see.  In fact, a common scenario is that you won't go to grad school for either subject.

I think someone said this already, but Cornell and Chicago are both great places for both math and physics, so you can't really go wrong.

[quote="yenlee"]On the other hand, college physics is not as qualitatively different from AP Physics. (However, imho nothing quite has the same beauty as mechanics. It's too bad one only gets to learn it once or twice at most.) But to make matters worse, college courses are still very different from research, with physics courses even more divorced from research-style thinking than math courses (especially experimental physics).[/quote]
Classical mechanics and electricity and magnetism is pretty similar to AP Physics C, although harder.  After that, it gets much more difficult pretty quickly.  Quantum mechanics is unlike any other physics course before it.
[quote="yenlee"]Real proof-based college math is absolutely nothing like high school calculus, and it's also very different from competition/AoPS-type stuff, so you'll have to wait and see how you like it.[/quote]
This is very true.  I found that high school math (especially AP Calculus), at my school at least, emphasized remembering a formula or technique and then applying it to very similar problems.  Techniques, not theorems, are emphasized.  For example, in most AP Calculus courses, the teacher will present the fundamental theorem of calculus or derivative tests and then say, "Here are some problems.  Solve them."  In college math, the professor will present the important theorems, prove them, and say "Here's a statement.  Prove it or disprove it," and all you have to work with are theorems.  It's really rewarding when you complete a proof, but usually pretty difficult.

Two questions:
Do grad admissions (in US)care much about social activities,music skills like undergrad admissions
I'm just a sophomore,and my school has'nt lots of famous proffesor,so how can I get a good recommendation?

[quote="nkht-tk14"]Do grad admissions (in US) care much about social activities,music skills like undergrad admissions[/quote]  No.

For the second question, "being famous" isn't a precondition for "writing a good recommendation."  "Knowing you fairly well" and "having worked with you in some more interactive context than just grading your problem sets" are much more important.

JBL is mostly right--what matters is having recommendations from people who can say something meaningful about you.  But it definitely matters who writes the recommendations.  This isn't for any evil reason.  It's just that people who read rec letters (not me) will always tell you that every letter says that this person is great and that person is great, so knowing the letter writer is one way of making meaningful distinctions.  And of course, I second the contention that no grad school cares about your social activities.

nkht-tk14, where do you go to school?  It matters because the most useful recommendations are ones that make comparisons.  If a prof says, "Oh, this guy is better than students I've had before who have gone on to Harvard grad school and done well," then that means something.  If you go to an international school that no one has heard of, it's always tough.  You'll essentially need people to announce that you are awesome to get into a top school.  But at lower-tier schools, a good rec and high GRE scores might see you through.  Or what happens occasionally is that random school X in some random country develops a mini-pipeline to some American grad school.

yenlee, I have a question for you: suppose the undergrad school does not offer chances to perform undergraduate research.  Then what do you look for in the letter of recommendation?

*I* don't look for anything, having never read graduate applications myself.  Even without research experience, recommendations can talk about classroom performance, or preferably independent study.  They can even give context for the student's entire body of work, for example, explaining the depth of the student's coursework overall.  But honestly, getting good recommendations (and writing them, too) is a tricky business.


Lo siento para muy ?s.. (Sorry for many questions.. I can't say that well in spanish)..

I just learned something called "indirect object pronoun" on my spanish class and I have NO clue. :(  I had to finish up my test so I didn't even get to participate in my class.. So, I'm hoping to understand it here.. :blush: Thanks for all your help!

[b]What I Do Understand[/b]

I get that indirect object pronouns are like "le" or "les" referring to the phrase that is in form of a + noun or a + pronoun. I get the concept but I"m not sure on the application. Or other pronouns..

[b]What I Do NOT Understand[/b]

1. Hoy mi madre ____ va a pedir al mesero los camarones al ajillo.

[hide="My Answer"]
[b]le[/b]
I think it's "le" because le describes the a el ajillo so isn't it should be "le" instead of "les"?[/hide]

2. Mi tio Timoteo siempre ____ pide un vaso de leche a la mesera.

[hide="My answer"]
[b]le[/b]
le describes a la mesera?[/hide]

3. El mesero ____ trae la sopa a nosotros.

[hide="My Answer"]
[b]nos[/b]

Nos = a nosotros?? I don't think "les" makes sense in this case.[/hide]

4. A mi hermana Elena _____ encanta el flan.

[hide="My Answer"]
[b]le[/b]

le describes for the el flan.[/hide]

5. El mesero va a traer ____ la cuenta a mi papa.

[hide="My Answer"]
[b]le[/b]

le describes a mi papa?[/hide]

6. A mi no ____ gustan las papitas.

[hide="My Answer"]

Pretty sure it's me.[/hide]

7. ¿A ti ___ gusta el jamon?

[hide=" My answer"]

Same reason as 6. Te.. A ti te makes sense and sounds right.[/hide]

Correcto? :? And can anyone give me more general explanation after checking this about the indirect object?

Muchas gracias.

4. le describes Elena, not el flan.

Slightly off-topic, this reminds me of two examples about direct/indirect pronouns that my teacher last year loved:
Direct Object: Yo cuelgo Jose (I hang Jose): Yo LO cuelgo.
Indirect Object: Yo cuelgo una cadena de flores a Jose (I hang a chain of flowers [b]to Jose[/b]): Yo LE cuelgo una cadena de flores.


AMC
AMC 8
AIME
AMC 10
AMC 10 A
USA(J)MO
USAMO

I just wondered... If you've been participated in any of AMC tests (only AMC8,10,12), can you tell the score and stuffs?
I'm interested in people taking tests and stuffs... And if you did well, could you share the tips with us?  Thanks!

Just look back at some old threads on this forum, namely the score polls created just after each exam this year.

or if he doesn't want to, I got 24 on the amc 8 (ive never gotten a 25 because of stupid mistakes), 120.5 on the first amc 12, 117 on the second, and 5 on the aime ...poop aime score was worse than last year...

not bragging or anything, but i got a 25 on amc 8 (not that it mattered).  then my scores started going down, down, down. 125 on amc10a and 136 on amc10b.  then, i really started dying and got a 3 on the aime (also got a 3 last year). anyone else?

144 AMC 12a

Didnt take AMC 12b

11 AIME

My brain literally shut down on USAMO, so I pulled off a 3 :(

hmm... well i got:

AMC 10A: 134.5
AMC 10B: 130.5

so many careless mistakes... shouldve  got 140.5 on A

AIME: 6 (Nooo... 1 from USAMO)

ya, thats all i did this year... but i'll be ready next year... yes i will

It seems there is almost no correlation between AMC/AIME and USAMO scores... I got a mere 124 on AMC12B and 10 on AIME2, but a 17 on USAMO...

Also, there are only 9 sophomore moppers... 3 of them are winners or honourable mentions.  Does that mean that there is only one of us that is not yellow?  If so, I hope it's me. lol

I went from a 130 AMC 10B last year and a 3 on AIME to a 137 10B and a 5 on AIME this year.

Might as well post my score...

AMC 12A: 132.5
AIME I: 8

Might as well post mine then too :-)

AMC 12 -- 130
AIME II -- 11
USAMO -- 10

AMC 10A--126.5

AIME I--6 (Gah!!! one from USAMO!!!)

Hopefully I can convince my school to offer both the A and B contests next year; I'd like to take the 12A and 10B.

Last year: 134.5 on AMC 12B (should have been 144, erk.)
This year: 121.5 on AMC 12B

First time taking AMC this year:

AMC 10B: 123
AIME: 3

Terrible, but I plan to improve for next year.

AMC12A- 97

TERRIBLE also

Hopefully next year I will have a AIME score and a USAMO score(very unlikely)

8th grade
4th place mathcounts

9th grade
AMC12A - 101.5
AMC12B - 109
AIME II - 4

10th grade
AMC12A - 116
AMC12B - 132.5
AIME II - 5 (just did bad)

AMC8 - back in 8th grade, I can't remember

AMC10A (9th gr) - 111 (?)
AMC10B (9th) - 108 (?)
AIME (9th) - 2

AMC 12A (10th) - 100 and something
AMC12B (10th) - 125 (actually sure on this one)
AIME (10th) - 3

Yup, I take both tests each year...Not exactly the best scores ever but still...

AMC 8- 22 (6th), 23 (7th), 22 (8th)
7th grade- AMC10B- 120.5 (just got into AIME) AIME- 3
8th grade- AMC10A- a pretty pathetic 105. didn't even make AIME...


logarithms
parameterization
ratio
probability
expected value
probability and stats

A stick of length 1 is broken at a random point that is uniformly distributed.  The right hand part is thrown away, and the left part is broken similarly (the breaking point being uniformly distributed).  Show that the length after n steps X(n) satisfies

X(n)^(1/n) converges to 1/e almost surely.

It's the law of large numbers. Show that the logarithm of a uniform 0-1 has expected value $ \minus{} 1$.
(Its negative is an exponential distribution with parameter 1)

What exactly are you applying the law of large numbers to?  The way I am trying to do it is to write each X(n) as a product of n random variables which are i.i.d. and then apply the strong law of large numbers to the log sequence.  The problem is I can't find the right expression.

Turn that product into a sum. That's why I mentioned a logarithm.

[quote="toad"]The problem is I don't see how to write X(n) as a product.  Any enlightenment would be appreciated.[/quote]

$ X(n)$ is a product of $ n$ uniform $ [0,1]$ rvs.  Think of them as ratios, not lengths.

So $ X(n)^{1/n} \equal{} \prod_{1\leq i\leq n} U_i^{1/n}$, where the $ U_i$'s are iid uniform $ [0,1]$ rvs, and hence

\[ \log(X(n)^{1/n}) \equal{} \frac{\sum_{1\leq i\leq n} \log U_i}{n} \to E[\log U_1] \equal{} \minus{}1
\]


algorithm
combinatorics unsolved
combinatorics

Let $ n\ge3$ be an integer. Each row in an $ (n\minus{}2)\times n$ array consists of the numbers $ 1,2,\ldots,n$ in some order, and the numbers in each column are all different. Prove that this array can be expanded into an $ n\times n$ array such that each row and each column consists of the numbers $ 1,2,\ldots,n$.

[hide]First we prove that the array must have $ n\minus{}2$ of each element by contradiction.  If it had more than that  by the pigeon hole principle, one of the $ n\minus{}2$ rows would have more than one of that element.  If it had less than $ n\minus{}2$ then at least one row would not have that element.  Thus we reach a contradiction a verify that each element appears $ n\minus{}2$.  Since no elements repeat in a column, we deduce that each element is missing from two distinct columns.

Now we present an algorithm to arrange the elements to form an $ n\times n$ array.  Pick some $ k$ between $ 1$ and $ n$.  Let the columns $ k$ is missing from be columns $ a,b$.  In the 2 rows being added, in column $ a$, place $ k$ in the bottom and the other element missing from $ a$, $ a_1$ in the top.  In column $ b$ do the opposite: place $ k$ in the top row and place the other element missing from $ b$, $ b_1$ in the bottom row.  Now there is some other column from which $ a_1$ is missing.  let this column be $ c_1$.  Then for this column, place $ a_1$ in the bottom row and place the other element missing from $ c_1$, $ a_2$, in the top row.  Continue on like this.  This assures that all the $ a_i$ are in different rows.  It ends once the 'missing element' from some column $ c_i$ is $ b_1$.  By the algorithm, $ b_1$ is placed in the top row in column $ c_i$.  Thus, it does not violate any rules since $ b_1$ is in the bottom row in column $ b$.

If there are still elements left then we perform this cycling algorithm again, chooosing an element that hasn't been filled in yet.  Since in the previous algorithms, both of any element were placed, there is no reason why the arrangement in this next algorithm should violate the fact that there is one of each element in each row.

Thus, we have an algorithm to create such an $ n\times n$ array as the problem describes. [/hide]


function
analytic geometry
graphing lines
slope

Find all functions $f : \mathbb R \to \mathbb R$ that have the following property: for any distinct points $A(p,f(p)), \ B(q,f(q))$, the line $AB$ intersects $Oy$ in $(0,-pq)$.

[hide]
The line's equation (considering its slope and point $A$) is:
$y = \left(\frac{f(q)-f(p)}{q-p}\right) (x-p)+f(p).$

So if it contains $(0,-p q)$:
$-pq = \left(\frac{f(q)-f(p)}{q-p}\right) (-p)+f(p),$

which, rearranging, becomes:
$\frac{p^{2}-f(p)}{p}= \frac{q^{2}-f(q)}{q}.$

This means each side is constant because any $p$ and $q$ will satisfy it (... or something like that). So if $k$ is some constant, then,
$f(x) = x^{2}-k x.$
[/hide]


trigonometry
geometry
3D geometry
pyramid

Hi,

I have been trying to create an equation to plot points (perspective projection style) for wireframe shapes (for Flash). I need to start with 4 points and go up to an unlimited number creating geodesic forms and need to do this dynamically. I have been trying various things to do this for me and am now at the stage of dreaming in sine and cosine wavelengths! :)

I have found co-ordinates for two forms and need an equation that would generate this kind of data.

For example a 3sided pyramid co-ordinate points are - (4 points)
-0.212331, -58.3687, -82.3175,
82.3334, 58.3686, 0.228231,
-82.7581, 58.3686, 0.228231,
0.212331, -58.3686, 82.774

and a square's are: (8 points)
-150, -150, -150,
-150, 150, -150,
150, 150, -150,
150, -150, -150,
150, -150, 150,
-150, -150, 150,
-150, 150, 150,
150, 150, 150

Is anyone familiar with perspective projection and plotting xyz co-ordinates on an xy plane - and may be able to shed some light on this equation? ;)

Thanks all!

you might want to check that page and download the chapter included..
http://www.macromedia.com/devnet/mx/flash/articles/samurai_ch2.html


I was reading the polynomials chapter and I was doing ok until I got to this "example" problem:

Find an equation whose roots are 3 greater than those of $x^{4}-3x^{3}-3x^{2}+4x-6.$

Mainly at this step:
[hide]$g(r_{i}+3)=f(r_{i}+3-3)=f(r_{i})=0$[/hide]

Apply a horizontal shift 3 units to the right.

so would that mean the answer is
[hide]
$(x-3)^{4}-3(x-3)^{3}-3(x-3)^{2}+4(x-3)-6$
[/hide]

 :huh:
i think I did it wrong
scratch that
I KNOW I DID IT WRONG :ninja:

Granted, it'll take a while to simplify, but I believe it's correct.


If S is a set of positive integers less than or equal to 100 with no two distinct elements of S summing to an element of (7,12,33,45,69,81) then the maxinum number of elements S can have is?

(Removed: Blah. Was thinking of something else. :) )

Hrm?  My suggestion:
[hide]Consider only one of the six numbers at a time.  In particular, what if [hide]you only have to avoid 81?[/hide][/hide]


[quote="cuong"]for a,b,c>0 and abc=1 prove

$\sum\frac{ab+bc}{\sqrt{b^2+1}}\geq 3\sqrt{2}$ :D

to NTTU:

why not give this ineq to MR Can??? :rotfl:  :ninja:  :mad:  :roll:[/quote]

  Mr Vu THien Can ? Hehe  :roll:  .
  [hide]I bet 100 dollars if Mr Can can solve this problem ![/hide]


If I am thinking of applying for Early Admissions, would it be a better choice to take AP Physics (i havent done Physics yet), instead of AP Biology or AP Chem? Thanks for your feedback!

It is hard to answer such a question.  It can depend on tons of factors.  Remember, your final application is a coherent picture of yourself.  Are you at the same time planning on getting involved in any of those corresponding olympiads? Taken by itself, those classes are just about equally impressive, (although since I like physics, I find getting double 5's on the Physics C is pretty awesome looking).

Either way, the important point is what JBL was saying.  Maybe your school has a particularily awesome teacher for one of those classes, or you are just dying to learn about maxwell's equations.  That should make a far larger impact on your decision.

What do you mean by "early admissions"? Do you mean applying for the early round of considering applications (submitting your paperwork before Thanksgiving rather than afterward), or do you mean going to college at a younger than usual age?

well, it does depend one which school you want to apply for (well actually, what major)

but also, AP physic tests are alot harder to get a 5 on (my opinion), so you should see if your school uses computer scans, like MIT, to read the applications or if real admistrators read them, like harvard. thus, if your school is computer-based, take the easier class, and if it is more real people based, take physics.

 yup, hope that helps  :|

[quote="JBL"]That question sounds analogous to the question,
"If I'm going to buy ice cream, should I wear stripes or plaid?"

What difference do you think it will make?  Take whichever class you think you'll like the most![/quote]

Vertical stripes. The lines going downward have the psychological effect of making one look thin.
Man, people need to think critically before replying.


hello, we compute $ \frac{1}{(1+\sqrt{x})^2}+\frac{1}{(1+\sqrt{y})^2}-\frac{2}{x+y+2}=
\frac{2+2x+2x^{\frac{3}{2}}+x^2-2x\sqrt{y}+2y-2\sqrt{x}y+2y^{\frac{3}{2}}+y^2-8\sqrt{xy}}{(1+\sqrt{x})^2(1+\sqrt{y})^2(2+x+y)^2}$
We have to prove that the numerator is always positive. We derive
(1)$ x^2-2x\sqrt{y}+y=(x-\sqrt{y})^2$
(2)$ y^2-2y\sqrt{x}+x=(y-\sqrt{x})^2$
(3)$ x+y\geq2\sqr{xy}$
(4)$ 2+2x^{\frac{3}{2}}+2y^{\frac{3}{2}}\geq6\sqrt{xy}$
By adding (1),(2),(3) and (4) we get the desired inequation.
Sonnhard.

When we're bored with easy problem, let find another complicated solution  :D  :D

Let $ x+y=2k$. Note that
\[ \frac{1}{(1+\sqrt{x})^{2}}-\left( -\frac{1}{\sqrt{k}\left( \sqrt{k}+1\right)^{3}}(x-k)+\frac{1}{\left( 1+\sqrt{k}\right)^{2}}\right) =
\]
\[ \frac{\left( k+x+2\sqrt{kx}+3\sqrt{k}+2\sqrt{x}+1\right) }{\sqrt{k}\left( \sqrt{k}+1\right) ^{3}\left( \sqrt{x}+1\right) ^{2}}\left( \sqrt{k}-\sqrt{x}\right) ^{2}\geq 0
\]

which means
\[ \frac{1}{(1+\sqrt{x})^{2}}\geq -\frac{1}{\sqrt{k}\left( \sqrt{k}+1\right)^{3}}(x-k)+\frac{1}{\left( 1+\sqrt{k}\right) ^{2}}
\]

where equality attained when $ x=k$. So we have
\begin{eqnarray*}\frac{1}{(1+\sqrt{x})^{2}}+\frac{1}{(1+\sqrt{y})^{2}}&\geq&-\frac{1}{\sqrt{k}\left( \sqrt{k}+1\right) ^{3}}(x+y-2k)+\frac{2}{\left( 1+\sqrt{k}\right)^{2}}\\&=&\frac{2}{\left( 1+\sqrt{k}\right) ^{2}}\end{eqnarray*}

Next, we have to prove that
\[ \frac{2}{\left( 1+\sqrt{k}\right) ^{2}}\geq \frac{2}{2k+2}=\frac{2}{x+y+2}
\]

but it's trivial since it's equivalent to $ \left( \sqrt{k}-1\right) ^{2}\geq 0$

(I'm sorry for the space wasted...)

Very nice proof, dyd!
U can avoid misuse of space by using single $ \$$ instead of $ \$\$$ while writing $ LATEX$ codes.
  :)

By applying CS Inequality in the Engel form, we get:
$ LHS\geq\frac{4}{x\plus{}y\plus{}2\sqrt{x}\plus{}2\sqrt{y}\plus{}2}$
Next, we'll prove $ \frac{4}{x\plus{}y\plus{}2\sqrt{x}\plus{}2\sqrt{y}\plus{}2}\geq RHS$, which is true from $ (\sqrt{x}\minus{}1)^2\plus{}(\sqrt{y}\minus{}1)^2\geq0$
Equality applies iff $ x\equal{}y\equal{}1$

Cauchy, we get :
$ LHS \ge \frac{1}{2(1\plus{}x)}\plus{}\frac{1}{2(1\plus{}y)}$
And then $ MHS\ge RHS$ is easily proved by HM-AM or Cauchy..

[quote="dyd"]Prove that for $ x,y\in\mathbb{R^ \plus{} }$,
$ \frac {1}{(1 \plus{} \sqrt {x})^{2}} \plus{} \frac {1}{(1 \plus{} \sqrt {y})^{2}} \ge \frac {2}{x \plus{} y \plus{} 2}$[/quote]

Use the obvious inequality $ 2(a^2 \plus{} b^2) \geq (a \plus{} b)^2$ and get:

$ LHS \geq \frac {1}{2}(\frac {1}{x \plus{} 1} \plus{} \frac {1}{y \plus{} 1})$

Now we have to prove that:

$ \frac {1}{2}(\frac {1}{x \plus{} 1} \plus{} \frac {1}{y \plus{} 1}) \geq \frac {2}{x \plus{} y \plus{} 2} \Leftrightarrow (\frac {1}{x \plus{} 1} \plus{} \frac {1}{y \plus{} 1}) \geq \frac {4}{(x \plus{} 1) \plus{} (y \plus{} 1)} \Leftrightarrow ((x \plus{} 1) \plus{} (y \plus{} 1))^2 \geq 4(x \plus{} 1)(y \plus{} 1)$

Which is obvious.

[quote="NickNafplio"]Use the obvious inequality $ 2(a^2 \plus{} b^2) \geq (a \plus{} b)^2$[/quote]
It is Cauchy after all!
 :)

easy solution  :D

we know $ (a\plus{}b)^2 \ge 4ab$

for $ a\equal{}x\plus{}1$ and $ b\equal{}y\plus{}1$, we get $ (x\plus{}y\plus{}2)^2 \ge 2(x\plus{}1)(y\plus{}1)$

then $ \frac{1}{2x\plus{}2}\plus{}\frac{1}{2y\plus{}2} \ge \frac{2}{x\plus{}y\plus{}2}$

it's obvious that $ 2 \sqrt{x} \le x\plus{}1$ and $ (\sqrt{x}\plus{}1)^2 \le 2x\plus{}2$

then $ \frac{1}{(\sqrt{x}\plus{}1)^2} \ge \frac{1}{2x\plus{}2}$

applying this to $ y$ will complete the proof  :D

By Titu's Lemma,
$\frac{1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} \ge \frac{4}{2+x+y+2\sqrt{x}+2\sqrt{y}}$
And
$\frac{4}{2+x+y+2\sqrt{x}+2\sqrt{y}} \ge \frac{2}{x+y+2} \iff x+y+2 \ge 2\sqrt{x}+2\sqrt{y}$,
which is true because $x+1 \ge 2\sqrt{x}$ and $y+1 \ge 2\sqrt{y}$.

$\color{red}{\textit{\textbf{Proof:}}}$ We have the function $f(x)=\frac{1}{(1+\sqrt x)^2}$ is a convex function as $f''(x)=\frac{4\sqrt x +1}{2(\sqrt x+1)^4 x^{\frac{3}{2}}}>0$. Therefore,
\begin{align*}
\frac {1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} &\ge \frac{2}{\left(1+\sqrt {\frac{x+y}{2}}\right)^2} \\ &\ge \frac{2}{x+y+2} \\ &\iff x+y+2\ge \left(1+\sqrt {\frac{x+y}{2}}\right)^2 \\ &\iff \frac{x+y}{2}+1\ge 2\sqrt{\frac{x+y}{2}}
\end{align*}
which is just AM-GM. $\quad \blacksquare$


inequalities
function
induction
LaTeX
vector
inequalities solved

Let $f$ be a convex function on an interval $I$.
Three numbers $a$, $b$, $c$ belong to $I$.

Prove that
$f\left(a\right)+f\left(b\right)+f\left(c\right)+3f\left(\frac{a+b+c}{3}\right) \geq 2\left(f\left(\frac{a+b}{2}\right)+f\left(\frac{b+c}{2}\right)+f\left(\frac{c+a}{2}\right)\right)$.

Popoviciu's inequality is like so:

f(x)+f(y)+f(z)+3f((x+y+z)/3)>=2f((x+y)/2)+2f((z+y)/2)+2f((x+z)/2)
where f is convex on an "I" interval, x,y,z are from I. If f is strictly convex we can only have equality if and only if x=y=z.

An idea to solve this is that you can violate the simetry with x>=y>=z and then apply Karamata's inequality.

cheers! :D

Hey just a thought .. Have you ever seen a proof of Karamata's Inequality ? I did and it's very difficult . You can't use it in problems like this (competition problems).

I have seen a solution of Karamata's inequality and it's just induction with a little help from Jensen's inequality...it doesn't look that hard.. :D

cheers!!

Can you post it here then ? I'd be quite interested .

[color=red][[b]EDIT by Darij:[/b] This proof of the Karamata inequality is, unfortunately, wrong. See below.][/color]

Alright here it is:

 Karamata inequality:

f(x1)+f(x2)+...+f(xn)>=f(y1)+f(y2)+..+f(yn), where f is convex on I interval, with y1>=y2>=...>=yn, x1>=x2>=...>=xn, with xn,yn from I , n>=2,
sum(xi)=xum(yi); i=1,n
x1>=y1;
x1+x2>=y1+y2;
...
x1+..+x_(n-1)>=y1+...+y_(n-1).
If f is strictly convex, then we have equality if and only if (x1,..,xn)=(y1,..,yn).

Proof:

I shall write a solution with indution.( my favourite method) :D :D
n=2
f(x1)+f(x2)>=f(y1)+f(y2), (this is what we have to prove)
we know that x1>=x2, y1>=y2, x1+x2=y1+y2 and x1>=y1. we get that x1>=y1>=y2>=x2 and then in the not trivial case x1>x2 we have that y1=px1+(1-p)x2, y2=(1-p)x1+px2, with p=(y1-x2)/(x1-x2), 1/2<=p<=1 because y1-x2>=(y1+y2)/2-x2=(x1+x2)/2-x2=(x1-x2)/2.

From Jensen we have that:
f(y1)=f(px1+(1-p)x2)<=pf(x1)+(1-p)f(x2) and
f(y2)=f((1-p)x1+px2)<=(1-p)f(x1)+pf(x2), and now we sum those two inequalities and we get what we want.

For f strictly convex and x1<x2 we have equality when p=1 which gives (x1,x2)=(y1,y2)

Now we consider that x1>=x2>=..>=x_(n+1) and y1>=y2>=..>=y_(n+1) with
sum(xi)=xum(yi); i=1,n+1
x1>=y1;
x1+x2>=y1+y2;
...
x1+..+xn>=y1+...+yn.

Let u>=0 be so that we have x1+x2+...+xn=y1+y2+..+y_(n-1)+(yn+u). we have that x_(n+1)+u=y_(n+1) and from the induction we have that:
f(x1)+f(x2)+..+f(x_(n+1))>=f(y1)+f(y2)+...+f(y_(n-1))+f(yn+u)+f(x_(n+1)).

[color=red][[b]EDIT by Darij:[/b] Alas, the proof is seriously wrong at this point. In fact, in order to be able too apply the induction assertion, we must check that the condition $y_{1}\geq y_{2}\geq ...\geq y_{n-1}\geq y_{n}+u$ is fulfilled. And actually, this condition is not always fulfilled.][/color]

what  is left for us to prove is that:
f(y1)+f(y2)+..+f(y_(n-1))+f(yn+u)+f(x_(n+1))>=f(y1)+f(y2)+..+f(x_(n+1)) which is equivalent to prove that f(yn+u)+f(x_(n+1))>=f(yn)+f(y_(n+1)).
Because yn+u>=yn>=y_(n+1)>=x_(n+1) and (yn+u)+x_(n+1)=yn+y_(n+1) we get that this ineq is true because the ineq is true for n=2 (we apply Karamata's ineq for n=2, which I proved to be true).

again if the function is strictly convex then we have equality only if (x1,x2,..,x_(n+1))=(y1,y2,..,y_(n+1)).

cheers! :D

hey Lagrangia, how do you prove popoviciu's inequality with karamata.
I prove the case (x+y+z)/3 >= (x+z)/2 but i can't prove the case
(x+y+z)/3 <= (x+z)/2 ??

And what about the first problem ??

Well like I said you violate simmetry with x>=y>=z. We have that x>=(x+y+z)/3>=y and we have to compare y with (x+y+z)/3.

We have two cases:

1) y<=(x+y+z)/3 <=> y<=(x+z)/2 you get the Popoviciu's inequality after you apply Karamata to the 6-uplets:
(x;(x+y+z)/3;(x+y+z)/3;(x+y+z)/3;y;z) and ((x+y)/2;(x+y)/2;(x+z)/2;(x+z)/2;(z+y)/2;(z+y)/2)

you just have to prove that the way I chose the uplets verifies Karamata.

2)y>=(x+y+z)/3 <=> y>=(x+z)/2 you get the Popoviciu's inequality after you apply Karamata to the 6-uplets:

(x;y;(x+y+z)/3;(x+y+z)/3;(x+y+z)/3;z) and ((x+y)/2;(x+y)/2;(x+z)/2;(x+z)/2;(z+y)/2;(z+y)/2)
you just have to prove that the way I chose the uplets verifies Karamata.

tschues!! :D

the karamata's inequality ,isnt it the same thing as hardy -littlewood-polya
majorization inequality :?

Hey .. galois you are right it's the same thing... dunno why they named it like that! :D:D

here is a generalization of the convex functions!

pf(x)+qf(y)+rf(z)+(p+q+r)f((px+qy+rz)/(p+q+r))>=(p+q)f((px+qy)/p+q) + (q+r)f((qy+rz)/(q+r)) + (r+p)f((rz+px)/(r+p)), where f is convex on an I interval, p,q,r>0 and x,y,z belong to I. (the proof is not that hard ... just don't try to use Karamata inequality... :D)

for a=b=c we get Popoviciu's inequality.

so almost all the convex functions have started from this one...

cheers! :D :D

Theorem of Popoviciu: f a convex function I--->R, then

f(x)+f(y)+f(z) + 3f((x+y+z)/3)  \geq

2[f((x+y)/2) + f((y+z)/2) + f((z+x)/2)]
-------------------------------------------------------------------------

Here the proof of the Romanian mathematicain Tiberiu Popoviciu
published in the romanian journal high-school
Revista Matematica Timisoara, N1, 1991

The case x=y=z is obvious, suppose at least two of x,y,z are distinct.
suppose x \leq y \leq z

If y \leq (x+y+z)/3 then

(x+y+z)/3 \leq (x+z)/2 \leq z
(x+y+z)/3 \leq (y+z)/2 \leq z

There exist s,t in [0;1]
(1)  (x+z)/2 = s.(x+y+z)/3 +(1-s)z
(2)  (y+z)/2 = t.(x+y+z)/3 + (1-t)z

add (1) & (2) (x+y-2z)/2 =(s+t)(x+y-2z)/3 => s+t = 3/2

f is convex so

(3) f((x+z)/2) = f(s.(x+y+z)/3 + (1-s)z)  \leq s.f((x+y+z)/3)+(1-s)f(z)

similarly

(4) f((y+z)/2) =f(t.(x+y+z)/3 + (1-t)z) \leq t.f((x+y+z)/3) + (1-t)f(z)

add (3)&(4) with f((x+y)/2)  \leq 1/2f(x)+1/2f(y) the conclusion follows.

The case (x+y+z)/3 < y can be treated in the same way.

I was asking in other thread for proof of Karamata inequality, but noone replyed. Now I have my proof, so please check it if it's ok.

Let's have convex function on interval $I$, positive integer $n$ and real numbers $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ in $I$ with restrictions:
$a_1\geq b_1\\
a_1+a_2\geq b_1+b_2\\
\ldots\\
a_1+a_2+\ldots+a_{n-1}\geq b_1+b_2+\ldots+b_{n-1}\\$
and
$a_1+a_2+\ldots+a_n= b_1+b_2+\ldots+b_n$

We will try to prove:
$f(a_1)+f(a_2)+\cdots+f(a_n)\geq f(b_1)+f(b_2)+\cdots+f(b_n)$

We use induction. Lagrangia showed how to prove it for $n=2$. Suppose it's true for $1,2,\ldots,n-1$ and we will prove it for $n$.

Denote
$X_1=a_1-b_1\\
X_2=(a_1+a_2)-(b_1+b_2)\\
\ldots\\
X_n=(a_1+a_2+\cdots+a_n)-(b_1+b_2+\cdots+b_n)$

At the beginning we know that $X_1,X_2,\ldots,X_{n-1}\geq0$ and $X_n=0$.
It's also easy to find out that $a_1\geq b_1\geq b_n\geq a_n$. From this we know that there exist some $k$ such that $a_{k-1}\geq b_n\geq a_k$ and such $m$ that $a_m\geq b_1\geq a_{m+1}$. Also cause of $a_m\geq b_1\geq b_n\geq a_k$ we can claim $m<k$.

Now we will use smoothing. Denote $min=minimum\{X_m,X_{m+1},\ldots,X_{k-1},a_m-b_1,b_n-a_k\}$. We know that $min\geq 0$. We make transformation: $a_m:=a_m-min$, $a_k:=a_k+min$. What has happened? (let old $a_m,a_k$ be $A_m,A_k$)

Arrays of $a$'s and $b$'s are still ordered because $a_{k+1}<A_k+min\leq A_k+(b_n-A_k)\leq a_{k-1}$ and also $a_{m-1}>A_m-min\geq A_m-(A_m-b_1)=b_1\geq a_{m+1}$. What about $X_1,\ldots,X_n$? Numbers $X_1,X_2,\ldots,X_{m-1}$ and $X_k,X_{k+1},\ldots,X_n$ didn't changed (because in the sum are both or none of $a_m$,$a_k$). Numbers $X_m,X_{m+1},\ldots,X_{k-1}$ has changed, but we know that each such $X_i$ was decreased by $min$, but $min$ is not bigger than $X_i$ so they stayed nonegative. And at last, the left side of our inequality has decreased cause of karamata inequality for $A_m,A_k$ and new $a_m,a_k$ (case $n=2$). So everything is all right, but we know something more. We know that after this transformation one of these things is true:

1)  One of $X_m,X_{m+1},\ldots,X_{k-1}$ is zero. This can happend if when we are choosing $min$, one of them is minimal in there. Let $X_i=0$. But then we have:
$a_1\geq b_1\\
a_1+a_2\geq b_1+b_2\\
\ldots\\
a_1+a_{2}+\ldots+a_{i-1}\geqq_1+b_{2}+\ldots+b_{i-1}\\
a_1+a_{2}+\ldots+a_i=b_1+b_{2}+\ldots+b_i\\$
And from now we can delete the last line from the others:
$a_{i+1}\geq b_{i+1}\\
a_{i+1}+a_{i+1}\geq b_{i+1}+b_{i+2}\\
\ldots\\
a_{i+1}+a_{i+1}+\ldots+a_{n-1}\geq b_{i+1}+b_{i+2}+\ldots+b_{n-1}\\
a_{i+1}+a_{i+1}+\ldots+a_n= b_{i+1}+b_{i+2}+\ldots+b_n$
So it's equivalent to prove Karamata inequality for $i$ and $n-i$ (we break the inequality into two, both that are true).

2) That $a_m=b_1$ (also the same reasons (being minimal)). But then we just use Karamata inequality for arrays $a_1,a_2,\ldots,a_{m-1},a_{m+1},\ldots a_n$ and $b_2,b_3,\ldots b_n$ and sum it with trivial $f(a_m)=f(b_1)$

3) If $a_k=b_n$ we use Karamata inequality for arrays $a_1,a_2,\ldots,a_{k-1},a_{k+1},\ldots a_n$ and $b_1,b_2,\ldots,b_{n-1}$ and sum it with trivial $f(a_k)=f(b_n)$

Hope it's correct, can someone please check it?
I also hope that it's understandable ;)

thank you.
but i can't understand something.
why$a_1\geq b_1\geq b_n\geq a_n $

$a_1\geq b_1$ and $b_1\geq b_n$ are necessary for majorization.
$b_n\geq a_n$ flows from:
$a_1+a_2+\ldots+a_n=b_1+b_2+\ldots+b_n$
$a_1+a_2+\ldots+a_{n-1}\geq b_1+b_2+\ldots+b_{n-1}$

I think i have known.
your approach have two step.
1,adjust $a$'s to satisfy $a_1+\cdots+a_i=b_1+\cdots+b_i$,and the sum will not increase.
2,use induction,
right?
nice solution.thank you,

Yes, by taking two terms of $a$ a little bit together I can decompose Karamata inequality to two Karamata inequalities for smaller vectors (less than $n$ terms). One possibility is that
$a_1+\cdots+a_i=b_1+\cdots+b_i$
happen. Or it can happen that $b_1$ or $b_n$ is equal to some $a_i$. Then we can delete from $a$ number $a_i$ and from $b$ $b_1$ or $b_n$ and one can show that after this step $a$ still majorize $b$.
My proof is not so detaily written, but I hope that understandable ;)

abel is a great man,because he made me come out the nice solution for kamarate inequality ;)
let $k_i=\frac{f(a_i)-f(b_i)}{a_i-b_i}$,$S_i=a_1+\cdots+a_i-b_1-\cdots-b_i$
then we have
$f(a_1)+\cdots+f(a_n)-f(b_1)-\cdots-f(b_n)=(a_1-b_1)k_1+(a_2-b_2)k_2+\cdots+(a_n-b_n)k_n=\sum_{i=1}^nS_i(k_i-k_{i+1})$(the last equality use abel vary)
and we have $k_i \geq k_{i+1},S_i \geq 0$
($\frac{f(a_i)-f(b_i)}{a_i-b_i} \geq \frac{f(a_i)-f(b_{i+1})}{a_i-b_{i+1}} \geq \frac{f(a_{i+1})-f(b_{i+1})}{a_{i+1}-b_{i+1}}$$
then we have proved the kamarate ineq. :)
is it nice??? :D

[quote="ondrob"]1: This proof is really nice, how did you find it?
2: Who is Abel?[/quote]

Abel's a norway mathematician many years ago. (maybe 100 or 200 years ago? i don't really remember). zhaobin makes use of the abel's identity. Very very nice proof.  :lol:
but i don't understand one thing:
[quote]$k_i \geq k_{i+1},S_i \geq 0$
($\frac{f(a_i)-f(b_i)}{a_i-b_i} \geq \frac{f(a_i)-f(b_{i+1})}{a_i-b_{i+1}} \geq \frac{f(a_{i+1})-f(b_{i+1})}{a_{i+1}-b_{i+1}}$
[/quote]
i don't understand this part. Can you elaborate? Thanks.

to siuhochung
for $a_i \geq a_{i+1},b_i \geq b_i+1$
I think it is easy to show $k_i \geq k_{i+1}$
do you agree :D ?
if there exist $a_i=b_i$,we can use induction.
I find it when I was think why should there have the condition $S_i \geq 0$

there's another proof here in chinese
http://episte.math.ntu.edu.tw/articles/mm/mm_06_4_02/page3.html
I think the idea used here is interesting.

This is ondrob's post #13, with typos fixed and (most importantly) LaTeX syntax corrected. (The original syntax used to be correct when the parser was different, but with the current version of the parser it just creates a horrible mess.)

If you are a mod and reading this: I think you can just copy the below quote into post #13.

Oh, and ondrob's proof of the Karamata inequality is correct (although case 3 at the very end wouldn't suffer from an additional explanation why Karamata can be applied to the arrays $a_1,a_2,\ldots,a_{k-1},a_{k+1},\ldots ,a_n$ and $b_1,b_2,\ldots ,b_{n-1}$).

[quote="ondrob"]I was asking in other thread for proof of Karamata inequality, but noone replied. Now I have my proof, so please check if it's ok.

Let's have a convex function $f$ on interval $I$, a positive integer $n$ and real numbers $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ in $I$ with restrictions:
\[
a_1\geq b_1\\
a_1+a_2\geq b_1+b_2\\
\ldots\\
a_1+a_2+\ldots+a_{n-1}\geq b_1+b_2+\ldots+b_{n-1}\ \ \ \ \ \ \ \ \ \text{and} \\
a_1+a_2+\ldots+a_n= b_1+b_2+\ldots+b_n
\]
as well as $a_1\geq a_2\geq ...\geq a_n$ and $b_1\geq b_2\geq ...\geq b_n$.

We will try to prove:
$f(a_1)+f(a_2)+\cdots+f(a_n)\geq f(b_1)+f(b_2)+\cdots+f(b_n)$.

We use induction. Lagrangia showed how to prove it for $n=2$. Suppose it's true for $1,2,\ldots,n-1$ and we will prove it for $n$.

Denote
\[
X_1=a_1-b_1\\
X_2=(a_1+a_2)-(b_1+b_2)\\
\ldots\\
X_n=(a_1+a_2+\cdots+a_n)-(b_1+b_2+\cdots+b_n)
\]

At the beginning we know that $X_1,X_2,\ldots,X_{n-1}\geq0$ and $X_n=0$.
It's also easy to find out that $a_1\geq b_1\geq b_n\geq a_n$. From this we know that there exist some $k$ such that $a_{k-1}\geq b_n\geq a_k$ and such $m$ that $a_m\geq b_1\geq a_{m+1}$. Also, because of $a_m\geq b_1\geq b_n\geq a_k$, we can assume that $m<k$.

Now we will use smoothing. Denote $min=minimum\{X_m,X_{m+1},\ldots,X_{k-1},a_m-b_1,b_n-a_k\}$. We know that $min\geq 0$. We make transformation: $a_m:=a_m-min$, $a_k:=a_k+min$. What has happened? (Let the old $a_m,a_k$ be $A_m,A_k$.)

Arrays of $a$'s and $b$'s are still ordered because $a_{k+1}\leq A_k+min\leq A_k+(b_n-A_k)=b_n\leq a_{k-1}$ and also $a_{m-1}\geq A_m-min\geq A_m-(A_m-b_1)=b_1\geq a_{m+1}$. What about $X_1,\ldots,X_n$? Numbers $X_1,X_2,\ldots,X_{m-1}$ and $X_k,X_{k+1},\ldots,X_n$ didn't change (because in the sum are both or none of $a_m$,$a_k$). Numbers $X_m,X_{m+1},\ldots,X_{k-1}$ have changed, but we know that each such $X_i$ was decreased by $min$, but $min$ is not bigger than $X_i$ so they stayed nonnegative. And at last, the left side of our inequality has decreased because of karamata inequality for $A_m,A_k$ and new $a_m,a_k$ (case $n=2$). So everything is all right, but we know something more. We know that after this transformation one of the following things is true:

1)  One of $X_m,X_{m+1},\ldots,X_{k-1}$ is zero. This can happen if when we are choosing $min$, one of them is minimal in there. Let $X_i=0$. But then we have:
\[
a_1\geq b_1\\
a_1+a_2\geq b_1+b_2\\
\ldots\\
a_1+a_{2}+\ldots+a_{i-1}\geq b_1+b_{2}+\ldots+b_{i-1}\\
a_1+a_{2}+\ldots+a_i=b_1+b_{2}+\ldots+b_i
\]
And from now we can subtract the last line from the others:
\[
a_{i+1}\geq b_{i+1}\\
a_{i+1}+a_{i+2}\geq b_{i+1}+b_{i+2}\\
\ldots\\
a_{i+1}+a_{i+2}+\ldots+a_{n-1}\geq b_{i+1}+b_{i+2}+\ldots+b_{n-1}\\
a_{i+1}+a_{i+2}+\ldots+a_n= b_{i+1}+b_{i+2}+\ldots+b_n
\]
So it's equivalent to prove Karamata inequality for $i$ and $n-i$ (we break the inequality into two, both that are true).

2) That $a_m=b_1$ (also the same reasons (being minimal)). But then we just use Karamata inequality for arrays $a_1,a_2,\ldots,a_{m-1},a_{m+1},\ldots , a_n$ and $b_2,b_3,\ldots , b_n$ and sum it with the trivial equality $f(a_m)=f(b_1)$.

3) If $a_k=b_n$ we use Karamata inequality for arrays $a_1,a_2,\ldots,a_{k-1},a_{k+1},\ldots ,a_n$ and $b_1,b_2,\ldots ,b_{n-1}$ and sum it with the trivial equality $f(a_k)=f(b_n)$.

Hope it's correct, can someone please check it?
I also hope that it's understandable ;)[/quote]


Points $P$ and $Q$ are outside $\triangle ABC$ such that $\triangle APB$ and $\triangle AQC$ are isosceles right triangles ($\angle P=\angle Q=90^\circ$).

If $\triangle APQ=10$, $\triangle BCP=16$ and $\triangle BCQ=14$, find the area of pentagon $APBCQ$.

This question is interesting and challenging.
I am not able to solve it.
Please give some hints.
 :)

[hide="Hint"]
Let $M$ be the midpoint of $BC$.
Then rotate $\triangle PBM$ and $\triangle QCM$.
[/hide]

Thanks a lot! This is a nice problem.  :)
I got an answer 65.
Is that correct?

Try again... the answer is less than that.
(I think I know how you got $65$, though.)

That's correct.

Can you post your solution now? :)

Better leave it for other people to enjoy! :)


evaluate this limit:

$ lim_{n\to\infty}\sqrt[2n]{(\cos(x))^{2n}\plus{}(\sin(x))^{2n}}$ with $ x\in{\mathbb{R}}$.

thanx.

Evaluate the limit $ \lim_{n\rightarrow\infty}(a^n\plus{}b^n)^{\frac{1}{n}}$ when $ a,b>0$.
Answer: $ \max\left\{a,b\right\}$.

$ (a^n\plus{}b^n)^{\frac{1}{n}}\equal{}a\left(1\plus{}\left(\frac{b}{a}\right)^n\right)^\frac{1}{n}$


it should be well-known;
Let $ \Delta ABC$ be a triangle and $ W$ its circumcircle.Let $ A',B',C'$ be three points on $ BC,CA,AB$ and $ W_a,W_b,W_c$ be circles that are tangent to $ BC,CA,AB$ at $ A',B',C'$ and to $ W$ internally such that their points of tangency with $ W$ aren't in the same side with $ A,B,C$ wrt $ BC,CA,AB$,respectively.Let $ (I)$ be the incenter of $ ABC$.Prove that there exists a fourth circle tangent to $ W_a,W_b,W_c,(I)$ if and only if $ AA',BB',CC'$ are concurrent.

nobody? :(
i edit the stupid mistake,it is:
 "there exists a circle tangent to...."
[quote="shoki"]...there exists a [u]fourth[/u] circle tangent to $ W_a,W_b,W_c,(I)$ if and only if $ AA',BB',CC'$ are concurrent.[/quote]

Dear Shoki and Mathlinkers,
see for example
http://forumgeom.fau.edu/FG2001volume1/FG200124.pdf
Sincerely
Jean-Louis


From CIA's report you know that Bin lives in one of 13 caves. These caves can be considered as 13 points on a circle.  Everyday, Bin move to one the two  caves that are adjacent to the case he lived in the day before. Each day, you can also visit 2 caves.

What is the optimal strategy for you to find out what cave  Bin is in?

Have A Great Day!

Label the caves 0-12.

Here is a strategy that might be optimal; I'm not sure.
[hide]On day $ n$, you check caves $ n$ and $ \minus{}n \mod{13}$.
To show that this strategy works:
On day 1, you check caves 1 and 12.  He is not there, or else you find him.
When he moves, you know he is not in cave 0.
On day 2, you check caves 2 and 11.  He is not there again, and not in cave 0.
So when he moves again, you know he is not in cave 1 or 12.
On day 3, you check caves 3 and 10.  He is not there and not in caves 1 or 12.
So when he moves, you know he cannot move into cave 2, 0 or 11.
Using the same reasoning 10 more times, you conclude that he is not in any of the caves.  Therefore, you must have found him on your search.[/hide]

sorry to revive, but the problem is not completely solved yet. furthermore, the solution above is not valid. suppose that he was in cave 2 on the day you checked 1 and 12. the day after he can go to 1 while you are checking 2 and 11.

So? Eventually, you will find him.  Try it with your situation: you will see that eventually you will check the same cave as he is in, even if it has to be after 13 days or so.

oh yes, i made a mistake :blush: . Well, can you prove that this is the optimal algorithm though?


I just noticed there is an aime problem series that must have just started, or started recently? Did I miss the entry date? How long has it been running?

Two classes have taken place.  If you call and order by phone, we can slip you into that class.


algorithm
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Diophantine equation
Euclidean algorithm
relatively prime

How do you solve Diophantine equations like $ 37x\plus{}21y\equal{}1$?

Thanks!

You want the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]extended Euclidean algorithm[/url].

but how do you find the roots of certain linear diophantine equation and generalize it?

for instance, i have no idea how to find x, y in $ 37x \plus{} 21y \equal{} 1$...

could you explain?

thank you  :)

EDIT: oh nvm i got it  :D

That's exactly what the extended Euclidean algorithm does.  What is giving you trouble in the Wikipedia article?

Applying the Extended GCD algorithm, you get gcd=1 (obvious, 37 is prime) and a particular solution of (4,-7).
We know that $ x\equal{}x_0\plus{}bn$ and $ y\equal{}y_0\minus{}an$ for any integer $ n$.
Thus, we get the forms of the solutions:
$ x\equal{}4\plus{}21n$
$ y\equal{}\minus{}7\minus{}37n$
And, to test it:
$ 37x\plus{}21y\equal{}1$
$ 37(4\plus{}21n)\plus{}21(\minus{}7\minus{}37n)\equal{}1$
$ 148\plus{}777n\minus{}147\minus{}777n\equal{}1$
$ 1\equal{}1$

i don't really understand this...
can someone please explain the wiki article? (t0rajir0 perhaps?)

I gave an example with larger numbers [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url], but I don't know if it's any clearer.

I'll try to explain it more clearly with another example. (I'm just picking random numbers)
Let's say that $ 5x \plus{} 3y \equal{} 12$, and we want the general form for the solutions $ x$ and $ y$.
For reference, this equation is of the form $ ax \plus{} by \equal{} c$. I'll refer to the numbers later on as $ a$,$ b$, and $ c$.
First, use the Extended Euclidean algorithm on $ a$ and $ b$. Working it out on paper, I find that the GCD is 1 (which is obvious because 5 and 3 are relatively prime) and that $ ( \minus{} 1,2)$ is a solution to the equation $ 5m \plus{} 3n \equal{} 1$, allowing the numbers we just found from the algorithm to be represented by $ (m,n)$. The equation we derive is $ am/gcd(a,b) \plus{} bn/gcd(a,b) \equal{} gcd(a,b)$. If you plug in the values for $ m,n,gcd(a,b)$ you'll end up with the equation.
Now, we multiply both sides by the number we would have to multiply by in order to acquire $ c$. In this case, $ c \equal{} 12$ and we must multiply 1 by 12 to get 12, so we now end up with: $ 5( \minus{} 1)(12) \plus{} 3(2)(12) \equal{} 12$, or $ 5( \minus{} 12) \plus{} 3(24) \equal{} 12)$. This means that $ ( \minus{} 12,24)$ is a particular solution to our first equation $ 5x \plus{} 3y \equal{} 12$.
You can check this by plugging in: $ 5( \minus{} 12) \plus{} 3(24) \equal{} \minus{} 60 \plus{} 72 \equal{} 12$.
Let this solution be denoted as $ (x_0,y_0)$, such that $ x_0 \equal{} \minus{} 12$ and $ y_0 \equal{} 24$. Now, as a rule, our solutions will be of the form $ x \equal{} x_0 \plus{} bt$ and $ y \equal{} y_0 \minus{} at$ for any integer $ t$.
Trying this with some random number like 3: $ x \equal{} \minus{} 12 \plus{} 3(3) \equal{} \minus{} 3$ and $ y \equal{} 24 \minus{} 5(3) \equal{} 9$.
Checking the solution $ ( \minus{} 3,9)$: $ 5( \minus{} 3) \plus{} 3(9) \equal{} \minus{} 15 \plus{} 27 \equal{} \boxed{12}$
To test the general solution, you can substitute for $ x$ and $ y$:
$ 5( \minus{} 12 \plus{} 3t) \plus{} 3(24 \minus{} 5t) \equal{} \minus{} 60 \plus{} 15t \plus{} 72 \minus{} 15t \equal{} 72 \minus{} 60 \equal{} \boxed{12}$
And thus, our generalized solutions are:
$ x \equal{} \minus{} 12 \plus{} 3t$
$ y \equal{} 24 \minus{} 5t$


Let $m$ and $n$ be natural numbers for which the equation
$(x+m)(x+n) = x+m+n$
has at least one integer solution. Show that $\frac{1}{2}<\frac{m}{n}<2$

$D=(m-n+1)^{2}+4n$. If m>n D is exactly square if and only if $D=(m-n+1+2k)^{2}, k\in N$ it give $m=n(1+\frac{1}{k})-k-1<2n-2$, therefore $\frac{m}{n}<2$. If $n>m$ we have (by symmetric) $\frac{m}{n}>\frac{1}{2}.$


calculus
integration
function
real analysis
real analysis unsolved

Let $ a<b$ and let $ f(x)$ be a function satisfying $ f'(x)$ is a continuous function in R such that:
 $ f(a)\equal{}f(b)\equal{}0$ and $ \int_{a}^{b}|f'(x)|dx\equal{}m$.
 Prove that:
$ |f(x)|\le \frac{m}{2} \forall x \in [a,b]$

Manifestly $ f(x) \equal{} \int _a^xf'(x)dx \equal{} \minus{} \int _x^bf'(x)dx\Longrightarrow$ $ \ (|f(x)| \leq \int _a^x|f'(x)|dx)\ \& \ (|f(x)| \leq \int _x^b|f'(x)|dx)$ $ \Longrightarrow |f(x)|\leq \min (\int_a^x|f'(x)|dx,\int_x^b|f'(x)|dx)$
Finally  $ \int_a^b|f'(x)|dx \equal{} \int_a^x|f'(x)|dx \plus{} \int_x^b|f'(x)|dx \geq$$ 2 \min (\int_a^x|f'(x)|dx),\int_x^b|f'(x)|dx) \geq 2|f(x)|$
 :cool:


real analysis
calculus
integration
inequalities
real analysis unsolved

If $ g\in L_p(x)$ and $ |f_n| \le g$ ,show that for each $ \epsilon >0$,there is a $ \delta(\epsilon)>0$,such that if $ E \subseteq X$ and $ m(E) < \delta(\epsilon)$,then
$ \int_E |f_n|^p dm \le \epsilon^p$ ,for all $ n \in \mathbb{N}$

Define $ \nu_n(E)\equal{}\int |f_n|^p dm$ and use the result that " (for positive measures) $ \nu\ll \mu$ iff for given $ \epsilon>0$, there exists a $ \delta>0$ such that $ \nu(E)<\epsilon$ whenever $ \mu(E)<\delta$

limsup,what do you mean by $ \nu\ll \mu$?

The symbol $ \nu\ll\mu$ means that the measure $ \nu$ is absolutely continuous with respect to $ \mu.$

But limsup doesn't quite have the proof, since he hasn't accounted for the $ \forall n$ quantifier yet.

More to the point would be noting that $ \mu(E)\equal{}\int_Eg^p$ is relatively continuous with respect to Lebesgue measure, and $ \nu_n\le \mu$ in the ordinary inequality sense that $ \nu_n(E)\le\mu(E).$


[quote="junggi"]Let $k-\frac{1}{2}\not=0$ and $f(x)=x^{2}-(k-1)x+2k-1$.
If $f(x)=0$ has 2 distinct real roots, is $f(\frac{k-1}{4k-2})$
A. $>0$
B. $=0$
C. $<0$
D. none of the above(?)[/quote]
[hide="Idea"] Plug in $\frac{k-1}{4k-2}$ for $x$ in the equation and set equal to $0$. Multiply by $(4k-2)^{2}$, then use the fact that the determinant is $>0$. [/hide]

You can use the quadratic equation on the original equation to find the two roots. Then see if k-1/4k-2 will, just by looking at it, yield a higher or lower answer. Not sure. :ninja:


$a,b,c in R^{+},abc=1$
Prove that: $a^{2}b+b^{2}c+c^{2}a\geq ab+bc+ca$


linear algebra
matrix
vector
geometry
geometric transformation
homothety
linear algebra unsolved

Suppose that V is finite dimensional T  V-->V
Prove if ST = TS for every S   then Tis scalar multiple of the identity

Let S be a matrix with lots of 0s and a single 1.  By left and right multiplying T by S, it's easy to find that the off diagonal entries are zero, and the on diagonal entries are equal, which is what you want.

blahblahblah   can u explain more , I need to understand . Pls bear with me .
Thank u in advance

Take any matrix you want that is all zeroes except for one entry, which is a 1.  Call this S.  Now take a matrix with arbitrary coefficients T, and calculate TS and ST.  You will get a lot of 0s, and a few nonzero entries - but the nonzero entries will be in different places in TS and ST.  So set the corresponding entries in TS and ST equal, and you will be able to conclude that many of the entries of T are zero, and that other entries are equal.

Repeat this with a new S, where the 1 is in a different place.

take $x$ an arbitrary vector and $p$ a projector on $\mathbb{K}x$.
Since $T(x)=TP(x)=PT(x)$, we know that $x$ and $Tx$ are colinear. This works for every $x$.
Then conclusion follows (very well known that if $Tx$ is always colinear to $x$, $T$ is an homothety)


find all integers $a$ s.t there exist two different sequences of natural numbers like:
$i_1,i_2,...,i_k ,,,,,,,,j_1,j_2,...,j_l$
satisfying:
$(a^{i_1}+1)(a^{i_2}+1)...(a^{i_k}+1)=(a^{j_1}+1)(a^{j_2}+1)...(a^{j_l}+1)$

isn't there a solution?$a=0,+1,-1,+2,-2$ works.

has anyone the solution?what about you dna?

can't anyone solve this,i thought on it two days but it failed to any result.

Must the terms of the sequence be pairwis distinct?
If yes.every $a$ except $-1,0,1$ don't have the poperty.

[quote="dashmiz"]two sequences are different if they have no elements in common.
and see that $a=2$ works cuz:
$2^3+1=(2+1)(2+1)$.[/quote]
yes,you are right.But what I said is whether the terms of one sequence can be equal.the example you give have the a sequence which has two equal number(THE second sequence,$j_1=j_2=1$)
If the terms of one sequence must be distinct .the problem came to easy.
otherwise ,It's really hard.

you are right Hawk Tiger,in the case you mentioned the problem is very easy.but as the example the elements of one of sequences need not be different.


It appears that different users count "reviving" a topic many different ways, so I really wish to know exactly what counts as a "revival" of a thread? Is it qualitatively (like as in people probably forgot it) or quantitatively (as in like, one month)? And also, why is "reviving" a topic considered a bad thing? If somebody needs an explanation for a concept they do not understand that was posted in a thread, shouldn't they revive it to ask for help?

Useless or stupid revivals are bad, not the cases you mention. And yes, that happens much too often.

[quote="r15s11z55y89w21"]It appears that different users count "reviving" a topic many different ways, so I really wish to know exactly what counts as a "revival" of a thread? Is it qualitatively (like as in people probably forgot it) or quantitatively (as in like, one month)?[/quote]

Thanks for the explanation, but I still need to know how much "gap-time" counts as a revival of a post.

Generally, if a thread is over a month old, and you aren't adding anything new, don't revive it. If you are asking a question (related to the thread), or answering an unanswered question, that's fine.

In more generality, don't answer if you have nothing to say. That's rather independent of time.
You can express thanks to someone that explained something to you (in fact, that's rather good instead of bad behaviour), but don't revive threads for telling people how much you liked that solutions five pages and two years before.

Short:
The only rule that applies is called "common sense".

If there is new and useful content added to a thread, reviving is fine. There is no associated time limit.

Sometimes revivals are ill-advised because they are confusing. Suppose there was a fairly substantial topic in 2008 discussing the XYZ contest of 2008 - maybe mostly posters were talking before the contest about what teams they would be on or what to expect, and then after the contest, there was some more discussion of the result. That topic is now strongly time-linked.  If, just before the 2009 XYZ contest, someone revives the topic by posting a comment about the 2009 contest, then people will see the topic, see that it seems to be a busy topic, and start responding to it - in some cases, responding to posts that were made in 2008 without realizing that that's what they're doing. The whole thing just gets confusing.

Clearly, anything about the 2009 version of the contest should go in a brand new topic. (But not seven different brand new topics, which is a different problem.)


Χρειάζομαι μια λύση σ αυτό το πρόβλημα γιατί έχω να παραδώσω paper. Είχε 9 ασκήσεις και μου μένουν 2 να λύσω. Μία δεν την έχω προσπαθήσει ακόμα αλλά αυτή με έχει παιδέψει πάρα πολύ...

Να αποδειχθεί ότι υπάρχει αμφιμονοσήμαντη απεικόνιση του $ A$ στο $ A x N$ όπου $ A$ είναι infinite set και $ N$ είναι οι φυσικοί.

Όταν είναι αριθμήσιμο είναι εύκολο αλλά τι γίνεται όταν είναι uncountable?

Akyro... Lythike telika... Otan vrw ligo xrono tha grapsw th lysh m na yparxei...

Νομίζω ότι όταν δίνεις ένα τέτοιο πρόβλημα, πρέπει να λες ποια κομμάτια της θεωρίας υποτίθενται ότι είναι γνωστά. Αυτό γιατί στη συγκεκριμένη περίπτωση έχεις μια πρόταση της στοιχειώδους συνολοθεωρίας η οποία λέει ότι για
δύο άπειρους πληθάριθμους $ \mathfrak{a}$, $ \mathfrak{b}$:

\[ \mathfrak{a}\cdot\mathfrak{b}\equal{}\max(\mathfrak{a},\mathfrak{b})\]

Συνεπώς όχι μόνο το $ A$ είναι ισοπληθικό με το $ A\times\mathbb{N}$ αλλά και με το $ A\times A$ ή και με το $ A\times A\times A$!

Για συγκεκριμένα σύνολα μπορείς συνήθως να χρησιμοποιείς τη συγκεκριμένη μορφή τους για να αποδείξεις την ύπαρξη τέτοιων απεικονίσεων. Για τυχαία σύνολα χρειάζεται γενικά να επιστρατεύσεις το λήμμα του Zorn και το θεώρημα Schroeder - Bernstein.

Δεν ξέρω και πολλά απο συνολοθεωρία... Τώρα τα κάνω αυτά αλλά είναι πολύ intensive το course αλλά Zorn και Scroeder-Bernstein είναι γνωστά.

Αν χωρίζαμε από το well-ordering το A σε disjoint countable subsets τότε υπάρχει απεικόνιση καθενός από τα subsets στο N... Αλλά αρκεί αυτό???

Αναλυτικά

Γράφω τα στοιχεία του Α ως z+n όπου z δεν έχει predecessor... Άρα υπάρχει απεικόνιση του A στο B x N όπου Β το σύνολο των z.

Άρα από το A x N πάμε αμφιμονοσήμαντα στο (B x N) x N = Β x (N x N) και από κει στο B x N (ως γνωστόν υπάρχει αντιστοιχία του N στο N^2)

Από το B x N όμως πάμε στο A όπως στην αρχή... Άρα τελειώσαμε...

Δεν ξέρω αν είναι σωστό αυτό...

Η ιδέα σου είναι πολύ όμορφη και σωστή. Με λίγη προσοχή μπορείς να αποφύγεις τη χρήση διατακτικών ως εξής:

Βήμα πρώτο: Με απλή (αλλά ενδιαφέρουσα) χρήση του λήμματος του Zorn δείχνουμε ότι κάθε απειροσύνολο μπορεί να διαμεριστεί σε απειραριθμήσιμα υποσύνολα. Αν $ A$ είναι το σύνολό μας και $ B$ είναι το σύνολο των υποσυνόλων αυτών, τότε:

\[ A\sim B\times \mathbb{N}\]

Βήμα δεύτερο(ίδιο ακριβώς με το δικό σου): Τώρα έχουμε:

\[ A\sim B\times \mathbb{N}\sim B\times(\mathbb{N}\times\mathbb{N})\sim (B\times\mathbb{N})\times\mathbb{N}\sim A\times \mathbb{N}\]

Ευχαριστώ πολύ για τη βοήθεια και τις επισημάνσεις!


[hide]N adds 1 each time you divide 1988 by 2, and you need to divide until you get 0. Since $2^{10}=1024$, you will need to divide 11 twos. So, the output is 11.[/hide]

[quote="Quevvy"]INT(X) always returns the greatest integer less than or equal to X. What is the out put of the following BASIC-language algorithm?

Step 1: N = 0 : X = 1988
Step 2: When X = 0 then output N and go to Step 6
Step 3: Change value of X to INT(X/2)
Step 4: Change the value of N to N + 1
Step 5: Go to step 2
Step 6: End[/quote]

from what i understand, it is asking for the exponent of the greatest power of 2 less than 1988. is it 11 ($2^{10}$ and the other 1 for the last time to get it to 0)? i think i'm not off by more than 1

You guys are both right.

Gosh! I wish I would have thought of that while we were doing this...


Find all odd positive integer $ n$ such that : $ n^2$ don't divide $ (n\minus{}1)!$

[quote="conan_naruto236"]Find all odd positive integer $ n$ such that : $ n^2$ don't divide $ (n \minus{} 1)!$[/quote]
[hide="Solution"]
Clearly, an odd prime $ p$ will work since $ (p \minus{} 1)!\equiv \minus{} 1\bmod p$ by Wilson's Theorem, so $ p$ is not a factor of $ (p \minus{} 1)!$, so $ p^2$ is not a factor of $ (p \minus{} 1)!$.  Now, let $ n$ be a composite odd number.  Let $ n \equal{} kl$, where $ k$ and $ l$ are positive integers above $ 1$.  Since $ n$ is odd, $ k$ and $ l$ are odd.  Assume that $ k\ne l$.  Then, $ k,l\ge 3$ and $ k, l < p$.  This means that $ l \equal{} \frac {n}{k}\le \frac {n}{3}$, so $ 2l\le \frac {2n}{3} < n$.  Similarly, $ 2k < n$, so $ 2l, 2k < n$.  Then, as $ 2l\ne l, k$ and $ 2k\ne l, k$, we have that $ 2l\cdot 2k\cdot l\cdot k|(n \minus{} 1)!$.  This means that $ 4n^2|(n \minus{} 1)!$, so $ n^2|(n \minus{} 1)!$.  This contradicts the problem statement, so we move on to the next case.

If $ k$ must equal $ l$, then $ n$ must be equal to $ p^2$ for some prime prime $ p$.  As long as $ 5\le p$, we have that $ p, 2p, 3p, 4p<5p\le p^2$, so $ p\cdot 2p\cdot 3p\cdot 4p|(p^2\minus{}1)!$, so $ 24n^2|(n\minus{}1)!$, so $ n^2|(n\minus{}1)!$, which contradicts the problem statement.  If $ n\equal{}1$, then we have a contradiction as $ 1|1\equal{}0!$ and if $ p\equal{}3\implies n\equal{}9$, then $ 9^2\equal{}81$ is not a factor of $ 8!\equal{}(9\minus{}1)!$.

It follows that the only $ n$ that work are primes and $ 9$.  [/hide]


inequalities
Euler
inequalities unsolved

If $ ABC$ is an acute-angled triangle then prove that $ \sum\frac{a(a^2\plus{}2bc)}{b\plus{}c}\le\frac{4(R\plus{}r)^2(2R\minus{}r)}{R}$ .

[quote="Mateescu Constantin"]If $ ABC$ is an acute-angled triangle then prove that $ \sum\frac {a(a^2 \plus{} 2bc)}{b \plus{} c}\le\frac {4(R \plus{} r)^2(2R \minus{} r)}{R}$ .[/quote]

this ineq holds for all triangle...

using s-R-r, we have

$ \sum\frac {a(a^2 \plus{} 2bc)}{b \plus{} c}\le\frac {4(R \plus{} r)^2(2R \minus{} r)}{R}$

$ \Longleftrightarrow 2{s}^{2}{R}^{2}r \minus{} R{s}^{4} \plus{} 6R{s}^{2}{r}^{2} \minus{} 2{R}^{2}{r}^{3} \minus{} 5R{r}^{4} \plus{} 4{R}^{3}{s}^{2} \plus{} 8{R}^{4}r \minus{} 2{r}^{3}{s}^{2} \minus{} 2{r}^{5}\ge0$

$ \Longleftrightarrow \minus{} R{s}^{4} \plus{} \left( 4{R}^{3} \plus{} 6R{r}^{2} \minus{} 2{r}^{3} \plus{} 2{R}^{2}r\right) {s}^{2} \minus{} 5R{r}^{4} \plus{} 8{R}^{4}r \minus{} 2{R}^{2}{r}^{3} \minus{} 2{r}^{5}\ge0$

let

$ f(t) \equal{} \minus{} Rt^2 \plus{} \left( 4{R}^{3} \plus{} 6R{r}^{2} \minus{} 2{r}^{3} \plus{} 2{R}^{2}r\right)t \minus{} 5R{r}^{4} \plus{} 8{R}^{4}r \minus{} 2{R}^{2}{r}^{3} \minus{} 2{r}^{5}$

where $ t \equal{} s^2$, by Gerretsen Inequality, we have

$ 4R^2 \plus{} 4Rr \plus{} 3r^2\ge s^2\ge 16Rr \minus{} 5r^2$,

i.e. $ t \in [16Rr \minus{} 5r^2,4R^2 \plus{} 4Rr \plus{} 3r^2]$

so we just need to prove that

$ f(16Rr \minus{} 5r^2)\ge0$ and $ f(4R^2 \plus{} 4Rr \plus{} 3r^2)\ge0$

but we have

$ f(16Rr \minus{} 5r^2) \equal{} 4r\left( R \minus{} 2r \right)\left(r^2(11R \minus{} r) \plus{} R^2(18R \minus{} 25r)\right)$

and

$ f(4R^2 \plus{} 4Rr \plus{} 3r^2) \equal{} 4{r}^{2}\left( R \minus{} 2r \right) \left( {R}^{2} \plus{} Rr \plus{} {r}^{2} \right)$

by Euler Inequality, easy to prove that $ f(16Rr \minus{} 5r^2)\ge0$ and $ f(4R^2 \plus{} 4Rr \plus{} 3r^2)\ge0$ holds, so, inequality holds.

[quote="hedeng123"][quote="kuing"][quote="Mateescu Constantin"]If $ ABC$ is an acute-angled triangle then prove that $ \sum\frac {a(a^2 \plus{} 2bc)}{b \plus{} c}\le\frac {4(R \plus{} r)^2(2R \minus{} r)}{R}$ .[/quote]

this ineq holds for all triangle...

using s-R-r, we have

$ \sum\frac {a(a^2 \plus{} 2bc)}{b \plus{} c}\le\frac {4(R \plus{} r)^2(2R \minus{} r)}{R}$

$ \Longleftrightarrow 2{s}^{2}{R}^{2}r \minus{} R{s}^{4} \plus{} 6R{s}^{2}{r}^{2} \minus{} 2{R}^{2}{r}^{3} \minus{} 5R{r}^{4} \plus{} 4{R}^{3}{s}^{2} \plus{} 8{R}^{4}r \minus{} 2{r}^{3}{s}^{2} \minus{} 2{r}^{5}\ge0$
[/quote]

$ \equal{} R( \minus{} s^4 \plus{} (4R^2 \plus{} 20rR \minus{} 2r^3)s^2 \minus{} r(4R \plus{} r)^3)$

$ \plus{} \frac {r(9R^2 \minus{} 3rR \plus{} r^2 \minus{} Rr^2)(R(4R \plus{} r)^2 \minus{} 2(2R \minus{} r)s^2)}{2R \minus{} r}$

$ \plus{} \frac {r^3(16R^4 \minus{} 29R^3 \plus{} 8rR^3 \minus{} 23rR^2 \plus{} r^2R^2 \minus{} r^2R \plus{} 2r^3)}{2R \minus{} r}\geq 0$[/quote]

how do you get the last ineq holds?
 :blush:


How can I show that $ f(x) \equal{} x^3$ and $ f(x) \equal{} |x|^3$ are linearly independent on all real $ x$? Using algebra and calculus and not just drawing the two graphs and stating it's common sense?

THanks

Let $ a,b$ be such that  $ \forall x\in R, ax^3 \plus{}b|x^3|\equal{}0$
Put the values $ x\equal{}1$ , $ x\equal{}\minus{}1$ .. and so on ..
 :cool:

I was wondering if there was a method using the wronskian matrix.

The wronskian is zero, which does [b]not[/b] imply that the functions are linearly dependent. This is the point.


trigonometry
inequalities
limit
inequalities proposed

Prove that the inequality $ \sin{x}\geq x\minus{}\frac{x^2}{\pi}$ holds for all real $ x.$

hello, defining $ f(x)=\sin(x)-x+\frac{x^2}{\pi}$ and differentiating with respect to $ x$ we get $ f^'(x)=\sin(x)-1+\frac{2x}{\pi}$. The equation $ f^'(x)=0$ has only the solutions $ x=0$ or $ x=\pi$, because $ f^{''}(x)=-\sin(x)+\frac{2}{\pi}$ we get $ f^{''}(0)=f^'(\pi)=\frac{2}{\pi}>0$ and $ f(x)$ has at the points $ x=0$ and $ x=\pi$ local minima. Because $ \lim_{x \to \pm \infty}f(x)=\infty$ and $ f(0)=f(\pi)=0$, we have $ f(x) \geq0$ for all real $ x$.
Sonnhard.


algebra
polynomial
geometry
3D geometry
quadratics
superior algebra
superior algebra unsolved

is every cubic field extention of the field $ \mathbb{Q}$ of the form :maybe:  $ \mathbb{Q}\{a^{\frac{1}{3}} \}$ ?
The answer is no.  An idea to solve this is to consider an irrudicible polynomial in $ \mathbb{Q}$  with irrational zero's, for example $ x^3\minus{}3x\plus{}1$. If every cubic field extention of the field $ \mathbb{Q}$ is of the form $ \mathbb{Q}\{a^{\frac{1}{3}} \}$. Then these field must be isomorphic.
But in the case of qubic equations. There are two possibilities:
1) There is exactly one real solution and 2 complex solutions.
2) There are three real solutions.
But I need to show that there will be a  contradiction by using some kind of isomorphism that lead to the contradiction..
Any one?

@-oo-: the problem is to show that no such $ a$ exists.

1. You can show that the splitting field of $ x^3\minus{}a$ has degree $ 2$ or $ 6$, whereas there are polynomials (those with discriminant being a square) such that the splitting field has degree $ 3$.

2. Another way is that some splitting fields (like of $ x^3\minus{}3x\plus{}1$) are purely real, whereas those of $ x^3\minus{}a$ aren't. A field extension $ K$ of $ \mathbb Q$ is called "purely real", iff any embedding of $ K$ into $ \mathbb C$ is already an embedding into $ \mathbb R$. If it's an extension by one polynomial's root, this happens iff that polynomial has only real roots [as the embeddings are given by sending a root of that polynomial to some (other or the same) root].

indeed I'm able to read ..
$ a$ is not specified so it could be $ \in \mathbb{C}$. probably it should be rational or even integer, but I wanted to know that from badida

sorry for my late reaction. So the meaning of the question that there are several cubic field extentions of $ \mathbb{Q}$ and not only the ones $ \mathbb{Q}\{ a^{\frac{1}{3}} \}$. So the idea was to give an example of such field extentions.

An example that definitely works: find a cubic polynomial with three real roots and a perfect square discriminant. Adjoining any one of the roots gets us the splitting field of that polynomial, which is of degree three over $ \mathbb{Q}$.

As ZetaX noted, [b]any[/b] rational cubic with all roots real allows you to construct one of these by adjoining a root. Solving a cubic with one real root is equivalent to taking a cube root, along with some quadratic operations. Solving a cubic with all three roots real is equivalent to trisecting an angle, along with some quadratic operations. The two are different, and there's no way to get one from the other.

In that light, here's my example of a cubic field that isn't generated by a cube root: $ \mathbb{Q}\{\cos(20^{\circ})\}$.