The $ n$th term of a sequence is $ a_n \equal{} (\minus{}1)^{n\plus{}1}(3n \plus{} 2)$. What is the value of $ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_{100}$?

[quote="GameBot"]The $ n$th term of a sequence is $ a_n \equal{} ( \minus{} 1)^{n \plus{} 1}(3n \plus{} 2)$. What is the value of $ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_{100}$?[/quote]

$ 3 \cdot 1\plus{}2\minus{}3 \cdot 2\minus{}2\plus{}3\cdot 3\plus{}2\plus{}\hdots$

THe twos cancel, so

$ 3\minus{}6\plus{}9\minus{}12\plus{}\hdots\plus{}300$
$ 3(1\minus{}2\plus{}3\minus{}4\plus{}\hdots\plus{}100)$
$ 3(49 \cdot \minus{}1\plus{}100)$
$ 153$.

[quote="Dr Sonnhard Graubner"]hello, we have $ \sum_{n \equal{} 1}^{100}a_n \equal{} \minus{} 150$.
Sonnhard.[/quote]

Are you sure?

Wow epic fail. 101 isn't even... the answer above is correct.

A solution is:

Consider $ a_{n\plus{}1}\plus{}a_n$, for odd n.

$ \minus{}(3(n\plus{}1)\plus{}2)\equal{}\minus{}3n\minus{}5$
$ 3n\plus{}2\minus{}3n\minus{}5\equal{}\minus{}3$
$ \minus{}3 \cdot 50\equal{}\minus{}150$.


Define S(n)=1/1+1/2+....+1/n=A(n)/B(n)   ((A(n),B(n))=1)
Prove that infinite n  lest  A(n)=p^t  (p is prime)

If you meant "prove that there an infinite number of integers $ n$ such that $ A(n)$ is not the power of a prime"
then, you may find a solution, in french here :
http://www.animath.fr/cours/arithm.pdf

See pb # 166

Moreover, I'm quite sure that this problem already appeared in this forum.

Pierre.

I don't know french.Can you write english solution?


Estamos trabajando aqui en Colombia en un proyecto de un libro de combinatoria dirigido a olimpicos de alto nivel y de competencia... Probablemente seria unico en su genero, por lo tanto queremos hacerlo lo mas completo posible, pero necesitamos su ayuda!

Que temas, problemas, tecnicas piensa que este libro deberia incluir? (O le gustaria en su defecto?)

Personalmente creo que hay algo que nunca habia visto y no faltara POSETS!, algunas otras ideas? Teselaciones, Invariantes, grafos,  Permutaciones, Identidades combinatoricas, Combinatoria Geometrica, Combiantoria y teoria de Numeros, Funciones Generatrices, Teoria de Conjuntos, Biyecciones, Numeros complejos en combi, Principio de Casillas, Principio Maximal, Algebra lineal en combi, Conteos, Coeficientes Binomiales, Ramsey, Van der Waerden, El metodo Probabilistico, que opinan ustedes? cuales de estalista les gustan y cuales no tanto, o que otros?

Pienso que va a ser un libro muy muy bueno! :)
Sabes si sera publicado antes de la proxima OIM?

Hmm.. pienso tambien en (es posible que sea en tu "post", pero, como no conozco las nombres en espanol.. :oops:) : Diamond theory (maybe it'll be discussed with tilings though?) + teorema de las cuatras colores + (much less useful for olympiads (I think.. so I may be wrong!)) "design theory" y "matroid theory"

Sorry for my poor spanish. :(

Buenas,
quiero saludar la iniciativa de Colombia por hacer
esta obra.

[url=http://mathworld.wolfram.com/topics/Combinatorics.html]here[/url]

revisar
Saludos,
Carlos Bravo ;)

I don't know if it'll help you but... :
http://www.google.com/search?q=bosons+combinatorics
:)

Talvez esta pagina ayude: http://www.combinatorics.org/


topology
real analysis
real analysis unsolved

it is true that, for $n \to \infty$

$f_{n}\to f$ uniform almost everywhere if and only $\parallel f_{n}-f\parallel_{\infty}\to 0$

what's a counterexample if $f_{n}$ does not converge uniformly, but converges normaly

What do you mean by 'almost everywhere'? I'll give it a shot anyway.

Let $A$ be a subset of a metric space on which $\{f_{n}\}$ takes its domain.
Suppose $f_{n}\rightarrow~f$ converges uniformly. Given any $\epsilon>0$; For all $x\in~A$ there exists an positive integer $N$ such that $\parallel ~f_{n}(x)-f(x)~\parallel <\epsilon$ for all $n\geq~N$ and for all $x\in~A$. So $\parallel ~f_{n}(x)-f(x)~\parallel_{\infty}=\sup_{x\in~A}|f_{n}(x)-f(x)|<\epsilon$. Since $\epsilon$ was arbitrary, $\parallel f_{n}-f\parallel_{\infty}\rightarrow~0$. Conversely, suppose $\parallel f_{n}(x)-f(x)\parallel_{\infty}\rightarrow~0$. Given any $\epsilon>0$; there exists an positive integer $N$ such that $\parallel ~f_{n}(y)-f(y)~\parallel \leq\parallel ~f_{n}(x)-f(x)~\parallel_{\infty}<\epsilon$ for all $n\geq~N$ and all $y\in~A$. Hence $f_{n}\rightarrow~f$ converges uniformly.

Consider $f_{n}: [0,1]\rightarrow\mathbb{R}$ defined by $f_{n}(x)=x^{n}$.
$\{f_{n}\}$ converges pointwise ('normally') to $0$ if $0\leq~n<1$ and to $1$ if $x=1$.
But $\{f_{n}\}$ does not converge uniformly.

The term 'almost everywhere' means all but a finite number of points in this context..


Hello

I'm returning to math after a long gap, so please be gentle!  :blush:

I am trying to calulate the [b]area[/b] and also determine the [b]radius[/b] of a circle given the following:

A sector AOB has an area of 250cm² and the angle at the centre, O, is 75°.

Many many thanks for your help!

[quote="LynnelleYe"]This would probably be more appropriate for 'getting started'.
Anyway,
[hide="hint"]If the angle is 75 degrees, the sector is what fraction of the full circle, which you can think of as a sector with an angle of 360 degrees?[/hide][/quote]

Thanks for your reply, (sorry I posted in the wrong forum)... I still don't get it.  :(  :blush: Could you/anyone expand on this??  Thanks!

the formula for an area of a circle is pi*r 2 . If thae sector is 75 degrees, then it is 75/360 of the entire circle.

So 250cm 2 =(75/360)*pi*r 2

Now solve

[quote="darkphoenix"]Thank you Xantos C. Guin.

Ok, I have the following answers:
[hide]
Area=1200 cm2
Radius = 19.5441 cm
[/hide]

Is this correct?  Thank you![/quote]

yes.

for tthe radius, leave it in terms  of pi.

so like:

Radius=$\sqrt{\frac{1200}{\pi}}$

You also may want to rationalize thedenominator and simplify the radical into

$\frac{20\sqrt{3\pi}}{\pi}$


geometry
probability
function
articles
number theory
\/closed

Summer Independent Study schedules have now been posted [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].

I suppose the self-evident question for many AoPSers is whether they will exasperate their online teachers by doing a residential summer program in the same summer as doing an AoPS independent study course. I'm loving the AoPS courses for my son, but he is not loving having his time to do those courses squeezed by his accelerated curricular math course, and we expect that he will submit an application to a residential summer program. How much will it snarl up your system to have a student enrolled in the summer course who might be gone for a month in the middle of the course?

This is up to the student.  It's fine to miss some time during any class or program here at AoPS.  Most of the weekly topics are different enough that continuity is not completely necessary.  Students can always make up work or time after returning from a program and can use more of their office hour time during one part of the schedule than another.  The Independent Study is fairly intense, but summer time might help students with time allocation allowing students who miss some time to still soak up the material thoroughly.

A question about Independent study: I am planning to enroll in Intermediate Independent Study A for the May-August term. The course description says I will have access to one subject class of my choice. However, I have already taken both Intermediate Algebra and Intro to Counting/Probability, and I believe Olympiad Geometry is beyond me. Would I be able to keep the credit to take a subject class for the next term, perhaps Intermediate Number Theory or Intro/Intermediate Geometry, which I haven't taken?

Yes, students sometimes hold the class credit until a course comes along that they would like to take.

[quote="plokoon51"]i'm assuming that the intro to counting & probability will have a different time schedule than the intro independent study, right?[/quote]

The Independent Study does not adhere to a strict class schedule like our regular classes.  Office Hours are scheduled at various times throughout the week allowing students to attend around 30 total hours for instruction.  The hours are even different from week to week to allow the greatest scheduling flexibility for students.  There is no need to work about scheduling conflicts.

What type of project are we talking about here? For the Final Project, that is.

The final project is up to you.  First, it's not mandatory.  Second, you can pick any subject you want to delve into.  We have [url=http://www.artofproblemsolving.com/Resources/Papers/LaurendiPartitions.pdf]joml88's paper[/url] posted as an example.  He enjoyed learning about partitions of integers and generating functions and decided to learn partitions of integers well enough to write a problem solving paper about the subject.

I suggest working through the Independent Study material for a couple of months and then asking yourself if there was any one part of the material that really sparked your interest.  You might then either write an educational article about it or pick a particular problem in that area to research and write a paper about your results.  The nice thing about doing that is that if you plan to write such a paper for a scholarship (the Compton Fellowship at Washington University in St. Louis requires such a paper) or one of the major math/science fairs/talent searches, then you have either a completed project or at least a firm base.

can you post another example? joml88's paper doesnt seem to open in my computer :(

We're only in the second semester of the Independent Study program.  Of the two papers turned in at the end of the first, one is still being polished as it is a candidate for publication.  We plan to post papers as they come in, but this is the first and only posted so far.

[quote="xxreddevilzxx"]Am I allowed to print this???

And what are the benefits except that it gives a complete project, for other purposes like scholarship??[/quote]

You can't publish or sell the paper, but you can certainly print it out.

The benefits of writing such a paper are severfold.  You gain experience researching an area of mathematics as well as communicating those ideas effectively.  I'm sure this has some positive externality for helping learn to write better proofs as well.

[quote="cj"][quote="MCrawford"]

 Instructors provide detailed feedback on students' work throughout the independent study. The feedback includes not only comments on the accuracy of solutions, but also hints for approaching missed problems, suggestions for further study or review, and commentary on solution-writing style.

Students with audited enrollments are not eligible to receive instructor feedback on the challenge sets, nor will they receive Certificates of Completion. In all other respects, the audited enrollment is the same as regular enrollment.[/quote]

Are students with audited enrollments eligible to receive instructor feedback on the weekly assignment?[/quote]

No.  That's the main difference between the two enrollment types - the audited-enrolled students don't send in their homework for us to review.

THEN HOW CAN

"Instructors provide detailed feedback on students' work throughout the independent study"

?

I think MCrawford was referring there to regular enrollments.


If $n$ is a positive integer then there exist polynomials $f$ and $g$ with integer coefficients such that
\[\frac{1-x^{2^{n}}}{1-x^{2}}= f(x)^{2}+g(x)^{2}\]

Very interesting fact :yup: good to know

:idea: Is there a question that goes along with it?

I was too terse :) The problem is to [i]prove[/i] that claim.

Can one say anything about
\[\frac{1-x^{{a^{n}}}}{1-x^{a}}\]
similar?

Not in the same vein, since it is not true that $(a^{3}+b^{3})(c^{3}+d^{3})$ is some sum of cubes, or anything similar.

I guess its not an intuitive result, just a problem where we tack on identities (thats why its not really generalizable)


what is the hardest geometry in test?
1. isl
2. apmo
3. fkmo
please answer the this question
i think answer is no.1
 :?:  (i am korean. so icannot speak english very well. please understand me. :) )


function
inequalities
induction
inequalities unsolved

Prove that all functions $ f: [ 0, \infty ) \rightarrow [ 0, \infty )$ such: $ f^{' '}(x)>0$ for all $ x \in [ 0, \infty )$ satisfies:
\[ f\Big(\frac{x_{1}+x_{2}+...+x_{n}}{n}\Big) \le \frac{f(x_{1})+f(x_{2})+...f(x_{n})}{n}\]

[quote="zgorkster"]This is a very commonly known theorem called Jensen's inequality.  Here is a good link:

[url]http://planetmath.org/encyclopedia/JensensInequality.html[/url][/quote]

thx zgorkster..  :lol:

I didnt find any proof in the above link,so here is a proof for jensen inequality:

first we know that $ f$ is convex iff for every $ \lambda \in [0,1]$ we have:

$ f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)$

i.e. $ f(w_{1}x+w_{2}y)\leq w_{1}f(x)+w_{2}f(y)$ where $ w_{1},w_{2}$ are psoitive numbers s.t $ w_{1}+w_{2}=1$...

now we want to prove:

[b][u]jensen inequality:[/u]for convex function $ f$ we have:

$ f(\sum_{i=1}^{n}w_{i}x_{i})\geq \sum_{i=1}^{n}w_{i}f(x_{i})$ where $ \sum_{i=1}^{n}w_{i}=1$[/b]

[b][u]proof:[/u][/b]we prove the above inequality with induction on $ n$...

first of all for $ n=2$ it follows immediately from the definition of convex functions mentiond above...

now assume that its true for $ n-1$ i.e. for every $ n-1$ numbers $ w_{1},...,w_{n-1}$ with sum equal to $ 1$ we have:

$ f(\sum_{i=1}^{n-1}w_{i}x_{i})\leq \sum_{i=1}^{n-1}w_{i}f(x_{i})$

now we want to prove that:

$ f(\sum_{i=1}^{n}w_{i}x_{i})\leq \sum_{i=1}^{n}w_{i}f(x_{i})$

we have:

$ RHS=w_{1}f(x_{1})+\sum_{i=2}^{n}w_{i}f(x_{i})$

now put $ w'_{i}=\frac{w_{i}}{1-w_{1}}$ for every $ 2\leq i\leq n$ so we get that $ \sum_{i=2}^{n}w'_{i}=1$

now we ge that:

$ RHS=w_{1}f(x_{1})+(1-w_{1})\sum_{i=2}^{n}w'_{i}f(x_{i})$

now according to the hypothesis of the induction we get that $ \sum_{i=2}^{n}w'_{i}f(x_{i})\geq f(\sum_{i=2}^{n}w'_{i}x_{i})$ so the
last inequality becomes:

$ RHS\geq w_{1}f(x_{1})+(1-w_{1})f(\sum_{i=2}^{n}w'_{i}x_{i})$

now according to jensen for$ n=2$ the above inequality becomes:

$ RHS\geq f(x_{1}w_{1}+(1-w_{1})\sum_{i=1}^{n}w'_{i}x_{i})=f(\sum_{i=1}^{n}x_{i}w_{i})$

and we are done...


linear algebra
matrix
linear algebra unsolved

Let $ n>2$ be an even integer, and $ A\in \mathcal{M}_n(\mathbb{Z})$.
Show that there exist $ B,C \in \mathrm{SL}_n(\mathbb{Z})$, s.t. $ A\equal{}B\plus{}C$.

Hello JC_math

using  I get that $ A$ is equivalent to $ B\plus{}C$ with $ B,C \in SL(n,Z)$

Moubinool ?
The result is valid for $ n$ even (in particular for $ n\equal{}2$).
Let $ M\in\mathcal{M}_n(\mathbb{Z})$. Using the Smith normal form
http://en.wikipedia.org/wiki/Smith_normal_form
there exist $ P,Q\in{S}L_n(\mathbb{Z})$ and a diagonal matrix $ D\in\mathcal{M}_n(\mathbb{Z})$ s.t. $ M\equal{}PDQ$. Thus, using block matrices, it remains to prove the result for $ n\equal{}2,M\equal{}\begin{pmatrix}\alpha&0\\0&\beta\end{pmatrix}$ where $ \alpha,\beta$ are given integers.
$ \begin{pmatrix}\alpha&\minus{}1\\1&0\end{pmatrix}\plus{}\begin{pmatrix}0&1\\\minus{}1&\beta\end{pmatrix}\equal{}\begin{pmatrix}\alpha&0\\0&\beta\end{pmatrix}$.

I am agree Loup Blanc

with smith normal $ M\equal{}PDQ$ working on  $ D$ we have  $ D$ is EQUIVALENT to  $ B\plus{}C$ with $ B,C \in SL(n,Z)$

Finally $ M\equal{}R(B\plus{}C)S$ with $ R,S \in GL(n,Z)$ and  $ B,C \in SL(n,Z)$

But JC_math said for n even and any $ M \in M(n,Z)$, then  $ M\equal{}E\plus{}F$ with $ E,F \in SL(n,Z)$

JC_math did not talk about equivalent matrices http://mathworld.wolfram.com/EquivalentMatrix.html

How do you prove the assertion of JC_math
Let $ n>1$ be an even integer, and $ A\in \mathcal{M}_n(\mathbb{Z})$. Show that there exist $ B,C \in \mathrm{SL}_n(\mathbb{Z})$, s.t. $ A\equal{}B\plus{}C$

Hi Moubinool,
using the the Smith normal form we know that $ M\equal{}PDQ$ where $ D$ is a diagonal matrix and $ P,Q$ are invertible matrices in $ \mathcal{M}_n(\mathbb{Z})$ that is $ det(P)\equal{}det(Q)\equal{}\pm{1}$. We may assume $ det(P)\equal{}det(Q)\equal{}1$ because $ diag(1,\cdots,1,\minus{}1,1,\cdots,1).diag(a_1,\cdots,a_n)\equal{}diag(a_1,\cdots,a_{i\minus{}1},\minus{}a_i,a_{i\plus{}1},\cdots,a_n)\equal{}diag(a_1,\cdots,a_n).diag(1,\cdots,1,\minus{}1,1,\cdots,1)$


I  am from the Philippines and I bought AOPS volumes 1 and 2 books recently. The book is really good. I hope they could produce more books of the same quality.

Yes, they are. Especially the Introduction to ______ books. They're really helpful for middle school competitions like the AMC 8 and MATHCOUNTS.

None of the books except AoPS Vol. 1 and maybe Intro to Prob. and Counting help with AMC 8 as far as I can tell. AoPS is generally aimed at more advanced competitions.


I posted this originally in Pre-Olympiad, but no one made alot of progress.

Find, with proof, all natural numbers $n$ such that $n^{4}+7^{n}+47$ is a perfect square.

If n is odd, then $n^{4}+7^{n}+47=7(mod \ 8)$ is not perfect square. If n is even, then $n^{4}+7^{n}+47=0(mod \ 8)$. It give $n=2m,16m^{4}+49^{m}+47=(7^{m}+k)^{2}, \ 4|7^{m}+k$. If $m\ge 4$, then $16m^{4}+47<2*7^{m}+1$, therefore had not solution with m>3. It is easy to chek, that m=1 give not solution. m=2 (n=4) give $n^{4}+7^{n}+47=52^{2}$. n=6 give not solution.

[quote]I posted this originally in Pre-Olympiad, but no one made alot of progress. [/quote]
Huh?  :roll:   The problem is solved there in that forum ... many days ago!


I'm definitely taking Math IIC and physics, but I can't decide on the third one. I had originally decided I would take Latin, but I have only had 2 years of latin and my school doesn't offer any more. I only have one more test date before I apply early action to some place, so I need to do well on this third one, and I don't know if I could do that on the Latin one. But if I did do well, would it help show colleges that require 3 years of foreign language that I know what I'm doing, or does it really matter at all?

Also, have any of you taken the SAT II latin?

under the new SAT with writing in itself, I heard you are required/recommended to take only two.

IIC+Physics is a good combination, as well as IIC + Latin in my opinion.
just take what you can do well on.

as far as I remember from last year's experience, most schools take your november SAT score as well if you apply EA/ED. so you have two chances.
It might be a good idea to take one on each month, just because if you take them both in one day and IF you did bad on latin while you did well on physics, you cannot cancel only latin.
but again, its only one pro for taking it separately, so it is totally up to you.

I've never took a foreign language SAT, so I cannot answer the last question tho.

I hope it helped a bit.


clearly x1y1+x2y2<= xy. (using the cauchy-shwarz  ineq)
so (x1+y1)^2+(x2+y2)^2<= x^2+y^2+2xy<= (x+y)^2<=4max(x^2,y^2)
with equality iff x=y and x1=yq x2=y2.


conics
parabola
ellipse
hyperbola
geometry
3D geometry
trigonometry

In school we're studying conics. On the first day we learned all about how parabolas, ellipses, and hyperbolas are what you get when you slice a plane through a cone. The next day we learned that a parabola is the set of all points that are equidistant from a line and a point. What I want to know is, how do these two ideas connect? How do 'they' know that such a set of points is a slice of a cone? which came first, the cone or the focus/directrix?
Then in the homework problems we had to do a bunch of stuff with parabloa-shaped suspension bridges. How do they know suspension bridges have every point the same distance from some point as from some line? why can't a suspension bridge just be an arc of a circle?
I asked my teacher and he said something vague that didn't sastisfy me. I get all the math with the parabolas, I just don't get how they knew the different things were related.

Actually, a hanging chain has a form of so called [i]catenary[/i] or [i]funicular curve[/i], which closely resembles a parabola albeit it differs from it. It's general equation is $y=\frac{a(e^{bx}+e^{-bx})}{2}$ or, which is the same, $y=a\cosh bx$ where $\cosh t$ is so called [i]hyperbolic cosine[/i]. As far as I know, though, civil engineers approximate the catenary by a parabola since the latter curve is easier to deal with and the results are close enough for all practical purposes.

Take a  look.

[url]http://www.komal.hu/cikkek/dandelin/dandelin.e.shtml[/url]


Ring Theory
algebra
function
domain
superior algebra
superior algebra solved

Is the set of reals Unique Factorization Domains (UFD)?
Is the set of reals Principal Ideal domain (PID)?
What is irreducible elements of reals?

Can someone help me like to very-very stupid one?

Best regards
Tanja
_________________
Tamplushka

To consider a ring, you have to have a set, with an addition and multiplication on it
I suppose you take the usual binary operations?

Well, in  that case it is a PID and UFD

You see, it is a field.  Every nonzero element is a unit.  An irreducible element is a nonzero nonunit not decomposable into two units.  Well that does not occur here.

Basically, fields are not very interesting PID, as in a field the only ideals are the zero ideal and the total ideal

:) [b]I am very grateful for your ansver.[/b]

Thank You [size=200]very[/size] much.
Best regards
Tanja :ddr:


Let $ a,b,c>0$. Prove that

$ \frac{a^2\minus{}bc}{4a^2\plus{}4b^2\plus{}c^2}\plus{}\frac{b^2\minus{}ca}{4b^2\plus{}4c^2\plus{}a^2}\plus{}\frac{c^2\minus{}ab}{4c^2\plus{}4a^2\plus{}b^2}\geq 0$.

I saw two solutions proposed, but none a nice one...

See http://www.mathlinks.ro/Forum/viewtopic.php?t=123742.
I dont' know why my nice solution was not posted on Mathematical Reflections.


Let $x_1,x_2,\ldots,x_n$ be distinct positive integers. Prove that
\[ x_1^2+x_2^2 + \cdots + x_n^2 \geq \frac {2n+1}3 ( x_1+x_2+\cdots + x_n). \]
[i]Laurentiu Panaitopol[/i]

If there exists $x_i>n$, this means that there is an element $j$ of the set $\{1,2,...,n\}$ not equal to any of the $x_i$. Now consider $x_i^2-\frac{2n+1}{3}x_i>j^2-\frac{2n+1}{3}j$. This implies that $x_1^2+x_2^2+\cdots + x_n^2-\frac{2n+1}3( x_1+x_2+\cdots+x_n)$ is minimized when $x_i$ is a permutation of $(1,2,..n)$. The conclusion follows then.

This problem is posted in section Solved problems and put in Old and New ineqs   ;)

Nttu could you post the topic of this ineq?

Let $x_1<x_2<\cdots<x_n$ without loss of generality. Set $x_i = y_i +i$. Check that since $x_{i+1} = y_{i+1}+i+1 \ge x_i+1 = y_i+i+1$, we have $0\le y_1\le \cdots \le y_n$. Equipped with this, we next observe
\[
\sum_{1\le i\le n}x_i^2 = \sum_{1\le i\le n}y_i^2 + 2\sum_{1\le i\le n}iy_i + \frac{n(n+1)(2n+1)}{6}.
\]
Likewise
\[
\frac{2n+1}{3}\sum_{1\le i\le n}x_i = \frac{n(n+1)(2n+1)}{6} + \frac{2n+1}{3} \sum_{1\le i\le n}y_i.
\]
Hence, it boils down proving
\[
\sum_{1\le i\le n}y_i^2+2\sum_{1\le i\le n}iy_i \ge \frac{2n+1}{3}\sum_{1\le i\le n}y_i.
\]
Now, using Chebyshev's sum inequality, we have
\[
\sum_{1\le i\le n}i y_i \ge \frac13 \left(\sum_{1\le i\le n}i \right)\left(\sum_{1\le i\le n}y_i\right).
\]
Hence
\[
\sum_{1\le i\le n}y_i^2+2\sum_{1\le i\le n}iy_i \ge\sum_{1\le i\le n}y_i^2+ \frac{n(n+1)}{6} \sum_{1\le i\le n}y_i.
\]
With this, it boils down checking
\[
\sum_{1\le i\le n}y_i^2 + \frac{(n-1)(n-2)}{6}\sum_{1\le i\le n}y_i \ge 0,
\]
which is obvious. The equality holds only when (a) either $n=1,2$ or (b) $n\ge 3$ and $y_i=0$, that is $(x_1,\dots,x_n)$ is a permutation of $(1,2,\dots,n)$.


calculus
geometry
induction
real analysis
AMC
USA(J)MO
USAMO

[quote="david0184"]
I have started reading the proofs section to the AoPS, but it isn't enough. Would it benefit me by going through as many sections as I can in volumes 1 and 2 before I get on campus or would my time investment be more beneficial elsewhere? I'm not too inclined to purchase the textbook Apostol Calculus right now since I don't believe it goes too much into proofs. But in your opinion should I do that? I'm intimidated that I might not be able to keep up with my peers if I go in unprepared.
[/quote]

There are plenty of proofs in Apostol.  Much of the material in that first problem set (induction, proof by contradiction) would be covered in a first discrete math textbook, or a first abstract algebra text, or a first analysis text, or the AoPS books.  It's replicated in many places, so there's no need to get it all out of a particular text.

However, later problemsets appear to be (unsurprisingly) completely on calculus/analysis type topics.  For these you should have a proper rigorous calculus text, and while there are some options here (Spivak, Apostol, Rudin (more difficult), etc), you may as well just buy Apostol now since you'll need it for the course anyways and it [i]is[/i] the recommended text.

David, first of all, be confident. Confidence goes a long way in helping you be more successful in mathematics!! If you have the confidence that you can do it, your mind will find a way to make it happen!

I agree with blahblahblah's advice that you should go ahead and get Apostol and start with the first chapter. In fact, it's advice you should follow all throughout college in all of your courses. Try to stay ahead in each of your classes by reading ahead in the book. That way you'll know where things are going and be able to ask better questions during class.

In addition to that, a book that helped me get ready for college mathematics was [u]Analysis: with an Introduction to Proof[/u] by Steven R. Lay. The first two chapters might be helpful to you.

Thank you both for your suggestions. I'm off to buy Apostol, which is surprisingly expensive.  :o

I'm glad I found this.  I'm a rising junior at Caltech, so I took math 1a a while ago.

Relax.  You're fine.  I repeat, relax.  Math 1a is supposed to wreck freshman.  It wrecked me, it wrecked most of my friends, it wrecked most everyone I ever met.  But guess what, you're on pass/fail!  It's ok to get wrecked.  Ok, wrecked is a strong word, but never before Ma 1a have I been proud to get a 60% on an exam.  I went into the class having no idea how to do proofs (as did most people FYI), and I did about average.  I've never qualified for USAMO, and neither have most Caltech students.  Be humble, but please don't be intimidated.  And for the record, all other core math classes are easier Ma 1a.

In fact, my advice is not to prepare at all.  You're supposed to use the first 2 terms of pass/fail to figure out how to adapt to the workload.  Preparing ahead of time just kind of ruins that.

I recommend you spend your last summer before Tech doing something fun.  And when you get here, don't spend your whole first week getting all your homework done early.  It's far more important you be involved in Rotation.

[quote="dauntless"]You're supposed to use the first 2 terms of pass/fail to figure out how to adapt to the workload.  Preparing ahead of time just kind of ruins that.[/quote]  Preparing ahead of time is one very sensible way to adapt to a heavy workload  :|  :)

[quote]Preparing ahead of time is one very sensible way to adapt to a heavy workload[/quote]

That's very true.  I just don't recommend it for Caltech freshmen.  Unless you're a bio or geo major, it gets much harder sophomore year, and you will be on grades then.  My reasoning is that while you're on pass/fail and grades don't matter, you shouldn't prepare ahead of time in order so that you kind of shock yourself into learning how to effectively manage your time studying subjects that are very difficult and taught at breakneck speed.  That way, you're used to whole "drinking from the firehose" thing before grades and the real classes start sophomore year.  Learning how to properly pull and recover from all nighters, and how to learn a subject you don't understand for a problem set in a day are useful skills you should learn early at Tech (though the all nighter thing really only applies if you have bad time management, or are a ChemE, EE, or physics major).

Preparing ahead of time for hard courses in your major after freshman isn't a bad idea, though it's generally frowned upon.

Again, it's just my recommendation as a student who's been there and done that.  The OP can obviously do whatever he wants, and it doesn't really matter to me.

All I can say is - I greatly disagree. I know [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=80077#80077]Richard Rusczyk would disagree also[/url].

Many people get better at writing proofs by reading proofs -- I know this worked for me -- so this may be a good place to start. Read over several proofs, and at each step, try to figure out:

1. Why that step is valid
2. How it helps the proof writer get to his/her desired conclusion
3. What might've motivated the proof writer to realize that step

From studying 1 and 2, you can get a feel of how to write an effective proof, and studying 3 might give you some ideas of things you can look for/try out when you're looking for the next step to take. When you've developed a sketch of your proof, go back through and make sure that at each step, 1, 2, and 3 seem clear, in that order of importance.

[u][b]Regarding time management in freshman year:[/b][/u]

I'm not sure if I understand [b]dauntless[/b]'s argument correctly, so I won't respond to it directly -- I'll just speak generally.

I don't think there can ever be a good reason to waste time. If you can prepare for your freshman year courses ahead of time, your transition into college life will be much smoother. Also, it'll open up a lot of free time that you can use to do other productive things, including (but not limited to):

- Join clubs/teams
- Explore research opportunities early on
- Get a job to make money and build your resume
- Personal projects
- Take more classes to get ahead or to experiment
- Meet faculty, especially helpful at small schools
- Network/socialize/make friends*
- Work out*
- Have fun*

Basically, if you take one step that lightens your semester, you can use the saved time to do other things that you would not have been able to do. So, it would ultimately, you can still prepare your time-management skills, just as if you didn't prepare your classes, but you'll end up getting much more out of your freshman year, and thus set yourself up for a richer overall college experience.

[i]* - Yes, these are productive![/i]

I know a lot of what I'm saying sounds irresponsible and unconventional.  I believe in having good time management skills, and keeping up with work.  I'm not advocating cramming, or falling behind, or being lazy.  In fact, if you're lazy or fall behind, you're pretty much screwed.  Cramming is something you'll have to learn how to do given our 2 day reading period before finals and our profs' tendency to still assign problem sets during midterms week, and up to the day before the reading period.

I have a lot more I could say on the topic, but I would just like to say that this is the same advice that most Caltech undergrads give to prefrosh, and that it is only specific to Caltech, due to it's unconventional nature.  I've given this same advice to prefrosh before, and they've never told me that I was wrong.  In fact, they mostly complain about how they tried to prepare and how it was mostly useless...  At any rate, the OP will soon come to understand what it is that I'm trying to say.

I don't disagree with any of the advice given, as I think it's all good advice.  I just think due to Caltech's style of curriculum that a lot of it becomes somewhat impossible to implement (though certainly not wrong).

Hey cool I'm a Caltech rising freshman too.

Sweet.

Anyways, I'm in ma1 also so I'll let you copy.

Ok no.

So here's my advice. While it is a fairly good idea to get on top of Apostol or some other similarly difficult calculus textbook, I suggest a different approach. I would instead purchase an introductory textbook on number theory (Number Theory by George E. Andrews is a good one I used) or a discrete math book instead. I think that this approach would work better because you have presumably taken calculus of some sort already, and your major weakness is proofs. These books are typically meant to feel as an introduction to proving statements, covering induction, pigeonhole, contradiction, etc., your standard proofing techniques. Plus they have good problems which you won't be bored over or resort to some previous snippet of calculus knowledge you knew previously. As blah^3 said earlier, pretty much any intro book will cover these, so choose a subject you're vaguely interested in.

If you feel like you want to specifically target calculus proofs, I'd suggest problems from high school calculus competitions such as [url=http://sumo.stanford.edu/smt/]SMT[/url] or HMMT. These typically test above your BC level and develop problem-solving skills, especially in the area of calculus. Sometimes you'll need the techniques you learned from proofs in handling these. Maybe. But if you want to focus on calculus-type proofs specifically, then apart from looking at baby Rudin on Google Reader (guilty), what you can do is take easy Putnam problems.

Yea I'm cheap so I don't like buying heavy textbooks when I can have problems for free. Hence the above emphasis on not buying books. But if you can (and it looks like we'll have to) go for it.

Rising junior at Caltech.

I agree with dauntless, in that ma1A should not be a reason for efforts on too much preparation. The course is designed so that you learn new skills, theorems and proofs in a very relaxed and collaborative environment. It is P/F so there is no stress. Going to your lectures, sections, collaborating with your friends and learning how to make better decisions with your time are much more important in the long run than just forcefully memorizing a couple of proofs before getting here.
(Note the exaggeration so that it's clear that I don't disagree with anyone above, just saying, you know.. the obvious :P )
What I would do is spend time on my favorite subjects/books be those number theory/ calculus/ puzzles etc.. (If it must really be studying/reading )
That way you learn and you enjoy your time, plus you will have more interesting topics to discuss with your friends(If you take the Putnam preparatory class, or actually even in Ma1 I know some TA's love talking about random math in their sections :P).

[quote="david0184"]Hi all,

I will be entering Caltech the fall of 2009. At my high school, the only class I've taken with proofs was geometry in 8th grade, but the proofs were neither rigorous or well presented, so they didn't really count. I have not had any other training in proofs. After looking at the first problem set for the math 1a, which I will be taking this upcoming term, I've realized that I have no foundation whatsoever in proofs. As a matter of fact, it's very difficult for me to even follow the solutions. I know that this is epically sad for a to-be caltech student, which is why I'm here to ask what I can do to prepare/brace myself for this upcoming course. The 1st problem set is found here: [url]http://www.math.caltech.edu/classes/ma1a/08Ma1aHWSol1.pdf[/url]

The link to all ma 1a problem sets can be found here: http://www.math.caltech.edu/classes/ma1a/index1.html

I have started reading the proofs section to the AoPS, but it isn't enough. Would it benefit me by going through as many sections as I can in volumes 1 and 2 before I get on campus or would my time investment be more beneficial elsewhere? I'm not too inclined to purchase the textbook Apostol Calculus right now since I don't believe it goes too much into proofs. But in your opinion should I do that? I'm intimidated that I might not be able to keep up with my peers if I go in unprepared.

Thank you so much in advance for your help. I look forward to hearing your comments. I would like to be as prepared as I can be when I get on campus.

Dave[/quote]

david0184
Did you [b]ever,ever consider[/b] what a mathimatical proof ,or in general a proof ,is???

Over the last couple of weeks τρικλινο's been making posts exploring the question of what constitutes a sufficient mathematical proof. To his credit, it is actually a legitimate question. But his opinion is in the extreme minority about what level of formality a proof needs in order to be considered legitimate. If you look through his last few posts you'll see what I mean.

[quote="david0184"]Hi all,

I will be entering Caltech the fall of 2009. At my high school, the only class I've taken with proofs was geometry in 8th grade, but the proofs were neither rigorous or well presented, so they didn't really count. I have not had any other training in proofs. After looking at the first problem set for the math 1a, which I will be taking this upcoming term, I've realized that I have no foundation whatsoever in proofs. As a matter of fact, it's very difficult for me to even follow the solutions. I know that this is epically sad for a to-be caltech student, which is why I'm here to ask what I can do to prepare/brace myself for this upcoming course. The 1st problem set is found here: [url]http://www.math.caltech.edu/classes/ma1a/08Ma1aHWSol1.pdf[/url]

The link to all ma 1a problem sets can be found here: http://www.math.caltech.edu/classes/ma1a/index1.html

I have started reading the proofs section to the AoPS, but it isn't enough. Would it benefit me by going through as many sections as I can in volumes 1 and 2 before I get on campus or would my time investment be more beneficial elsewhere? I'm not too inclined to purchase the textbook Apostol Calculus right now since I don't believe it goes too much into proofs. But in your opinion should I do that? I'm intimidated that I might not be able to keep up with my peers if I go in unprepared.

Thank you so much in advance for your help. I look forward to hearing your comments. I would like to be as prepared as I can be when I get on campus.

Dave[/quote]

Hey Dave.
I'm a rising senior at Caltech. Well, in fact, according to Caltech traditions, I'm already a senior, but that's another story.
Dave, I understand your desire to prepare for something that seems very foreign. It's normal.  When I came to Tech, proofs were pretty new to me too. I had a bit of a hard time getting used to the whole "deriving from axioms" approach of Ma1a and Tommy (Apostol).  But I read, and I tried, and I read, and I tried, until I felt I understood.
By no means did I get 100% of the full score in Ma 1a, but I did well;  later I went on to take harder Math classes feeling pretty good about proofs.

If you _really_ want to, then by all means do start preparing for Ma 1a. I personally liked Tommy and still own both volumes.
However, what I urge you to do this summer is actually to think about all the reasons why you find science exciting. What is it that you want to do with science, or in science?
Read stuff on it, think about it, dream about it!

Now let me explain why I think this may turn out to be a good idea in the long-run: After the firehose starts pouring (ie. when school starts), you will sooner or later find yourself trying to play catch up with classes, life, sleep etc.  It'll become very easy to loose the broader perspective; you may forget why you're doing this and how does this fit into your bigger plans, you may (as I have) make decisions that are locally optimal, but broadly not.
To prevent this from happening, I believe it is very important that each Techer remember why they're in this. The reason is (or are), ideally, something that deeply resonates with you.
[i]This is when you should think back on your dreams and remember your bigger goals.[/i]

Some may say that whatever goals or aspirations one can have before even starting college will turn out to be silly, unrealistic or otherwise useless.
They're wrong. Having a bigger perspective and a plan is _always_ good, even if you end up changing it often.

Whatever you end up doing, relax and don't stress out. You'll have plenty of time for that over the next four years.

[quote="AndersoN"]

Hey Dave.
I'm a rising senior at Caltech. Well, in fact, according to Caltech traditions, I'm already a senior, but that's another story.
Dave, I understand your desire to prepare for something that seems very foreign. It's normal.  When I came to Tech, proofs were pretty new to me too. I had a bit of a hard time getting used to the whole "deriving from axioms" approach of Ma1a and Tommy (Apostol).  But I read, and I tried, and I read, and I tried, until I felt I understood.
By no means did I get 100% of the full score in Ma 1a, but I did well;  later I went on to take harder Math classes feeling pretty good about proofs.

If you _really_ want to, then by all means do start preparing for Ma 1a. I personally liked Tommy and still own both volumes.
However, what I urge you to do this summer is actually to think about all the reasons why you find science exciting. What is it that you want to do with science, or in science?
Read stuff on it, think about it, dream about it!

Now let me explain why I think this may turn out to be a good idea in the long-run: After the firehose starts pouring (ie. when school starts), you will sooner or later find yourself trying to play catch up with classes, life, sleep etc.  It'll become very easy to loose the broader perspective; you may forget why you're doing this and how does this fit into your bigger plans, you may (as I have) make decisions that are locally optimal, but broadly not.
To prevent this from happening, I believe it is very important that each Techer remember why they're in this. The reason is (or are), ideally, something that deeply resonates with you.
[i]This is when you should think back on your dreams and remember your bigger goals.[/i]

Some may say that whatever goals or aspirations one can have before even starting college will turn out to be silly, unrealistic or otherwise useless.
They're wrong. Having a bigger perspective and a plan is _always_ good, even if you end up changing it often.

Whatever you end up doing, relax and don't stress out. You'll have plenty of time for that over the next four years.[/quote]

Thanks for your reply Anderson. :)

[quote="david0184"]Well, a proof is a formal series of statements showing that if one thing is true something else necessarily follows from it. Why ask such an obvious question?  :roll:[/quote]

What is "a formal series of statements" ,can you give an example ,if you like, please?

What is the meaning of "necessarily follows"?

I have two not very helpful and kind of obvious things to say that are not specific to Caltech.

1)  Unless your earlier preparation was inadequate*, it is neither expected nor required that you spend any time whatsoever studying in advance for a math class, so you shouldn't feel pressure to do so.

2)  It never hurts to study anything on your own.

You can study ahead if you want to improve your performance in the class (which is true for any class) or just for fun, or you can spend your time doing something else that you feel has more value to you.  Specifically, if your main motivation for studying ahead was that you're afraid that you'll flunk out of the class simply because of your inexperience with proofs, don't bother.  As mentioned above, most of your peers will have no proof experience either.

* which I'll arbitrarily think of as skipping prereqs and/or having a math background at a level far below your peers.

(Biographical note:  I honestly don't remember doing anything the summer before college besides sleeping and watching anime with friends.  I also learned the matrix definition of the derivative.  I think I must have done other things, but I can't remember.  Maybe reading the course catalog.)

@τρικλινο, please don't hijack threads.  It's not polite.

[quote="yenlee"]

@τρικλινο, please don't hijack threads.  It's not polite.[/quote]

When and where

The OP's question had to do with concerns about preparing for college. You've been trying to force-start a conversation about what constitutes a formal proof.

There is no evidence that people in this thread were, at any point, interested in abandoning the original topic to discuss your question (and there is evidence to the contrary). So while the question is an interesting one, it doesn't really belong here.

I see that you've created previous threads that closely revolved around the topic of formal proofs [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=291701]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293832]here[/url]. If you want to discuss it, it would be more appropriate to do so in those threads. Although, I think it's clear what everyone thinks, so there's not much left to discuss anyway. Thanks!  :)


Let ABC be a triangle. Find the minimum value of
\[\frac{1}{r}.\sqrt [6]{\frac{s}{r}}\left[\left(l_{a}+l_{b}\right)\sqrt{\tan\frac{C}{2}}+\left(l_{b}+l_{c}\right)\sqrt{\tan\frac{A}{2}}+\left(l_{c}+l_{a}\right)\sqrt{\tan\frac{B}{2}}\right] \\ \\l_{a}\ angle\ bisector \\s\ semiperimeter \\r\ radius\]

$\left( l_{a}+l_{b}\right)\sqrt{\tan \frac{C}{2}}+\left( l_{b}+l_{c}\right)\sqrt{\tan \frac{A}{2}}+\left( l_{c}+l_{a}\right)\sqrt{\tan \frac{A}{2}}=K \\By\ AM-GM\ we\ have\ K\geq 3.\sqrt[3]{ \left( l_{a}+l_{b}\right)\left( l_{b}+l_{c}\right)\left( l_{c}+l_{a}\right) .\sqrt{\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}}}\\\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}=\frac{r^{3}}{(s-a)(s-b)(s-c)}=\frac{A}{s^{2}}\\ l_{a}+l_{b}\geq2\sqrt{l_{a}.l_{b}}\\ l_{a}+l_{b}\geq2\sqrt{l_{a}.l_{b}}\\ l_{a}+l_{b}\geq2\sqrt{l_{a}.l_{b}}\\\left( l_{a}+l_{b}\right)\left( l_{b}+l_{c}\right)\left( l_{c}+l_{a}\right)\geq 8.l_{a}.l_{b}.l_{c}\geq8.h_{a}.h_{b}.h_{c}= 8abc\sinh\sinh\sinh=\frac{16A^{2}}{R}\ \ \ \ (*) \\\$$

Can we multiply inequalities and is (*) correct? And if it's correct, what to do next?

Why $h_{a}h_{b}h_{c}=abc$ and why $abc=\frac{2A^{2}}{R}$

sorry my mistake,I didnt read the post carefuly,I edited my post,you can see the corrected result in my post above...

[quote="night^wish"][color=darkred]Let $ABC$ be a triangle. Find the minimum value of
\[K=\frac{1}{r}.\sqrt [6]{\frac{s}{r}}\left[\left(l_{a}+l_{b}\right)\sqrt{\tan\frac{C}{2}}+\left(l_{b}+l_{c}\right)\sqrt{\tan\frac{A}{2}}+\left(l_{c}+l_{a}\right)\sqrt{\tan\frac{B}{2}}\right] \\ \\ l_{a}\ angle\ bisector \\ s\ semiperimeter \\ r\ radius \][/color]
[/quote][color=darkblue]
[b]Proof.[/b] We have: \begin{eqnarray*}\frac{1}{r}\cdot\sqrt[6]{\frac{s}{r}}\cdot\left[\sum\left(l_{b}+l_{c}\right)\sqrt{\tan\frac{A}{2}}\right]&\ge& \frac{1}{r}\cdot\sqrt[6]{\frac{s}{r}}\cdot 3\sqrt[3]{\prod\left(l_{b}+l_{c}\right)\cdot\sqrt{\prod\tan\frac{A}{2}}}\\ &\ge& \frac{1}{r}\cdot\sqrt[6]{\frac{s}{r}}\cdot 3\sqrt[3]{\prod 2\sqrt{l_{b}l_{c}}}\cdot\sqrt[6]{\prod\frac{r}{s-a}}\\ &\ge&\frac{1}{r}\cdot\sqrt[6]{\frac{s}{r}}\cdot 6\sqrt[3]{\prod h_{a}}\cdot\sqrt[6]{\frac{r}{s}}\\ &=&6\cdot\frac{1}{r}\cdot\sqrt[3]{\prod h_{a}}\end{eqnarray*}

On the otherhand, \[\prod ah_{a}=8s^{3}r^{3}\Longrightarrow\prod h_{a}\cdot 4Rrs=8s^{3}r^{3}\\ \Longrightarrow \prod h_{a}=\frac{2s^{2}r^{2}}{R}\Longrightarrow\sqrt[3]{\prod h_{a}}=r\cdot\sqrt[3]{\frac{2s^{2}}{Rr}}\ge 3r\]

Hence $K_{\min}=18$[/color]

Thanks very much for your help. But why $\sqrt[3]{\frac{2s^{2}}{Rr}}\ge3$?

It's well-known and not very hard  :)
We have: $abc\le\frac{1}{27}(a+b+c)^{3}\implies 27\cdot 4Rrs\le 8s^{3}\implies 27Rr\le 2s^{2}\implies\boxed{\sqrt[3]{\frac{2s^{2}}{Rr}}\ge 3}$


topology
function
integration
geometry
3D geometry
sphere
Functional Analysis

Prove that if $F$ is a subspace of $E=C([0,1],\mathbb{R})$ such that the norms induced on $F$ by the $||.||_\infty$ and $||.||_2$ norms are equivalent then $F$ is of finite dimension. What happens if there will be $||.||_\infty$ and $||.||_1$ norms ?

First of all, notice that the second problem can be reduced to the first one: on $\mathcal C([0,1]),\ \|\cdot\|_{1}\le\|\cdot\|_{2}\le\|\cdot\|_\infty$, so if the supremum norm and the $\|\cdot\|_{1}$ norm induce the same topology on $F$, then the supremum norm and the $\|\cdot\|_{2}$ norm also induce the same topology on $F$.

Now for the first problem:

We assume that $F$ is infinite-dimensional, and seek a contradiction.

The closure of $F$ in $\mathcal C([0,1])$ with respect to the $\sup$ norm coincides with its closure in $\mathcal L^{2}([0,1])$ with respect to $\|\cdot\|_{2}$, since $F$ has the same Cauchy sequences in the two norms. We may thus assume WLOG that $F$ is closed. Being an infinite-dimensional subspace of the separable Hilbert space $\mathcal L^{2}([0,1]),\ F$ has an infinite orthonormal basis $(e_{n})_{n\in\mathbb N}$ consisting of continuous uniformly bounded functions on $[0,1]$.

We cannot have $e_{n}\to 0$ pointwise, because then Lebesgue’s Dominated Convergence Theorem would imply that $1=\int e_{n}^{2}\to 0$, which is absurd. This means that we can find $x\in[0,1]$ such that $e_{n}(x)$ does not converge to $0$. For our purposes, we can assume WLOG that $|e_{n}(x)|>t,\ \forall n\in\mathbb N\ (*)$ for some $t>0$. For every $\ell^{2}$ sequence $(\alpha_{n})_{n},\ \sum\alpha_{n}e_{n}$ converges in $\|\cdot\|_{2}$ and hence in $\|\cdot\|_\infty$ to an element of $F$. In particular, $\sum\alpha_{n}e_{n}(x)\ (**)$ is a convergent series. Take $\alpha_{n}=\frac 1{ne_{n}(x)}$. $(*)$ shows that this is indeed a square-summable sequence, but $(**)$ is clearly divergent (it’s the harmonic series). We have our contradiction.

A more general result (I was reminded of it by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=225739]this other thread[/url]):

Let $ F$ be a subspace of $ C([0,1],\mathbb R)$ on which the supremum norm and the $ L^p$ norm are equivalent for some $ 1\le p < \infty$. Then $ F$ is finite-dimensional.

Since the norms $ \|\cdot\|_p$ and $ \|\cdot\|_\infty$ are equivalent, we can complete the space with respect to the supremum norm, so we assume WLOG that it was complete to begin with. Also, I'll assume that $ 1 < p$, since $ p \equal{} 1$ was treated above. Being a closed subspace of $ L^p$ for $ 1 < p < \infty,\ F$ is reflexive and separable. It follows that its unit ball is a compact metrizable space in its weak topology. Furthermore, the unit sphere is dense in the unit ball. Hence we can find a sequence $ (f_n)_n$ in the unit sphere of $ F$ with respect to $ \|\cdot\|_p$ which tends weakly to $ 0$. But the weak topologies generated by $ \|\cdot\|_p$ and $ \|\cdot\|_\infty$ coincide, so $ f_n$ tends to $ 0$ pointwise. Also, $ (f_n)_n$ is bounded in the supremum norm. By the Dominated Convergence Theorem it follows that $ \|f_n\|_p\to 0$, contradicting the fact that the functions $ f_n$ were chosen in the unit sphere of $ L^p$.


Does anyknow know how to write colored latex formulas ?

for istance if we want to type $\frac{a+b+c}{3}$ , it will be typed as black. Can we change it to red?

Note that the normal color code [code][color=red]\frac{a+b+c}{3}[/color][/code] doesnt work.Below , you can see the differnt places of $\$$ signs and Color command. look what happends to it :D

$[color=red: 3ed8164561]\frac{a+b+c}{3}[/color: 3ed8164561]$
____________________________________________________________
$\frac{[color=red: 3ed8164561]a+b+c[/color: 3ed8164561]}{3}$

LaTeX is basically an image that you create. You can't change it's color, it's always black (as far as I know).

As i_like_pie says $\frac{a+b+c}{3}$ is an image so can't be coloured by using the [color ] tag which is for text . But you can change colour in LaTeX itself by using the color package with
\color{red}\frac{a+b+c}{3}
However, it won't work here unless you manage to persuade the administrators to add the color package to the forum.


Let $a,b,c\ge 0$ such that $a+b+c=3$. Prove that
\[a^{2}+b^{2}+c^{2}\ge \frac{3}{2}(ab+bc+ca-abc) \]

Which one is true $a+b+c=1$ or $a+b+c=3$ ??

If $a+b+c=3$

So the problem equivalent with

$2(a+b+c)(a^{2}+b^{2}+c^{2})\geq 3(a+b+c)(ab+bc+ac)-9abc$
$\leftrightarrow 2(a^{3}+b^{3}+c^{3})\geq a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b$

Which true by Muirhead


[hide="Hint 1"]
Complementary counting. In other words, find the [b]number of cases in which the conditions are NOT satisfied[/b], and then subtract that from the [b]total possible number of cases[/b].
[/hide]

[hide="Hint 2"]
Treat the applicants that can't sit next to each other as a single person. Remember that there are 2 ways to arrange them in this "hyper-person" (i.e. if the two people are A,B, and it can be AB or BA if they are sitting next to each other).
[/hide]

Imagine that there are actually 4 people and 4 chairs. (The people that can't sit next to each other are the same person.) There are 4! = 24 ways to seat them. Multiply by 2 to get 48, because the two people could be AB or BA. The total number of ways you could seat them is 5! = 120 ways. 120 - 48 = 72 ways to seat them.

The easiest way to solve this problem is by complimentary counting, which just means, find the total number of arrangements, and subtract the number of arrangements that are not aloud.

There are 5!, or 120 arrangements, and there are 48 ways to arrange them, such that two are sitting next to each other.

120 - 48, equals 96.

oops, sorry about that  :oops:

Thanks for fixing it AIME15


[quote="archimedes1"]When looking over my class transcript for my current class Int. Trig/Complex Numbers, I sometimes see stuff like

[b]rrusczyk[/b] (20:26:13)
This is DeMoivre. Now what? How does this help?""MCP","3999","2007-08-02 20:26:13","Classroom1","joshuazucker","[RM]","L/7==R/7 (mod 5)""MCU","4000","2007-08-02 20:26:14","Classroom1","plumbst","","the mod stays the same since 7 shares no common factors with 5""MCU","2322","2007-08-02 20:26:15","Classroom3","Nexmus","","L/7==R/7 (mod 5)

Weird...  Intro number theory showing up in the middle of De Moivre's...  It seems like part of another class is getting mixed in.

Also, during class when I type something like
[code]"negative" root of i[/code]
the class transcript adds more """"s.[/quote]

We have just updated the transcripting program which may alleviate these happenings.

That being said, since it's a new upgrade, there are bound to be some new bugs (or even some old ones) in the system.  Please submit any bugs you find to me through PM.

Thanks


Pls help.

This problem requires the use of calculus.) Con-
sider a Cobb–Douglas production function with
three inputs. Kis capital (the number of ma-
chines),Lis labor (the number of workers),and H
is human capital (the number of college degrees
among the workers).The production function is
Y=K^1/3L^1/3H^1/3.
a. Derive an expression for the marginal product
of labor.How does an increase in the amount
of human capital affect the marginal product of
labor?
b. Derive an expression for the marginal product
of human capital.How does an increase in the
amount of human capital affect the marginal
product of human capital?
c. What is the income share paid to labor? What
is the income share paid to human capital?
In the national income accounts of this
economy, what share of total income do you
think workers would appear to receive? (Hint:
Consider where the return to human capital
shows up.)
d. An unskilled worker earns the marginal prod-
uct of labor,whereas a skilled worker earns the
marginal product of labor plus the marginal
product of human capital.Using your answers
to (a) and (b),ﬁnd the ratio of the skilled wage
to the unskilled wage.How does an increase in
the amount of human capital affect this ratio?
Explain.
e. Some people advocate government funding of
college scholarships as a way of creating a more
egalitarian society. Others argue that scholar-
ships help only those who are able to go to
college. Do your answers to the preceding
questions shed light on this debate?

Do not double post! It does not increase your chances of getting an answer, just irritates people. ;)

If you feel that your post was placed in a wrong forum, just PM a moderator asking to move it. Note that he may disagree with you.

This particular post is fine for both economics and calculus. Since the discussion has already started in the calculus subforum, please, continue there. Having two parallel threads devoted to exactly the same question just creates a lot of confusion.


modular arithmetic
number theory unsolved
number theory

If p is a prime not equal to 3, then why is it true that there exists some r such that $3r \equiv 1 \pmod{p}$?

For any $(a,b)=1$ exist $q,r$, suth that $br-qa=1$ or $br=1(mod a)$.
If p is prime $p\not =3$, then $(p,3)=1$, $b=3, a=p$ get your result.


function
real analysis
real analysis unsolved

Hi everybody

Let f be a function W-->(E,G) Where G is Sigma Algebra
Show that
f ={A belong to W/ ( Exist)  B belong to G such that A = f^-1(B)) IS A SIGMA ALGEBRA

NOTE SIGMA ALGEBRA = BOREL SET.

THANK U IN ADVANCE

There's really no reason why you wouldn't be able to do this ... where are you stuck ?


In a non-decreasing sequence $a_{1},a_{2},...$ all values are positive integers and exactly k values equal to k (for all $k \in Z^{+}$).Find all the primes of the form $a_{1}+a_{2}+...+a_{n}$

Clearly, the sequence is $1,2,2,3,3,3,4,4,4,4,5 \cdots .$
Now, let $p$ be a prime. Let $n$ be maximal such that the sum of the first $n$ squares, namely $\frac {n(n+1)(2n+1)} 6 $, is not greater than $n$. Thus, $p$ can be written as the desired form if and only if there exists $k \in \{0,1, \cdots , n \}$ (1) such that $p = \frac {n(n+1)(2n+1)} 6 + k(n+1).$
This reduces to $6p = (n+1)(2n^2+n+6k).$

Note that if $n \geq 6$ then each of the two factors of the rhs is greater than $6$, which contradicts that this product has to be $6p.$
Thus, it leaves $n \leq 5.$

Direct checking, using (1), gives that the solutions are $p=3, 5, 11, 61,67,73,79.$

Pierre.


$ f(x,y): R^2 \minus{} > R$ satisfying:

i) for any fixed $ x_0$, $ f(x_0,y)$ is measurable for y.
ii) for any fixed $ y_0$, $ f(x,y_0)$ is continuous for x.

Prove that f(x,y) is measurable on $ R^2$

This is false. Assuming AC, one can construct a non-measurable set that intersects each line by just 2 points. I believe [b]grobber[/b] posted the construction on AoPS once. I have too little time right now to repost it or to find the link :(.

Thank you for your reply. I have modified the problem a little and I think it is true now. Lookin' forward to a proof. :D

this post has sunk...I really need s solution...nobody?

Take a fine grid $ x_k\equal{}k\delta$ ($ k\in\mathbb Z$, $ \delta>0$) on the $ x$ axis and consider the function $ f_\delta(x,y)\equal{}f(x_k,y)$ for $ x_k\le x<x_{k\plus{}1}$. This function is measurable (why?). Now make the grid finer and finer. You'll get a sequence of measurable functions that converges to $ f$ pointwise. Finally, just recall that a pointwise limit of a sequence of measurable functions is measurable.


Here's a problem that was given by one of my students.

Suppose you have a straight picket fence with n boards to be painted using the colors blue and white. How many ways are there to paint all the boards of the fence, if no two adjacent boards are to be painted blue?

We consider the double recursion w_(n+1)=w_n+b_n, b_(n+1)=w_n, with w_1=b_1=1. w_n counts the number of n-fences ending with a white one. b_n counts the number of n-fences ending with a blue one. Then b_n=w_(n-1), so w_(n+1)=w_n+w_(n-1), which is the Fibonacci recurrence. Thus w_n=F_(n+1) and b_n=F_n, so w_n+b_n=F_(n+2).

I'm kind of confused.  What's the underline thing mean?

_ here just is to avoid confusion with multiplication, ie w_n is w subscript n, w being a function, not w times n. I see it was missed out once though  :lol:


Inspired from Magic Square Sum and Magic Square Product:

Is it possible to have a magic rectangle sum (yes, I know a square is a rectangle, but let's consider rectangles that aren't squares), where all the rows add to a number n, and all the columns add to the same number n.

Is this a challenge or do you need help with it? Cause if you need help I'll give you a hint otherwise I'll leave it for others.

You're allowed using any numbers you want, except n cannot equal 0.

[quote="SnowStorm"][hide]
I don't think it's possible without reusing numbers
Say we have
a b
c d
e f
Then a + b = a + c + e
So b = c + e

This means that c + e + d + f would have to be equal to c + d which isn't possible if they are distinct positive integers or negative integers.
[/hide][/quote]

It doesn't say that it has to be a 2x3 rectangle.


Solve for $ x$: $ x^{x\plus{}1}\equal{}(x\plus{}1)^x$ in the reals.


please choose one option and name the number. better not use google or some other searching tools.

Two interesting sources:

[url=http://en.wikipedia.org/wiki/United_States_casualties_of_war]US Casualties of War[/url]

[url=http://en.wikipedia.org/wiki/Casualties_of_the_conflict_in_Iraq_since_2003]Casualties of the Conflict in Iraq since 2003[/url]

Well I was convinced that it was gonna be about the Iraqi population, because the USA Army casualties are very well documented compared to the Iraqi casualties, which is a source of much debate.

I was gonna say : more than 100.000 .  Well that is what Robert Fisk said.
He claims that if you look at the casualties in just one city in one month, it has to be that much.

It's interesting that when one talks about casualties in the army, they sometimes forget there is a large number of UK troops in Iraq too, and more than hundred of them have died as well.

[quote="Buffalo"]For the number, I mean the rough magnitude, i.e. 2000 is the same as 5000, they both mean thousands. I noticed that there are three votes on the second, pieterminate, I guess you are one of them, what's the 50000 referring?
[b]Without[/b] the help of searching machine, do you have a rough idea of the numers?[/quote]
I said [b][i]total[/i][/b].  And um, yeah, that number was by Google News a few months back, so I don't know what it would be now; I haven't checked.


Find all $x_0,x_1$ such that the sequence defined by $x_{n+1}=x_{n-1}\frac{x_n}{3x_{n-1}-2x_{n}}$ contains infinitely many positive integers.


In the United States, coins have the following thicknesses: penny, 1.55 mm; nickel,
1.95 mm; dime, 1.35 mm; quarter, 1.75 mm. If a stack of these coins is exactly 14 mm high,
how many coins are in the stack?

We cheat and note that $ \boxed8$ quarters gives us a stack $ 14$ mm high.

I noticed that the sum of 1.35 and 1.75 was .10 over an even dollar amount and that 1.95 was .05 under so I tried 1.35, 1.75 and 2 1.95.  That summed to 7.00 so I just took 2 sets to get 14.

[quote=Lamboreghini]I used the floor function to solve this problem.[/quote]

Lamboreghini's method seems efficient, can someone explain how?

Using the Floor Function? The quesstion says "EXACTLY 14 mm high"
I think that, as math154 did, we should check each coin individually before trying combinations


Calculati urmatoarea limita fara a folosi L'hospital:\[
\mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{x^3 }} = ?.
\]

Am reusit sa rezolv aceasta peoblema.Notam limita ceruta cu l.
Avem:\[
\mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{x^3 }} = \mathop {\lim }\limits_{x \to 0} \frac{{2(\frac{x}{2} - \sin \frac{x}{2})}}{{x^3 }} + \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \frac{x}{2}(1 - \cos \frac{x}{2})}}{{x^3 }} = \frac{l}{4} + \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \frac{x}{2}}}{{x^2 }} = \frac{l}{4} + \frac{1}{8} \Rightarrow l = \frac{1}{6}.
\]

Metoda 2:
Notam x=3t si calculam limita din f(x).Rezulta:f(x)=$\frac{3t-sin3t}{(3t)^3}=\frac{3(t-sint)+4(sint)^3}{27t^3}=\frac{f(t)}{9}+\frac{4}{27} $
Deci l=$\frac{1}{6}$.

scuze.mai greu cu latex. sunt incepator si daca nu pot scrie ceva o las balta. normal era
$\sin x=x-\frac{x^3}{6}+o(x^5)$

nu te ajuta la rezolvare dak o scrii asa pt ca o sa ai $\frac{\cdots+o(x)}{x^3}$ ... scrii $o(x^4)$
inainte era bine doar ca trebuia sa fie $o(x^5)$ in loc de $o(x)$. :D
$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+o(x^{2n+2})$ cum urmatorul termen in dezvoltarea asta asimtotica ar fi $(-1)^{n+1}\frac{x^{2n+3}}{(2n+3)!}=O(x^{2n+3})=o(x^{2n+2})$ (*)deci in loc de $o(x^{2n+2})$ ai putea sa scrii si $o({x^k})$ cu $k\le 2n+1$ dar asta nu e asa de "tare" :) in schim nu poti sa scrii in loc de $o(x^{2n+2})$ pe $o(x^{2n+3})$ (asa cum ai scris in ultimul post) pentru ca (vezi (*) ). :)


A piece of string is $ 210$ feet long and is cut into two pieces. If the length of one piece is $ \frac25$ the length of the other, what is the number of feet in the shorter piece?

7x/5=210
x=150

The longer piece is 150 feet, so the shorter is 2*150/5=60 feet.


i like the Lamborghini Gallardo :D
it looks cool and it has a vicous engine.

Eh. Aston Martin Vanquish, but the Mercedes-Benz SLR McLaren is a close second. My only thing with the SLR is that it costs about a half million and is nowhere as comfortable as the Vanquish. We'll overlook its supremacy in engine power for now.

Geez I love this thread. Heck, this deserves a sticky. *nods*

+1 for sticky

1990-96 Nissan 300ZX, 350Z, RX7's, Supra's, Subaru WRX STi, pretty much all Japanese sports cars. 300zx is my favorite.

[quote="dinoman"]McLaren F1....:love:[/quote]

-.- That doesn't count. Hmph.

[quote="furious"][quote="GoBraves"]How about a nice Civic?[/quote]
if you're trying to get great gas milage....awesome car
if you're trying to look cool or go fast...not so much.[/quote]

Well I guess that rules out the hybrids.  :D

see.. what i dont understand is why nice cars have to cost 300 grand.. get a nice Mustang Convertible, spend like 20-30 grand only, look awesome in it...  :lol:  - besides if you own like.. ANY of the above cars (besides the civic) its much more likely to be stolen.

In SUV style, i love the idea of a Maybach....

[quote="ToBe"]In SUV style, i love the idea of a Maybach....[/quote]

Maybach...um...57 or 62? Neither resembles an SUV remotely... O.O

But I like the 62. Very nice for limos and whatnot. And uh...yeah. It's not just the "fast" or "cool". It's the fact that you know your car is worth 10 times what the ones next to you are and has 10 times the acceleration. It's the fact that you can roll down your windows, put on your shades, look at the guy in the convertible next to you (hope it's not a Carrera GT convertible) and burn him when the light turns green. :D And then it's the fact that who you are is what you drive. So uh...onwards with the Vanquish.

But to be honest, a nice F430 would be okay too. And uh...can't rule out a Modena...eh some 360 or whatnot...and uh... :)

I have this thing against Lamborghinis though. They're just...I don't know. They annoy me. I mean what is up with this company and light, blinding colors! Come on their cars are all in yellow, light green, orange...I mean sheesh.

[quote="hwenterprise"]What is your favorite car?

I personally like the BMW M3 GTR, or the BMW 760Li Sedan.

I'm just into BMWs. :D[/quote]

'69 Chevelle SS with 454 engine :)

[quote="MithsApprentice"][quote="K81o7"]I say go for a 787B...
Or an Escudo Pikespeak  :D  and tweak it up to 3khp or so...[/quote]

The uh...not street legal one? -.-[/quote]

Yeah...the rally car
Imagine taking an F1 car and driving around on the street...going under stuff, etc...

[quote="furious"][quote="K81o7"]Yeah...the rally car
Imagine taking an F1 car and driving around on the street...going under stuff, etc...[/quote]

then imagine getting pulled over, getting a ticket, and whatever other penalties there are for driving a non-street legal car. what fun.[/quote]

Need for speed man! jk :P  If it were only a ticket, it might be worth it...but they would revoke your license most likely which would not be fun :(

[quote="K81o7"]Well I mean a situation where you wouldn't be pulled over for it.  And if they try to pull you over, you could flee.  In which case...bad.  Just assume you're invisible or something.  That would be awesome.
But seriously, as for street-legal cars, I like Skylines.  But then you might get pulled over for excessive driftin and skidding.  :D
Heck, I don't actually drive, so I can't say much as yet.[/quote]
trying to outrun the cops =  bad idea. they have radios and helicopters for a reason.
also, skylines are not street legal in the US. There is a company that does a conversion to make it legal and then import it to America but then your skyline costs 100k

No people come on...seriously. What's wrong with just a plain old SLR? or a Carrera GT?

what's wrong with plain old sunglasses, miths?  *coughs*

Ok whatev.  I do agree that obsessing over cars is pretty much useless.  Most boys in my town seem to do so... so whatev.

[quote="bubala"]what's wrong with plain old sunglasses, miths?  *coughs*[/quote]

The Oakleys looked better! Besides, they weren't that expensive. And I wear them a lot so it's worth it. -.-

And just fyi, the SLR and Carrera GT aren't exactly "plain old". :P

[quote="hwenterprise"]No one likes the Corvette?[/quote]

Corvette is nice...it's just...american... :D
Yeah it's good...

Nothing wrong with getting like an old Corvette (60's and 70's?) and doing something or other to it...

It's not not really at the top of most people's lists these days...

I don't know anything about cars because I'm a girl :D

but I'd like one of those flashy silver Porcsche-y convertibles... but in light pink, orange, lime green, yellow or baby blue. most likely lime green. :)

or one of those volkswagon (sp? lol) beetles... they're pretty cute and girly.

[quote="Chigr"]How hasn't anyone mentioned the Saleen S7? If that isnt the nicest car currently in production then i really dont know what is....[/quote]

Um...yeah that's one of my least favorite cars. Have you ever tried to drive one?

Almost everyone who likes one hasn't :P Horsepower and speed isn't everything in a car. The interior of those things...man...I think every review ever done on the car says it's impossible to handle and feels like a metal rack inside.

Well the fact is cars of that speed are not going to be easy to drive, then again, do you want just anybody driving a car at 200mph. And many of the amenities people are used to are just not worth the added weight in such a car.  Sure speed and horsepower arent everything in a car, but it is probably the most important thing in car like this.  And the most important thing in car you spend half a million bucks on is that it looks fabulous and turns heads, its about how it looks, not the ammenities your low-end can minivan provide.
btw I dont think you've driven one either  :P

I strongly disagree. There's plenty of excellent cars with high-end price tags (Mercedes SLR, Ferrari Enzo, Aston Martin Vanquish, Porsche Carrera GT, Maybach 62, Rolls Phantom, etc...). All of them have far better interiors. And to be honest, if you want to turn heads...get one of those. People actually know what those are. On the other hand, only people actually interested in high-speed cars would know what a Saleen is (although there are probably quite a few who have heard of the founder).

Just for the record, critics claim that the fastest car "just about anyone" can drive is the Mercedes SLR McLaren, which can hit upwards 205 mph.

I'd say if you were going to spend a half million on a car, get one that's nice to ride and drive. You're obviously not going to be hitting up 200 mph on the freeway anyways. They've got racetracks for that purpose. I would be very satisfied with anything that could do 160 and have something like a 4.0 sec 0-60 with those luxuries. I assure you, there are plenty of things in a Vanquish that you don't see in a low-end minivan...or a high-end minivan for that matter.

But if all you're looking for is something "exotic", then yeah the Saleen would be the right choice.

if you want a really fast car and you have 1.2 million dollars, the Bugatti Veyron is for you. i believe its the fastest production car, at 250 mph.
http://www.autoweek.com/article.cms?articleId=103261

Has everybody forgotten about Ferrari? The new Superamerica 575M is the fastest covertible in the [b]world[/b]  :ninja:

[quote="slamchess"]whats the point of extra horsepower if u cant use it on the streets as miths said it is the interior that counts[/quote]
I agree, theres no point in getting a car with tons of horsepower if you can't use it.  I guess they just have more horsepower to make you buy it.

my favorite muscle car was the '69 Charger Hemi
the best supercar is the next-gen Murcielago (will have 1000 hp!)
best sports car- Lotus Exige or Cayman S
luxury car- Phantom (Rolls)
outrageously expensive car- Maybach Exelero
SUV- G-class AMG


prove that if $|a|+|b|+|c|\leq \sqrt{2}$ ,then the equation $x^{4}+ax^{3}+bx^{2}+cx+1=0$
does not have a real root.
answer Aviatorpilot its an order :D


Let a,b,c be nonnegative real number such that

a+b+c >= abc.

Prove that:        a^2 + b^2 +c^2 >= abc.

The first problem from the Balkan olympiad in 2001 states that in the given conditions $a^2+b^2+c^2\ge \sqrt 3 abc$, so it's stronger. We can make it even stronger: $ab+bc+ca\ge \sqrt 3 abc$. In order to show this:

Apply the inequality $x+y+z\ge \sqrt{3(xy+yz+zx)}$ for $x=bc,y=ca,z=ab$ to get $ab+bc+ca\ge\sqrt{3abc(a+b+c)}\ge \sqrt 3abc$, QED.


inequalities
LaTeX
inequalities unsolved

let $ a,b,c>0$ such that $ abc\equal{}1$; show that:
$ a^2b\plus{}b^2c\plus{}c^2a \ge \sqrt{( \frac{1}{a}\plus{} \frac{1}{b}\plus{} \frac{1}{c})(a\plus{}b\plus{}c)} \ge a \sqrt{c} \plus{} b \sqrt{a} \plus{}c \sqrt{b}$

Any ideas? Personally I found it easy to medium.....

[quote="wallace"]let $ a,b,c > 0$ such that $ abc \equal{} 1$; show that:
$ a^2b \plus{} b^2c \plus{} c^2a \ge \sqrt {( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(a \plus{} b \plus{} c)}$
[/quote]
It's equivalent to $ \sum_{cyc}\left(\frac{a^2}{c^2}\minus{}\frac{a}{c}\plus{}\frac{a}{b}\minus{}1\right)\geq0,$ which true by AM-GM.

[quote="wallace"]let $ a,b,c > 0$ such that $ abc \equal{} 1$; show that:
$ a^2b \plus{} b^2c \plus{} c^2a \ge  a \sqrt {c} \plus{} b \sqrt {a} \plus{} c \sqrt {b}$

[/quote]
AM-GM gives:
$ a^2b \plus{} b^2c \ge 2b \sqrt {a}$,do the same thing for the others then sum.

to prove
the inequality between the two expression to the right you have to use AM-HM and cauchy scwarz
If you wan't more details ask, I don't post it because I'm not used to Latex :P
Daniel

Rachid , if you want to prove that $ a\geq b\geq c$ it is not necessary to prove $ a\geq c$  :P

[quote="wallace"]let $ a,b,c > 0$ such that $ abc \equal{} 1$; show that:
$ \sqrt {( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(a \plus{} b \plus{} c)} \ge a \sqrt {c} \plus{} b \sqrt {a} \plus{} c \sqrt {b}$[/quote]
by cauchy scwarz :  $ a \sqrt {c} \plus{} b \sqrt {a} \plus{} c \sqrt {b} \le \sqrt{ab\plus{}bc\plus{}ca} \sqrt{a\plus{}b\plus{}c}\equal{}\sqrt {( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(a \plus{} b \plus{} c)}$


How to prove this?
For all $ n \in \mathbb{N}$
\[ \frac{1}{\binom{n}{1}}\plus{}\frac{1}{2\binom{n}{2}}\plus{}\frac{1}{3\binom{n}{3}}\plus{}...\plus{}\frac{1}{n\binom{n}{n}} \equal{} \frac{1}{2^{n\minus{}1}}\plus{}\frac{1}{2\cdot2^{n\minus{}2}}\plus{}\frac{1}{3\cdot2^{n\minus{}3}}\plus{}...\plus{}\frac{1}{n\cdot2^{0}}\]

Nice identity! My solutions is very messy. :blush:

[hide]We will proceed by induction. The base case is trivial. Now assume the result is true for $ n$. We shall prove that
\[ \frac{1}{\binom{n + 1}1} + \frac1{2\binom{n + 1}2} + \cdots + \frac1{(n + 1)\binom{n + 1}{n + 1}} = \frac1{2^n} + \frac1{2\cdot 2^{n - 1}} + \frac1{3\cdot 2^{n - 2}} + \cdots + \frac1{(n + 1)\cdot 2^0}
\]
But the RHS equals
\[ \frac12\left(\frac{1}{2^{n - 1}} + \frac{1}{2\cdot 2^{n - 2}} + \frac{1}{3\cdot 2^{n - 3}} + ... + \frac{1}{n\cdot 2^{0}}\right) + \frac1{n + 1} = \frac12\left(\frac{1}{\binom{n}1} + \frac1{2\binom{n}2} + \cdots + \frac1{n\binom{n}{n}}\right) + \frac1{n + 1}\]
(by the induction hypothesis).
So it is sufficient to prove that
(1) \[ \frac{1}{\binom{n}1} + \frac1{2\binom{n}2} + \cdots + \frac1{n\binom{n}{n}} = \frac{2}{\binom{n + 1}1} + \frac2{2\binom{n + 1}2} + \cdots + \frac2{n\binom{n + 1}{n}}\tag{1}\]
Let's assume $ n$ is even (if $ n$ is odd, we can prove analogously). Note that
$ \frac1{k\binom{n + 1}{k}} + \frac1{(n - k + 1)\binom{n + 1}{n - k + 1}} = \frac1{k\binom{n}{k}}$ (this is easy to verify). So we do pairings on the RHS of (1). Simplify the equation again (let $ n = 2m$), it becomes
\[ \frac1{\binom{2m}1} + \frac1{2\binom{2m}2} + \cdots + \frac1{m\binom{2m}m} = \frac1{(m + 1)\binom{2m}{m + 1}} + \frac1{(m + 2)\binom{2m}{m + 2}} + \cdots + \frac1{2m}.\]
We can easily check that the the $ k$-th term of LHS equals the $ n + 1 - k$-th term of RHS. So we are done.[/hide]

I have just found this problem, with Johan Gunardi's solution (which I don't find messy at all...), in a Bulgarian problem book, with "USA 2001" as the source. But it's not from the USAMO. Can anyone find the true source?

  darij

I remember seeing it in homework problems of MOP, maybe 2005 or something.


[hide] Since $log_{b}(a)=c, b^{c}=a$.  Squarerooting both sides, we get $(\sqrt{b})^{c}=\sqrt{a}\Rightarrow log_{\sqrt{b}}(\sqrt{a})=c$.  So, the numerator is c.  Now the denominator: since $b^{c}=a, b^{(2)(\frac{c}{2})} a\Rightarrow log_{b^{2}}{a}=\frac{c}{2}.$  So, the denominator is c/2.  $\frac{c}{\frac{c}{2}}=2$, which is our answer. [/hide]

[quote="Ignite168"]If $log_{b}(a)=c$, find the numerical value of $\frac{log_{\sqrt{b}}(\sqrt{a})}{log_{b^{2}}(a)}$.[/quote]

[hide]$log_{\sqrt{b}}\sqrt{a}=\frac{log \sqrt{a}}{log \sqrt{b}}=\frac{log a}{log b}$

$log_{b^{2}}a=\frac{log a}{log b^{2}}=\frac{1}{2}\frac{log a}{log b}$

therefore the answer is $\boxed{2}$.[/hide]


geometry
trigonometry
geometric transformation
rotation
invariant

Let ABC an equilateral triangle . Prove that $ ab\plus{}bc\plus{}ca\equal{}a^2\plus{}b^2\plus{}c^2$ , where $ a,b,c\in \mathbb{C}$ are the affixes of the points $ A,B,C$ .

[quote="mrmohamed"][quote="Pain rinnegan"]Let ABC an equilateral triangle . Prove that $ ab \plus{} bc \plus{} ca \equal{} a^2 \plus{} b^2 \plus{} c^2$ , where $ a,b,c\in \mathbb{C}$ are the affixes of the points $ A,B,C$ .[/quote]
 :huh:  :huh:  :huh:
[img]http://www.mathmontada.net/vb/uploaded/17_1257842517.gif[/img]
with my best wishes
mrmohamed[/quote]

I don't think that $ AB\equal{}c$ , i think that $ AB\equal{}|a\minus{}b|$ . By $ c$ i reffered to the affix of the point $ C$ , not the length of $ AB$

[quote="Pain rinnegan"][quote="mrmohamed"][quote="Pain rinnegan"]Let ABC an equilateral triangle . Prove that $ ab \plus{} bc \plus{} ca \equal{} a^2 \plus{} b^2 \plus{} c^2$ , where $ a,b,c\in \mathbb{C}$ are the affixes of the points $ A,B,C$ .[/quote]
 I don't think that $ AB \equal{} c$ , i think that $ AB \equal{} |a \minus{} b|$ . By $ c$ i reffered to the affix of the point $ C$ , not the length of $ AB$[/quote][/quote]
sorry for that sir
i will tray again
thanks

[quote="Pain rinnegan"]Let ABC an equilateral triangle . Prove that $ ab \plus{} bc \plus{} ca \equal{} a^2 \plus{} b^2 \plus{} c^2$ , where $ a,b,c\in \mathbb{C}$ are the affixes of the points $ A,B,C$ .[/quote]

Triangle ABC is equilateral, so $ \triangle ABC\sim \triangle BCA$ .

$ \implies \left\|\begin{array}{cc}
\frac{AB}{BC}\equal{}\frac{BC}{CA} \\
 \\
\angle ABC\equal{}\angle BCA \end{array}\right\| \Longleftrightarrow$ $ \left\|\begin{array}{ccc}
\frac{|a\minus{}b|}{|c\minus{}b|}\equal{}\frac{|b\minus{}c|}{|a\minus{}c|} \\
 \\
\arg\frac{a\minus{}b}{c\minus{}b} \equal{} \arg\frac{b\minus{}c}{a\minus{}c}\end{array}\right|\ \implies$ $ \frac{a\minus{}b}{c\minus{}b}\equal{}\frac{b\minus{}c}{a\minus{}c}$

$ \Longleftrightarrow\ a^2\minus{}ac\minus{}ab\plus{}bc\equal{}bc\minus{}b^2\minus{}c^2\plus{}bc\ \Longleftrightarrow\ a^2\plus{}b^2\plus{}c^2\equal{}ab\plus{}bc\plus{}ca$ .

I have found another solution:

$ \triangle ABC$ is equilateral $ \implies\ c\minus{}a\equal{}\omega (b\minus{}a)$ or $ b\minus{}a\equal{}\omega(c\minus{}a),\ \omega\equal{}\cos\frac{\pi}{3}\plus{}\mathrm{i} \sin\frac{\pi}{3}$ and $ \omega^3\equal{}\minus{}1,\ \omega^2\minus{}\omega\plus{}1\equal{}0$ .

$ \implies (c\minus{}b\omega\plus{}a\omega^2)(c\minus{}a\omega\plus{}b\omega^2)\equal{}0\ \Longleftrightarrow\ a^2\plus{}b^2\plus{}c^2\equal{}ab\plus{}bc\plus{}ca$ .

[quote="Mateescu Constantin"]I have found another solution:

$ \triangle ABC$ is equilateral $ \implies\ c \minus{} a \equal{} \omega (b \minus{} a)$ or $ b \minus{} a \equal{} \omega(c \minus{} a),\ \omega \equal{} \cos\frac {\pi}{3} \plus{} \mathrm{i} \sin\frac {\pi}{3}$ and $ \omega^3 \equal{} \minus{} 1,\ \omega^2 \minus{} \omega \plus{} 1 \equal{} 0$ .
[/quote]

Can you explain why ? Thanks .

Let the length of each side be $ k$. Now, $ \frac 1 {a\minus{}b} \equal{} \frac {a'\minus{}b'}{(a\minus{}b)(a'\minus{}b')} \equal{} \frac {a'\minus{}b'}{|a\minus{}b|^2} \equal{} \frac {a'\minus{}b'} {k^2}$   ($ a'$ denotes conjugate of $ a$)

$ \implies \frac 1 {a\minus{}b} \plus{} \frac 1 {b\minus{}c} \plus{} \frac 1 {c\minus{}a} \equal{} 0$ which on simplification gives $ ab\plus{}bc\plus{}ca\equal{}a^2\plus{}b^2\plus{}c^2$

Note that the equation is invariant under:
1. Scaling and rotation w.r.t. origin (they are basically the same if you allow for complex scale factor)
2. Translation
Thus, you can just prove the equation for any single case (for example, suppose a=0, b=1).

[quote="Pain rinnegan"]Can you explain why ? Thanks .[/quote]
[quote="Mateescu Constantin"]I used the formula of rotation in complex numbers (rotation with 60 degrees angle, because triangle ABC is equilateral) .[/quote]
Note that Mateescu assumed that C is 60 degrees counterclockwise of B around A, where A, B, C are the points in complex plane representing the numbers a, b, c. Because the final equation is symmetric, this assumption does not matter.


[quote="tmathst"]Cho một hình chữ nhật m*n.Mỗi ô có hai mặt trắng đen.Khi lật một ô(i,k) thì 4 ô xác định (i,k+a),(i,k-a),(i-a,k),(i+a,k) đổi màu theo(nếu một trong các tọa độ không tồn tại thì bỏ qua). Lúc đầu tất các ô đều có màu đen.Tìm số ô màu trắng lớn nhất có thể lật được (đặc biệt nếu lật được m*n ô thành trắng)[/quote]
bài này hình như hôm nọ cũng đăng trên Diễn đàn VMF thì phỉa, nhưng đếm mãi ko đủ!


Let $I$ be the incenter of a triangle $ABC.$ The circle with diameter $[AI]$ meets the bisectors of $\angle B$ and $\angle C$ in $M,\,N$ respectively. Prove that $M$ and $N$ lie on the line joining the midpoints of $[AB]$ and $[AC].$

[hide]We can restate the problem. Let $P,Q$ be the midpoints of $AB$, $AC$ and angle bisectors from $B$ and $C$ intersect line $PQ$ at $M$ and $N$. Prove that $A,I,M,N$ lie on a circle with diameter $AI$.
As $PQ$ is paralel to $BC$ $\angle BMP=\angle MBC$ and this shows that $\triangle BMP$ is isosceles $BP=PM$. But then in $\triangle BMA$ $AP=BP=MP$ so $\triangle BMA$ is right $\angle AMB=90^{0}=\angle AMI$.[/hide]


algebra
polynomial
function
algebra proposed

Let $ \mathcal F$ be a field and $ P : \mathcal F \to \mathcal F$ be a function such that $ \forall x_0 \in \mathcal F$, $ P \left( x_0, y \right) \in \mathcal F[y]$ and $ \forall y_0 \in \mathcal F$, $ P \left( x, y_0 \right) \in \mathcal F[x]$.

Does the above imply that $ P \in \mathcal F[x,y]$, when:
(a) $ \mathcal F \equal{} \mathbb Q$?
(b) $ \left| \mathcal F \right| < \infty$?

I haven't found it, but I do have a feeling that it has been discussed before.

Err... <bump>  :blush:

For this not to seem just a useless post, let me point out that (b) is "hyper-classical": If $ \mathbb F$ is a finite field and $ f : \mathbb F^k \to \mathbb F$ is function, then there is a polynomial $ p \in \mathbb F [ x_1, \ldots, x_k ]$ that is equal to $ f$ in every point. Indeed, let $ a_1, \ldots, a_n$ be the elements of $ \mathbb F$, so that ${ f ( x_1, \ldots, x_k ) = \sum_{i=1}^n f ( x_1, \ldots, x_{k-1}, a_i} ) \cdot \prod_{j \neq i} \frac{x_k - a_j}{a_i - a_j}$. Hence the claim follows by induction.

I think that (a) is much more interesting, but equally well-known. However, it seems to be over my powers. Hints, anyone?


[quote="dondigo"]Positive numbers $a,b,x,y$ satisfy following conditions:
$a^5+b^5\leq1$
$x^5+y^5\leq1$
Prove that:
$a^2x^3+b^2y^3\leq1$[/quote]
Very easy. We have
$2a^5+2b^5\le 2$
$3x^5+3y^5\le 3$
And $2a^5+3x^5\ge 5a^2x^3$
       $2b^5+3y^5\ge 5b^2y^3$


Do there exist such linear transformation $ \phi : R^4 \rightarrow R^3$ that: $ ker(\phi)\equal{}span\{(1,1,1,1), (1,1,1,0)\}$ and $ im(\phi)\equal{}span\{(1,1,1), (1,1,0)\}$ I'm asking for explanation.

It may be easier to prove a more general statement: for any subspace $ K\subset \mathbb R^n$ and any subspace $ L\subset \mathbb R^m$ such that $ \mathrm{dim}(K) \plus{} \mathrm{dim}(L) \equal{} n$, there exists a linear transformation $ \phi\colon \mathbb R^n\to\mathbb R^m$ such that $ \mathrm{ker}(\phi) \equal{} K$ and $ \mathrm{im}(\phi) \equal{} L$.

Hint: consider the quotient space $ \mathbb R^n/K$.

aaa I see now!
I've tried to prove your statement:
Let $ dim(K)\equal{}a$ and $ dim(L)\equal{}b$. $ a\plus{}b\equal{}n$ Let base of $ K$ be $ (k_1,k_2, \cdots, k_a)$, base of $ L$ be $ (l_1,l_2,\cdots, l_b)$
We take $ U$ -  subspace of $ R^n$ such that $ K\oplus U\equal{}R^n$. From $ a\plus{}b\equal{}n$ we get $ dim(U)\equal{}b$, so let $ (u_1,u_2,\cdots,u_b)$ be base of $ U$. We define $ \phi$ as follows:
\[ \phi(k_i)\equal{}0 \quad \forall i\in\{1,\cdots,a\}\]
\[ \phi(u_i)\equal{}l_i \quad \forall i\in\{1,\cdots,b\}\]

So in my first exmple it suffice to take $ f$ such that:
\[ f(1,1,1,1)\equal{}(0,0,0)\]
\[ f(1,1,1,0)\equal{}(0,0,0)\]
\[ f(1,1,0,0)\equal{}(1,1,1)\]
\[ f(1,0,0,0)\equal{}(1,1,0)\]

Thanks for your help!

Computationally:

Take the two vectors that span the desired null space and extend them to a basis. This gives us an invertible $ 4\times 4$ matrix, two of whose columns are those two vectors. Now created a $ 3\times 4$ matrix $ B$ as follows: Two of the columns of $ B$ are zero, and they are in the same positions as the two columns of $ P$ that span the null space. The other two columns of $ B$ are the basis for the desired image.

We want to find $ A$ such that $ AP\equal{}B.$ We can do that by letting $ A\equal{}BP^{\minus{}1}.$

There are choices along the way - which vectors to put in which columns, and how to write the other two vectors when we extend to a basis to create $ P.$ For that reason, our answer won't be unique.

In this particular case, I'll make the choices as follows:

Let $ P\equal{}\begin{bmatrix}1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}1&1&0&0\\ 1&1&0&0\\ 0&1&0&0\end{bmatrix}.$

Then $ A\equal{}BP^{\minus{}1}\equal{}\begin{bmatrix}1&1&0&0\\ 1&1&0&0\\ 0&1&0&0\end{bmatrix} \begin{bmatrix}1&\minus{}1&0&0\\ 0&1&\minus{}1&0\\ 0&0&1&\minus{}1\\ 0&0&0&1\end{bmatrix}$

$ \equal{}\begin{bmatrix}1&0&\minus{}1&0\\ 1&0&\minus{}1&0\\0&1&\minus{}1&0\end{bmatrix}.$

This matrix $ A$ is one answer to the problem, among many possibilities. As required, it is a matrix of rank 2.

Added in edit: This differs from the choices dondigo has just made only in the order of the vectors. If he pursues what he has started, he should end with a permutation of this answer.


algebra
polynomial
quadratics
quadratic formula
complex numbers

[b]Let P be the product of those roots of $ z^6\plus{}z^4\plus{}z^3\plus{}z^2\plus{}1\equal{}0$ that have a positive imaginary part. Find the argument of P. [/b]

[hide="Hint"]Divide by z^3[/hide]

[hide="Hint2"]Let y=z+1/z[/hide]

[hide="Solution"]Divide by z^3 and let y=z+1/z. Then we get y^3-2y+1=0. y=1 is a solution, so divide by y-1 to get y^2+y-1 = 0. By the quadratic formula, the remaining solutions are $ y\equal{}\frac{\minus{}1\pm \sqrt{5}}{2}$. Then you solve for z.[/hide]

[hide="Solution"]
$ z^6\plus{}z^4\plus{}z^3\plus{}z^2\plus{}1$
$ \equal{} z^6 \minus{} z \plus{} z^4 \plus{} z^3 \plus{} z^2 \plus{} z \plus{} 1$
$ \equal{} z(z^5 \minus{} 1) \plus{} \frac{z^5 \minus{} 1}{z\minus{}1}$
$ \equal{} (z^5 \minus{} 1)( z \plus{} \frac{1}{z\minus{}1} )$
$ \equal{} (z^5 \minus{} 1)( z^2 \minus{} z \plus{} 1)\left(\frac{1}{z\minus{}1}\right)$
$ \equal{} (z^5 \minus{} 1)\left(\frac{z^3\plus{}1}{z\plus{}1}\right)\left(\frac{1}{z\minus{}1}\right)$
$ \equal{} \frac{(z^5 \minus{} 1)(z^3 \plus{} 1)}{(z\plus{}1)(z\minus{}1)}$

Thus $ z^5 \minus{} 1 \equal{} 0$ or $ z^3 \plus{} 1 \equal{} 0$, except $ z\neq 1$ and $ z\neq \minus{}1$.

Thus, using roots of unity, the arguments for the roots with a positive imaginary part are $ \frac{2\pi}{5}$ and $ \frac{4\pi}{5}$ for the first equation and $ \frac{\pi}{3}$ for the second equation. The product of the roots, $ P$, has an argument that is a sum of the individual arguments: thus, the argument of $ P$ is $ \frac{23\pi}{15}$.
[/hide]

[quote="pythag011"][hide="Hint"]Divide by z^3[/hide]

[hide="Hint2"]Let y=z+1/z[/hide]

[hide="Solution"]Divide by z^3 and let y=z+1/z. Then we get y^3-2y+1=0. y=1 is a solution, so divide by y-1 to get y^2+y-1 = 0. By the quadratic formula, the remaining solutions are $ y \equal{} \frac { \minus{} 1\pm \sqrt {5}}{2}$. Then you solve for z.[/hide][/quote]
I was actually looking for a good clean complex number argument. Though that works, it's extremely algebraic, and inelegant.  :)
What's $ \frac{\minus{}1\pm \sqrt{5}}{2}$?


I've learnt that a limit point of a sequence is a point for which the sequence converges to.

There is a formal definition of limit:
Let $A\subseteq X$. An element $x\in X$ is said to be a [i]limit point[/i] of $A$ if every open set containing $x$ contains at least one point of $A$ distinct from $x$.

I don't quite see the connection between my understanding of the limit point and the definition given.

Please help.

Your understanding needs to have "some subsequence converges to" to work. Also, that's only the same if $A$ is countable or if $X$ is nice.

A sequence converges to $x$ if it is eventually arbitrarily close to $x$.
A point $x$ is a limit point of a set if the set is arbitrarily close to $x$ infinitely often. This is not quite the same as "there exists a subsequence of elements converging to $x$", and even this statement assumes that the space has some separation property.

You can safely replace "at least one" with "infinitely many" as long as for each $a,b$, there exists an open set containing $a$ but not $b$.

The purpose of the formal definition : make things easier to prove, and apply universally. There's nothing wrong with using informal definitions, as long as you understand their limits and can make them rigorous if you need to.

Thank you for your reply.

I am looking for some explanation for the connection (that was the reason I posted this question). The definition does not mention about convergence at all.

Another question why chose $x \in X$ while what is interested is about limit that is in $A$?

I am new to this topic, but I would like to thank you in advance for being patience with me.


number theory proposed
number theory
httpbit.ly3vjUcdg

Suppose that $ a,b,$ and $ \sqrt[3]{a}\plus{}\sqrt[3]{b}$ are rational numbers. Prove that $ \sqrt[3]{a}$ and $ \sqrt[3]{b}$ are also rational.

hmmm.... this seems a very nice problem.this is what i got so far:
(this may be kinda obvious.nevertheless, i think this is the right track.)
$ (a^{1/3} \plus{} b^{1/3})^{3}$ is rational , so we can conclude that $ (ab)^{1/3}$ is also a rational.
this is little information... :( Can anyone post a asolution for this problem?
if i solve it, i´ll try to post it.

it isnt as simple as it seems..
it has a deep meaning where we have to prove very basic identities which are used axiomatically..

[quote="moldovan"]Suppose that $ a,b,$ and $ \sqrt [3]{a} \plus{} \sqrt [3]{b}$ are rational numbers. Prove that $ \sqrt [3]{a}$ and $ \sqrt [3]{b}$ are also rational.[/quote]

I proved that $ a^{2/3}$ is rational. (it isn´t hard, all you have to do is manipulate it a little.)But i´ve done nothing more more in this problem since them.

Not true as such, take $a=1/2, b=-1/2$. The extra condition $\sqrt[3]{a}+\sqrt[3]{b}\not=0$ is required. Let's solve the problem with this extra condition. Write $p:=\sqrt[3]{a},q:=\sqrt[3]{b}$ then $p^3,q^3,p+q\in Q$ and $p+q\not=0$.
Then also $p^3+q^3+3pq(p+q)\in Q\implies pq\in Q$. Write $c:=p+q,d:=pq$ then $c,d\in Q,c\not=0$ and $p,q$ satisfy $x^2=cx-d$.
Hence $p^3=p(cp-d)=c(cp-d)-pd=(c^2-d)p-cd\in Q\implies (c^2-d)p\in Q\implies d=c^2 \vee p\in Q$.
If $d=c^2$ then $pq=(p+q)^2\iff p^2+pq+q^2=0\iff p=q=0$. Since $p+q\in Q$ we have $p,q\in Q$ in all cases.


all wheels r on ground

http://photos1.blogger.com/blogger/5639/2020/1600/optical%20illusion%20stroller%2001.jpg

[/hide]

[quote="http://photos1.blogger.com/blogger/5639/2020/1600/optical%20illusion%20stroller%2001.jpg
"]Forbidden

Error 403[/quote]

(From the point of view of sitting in it) The front left wheels do not look to be touching the ground from how I see it.

actually, based on some shadows, the front-right wheels (from looking at the picture - really the front-left wheels of the stroller) are not on the ground because they have no shadow

Hmmm, I don't see anything unusual with the picture. :huh:

However, what's really amazing is that now the forum tables are stretched out because there were too many exclamation points :mad: :lol:

Weird, I only put 3 smileys and it thinks there are 4...


abstract algebra
combinatorics open
combinatorics

A permutation $ p \equal{} (p_1, p_2, \ldots, p_n)$ of $ {\bf Z}/n\bf Z$ is said to be [i]good[/i] if the first differences $ (p_2 \minus{} p_1, p_3 \minus{} p_2, \ldots, p_n \minus{} p_{n \minus{} 1})$ are a permutation of $ \left({\bf Z}/n{\bf Z}\right) \setminus \{0\}$.

For $ n$ even, compute the number $ a_n$ of good permutations of $ {\bf Z}/n\bf Z$.

So far as I can tell, this is open.  For $ n \equal{} 0, 2, 4, \ldots, 16$, the associated values are $ 1, 2, 8, 24, 192, 2880, 46272, 1250592,$ and $ 44095488$.  Sequence [url=http://www.research.att.com/~njas/sequences/A067601]A067601[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url] is related.

Easy exercises associated with this question:
1) Show that if $ n > 1$ is odd then there are no good permutations of $ {\bf Z}/n\bf Z$.
2) Show that if $ n \geq 2$ is even then $ n$ divides $ a_n$.
2) Show that if $ n \geq 4$ is even then $ 2n$ divides $ a_n$.
3) Show that if $ n \equal{} 2m$ then any good permutation $ p$ satisfies $ p_1 \minus{} p_n \equiv m$.

Generalization: for an arbitrary finite group $ G \equal{} \{g_1, g_2, \ldots, g_n\}$ with identity $ e$, how many permutations $ p \equal{} (p_1, \ldots, p_n)$ of $ G$ are there so that the first differences $ (p_2p_1^{ \minus{} 1}, p_3p_2^{ \minus{} 1}, \ldots, p_np_{n \minus{} 1}^{ \minus{} 1})$ are a permutation of $ G \setminus \{e\}$?

Comments: By analogy with exercise (1), the answer is $ 0$ if $ G$ is abelian and the product of the elements of $ G$ is $ e$ (and so in particular if $ G$ has odd order).  For any group, the enumerative problem is independent of whether we take first differences as indicated or in the reverse order, i.e. $ (p_1^{ \minus{} 1}p_2, p_2^{ \minus{} 1}p_3, \ldots, p_{n \minus{} 1}^{ \minus{} 1}p_n)$, since reversing permutations and inverting them element-wise is a bijection between good permutations in one case with good permutations in the other.

This might be more appropriate [url=http://www.artofproblemsolving.com/Forum/index.php?f=520]here[/url].  The group generalization, in particular, looks pretty interesting.

Your wish is my [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199166]command[/url].


Construct triangle $ABC$, given the midpoint $M$ of $BC$, the midpoint $N$ of $AH$ ($H$ is the orthocenter), and the point $A'$ where the incircle touches $BC$.

Let c be the circle of diametr MN and let I= c^MA'.
I is the foot of the altitude through A to BC. The circumcircle (MNI) c is the NP circle. Knowing that the NP circle and the incircle i of ABC are tangent we can construct i, using the fact that it is tangent to MI at A'. Now if A" is the simmetric of A' wrt M and A"' is the antipode of A' wrt i, we know that A=A"A"' ^ NI. And finally B and C are the intersection of the tangents to i from A with the line MA'.


function
trigonometry
real analysis
real analysis unsolved

Consider $ \sum_{n\equal{}1}^{\infty}(\minus{}1)^n\frac{\sin{nx}}{n}$.  Show that for $ |x|<1$, the sum uniformly converges to a continuous function.

im not sure, but can we apply drichlet's test? :maybe:

Fourier Sine Series?

cf.
[quote="Sugaku Koshiki II (Mathematical Formulas II) (Iwanami Shoten, Tokyo, 1987) [in Japanese]"]$ \sum_{n\equal{}1}^\infty (\minus{}1)^{n\minus{}1}\frac{\sin nx}{n} \equal{} \frac{x}{2} \ \ \ \ \ [\minus{}\pi<x<\pi]$[/quote]

I think it's pretty easy to use the Cauchy criterion to show it converges uniformly, and any sequence of continuous functions that converges uniformly converges to a continuous function.

[quote="calculuslover800"]I think it's pretty easy to use the Cauchy criterion to show it converges uniformly,....[/quote]
Easy? Cauchy criterion? How so?

Let me ask this: would the argument you have in mind show that this series converges uniformly on $ [\minus{}\pi,\pi]$ or even $ (\minus{}\pi,\pi)?$ Because any argument that would show that is dead flat wrong; the convergence is not uniform on those larger intervals.

"Easy" would be something like the Weierstrass M-test, which would show both absolute and uniform convergence. But this series does [i]not[/i] converge absolutely. We must use more delicate means. There is no way to avoid some summation-by-parts based argument (such as Dirichlet's test, which henry23 suggested).


There are many opposition societies in the city of N.
Each society consists of $10$ members. It is known that for every $2004$ societies there is a person belonging to at least $11$ of them.
Prove that the government can arrest $2003$ people so that at least one member of each society is arrested.

[i]Proposed by V.Dolnikov, D.Karpov[/i]

Replace $2003$ with $m$ and $10$ with $k$. So we want to show that given any number of societies with $k$ members such that any $m+1$ have $k+1$ with a common member, then there are $m$ members which cover all the societies.

Call a family of societies "good" if each member appears at most $k$ times in the family. Take the maximal good family $\mathcal{F}$, and suppose it has $x$ societies in it. Clearly $x \le m$ by the problem condition. Now call a member "common" if it appears in exactly $k$ societies in $\mathcal{F}$, suppose there are $y$ common members. Now each society not in $\mathcal{F}$ must contain some common member of $\mathcal{F}$, otherwise we could add it to $\mathcal{F}$ which would contradict its maximality. Hence, if we pick the $y$ common members of $\mathcal{F}$, we cover all societies not in $\mathcal{F}$. Furthermore, each common member appears $k$ times in $\mathcal{F}$, so they cover altogether at least $y$ sets in $\mathcal{F}$. So if we take an arbitrary member of each of the remaining $x-y$ sets in $\mathcal{F}$, we will cover $\mathcal{F}$ as well. Hence we can use exactly $x\le m$ members to cover all the societies, as desired.