calculus
derivative
algebra
polynomial
inequalities
real analysis
real analysis solved

Let $ Q(x)=x^6-3x^5-3.75x^4+Ax^3+Bx^2+Cx+D $   having only real roots $ x_j ,\;  1\le j\le  6 . $
  It is assumed that
  \[ x_1\ge x_2 \ge  x_3 \ge x_4 \ge x_5 \ge x_6 \hspace*{0.5cm}
\mbox{\rm and}\hspace*{0.5cm}   x_5= 0.5 - \sqrt{5} .\]

If $ \varphi =\displaystyle \frac{1+\sqrt{5}}{2} $ ,  prove or disprove  that  $ \; \displaystyle Q'(\varphi)=0 .$

Here is a solution by proposer .
[b]Solution.[/b] Let us denote $\mu=\displaystyle \frac{1}{6}\sum\limits_{k=1}^6 x_k$ and $\; \Delta=6\sum\limits_{k=1}^6(x_k-\mu)^2\; .$ In our case we have $\; \mu =\displaystyle \frac{1}{2}$,$\Delta=90$. Because $\sum\limits_{k=1}^6(\mu-x_k)=0$ , and  \[ x_6-\mu \le x_5-\mu =- \sqrt{5}<0\le \ldots \le x_1-\mu \; ,  \] it follows that there exists a natural number $k_0\in \{4,5,6\}$ such that
\[ x_1-\mu\ge x_2- \mu \ge \ldots \ge x_{k_0}-\mu \ge 0>x_{k_0 +1}-\mu \ge \ldots \ge x_6-\mu\; .  \] Let us prove that for $k\in \{ 2,...,6\}$ we have
\[ (*)\; \; \; x_k \ge \mu-\frac{1}{6}\sqrt{\frac{k-1}{7-k}\Delta}= \frac{1}{2}\left(1-\sqrt{\frac{10(k-1)}{7-k}}\right)\; .  \]
If $j\in \{1,...,k_0\}$ then it's clear that $\; x_j-\mu\ge \displaystyle -\frac{1}{6}\sqrt{\frac{j-1}{7-j}\Delta} .$ In other words  (*) is verified for $\; 1\le k\le k_0 .$ Suppose further that $p$is a natural number such that $2\le 1+k_0\le p\le 6.$ Then
\[ (**)\; \; \; \; \; \; \; m_p: =\sum\limits_{i=p}^{6}(x_i-\mu)\le (7-p)(x_p-\mu)\le 0  \]
\[ (***)\; \; \; \; \Delta=\frac{6}{p-1}m_p^2+6\sum\limits_{i=1}^{p-1}\left(x_i-\mu+\frac{m_p}{p-1}\right)^2+ n\sum\limits_{i=p}^{6}(x_i-\mu)^2\ge  \]
\[ \ge \frac{6(7-p)^2}{p-1}(x_p-\mu)^2+6(7-p)(x_p-\mu)^2  \]
that is $\; |x_p-\mu |= \mu-x_p \le \displaystyle \frac{1}{6}\sqrt{\frac{p-1}{7-p}\Delta}\; .$ This completes the proof of (*). From (**)-(***) it is seen that the equality case in (*) holds, iff
\[ x_1=x_2=\ldots=x_{p-1}=a+\mu \; ,\; \; (a\ge 0) \]
\[ x_p=x_{p+1}=\ldots=x_6=b+\mu \; , \; \; \; (b<0). \]
But this implies that the extremal polynomial, in order to have equality in (1),  must be
\[ Q^*(x)=Q_k^*(x)=\left(x-\mu+\frac{1}{6}\sqrt{\frac{k-1}{7-k}\Delta}\right)^{7-k} \left(x-\mu+\frac{1}{6}\sqrt{\frac{7-k}{k-1}\Delta}\right)^{k-1}=  \]
\[ =\left(x-\frac{1}{2}+\frac{1}{2}\sqrt{\frac{10(k-1)}{7-k}}\right)^{7-k} \left(x-\frac{1}{2}-\frac{1}{2}\sqrt{\frac{10(7-k)}{k-1}}\right)^{k-1}\; .  \]
Further consider $k=5$. Then (*) becames $\; \; \; (*') \; \; \;\; x_5 \ge 0.5-\sqrt{5}\; .$ But in problem is stated that $x_5=\displaystyle 0.5-\sqrt{5}.$ Therefore $Q$ coincides with extremal polynomial for $k=5$ , i.e.
\[ Q(x)=Q_5^*(x)=(x-0.5+\sqrt{5})^2(x-\varphi )^4\; .  \]
In conclusion $Q(\varphi)=Q'(\varphi)=Q\prime\prime (\varphi)=Q^{(3)}(\varphi)=0.$
Possible references are :

[1]A.Lupas , [i]Problem 246[/i] Mat.Vesnik 8(23) (1971).

[2]A.Lupas , [i]Problem 372[/i] Mat.Vesnik 13(28) (1976) 237-238.

[3]A.Lupas , [i]Inequalities for the roots of a class of polynomials.[/i] Univ.Beograd Publ.Elektrotehn.Fak.Ser.Mat.Fiz. No.577-598 (1977) 79-85 .

[4] G.V. Milovanovic , D.S. Mitrinovic, Th.M.Rassias , [i] Topics in polynomials: Extremal problems, Inequalities, Zeros.[/i] World Scientific , Singapore,London, 1994, (see pages 101-106, Theorem 1.3.7).


calculus
derivative
real analysis
integration
function
trigonometry
topology

Let $ g$ be absolutely continuous on $ [a,b]$ and let $ E\equal{}\{ x: g'(x)\equal{}0 \}$. Show that $ m(g[E])\equal{}0$.

With regard to this problem, is $ g$ constant on $ E$?

$ m(g[E])$ is the Lebesgue measure of the image of $ E$ under $ g$. Sorry for the confusion.

$ \mu_g (E) \equal{} \int_E \, d\,g \equal{} \int_E g' \, dx \equal{} 0$
By definition of $ \mu_g$ and lebesgue measure we could not have $ m(g(E)) > 0$.

Of course $ g$ is not constant on $ E$, why would you expect that? This is not true evn if $ g$ is $ C^1$. Many counterexamples...

It's the stieltjes measure induced by the set function $ \lambda_g([c,d]) \equal{} g(d) \minus{} g(c)$. If you don't know about this, you're looking for a more elementary solution.

Ok, let's do it less abstractly. Do you know this theorem:
There is an integrable function such that
$ g(x) \equal{} g(a) \plus{} \int_a^x f(t) \, dt$ and $ g' \equal{} f$ a.e.?

Well we also have $ g \equal{} g_1 \minus{} g_2$ where $ g_1$ is increasing, $ g_2$ decreasing and both absolutely continuous. So $ g'(x) \equal{} 0$ (almost) if and only if $ g_1'(x) \equal{} 0$ and $ g_2'(x) \equal{} 0$.

I think I need more help here because I still cannot see the link between these results and the one I wish to prove.

Ok, so basically we're looking at the measure of $ G \equal{} \{\int_a^x f(t) \, dt | x \in f^{ \minus{} 1}(0)\}$ where $ f(t) \geq 0$. I will also assume $ f$ is bounded by $ M$. You have to do some approximation argument in general case. Denote $ g(x) \equal{} \int_a^x f(t) \, dt$. $ G \equal{} g(f^{ \minus{} 1}(0))$

Cover $ f^{ \minus{} 1}(0)$ by a disjoint union of intervals $ H \equal{} \cup_{i \equal{} 0}^\infty (a_i, b_i)$ such that $ \sum c_i < \epsilon$ where $ c_i \equal{} m( (a_i, b_i) \minus{} f^{ \minus{} 1}(0) )$.

Since $ g$ is increasing, the minimum of $ g$ on $ (a_i, b_i)$ is $ g(a_i)$ and the maximum is $ g(b_i) \equal{} g(a_i) \plus{} \int_{a_i}^{b_i} f(t) \, dt \equal{} g(a_i) \plus{} \int_{(a_i, b_i) \minus{} f^{ \minus{} 1}(0)} f(t) \,dt \leq g(a_i) \plus{} c_iM$.

It follows that $ g( (a_i, b_i) ) \subset (g(a_i), g(a_i) \plus{} c_iM)$

So $ m(G) \leq m(g(H)) \leq \sum m(g( (a_i, b_i) )) \leq \sum c_i M < \epsilon M$

There are holes to fill in about my assumptions in the beginning ($ 0 \leq f(t) \leq M)$

Note that the whole argument is basically just about approximating $ f^{ \minus{} 1}(0)$ with an open set, and compare this with the new post you made about absolute continuity. Basically, my first response in this thread is using what you stated in your other post + "oh well it's probably true for all sets since they can be approximated by open sets". Note that any open set $ H$ can be written a countable union of the connected components which gives $ H \equal{} \cup_{i \equal{} 0}^\infty (a_i, b_i)$

Also note that integration theory sucks, because there's so much to write to express simple ideas. If you're going to fill in the holes it will probably be at least twice as long.

Looking at your proof, I was thinking about whether it is possible to use the result that if $ g$ is monotone increasing, then $ f$ is nonnegative and the integral of $ f$ over a set of small measure will be small, so as to avoid the assumption that $ f$ is bounded. A problem I encountered was that I ended up bounding each of the integrals (on $ (a_{i},b_{i})\minus{}f^{\minus{}1}(0)$) by the same $ \varepsilon$, which doesn't make the sum small. What do you think?

I don't know if it will be so easy to fix directly in the proof. And I think if you want to do it you should not use absolute continuity, but the property that $ m( \{x : f(x) > t\} ) \to 0$ as $ t\to \infty$. But what I had in mind a standard type of argument like this:

If we have a general $ f$ (still $ \geq 0$), then we let $ f_n \equal{} \min(f, n)$. We know the theorem is true for the $ f_n$, so by passing a limit we get it for $ f$.

There was a previous part to the question as follows:
Let $ F$ be absolutely continuous on $ [c,d]$ and $ g$ be absolutely continuous on $ [a,b]$ with $ c \leq g \leq d$. Show that $ F \circ g$ is absolutely continuous on $ [a,b]$.

Two things I would like to ask:
1. Does $ g$ have to be monotone for the above to be true?
2. Is there any relation between this part and my original post? (To show that $ m(g(\{ x: g'(x)\equal{}0 \}))\equal{}0$)

1 follows from the basic definition of absolutely continuous (the one that expresses that $ \mu_g << m$). Monotonicity is not necessary.

2 I suppose there has to be a relationship if it's the same question, but I don't know it

I am a little confused. I saw the following example where $ F$ and $ g$ are absolutely continuous but not $ F \circ g$:

$ F$ is defined by $ F(x)\equal{}\sqrt{x}$ for $ x \in [0,1]$, $ g$ is defined by $ g(0)\equal{}0$ and $ g(x)\equal{}(x \sin x^{\minus{}1})^{2}$ for $ x \in (0,1]$. So $ (F \circ g)(0)\equal{}0$ and $ (F \circ g)(x)\equal{}x \sin x^{\minus{}1}$ for $ x \in (0,1]$.

Please pardon my lack of understanding. Can you please show me how the proof goes, using the $ \varepsilon\minus{}\delta$ definition for absolute continuity of functions? At this point in time, I have not gotten to the absolute continuity of measures. When I did it with the $ \varepsilon\minus{}\delta$ definition, I started with a finite collection $ \{ (x_{i},x_{i}') \}$ of nonoverlapping intervals with $ \sum |x_{i}'\minus{}x_{i}|<\delta$ and got $ \sum |g(x_{i}')\minus{}g(x_{i})|<\varepsilon'$. I wanted to then use this with $ F$ but the intervals $ (g(x_{i}),g(x_{i}'))$ may not be pairwise disjoint.

[quote="Kalle"]Well, that is not a true example (because composition does preserve absolute continuity. It's a trivial proof!). Is $ \sqrt {x}$ absolutely continuous?[/quote]
$ \sqrt x \equal{} \int_0^x{1/(2\sqrt {t})\;dt}$, so that it's absolutely continuous.

another direct prove is by using $ \sqrt {y \plus{} b} \minus{} \sqrt {y} < \sqrt {x \plus{} b} \minus{} \sqrt {x}$ for $ x < y$.

You guys are right, the composite of abs cont. functions need not be abs. cont. Sorry. To "prove" this is also an exercise in a different analysis book I have, so it seems to be a widespread myth.


quadratics
algebra
polynomial
geometry
3D geometry
quadratic formula

If $a^3 - b^3 = 24$ and  and $a - b = 2$, then find all possible values of $a + b$

[hide="not the complete thing"]Put $a$ in terms of $b$ after expanding $a^3-b^3$ and solve. Will show working later,in a hurry :) [/hide]

[hide]

$a^3-b^3$ factors into $(a-b)(a^2+ab+b^2)$.  you already know a-b=2, so

$(2)(a^2+ab+b^2)=24$

so then divide by 2

$a^2+ab+b^2=12$

if you solve $a-b=2$ for a, you get $a=b+2$  so substitute that in for a

$(b+2)^2+(b+2)b+b^2=12$

and eventually you can solve that for b, and get an answer, which i don't particularly feel like figuring out right now... :?

okay fine i'll do it (i don't like leaving math problems unsolved!!!!!)

$b^2+4b+4+b^2+2b+b^2-12=0$

$3b^2+6b-8=0$

using the quadratic formula

$b=\frac{-6\pm\sqrt{36+96}}{6}$

$b=\frac{-6\pm2\sqrt{33}}{6}$

$b=\frac{-3\pm\sqrt{33}}{3}$

so, since a=b+2, $a=\frac{3\pm\sqrt{33}}{3}$

so a+b is $\frac{-3+\sqrt{33}}{3}+\frac{3+\sqrt{33}}{3}=\frac{2\sqrt{33}}{3}$  or $\frac{-3-\sqrt{33}}{3}+\frac{3-\sqrt{33}}{3}=\frac{-2\sqrt{33}}{3}$[/hide]

[hide="Faster solution"]
Substitute the symmetric polynomials $x = a + b, y = ab$.  We can rewrite

$a - b = 2 \implies a^2 - 2ab + b^2 = 4 \implies x^2 - 4y = 4$
$(a-b)(a^2 + ab + b^2) = 24 \implies x^2 - y = 12$

We wish to find all possible values of $x$.  Now, from the second relation we can derive

$4x^2 - 4y = 48$
$3x^2 = 44$
$\boxed{x = \pm \frac{ 2\sqrt{33} }{3}}$
[/hide]

[hide]Our main goal is to find ab and a^2+b^2
$\text{First cube (a-b) to find ab and we get....}$
$a^3-3a^2b+3ab^2-b^3=a^3-b^3-3ab(a-b)$
We know a-b, and a^3+b^3 so we plug those values in
$24-3ab(2)=8$
$ab=\frac{8}{3}$
Now we square (a-b) to find a^2+b^2
$a^2-2ab+b^2=4, a^2+b^2-2(\frac{8}{3})=4$
$a^2+b^2=28/3$
$\text{Finally, we square a+b}$
$a^2+b^2+2ab=(a+b)^2$
$\frac{28}{3}+\frac{8}{3}=(a+b)^2$
$a+b=\boxed{\pm\frac{2\sqrt{33}}{3}}$

[/hide]


group theory
combinatorics proposed
combinatorics

$ Problem$

Let $ \pi \in S_n$ where $ S_n$ is the [url=http://en.wikipedia.org/wiki/Permutation_group]permutation group[/url] and $ \pi^{\minus{}1}$ be the inverse of $ \pi$. Enumerate the number of permutations with $ \pi\equal{}\pi^{\minus{}1}$.

Identity mapping is a trivial example.

$ \pi^2 \equal{} e$, identity.  Decompose $ \pi$ into disjoint cycles and note that each cycle must be a transposition.


calculus
integration
algebra
function
domain
polynomial
Ring Theory

These are some of the practice problems for the test I have tomorrow. I have done most of the problems so far. I would love to see perhaps shorts proofs or some ideas that may help me tomorrow on the test:

1. Show that there are exactly two rings of order 3, up to isomorphism

2. Prove that if R is a commutative ring with unity, so is R/I for any ideal I of R. Show that R/I need not be an integral domain.

3. Prove that F[x] is an integral domain for any field F.

1. The additive group is cyclic: $\{0,a,-a\}$. Consider the possible values of $a^{2}$: it is either $0$ (and then all products are zero), or $a$ (in which case $(-a)^{2}=a^{2}=a$ and the ring is $\mathbb Z/(3\mathbb Z)$), or $-a$ (which is the same as the preceding case with $a$ and $-a$ interchanged).

2. For the first part, you just need to check the axioms one by one. For the second, consider $\mathbb Z/(n\mathbb Z)$ where $n$ is a composite number.

3. The leading coefficient of the product of two nonzero polynomials $p$ and $q$ is the product of the leading coefficients of $p$ and $q$, which is not zero.

Thank you mlok.
And for number 1): How does one exhibit the rings you have found above as quotients of $\mathbb Z_{9}$ ?


[quote="quckqhelpboogyneedz"]what is the quick way to finding out all the different ways these numbers can be arranged:

11, 51, 2, 5

so I mean like

51/11/2/5
2/5/11/51
5/2/11/51

no numbers are used twice, and if I am correct, there are twenty-four different combinations.[/quote]

I think it would be 4! (4*3*2*1=24)

Yes it would be 4! (and no we are not being enthusiastic. It means multiplying all the numbers before it like nonie said.

alright thanks for the help

i have another q though!

I have a brick wall, there are eight colummns at the bottom, in each is a letter, in colum one, there is A, colum two B, so on to collum eight which has H, above these eight I have seven, then six, then five, and so on to one. I have to add A + B and put the outcome in the colum above, outcome is A+B, so on for B + C outcome is B+ C and goes in one above, C+D so on... then I get A + B, and B  + C, which is A + 2B + C, it goes on like this for all collumns to the top, what do I get in the end all the way in the top collumn?

[quote="quckqhelpboogyneedz"]what is the quick way to finding out all the different ways these numbers can be arranged:

11, 51, 2, 5

so I mean like

51/11/2/5
2/5/11/51
5/2/11/51

no numbers are used twice, and if I am correct, there are twenty-four different combinations.[/quote]
[hide]you would do 4*3*2*1 which is 24[/hide]

[hide]Yup, there's 24 combinations, which is 4! or 4*3*2*1. There's 4 options for the first number, then 3 for the second number, 2 for the third number, and only 1 choice left for the last number.[/hide]


search MATHCOUNTS AMC AIME USA(J)MO USAMO email

I think it might...I mean, I took the AMC 10 and AMC 8 this year, and...on some of the practice competitions there was a little modular arithmetic that I (obviously) couldn't solve...so I suggest you read Intro to Number Theory.

[quote="batteredbutnotdefeated"]Currently, I have been reading AoPS volume 2, intro to geo, and intermediate algebra... I have read most of the intros, but I have NOT read intro to number theory... Will this put me at a disadvantage when I take the AMC8/AMC10/AMC12/AIME/MC?[/quote]

Uh, well, a lot depends on how much number theory you know. If you don't know anything about modular arithmetic, bases, etc., then yes, it definitely will. If you do know stuff like that and find you're not getting the easier number theory questions wrong, then you should be okay. Though, for AMC12/AIME you might want to take the Intermediate Number Theory Seminar.

Is reading intro to NT o.k. after reading a book like ACoPS? Or will the drastic decrease in difficulty throw me off?

I'm reading Intermediate Algebra but when I try to do 1+1 I'm fine, so I think you should be as well. Though you may want to consider doing Intro to NT before ACoPS, because some concepts in ACoPS may be a little difficult without the basic foundations.

I'm assuming izzy is partially joking but she is right, it won't throw you off, it might get you to think in a way which you don't want to. For example, after doing AIME's for a long time, you wouldn't want to go back to AMC 8's and get comfortable with them, and not want to challenge yourself with AIME's. But I don't think that applies here.

No; it is the "Art and Craft of Problem Solving". Remember that Google is your friend if you wish to know more information.

EDIT: I meant searching "art and craft of problem solving".  :P

Look what I get when I search ACoPS
[url]http://lmgtfy.com/?q=ACoPS[/url]
Not incredibly useful.

Try searching "The Art and Craft of Problem Solving"

The author is Paul Zeitz btw.
Here is a link if you wish to buy it.
http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471135712

I suggest you look somewhere else though, Amazon is rather expensive.
 :o

$ \$$60.00!!!!
ZOMG!!!!

Some sites have it for free download, but I don't know if that is legal...

Not legal; you can take your chances, but I'm sure you can find a cheap used copy somewhere...

Yeah, I think I'll first work on getting AoPS Vol. 1 first.

[quote="PowerOfPi"]You should generally go in order, so master the introduction series (this doesn't necessarily mean read all the books, just know the material) before starting the intermediate series. A book like ACoPS would probably be best with the Intermediate Series.[/quote]
Should I start with the intro series or AOPS v 1?

It doesn't really matter. Either way works because both are  basics. The only difference is that the Introduction series goes into way more depth than volume 1, so if you go through that, Volume 1 would serve as a nice review. I myself did something like Volume 1, then the Intro series, then reviewed Volume 1, then Intermediate Algebra, and now I'm reviewing Vol. 1 one more time before moving on to Volume 2.

Take a look at this:

http://www.amazon.com/exec/obidos/ASIN/0471789011/artofproblems-20

Is the NT in AoPS Volume 1 and 2 sufficient for ACoPS? I'm really considering whether or not to buy intro to NT...

I recommend buying intro to NT. It really helps in AMC 8, AMC 10, Mathcounts, etc...

Is the intro series necessary or would AoPS 1, 2, ACoPS, PSS good enough to handle things on the AIME and some stuff on the USAMO?

This is probably not the right forum to talk about what to read (unless you want to relate it to a competition), but I don't know which forum. You could also email books@artofproblemsolving.com for suggestions.

I suggest that you buy the books in the intro series that specialize on your weak points.

I too am trying to get to USAMO level like you AIME15USAMO.

you could try books besides AoPS, you know. try problem solving strategies by engel, you will be able to solve all and i mean all olympiad problems after you read engel.

@Sephiroth:

PSS=problem solving strategies
ACoPS=Art and craft of problem solving

and you will not be able to do [b]all[/b] the USAMO problems after reading 1 book, it takes practice and some skill.

i have found that PSS is a miracle to me, but i have never read ACoPS or any introduction AoPS books.

I don't really have any goals beyond doing OK on AMC10 and AMC8 and making national Mathcounts.  Is AcOPS necessary for me or could I wait?

Personally, I did not get Intro to Number Theory, because it seemed to easy to me when I looked at the topics that it covered. However, I did get Intro to Algebra, and that was pretty easy for me. Most of the topics were things that I could already do. If you understand all of the Number Theory in AoPS Volume II, then you don't need the Intro to Number Theory. In other words, I don't think you are at much of a disadvantage. Intermediate Counting/Probability is a good read though. That is my only suggestion. However, if you haven't mastered at least 3/4 of the counting/probability problems that show up during a MathCounts test, then the Intro text might be a good book to get instead. That is your descision though.


Let a cylinder with a diameter of 8 and a height of 6 be sliced in half, and on top of that a trianglular prism is put on, with a height of 4. {And a base of 8.} Find the volume of the new solid in terms of pi.

2.) If a right circular cylinder is inscribed in a cube with an edge of 4, what is the ratio of the volume of the cylinder to the volume of the cube?

[hide="2"]The area of the cube is $ 4\cdot4\cdot4\equal{}64$. The cylinder's diameter is $ 4$ and so is the height so the volume is $ 2\cdot2\cdot\pi\cdot4\equal{}16\pi$. Hence our answer is $ \frac{16\pi}{64}\equal{}\frac{\pi}{4}$.[/hide]

[hide="1)"] The volume of the half cylinder is $ \frac{1}{2}\left(\frac{8}{2}\right)^2(6)\pi\equal{}48\pi$. The volume of the (equilateral) triangular prism is $ \frac{8^2\sqrt{3}}{4}(4)\equal{}64\sqrt{3}$. Summing these two volumes gives $ \boxed{48\pi\plus{}64\sqrt{3}}$ [/hide]


Merge cineva anul acesta la Pitesti? Eu nu stiu de ce nu m-am calificat ca am trimis :( ...anu trecut am fost.
Parca se putea merge si in afara concursului .Responsabili de acest concurs sunt domnii profesor: P. Radovici, M. Trifu, M. Ţena.Daca stie cineva un numar de telefon sau mail sa pot lua legatura cu dansii sa ii intreb daca pot merge in afara concursului

Poti merge oricum chiar daca nu te-ai calificat.(e posibil sa platesti cazarea) . Cel putin anul trecut au mai mers asa.

Stie cineva de unde pot lua subiectele de la clasele 9-10?

$ 1$ Fie $ a_{n}\equal{}11^{n}\minus{}21^{n}\minus{}31^{n}\plus{}41^{n}$,unde $ n\in N$.Demonstrati ca exista o infinitate de numere naturale $ n$,astfel incat $ a_{n}$este multiplu de $ 1000$. (Liviu Parsan,Bucuresti)
$ 2$ Fie $ p>0$ un numar prim dat.Sa se rezolve in $ Z^{2}$ ecuatia $ x^{4}\plus{}x^{2}y^{2}\plus{}y^{4}\equal{}p$. (Honoriu Andreescu,Alexandria)
$ 3$ Sa se arate ca exista un unic punct in interiorul unui triunghi ascutitunghic,astfel incat proiectiile sale pe laturi sa fie varfurile unui triunghi echilateral. (***)
$ 4$ Fie $ a\geq 2$ si $ x_{1},x_{2},...,x_{n}\in R$.Sa se arate ca numerele
\[ A\equal{}\cos x_{1}\plus{}\frac{\cos x_{2}}{a}\plus{}\frac{\cos x_{3}}{a^{2}}\plus{}...\plus{}\frac{\cos x_{n}}{a^{n\minus{}1}}\]
\[ B\equal{}\sin x_{1}\plus{}\frac{\sin x_{2}}{a}\plus{}\frac{\sin x_{3}}{a^{2}}\plus{}...\plus{}\frac{\sin x_{n}}{a^{n\minus{}1}}\]
nu pot fi simultan nule.
(Gheorghe Szollosy,Sighetul Marmatiei)
In prima zii au dat o fel de proba pe clase, si am decat subiectele de a IX-a.Daca vrei, le postez si pe alea.:D

[quote="Vrajitoarea Andrei"]
$ 3$ Sa se arate ca exista un unic punct in interiorul unui triunghi ascutitunghic,astfel incat proiectiile sale pe laturi sa fie varfurile unui triunghi echilateral. (***)
[/quote]

Interesant ca au dat o problema destul de specializata in concurs. Aceasta teorema este cunoscuta ca o proprietate a punctelor isodinamice intr-un triunghi. In orice caz, [u]in plan[/u], exista 2 puncte inverse unu fata de celalalt in raport cu cercul circumscris (punctele isodinamice) pentru care triunghiul lor pedal este echilateral.

Totusi, pentru cazul triunghiului ascutitunghic, pentru a se arata ca exista un punct in interior pentru care triunghiul lui pedal sa fie echilateral nu era nevoie de aceste aspecte teoretice.

Fie $ \triangle{XYZ}$ triunghiul pedal al lui $ P$, pentru care $ X\in BC$, $ Y\in CA$, $ Z\in AB$.

Avem $ \angle{BPC}\equal{}\angle{BPX}\plus{}\angle{CPX}$. Dar $ \angle{BPX}\equal{}\angle{BZX}$ si $ \angle{CPX}\equal{}\angle{CYX}$.

Deci $ \angle{BPC}\equal{}\angle{BZX}\plus{}\angle{CYX}\equal{} 180\minus{}\angle{AZX}\plus{}180\minus{}\angle{AYX}\equal{}\angle{A}\plus{}\angle{YXZ}$.

Asadar, problema se reduce a arata faptul ca exista un punct $ P$ in interiorul unui ascutitunghic pentru care $ \angle{BPC}\equal{} 60\plus{}\angle{A}$, $ \angle{CPA}\equal{} 60\plus{}\angle{B}$ si $ \angle{APB}\equal{} 60\plus{}\angle{C}$, ceea ce este evident.

Ultima e draguta dar e clasica.Fie $ \omega_{i}=\cos{x_{i}}+i\sin{{x_{i}}}$.Sa presupunem prin absurd ca ambele sume ar putea fi 0.Aceasta ar fi echivalenta cu $ a^{n-1}\omega_{1}+...+a\omega_{n-1}+\omega_{n}= 0$.O rescriem ca $ |a^{n-1}\omega_{1}| = |a^{n-2}\omega_{2}+...+\omega_{n}|\leq |a^{n-2}\omega_{2}|+...+|\omega_{n}|$ de unde avem $ a^{n-1}\leq a^{n-2}+...+a+1$ de unde contradictia cu $ a\geq 2$. :)


\[ log{}_2{}_n (1944) = log_n (486 sqrt(2))\]
I can't seem to figure out where to start without using the bruteforce method (converting to base 10 and using calculator).

Thanks.

Let $ \log_{2n}1944 \equal{} x$. Then, by the properties of logarithms,

$ (2n)^x \equal{} 1944$ and $ n^x \equal{} 486\sqrt {2}$.

Now, $ (2n)^x \equal{} 2^xn^x$, so

$ 2^x(486\sqrt {2}) \equal{} 1944\implies2^x \equal{} 2\sqrt {2}\implies2^{x \minus{} 1} \equal{} \sqrt {2} \implies x \equal{} \frac {3}{2}$.

Thus,

$ n^{\frac {3}{2}} \equal{} 486\sqrt {2} \implies n \equal{} (486\sqrt {2})^{\frac23} \implies n \equal{} \boxed{54\sqrt[3]3}$.

[hide="Somewhat simpler solution"]$ \frac{3\ln 2\plus{}5\ln 3}{\ln 2\plus{}\ln n}\equal{}\frac{\ln 2\plus{}5\ln 3\plus{}\frac 12\ln 2}{\ln n}$

Solving yields $ \ln n\equal{}\frac{3\ln 2\plus{}10\ln 3}{3}\equal{}\ln (2\cdot 3^3\cdot 3^{1/3})\equal{}\ln (54\cdot 3^{1/3})$

$ n\equal{}\boxed{54\sqrt[3]3}$[/hide]

[quote="grn_trtle"][hide="Somewhat simpler solution"]$ \frac {3\ln 2 \plus{} 5\ln 3}{\ln 2 \plus{} \ln n} \equal{} \frac {\ln 2 \plus{} 5\ln 3 \plus{} \frac 12\ln 2}{\ln n}$

Solving yields $ \ln n \equal{} \frac {3\ln 2 \plus{} 10\ln 3}{3} \equal{} \ln (2\cdot 3^3\cdot 3^{1/3}) \equal{} \ln (54\cdot 3^{1/3})$

$ n \equal{} \boxed{54\sqrt [3]3}$[/hide][/quote]

Thanks, but how did you arrive at the first step?

[quote="qszwdxefc"]Thanks, but how did you arrive at the first step?[/quote]

Using the change-of-base formula:

$ \log_ab\equal{}\frac{\log_cb}{\log_ca}$.

[hide="Proof"]
$ \log_ab\equal{}x$

$ a^x\equal{}b$

$ \log_c a^x\equal{}\log_cb$

$ x\log_c a\equal{}\log_cb$

$ x\equal{}\log_ab\equal{}\frac{\log_cb}{\log_ca}$ $ \Box$
[/hide]

In this particular case, I used $ c\equal{}e$ for no particular reason. It might have been a tiny bit quicker to use $ c\equal{}2$ or $ c\equal{}3$, but whatever...


[quote="dwx314"][hide]The area of a rectangle is LW=A
The areas are in ratio of 2:9
The lengths are in ratio of 4:27

So what you're arsking for is the other two sides.
So make a proportion: 2/9=4a/27b

b has to be 2, otherwise you cannot reduce the 4 into a 2
a has to be 3, otherwise you cannot reduce the 27 into a 9

12/54 --> 2*2*3/3*3*3*2 --> 2/3*3 (cancel some factors) --> 2/9

You want the sum of a+b.
3+2=5

>>5<<[/hide][/quote]

I dont see it asking the sum anywhere.....

[quote="charikaar"]i have maths question on ratio with shapes. this is wat it says. the ratio of the areas of these two rectangles is 2:9 Find the values of a and b. it gives 4 on 1 rectangle and 27 on the other and u hav to work out a and b.

can you do this using algebra or other basic method. thanks for your help.

Charikaar[/quote]

[hide]Assuming that a is on the smaller rectangle and that b is on the larger one, all you have to do is factor out the numbers. When you factor 4, you get 2*2, and when you factor out 27 you get 3*3*3.  Now if you want the ratio of 2:9, you need to eliminate one of the twos and one of the threes, so all you have to do is set a at three and b at two, so that they can be reduced. In theory, this question is somewhat vague, because there are an infinite jumber os solutions, if you multply the basic values of a and b by a common number. If the problem asked for the ratio of a to b, the answer would be 3/2, but assuming that you are asking for the simplest answer, A=3, B=2[/hide]

This may seem like a lot of work but it all really is mental, and you really don't have to write down anything on paper.

thanks for your replies. that really helped.

charikaar


Which is the larger number, $ 2^{100}$ or $ 3^{75}$?

plugging this into a TI-89 we get tat $ 3^{75}$ is bigger.

hehe....

but really, we see that $ 3^3>2^4$, so $ 3^{3\cdot25}>2^{4\cdot25}$


I have a weird solution:

note that:

\begin{eqnarray*}a^3 + b^3 + c^3 - 3abc = (a + b + c)\left(a^2 + b^2 + c^2 - (ab + bc + ca)\right) = (a + b + c)\left(1 - (ab + bc + ca)\right)\end{eqnarray*}

thus we have to prove that:

$ \left|(a + b + c)\left(1 - (ab + bc + ca)\right)\right|\leq 1$

$ \iff (a + b + c)^2\left(1 - (ab + bc + ca)\right)^2\leq 1$

now let $ S = ab + bc + ca$ so we must prove that:

$ (1 + 2S)(1 - S)^2\leq 1$

$ \iff 2S^3\leq 3S^2$ (*)

now $ S^2\geq 0$ so (*) is equivalent to:

$ S\leq \frac 32$

which is obviously true since $ ab + bc + ca\leq a^2 + b^2 + c^2 = 1$ :maybe:

so equality can't occur...

Let $ |a\plus{}b\plus{}c|\equal{}x$ then $ 0\le x\le\sqrt{3}$.
$ |a^3\plus{}b^3\plus{}c^3\minus{}3abc|\equal{}|(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)|\equal{}x\times\frac{1}{2}\times (3\minus{}x^2)\le 1$

$ \leftrightarrow x^3\plus{}2\ge 3x\leftrightarrow(x\minus{}1)^2(x\plus{}2)\ge 0$

[quote="BaBaK Ghalebi"]
so equality can't occur...[/quote]

Equality occurs when $ a\equal{}1, b\equal{}c\equal{}0$


quadratics
number theory unsolved
number theory

Prove that for every prime $ p$, there exists an integer $ a$, $ 1\le a \le 9$, such that $ a$ and $ a+1$ are squares modulo $ p$.

If $ 2$ is a quadratic residue, take $ a=1$.
If $ 5$ is a quadratic residue, take $ a=4$.
Otherwise, $ 10$ is a quadratic residue, then take $ a=9$.


[quote="AndrewTom"](Based on others I've seen but slightly different.) Solve the following simultaneous equations where a, b and c are whole numbers:
ab + 4 = 2(a+b)
bc + 1 = 2(b+c)
ca + 4 = 2(c+a).[/quote]

[hide]
The first equations can all be rearranged to

$ (a\minus{}2)(b\minus{}2)\equal{}0$
$ (b\minus{}2)(c\minus{}2)\equal{}3$
$ (c\minus{}2)(a\minus{}2)\equal{}0$

The first and second imply that $ a\equal{}2$

Therefore we seek $ (b\minus{}2)(c\minus{}2)\equal{}3$

By considering prime factors of 3

$ (a,b,c)\equal{}(2,5,3),(2,3,5),(2,1,\minus{}1),(2,\minus{}1,1)$[/hide]


For x not equal to -1 or 1, simplify

$\displaystyle{\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{\frac{x+1}{x-1}+\frac{x-1}{x+1}}}$. Express your answer as a common fraction in terms of x.

I don't want to find common denominators, although that's what i had to do.
I can do it in about 20 seconds through quadratic properties, but I want a faster way  ;)

[hide="is it"]$\frac{2x}{x^2+1}$[/hide]

You could try substituting 2 and 3, for example, to get

$x=2 \Rightarrow \frac{3-\frac{1}{3}}{3+\frac{1}{3}}=\frac{\frac{8}{3}}{\frac{10}{3}}=\frac{4}{5}$

$x=3 \Rightarrow \frac{\frac{4}{2}-\frac{2}{4}}{\frac{4}{2}+\frac{2}{4}}=\frac{\frac{3}{2}}{\frac{5}{2}}=\frac{3}{5}$

You might notice a pattern, you might not... :P

Think you can notice a pattern quickly? Where are you going to figure out it's $\frac{2x}{x^2+1}$

Very interesting approach. now we have three different approaches here.


How many integers in the list 1,2,3,...,10000 have exactly two consecutive 9's, such as 9929 or 1992 but 1999?

[hide="Solution"]
Possibilities:
* _ 9 9
_ 9 9 _
9 9 _ *

Where _ means 9 can't appear and * means it can appear in that position. 9 * 10 + 9 * 10 + 9 * 9 = 261
[/hide]


sometimes i have 8 subjects in my report and sometimes 12, and often i cant see some videos about the subjects :(

Sorry to hear about your problems viewing the topics and reports. There may be times while the server is under maintenance that not all your subjects will appear. If you try back in an hour or two they should re-appear. It is also possible that you are falling out of the "ready", "passed", or "mastered" state of the subjects and thus no longer appear. This can happen if you start answering problems in a topic or its associated topics incorrectly.

Please provide additional details on videos, such as which topics you feel you are not seeing all the videos.

There's another possibility here: if you are on the borderline of passing a topic, you might go from "Passed" back to "Not Passed" as a result of missing a couple review problems.  You will then be limited more in the topics and videos you can access until you re-pass the subject, and thereby regain access to the topics that depend on this one.


i don't understand why you would want to do that..its an invalid and false statement in mathematics. like that famous proof that 2=1.

*p.s.-isn't your username just a replica of [b]youdonotknowme[/b]?


geometry
group theory
abstract algebra
linear algebra
matrix
analytic geometry
algebra

In ''Joe Harris, Algebraic Geometry, a first course (GTM, Springer), page 4 '' has following text

...The group $ \text{PGL}_{n+1}K$ acts on the space $ \mathbb{P}^{n}$...

But i don't understand , what is $ \text{PGL}_{n+1}K$ ?

The quotient group $ \mathrm{GL}_{n+1}(K)/Z$ where $ Z$ is subgroup consisting of the multiples of the identity matrix (so $ Z\cong K^\times$). Actually, "P" stands for "projective". owk

Thank you very much, owk  :wink: But in that page , I don't understand. What is $ \text{Sym}^{d}(V^{*})$ ? . If you (or who) haven't that book , i will post that text here. Thanks , again.

Sym denotes symmetric powers. If you choose a basis $ X_{1},\ldots,X_{n}$ of $ V^{*}$ (consider coordinates as linear functionals on $ V$), you may identify $ \mathrm{Sym}^{d}(V^{*})$ with homogeneous polynomials in $ X_{1},\ldots,X_{n}$ of degree $ d$. owk


What is the largest number that can be made by inserting two multiplication signs between the digits $ 2, 2, 3, 3,$ and $ 3$ in that order? [For example $ 223 \times 3 \times 3 \equal{}2007$.]

[hide="answer"]

$ 2178 \equal{} 22 \times 3 \times 33$

[/hide]


When pruning your gallery, it says "Moderator Control Panel" at the top. That seems to give the impression that one is a gallery moderator. It's not a big deal, but...

Next time please put support issues in support. This is neither a question or suggestion, but basically a bug.

Hm, I thought it was more of a suggestion rather than a bug (I almost posted in support originally, but decided against it), but I will next time.


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

I am having trouble with this problem. I have tried quite a few approaches but nothing seems to work

Prove that the commutator subgroup of $ GL_{2}(\mathbb R)$ is $ SL_{2}(\mathbb R)$


SMO 2007, Open #10

Let $x_{1},x_{2},\ldots,x_{1970}$ be positive integers satisfying $x_{1}+x_{2}+\cdots+x_{1970}= 2007$. Determine the largest possible value of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{1970}^{3}$.

[hide]Since they're positive integers, the greatest possible term is 2007-1969=38.
Next, since we want the maximum, we make it so 38 can exist as a term.
Thus, the maximum of the second expression is $38^{3}+1969=56841$[/hide]


integration
inequalities
real analysis
logarithms
limit

Είχα βάλει μία παρόμοια στο παρελθόν, αλλά δε θυμάμαι τί απέγινε και αν είχε λυθεί. At any rate, here goes nothing :)

Έστω θετική συνάρτηση $ f: [0,1]\to\mathbb{R}$, όχι απαραίτητα συνεχής αλλά ολοκληρώσιμη (ας πούμε $ \int_a^b f(x) dx \equal{} 1$ για απλότητα).

Είναι αλήθεια ότι $ \int_0^1 f(y)\cdot\big| \log|x \minus{} y| \big| dx < \infty$ σχεδόν για όλα τα $ y\in [0,1]$?

Αν και εφόσον η απάντηση στο προηγούμενο ερώτημα είναι καταφατική, είναι αλήθεια ότι $ \int_0^1\int_0^1 f(x) f(y) \cdot\big|\log|x \minus{} y|\big| dy dx < \infty$?

Cheerio,

Durandal 1707

Edit: έβαλα το πρώτο ερώτημα...

Να βάλω ένα hint... αν εξακολουθεί να μην υπάρχει ενδιαφέρον, θα βάλω τη λύση μετά από μερικές μέρες :)

Από ανισότητες [url=http://en.wikipedia.org/wiki/Holder%27s_inequality]Hölder[/url] και [url=http://en.wikipedia.org/wiki/Young_inequality#An_inequality_for_Lp_norms]Young[/url], δείξτε ότι αν υπάρχει $ \varepsilon > 0$ ώστε $ \int |f|^{1 \plus{} \varepsilon} < \infty$, η απάντηση και στις δύο ερωτήσεις είναι καταφατική.

Άρα, για μία πυκνότητα $ f(x)$ που να κάνει το διπλό ολοκλήρωμα να απειρίζεται, η απάντηση δένει αρκετά καλά με [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262834]αυτό το topic[/url].

Αν και μπορεί να φαίνεται εντελώς αυθαίρετο, το πρόβλημα αυτό έχει αρκετά μεγάλη φυσική σημασία: το διπλό ολοκλήρωμα απλώς εκφράζει την ενέργεια μίας γραμμικής κατανομής φορτίου σε δισδιάστατο χώρο :)

Cheerio,

Durandal 1707

Edit: o gpapargi μου έδειξε ότι υπάρχει ένα typo στο πρώτο ερώτημα όπου το ολοκλήρωμα θα έπρεπε να είναι: $ \int_0^1 f(x)\cdot \big|\log|x \minus{} y|\big| dx$. Sorry αν παιδεύτηκε κανείς άδικα... θα το διόρθωνα απευθείας στο post, αλλά δυστυχώς δεν μπορώ πια :(

Meeeh, δε βλέπω να υπάρχει ενδιαφέρον οπότε ας το σκοτώσω για να μη μένουν άλυτα προβλήματα στο forum :(

Αρχικά, για συμβολισμό ας θέσουμε $ \|g\|_p \equal{} \left(\int_0^1 |g(x)|^p dx\right)^{1/p}$. Αυτή είναι η $ L^p$ νόρμα και είναι πάρα πολύ σημαντική στην ανάλυση και στη φυσική (π.χ. η κβαντομηχανική γίνεται σχεδόν εξ'ολοκλήρου πάνω στο χώρο των τετραγωνικά ολοκληρώσιμων κυματοσυναρτήσεων, δηλαδή αυτών που έχουν πεπερασμένη $ L^2$ νόρμα). Αυτό που δίνει στη νόρμα αυτή πολύ ισχυρές ιδιότητες είναι η ανισότητα Hölder: $ \|gh\|_1\leq \|g\|_p \|h\|_q$ οποτεδήποτε $ 1/p \plus{} 1/q \equal{} 1$ (τότε οι εκθέτες p,q λέγονται συζυγείς; παρατηρείστε ότι για $ p \equal{} q \equal{} 2$ η Hölder παίρνει τη μορφή της CBS και είναι ακριβώς αυτή που επιτρέπει στο εσωτερικό γινόμενο της κβαντομηχανικής να δουλέψει).

Τα παραπάνω μας είναι χρήσιμα μόνο από συμβολισμό, όπως είχα πει και στο δεύτερο post, οι ανισότητες είναι που κάνουν όλη τη δουλειά εδώ.

Παρατηρείστε λοιπόν ότι $ \|\log|x \minus{} y|\|_p^p \equal{} \int_0^1\log|x \minus{} y|^p dx < \infty$ για κάθε $ p > 1, x\in[0,1]$. Τότε, το ολοκλήρωμα $ \int f(y) \log|x \minus{} y|dy$ θα είναι απλώς η συνέλιξη (convolution) των $ f$ και $ \log|\cdot|$. Έτσι, από την ανισότητα Young προκύπτει ότι το ολοκλήρωμα θα είναι πεπερασμένο σχεδόν για κάθε $ x$ και, επιπλέον θα ισχύει: $ \|f*\log\|_{1 \plus{} 1/\varepsilon}\leq \|f\|_1 \|\log\|_{1 \plus{} 1/\varepsilon}$.

Τότε, το αρχικό διπλό ολοκλήρωμα θα είναι απολύτως φραγμένο από την παράσταση:
$ \|f\cdot(f*\log)\|_1\leq\|f\|_{1 \plus{} \varepsilon}\|f*\log\|_{1 \plus{} 1/\varepsilon} \leq \|f\|_1\|f\|_{1 \plus{} \varepsilon} \|\log\|_{1 \plus{} 1/\varepsilon} < \infty$.

Άρα, για να μην είναι πεπερασμένο το ολοκλήρωμα οι μόνες πιθανές υποψήφιες είναι συναρτήσεις $ f$ που να μην είναι σε κανέναν $ L^p$ για $ p > 1$ (αλλά να είναι παρολ'αυτά ολοκληρώσιμες). Αν βάλουμε $ f \equal{} \sum_n \frac {n^2}{2^n}1_{[2^{ \minus{} n},2^{ \minus{} n \plus{} 1})}$ τελειώσαμε :D

Looking back, μάλλον παραήταν εξειδικευμένο το πρόβλημα... Apologies for that, απλώς ήταν που είχε πολύ σημαντική φυσική ερμηνεία και παρασύρθηκα... μου είχε φανεί πολύ εντυπωσιακό ότι μπορείς σε δύο διαστάσεις να φτιάξεις συνεχείς κατανομές φορτίου με άπειρη ενέργεια ;)

Cheerio,

Durandal 1707

Edit: O George (gpapargi) είχε μία διαφορετική προσέγγιση όταν οι συναρτήσεις $ f$ είναι κάπως πιο ομαλές (π.χ. συνεχείς στο $ (0,1)$ αλλά όχι απαραίτητα φραγμένες) και η οποία πιθανώς να οδηγεί σε διαφορετική λύση... George, όταν βρεις χρόνο :)

Στέλνω κι εγώ μια λύση που σκέφτηκα μετά από hint του Παναγιώτη ότι το μέσα ολοκλήρωμα είναι το convolution 2 συναρτήσεων. Το convolution είναι μια πράξη με μεγάλη εφαρμογή στη θεωρία σημάτων και συστημάτων και όσοι έχουν ασχοληθεί με τηλεπικοινωνίες την ξέρουν καλά. Είναι μια λύση χωρίς απαιτήσεις στην ανάλυση και νομίζω παρακολουθείται εύκολα από κάποιον με στοιχειώδεις γνώσεις.

Ξεκινάω από τη γεωμετρική ερμηνεία του convolution για να εστιάσω στα κρίσιμα σημεία και να γλιτώσω πράξεις.

Όταν έχεις να βρεις το convolution 2 συναρτήσεων αφήνεις τη μια ακίνητη. Την άλλη την καθρεφτίζεις (φέρνεις το δεξιά αριστερά) και το ολισθαίνεις από το άπειρο μέχρι το -άπειρο. Σε κάθε σχετική θέση βρίσκεις το εμβαδό του γινομένου. Αυτό σημαίνει ότι όταν η μια είναι έξω από την άλλη τότε η τιμή του convolution είναι 0. Όταν η μια μπει μέσα στην άλλη τότε στην κοινή περιοχή βρίσκεις το γινόμενο και μετά το εμβαδό κάτω από αυτό.

Για να είναι πεπερασμένη τιμή το convolution θα πρέπει να είναι πεπερασμένο το εμβαδό ακόμα και στην πιο κρίσιμη περιοχή. Αυτή είναι στη σχετική θέση που ο χειρότερος απειρισμός της f ταυτίζεται με το χειρότερο απειρισμό της lnx δηλαδή το 0.
Για να μην μπλέκω με όλους τους όρους παίρνω πάντα τον μεγαλύτερο... τον dominant στην περιοχή του απειρισμού και ελέγχω τι γίνεται σε αυτόν. Αν αυτός είναι πεπερασμένος οι υπόλοιποι δε δημιουργούν πρόβλημα.

Αρχικά ασχολούμαι με συναρτήσεις της μορφής $ f(x)\equal{}\frac{1}{x^n}$
Εξετάζω λοιπόν μόνο το στιγμιότυπο στο οποίο οι απειρισμοί έχουν ταυτιστεί. Ελέγχω το ολοκλήρωμα του $ lnx\cdot \frac{1}{x^n}$
Αγνοώντας τις σταθερές βρίσκω   $ \frac{lnx}{x^{n\minus{}1}} \minus{} \frac{1}{x^{n}}$
Στην περιοχή του 0 ο dominant όρος είναι ο πρώτος. Αυτός πρέπει να φάει και τη δεύτερη ολοκλήρωση στην ίδια περιοχή και να δούμε τι συμβαίνει.
Μετά τη δεύτερη ολοκλήρωση (και αγνοώντας τις σταθερές) έχουμε:
$ \frac{lnx}{x^{2n\minus{}2}}\minus{}\frac{1}{x^{2n\minus{}1}}$
Dominant όρος είναι ο πρώτος. Βρίσκω το όριο στο 0. Με de l hospital βγαίνει ότι συμπεριφέρεται όπως ο  $ \frac{1}{x^{2n\minus{}2}}$

Το συμπέρασμα είναι ότι για συναρτήσεις τις μορφής $ \frac{1}{x^n}$ αν $ n<1$ τότε η ενέργεια είναι πεπερασμένη. Αν  $ n\geq 1$ τότε η ενέργεια είναι άπειρη. Το όριο για τις συναρτήσεις αυτής της μορφής είναι η $ f(x)\equal{}\frac{1}{x}$

Πάμε τώρα να δούμε τι γίνεται με τις υπόλοιπες συναρτήσεις.
Πρέπει να τις συγκρίνουμε ως προς τον απειρισμό τους με το $ \frac{1}{x}$ . Ότι απειρίζεται χειρότερα από την   $ \frac{1}{x}$ το εμβαδό είναι άπειρο.

Θέλω να συγκρίνω πολύ κοντά στο 0 την f με κάποια από την οικογένεια $ \frac{1}{x^n}$

Άρα η σκέψη μου είναι να πάρω το όριο της $ \frac{f(x)}{\frac{1}{x^n}}$   στο 0 (δηλαδή να βάλω τη μια στον αριθμητή και  την άλλη στον παρονομαστή) και να δω τι γίνεται.
Αν υπάρχει αριθμός ε τέτοιος ώστε το όριο του $ \frac{f(x)}{\frac{1}{x^{1\minus{}\epsilon }}}$  να είναι 0, αυτό σημαίνει ότι από κάποιο σημείο και μετά η $ \frac{1}{x^{1\minus{}\epsilon }}$ γίνεται μεγαλύτερη από την f και την φράσει από πάνω (η γραφική της παράσταση σκεπάζει αυτή της f). Αφού η «πάνω» έχει πεπερασμένο εμβαδό άρα και η κάτω έχει πεπερασμένο εμβαδό.

Για την περίπτωση που θέλω να δείξω ότι η f έχει άπειρο εμβαδό αρκεί να δείξω ότι φράσσεται από κάτω από μια συνάρτηση ανάλογη της $ \frac{1}{x}$

Αν το όριο της $ \frac{f(x)}{\frac{1}{x}}\equal{}xf(x)$ είναι άπειρο τότε το εμβαδό μας φράσσεται από κάτω από το εμβαδό της $ \frac{1}{x}$ γιατί η f κοντά στο 0 είναι μεγαλύτερη από την $ \frac{1}{x}$ . Δηλαδή απειρισμός.

Αν το όριο είναι κάποια μη μηδενική σταθερά τότε μπορώ να φράξω από κάτω με κάποια συνάρτηση της μορφής $ \frac{1}{cx}$ με c σταθερά. Δηλαδή έχω απειρισμό της f ίδιο με αυτόν της $ \frac{1}{x}$ και το ζητούμενο εμβαδό είναι άπειρο.

Ένα ερώτημα που μένει να απαντηθεί είναι το τι γίνεται αν το όριο στο 0 του $ xf(x)$ είναι 0. Είναι συνθήκη ικανή για να έχουμε πεπερασμένη ενέργεια διπλό ολοκλήρωμα); Αν ναι χρειάζεται μια απόδειξη. Αν όχι θέλουμε ένα αντιπαράδειγμα που να έχει όριο του   $ xf(x)\equal{}0$ άλλα άπειρη ενέργεια.
Τελικά φαίνεται πως ισχύει το δεύτερο. Η βασική ιδέα για την κατασκευή μιας τέτοιας f είναι να γλυκάνουμε λίγο τον απειρισμό της $ \frac{1}{x}$ έτσι ώστε να είναι κάτω από την $ \frac{1}{x}$ και άρα να έχει όριο της $ xf(x)\equal{}0$ αλλά να είναι πάνω από όλες τις συναρτήσεις της μορφής $ \frac{1}{x^n}$ με $ n<1$
Άρα πρέπει να «αφαιρέσουμε» κάτι από τον εκθέτη του x στον παρονομαστή. Μια τέτοια επιλογή είναι η $ lnx$
Η πρώτη υποψήφια συνάρτηση είναι η $ f(x)\equal{}\frac{1}{xlnx}$ και αν δεν έκανα κάποιο λάθος στις πράξεις αποδεικνύεται ότι απειρίζει το διπλό ολοκλήρωμα.

Επίσης αξίζει να σημειωθεί ότι αν «γλυκάνει» λίγο ακόμα ο απειρισμός με την $ f(x)\equal{}\frac{1}{x\cdot ln^2x}$ καταλήγουμε σε μια κατανομή φορτίου με πεπερασμένο συνολικό φορτίο και άπειρη ενέργεια γεγονός που σημαίνει ότι έχουμε «ζουλήξει» πάρα πολύ τα φορτία μας στην περιοχή του 0. Εννοείται πως άπειρο φορτίο στο πεπερασμένο χώρο δίνει άπειρη ενέργεια (αλλιώς παραβιάζεται ο νόμος του Gauss).
Φαίνεται λοιπόν πως αν ξεκινήσουμε από την $ \frac{1}{x}$ και σταδιακά κατεβάζουμε τον απειρισμό της τότε πρώτα θα μηδενιστεί το όριο της $ xf(x)$ μετά θα γίνει πεπερασμένο το φορτίο (με άπειρη ενέργεια) και τέλος θα γίνει πεπερασμένη και η ενέργεια.

[quote="Durandal"]
Edit: O George (gpapargi) είχε μία διαφορετική προσέγγιση όταν οι συναρτήσεις $ f$ είναι κάπως πιο ομαλές (π.χ. συνεχείς στο $ (0,1)$ αλλά όχι απαραίτητα φραγμένες) και η οποία πιθανώς να οδηγεί σε διαφορετική λύση... George, όταν βρεις χρόνο :)[/quote]

Νομίζω Παναγιώτη ότι η προσέγγισή μου πρέπει να δουλεύει και για ασυνεχείς συναρτήσεις. Ίσως να μην είναι Riemman ολοκληρώσιμες κάποιες αλλά μπορείς να δουλέψεις με τη μέγιστη τιμή της συνάρτησης σε κάθε διαμέριση. Θέλω να πω ότι αν η συνάρτηση είναι ασυνεχής και μη ολοκληρώσιμη (λόγω διαφορετικών ορίων της άνω και κάτω διαμέρισης) αλλά με πεπερασμένες τιμές σε κάθε ασυνέχεια, δε βλέπω κάποιο πρόβλημα. Οι απειρισμοί είναι αυτοί που μπορούν να κάνουν άπειρη την ενέργεια και όχι οι πεπερασμένες ασυνέχειες. Χάνω κάτι άλλο;

ΥΓ
Στο αρχικό μου post είδα ένα προβληματάκι στις 2 συναρτήσεις που έγραψα με το λογάριθμο στον παρονομαστή. Δεν έχουμε απειρισμό μόνο στο 0 αλλά και στο 1 που δε μας ενδιαφέρει. Για να το παρακάμψουμε αυτό μπορούμε απλά να απλώσουμε για λίγο το το λογάριθμο για να μη μηδενίζεται στο 1 πχ $ \frac{1}{xln\frac{x}{2}}$

[quote="gpapargi"]Νομίζω Παναγιώτη ότι η προσέγγισή μου πρέπει να δουλεύει και για ασυνεχείς συναρτήσεις. Ίσως να μην είναι Riemman ολοκληρώσιμες κάποιες αλλά μπορείς να δουλέψεις με τη μέγιστη τιμή της συνάρτησης σε κάθε διαμέριση. Θέλω να πω ότι αν η συνάρτηση είναι ασυνεχής και μη ολοκληρώσιμη (λόγω διαφορετικών ορίων της άνω και κάτω διαμέρισης) αλλά με πεπερασμένες τιμές σε κάθε ασυνέχεια, δε βλέπω κάποιο πρόβλημα. Οι απειρισμοί είναι αυτοί που μπορούν να κάνουν άπειρη την ενέργεια και όχι οι πεπερασμένες ασυνέχειες. Χάνω κάτι άλλο;[/quote]

Έστω ότι υπάρχει μόνο ένας απειρισμός στο $ x_0\in [0,1]$. Πρακτικά, εσύ λες: αν $ \lim_{x\to x_0} \frac {f(x)}{1/(x \minus{} x_0)^p} \equal{} 0$ για κάποιο $ p > 1$, τότε υπάρχει περιοχή του $ x_0$ όπου $ f(x) < \frac {\epsilon}{(x \minus{} x_0)^p}$ κλπ, κλπ. Αυτή η περιοχή είναι που κάνει το επιχείρημά σου να δουλεύει.

Ωστόσο, αυτήν την περιοχή την εξασφαλίζεις εύκολα μόνο αν η $ f$ είναι κάπως ομαλή κοντά στο $ x_0$. Aν έχεις πεπερασμένο πλήθος από απλές jump discontinuities, η διαίσθησή σου παραμένει σωστή, δε θα έχεις πρόβλημα (απλώς μίκρυνε την περιοχή σου ακόμα περισσότερο). Easy food for thought: ακόμα και αριθμήσιμο πλήθος από διορθώσιμες ασυνέχειες απείρου ύψους να έχεις που να είναι και πυκνές κοντά στο $ x_0$, πάλι θα μπορείς να σώσεις το επιχείρημά σου (αν και θα έχεις αρχίσει να χάνεις την ολοκληρωσιμότητα κατά Riemann).

Ωστόσο, τέτοιου τύπου ασυνέχειες είναι εντελώς τετριμμένες και φυσικά δε μας ενδιαφέρουν. Τι γίνεται λοιπόν με συναρτήσεις που έχουν ουσιώδεις (μη διορθώσιμες) ασυνέχειες όπως π.χ. η $ f(x) \equal{} \frac {1}{x^p} \plus{} \sum_{n \equal{} 1}^\infty \frac {2^{ \minus{} n}}{(x \minus{} 1/n)^p}$? Αυτές χρειάζονται λίγο περισσότερη δουλειά και λίγο δυνατότερη θεωρία.

Cheerio,

Durandal 1707

Ίσως δεν έχω καταλάβει κάτι…  :maybe:
Επειδή το ολοκλήρωμα του αθροίσματος είναι ίσο με το άθροισμα των ολοκληρωμάτων μπορώ να εξετάσω κάθε όρο χωριστά. Με ενδιαφέρει ο χειρότερος. Αν κάποιος από τους 2 απειρίζει το διπλό ολοκλήρωμα τότε απειρίζεται και το άθροισμα. Αν και τα 2 δίνουν πεπερασμένο ολοκλήρωμα τότε το ίδιο ισχύει και για το άθροισμα.
Δηλαδή στην συνάρτηση που αναφέρεις θα εξέταζα κάθε όρο χωριστά.
Για τον πρώτο όρο όλα καλά ως προς το τι θα κάνουμε.
Ας μιλήσουμε για τον δεύτερο. Είναι ένα άθροισμα απείρων συναρτήσεων, ας το συμβολίζουμε με Σ.
Γενικά η προσέγγισή μου είναι να φράξω από κάτω με κάτι που αποκλείνει και άρα το Σ να αποκλείνει ή να φράξω από πάνω με κάτι που συγκλίνει και άρα το Σ συγκλίνει.
Κατέληξα λοιπόν πως αν υπάρχει ε τέτοιο ώστε το όριο στο 0 του $ \frac {f(x)}{\frac {1}{x^{1 \minus{} \epsilon }}}$ να είναι 0 τότε και το διπλό ολοκλήρωμα είναι 0.
Αν το όριο της $ xf(x)$ είναι πάνω από 0 τότε το διπλό ολοκλήρωμα απειρίζεται.
Τέλος αν το όριο της $ xf(x)$ είναι 0 τότε δεν μπορούμε να βγάλουμε συμπέρασμα. Άλλοτε έχουμε άπειρο ολοκλήρωμα και άλλοτε πεπερασμένο όπως δείχνουν τα 2 συγκεκριμένα παραδείγματα.

Στο παράδειγμα που αναφέρεις θέλω να ελέγξω τι κάνει το Σ αν μπει στη θέση της f στα κριτήρια σύγκλισης-απόκλισης που αναφέρω.

Πρόσεξε: άλλο το αν μπορώ να βρω το όριο και άλλο αν ισχύει το κριτήριο που θέτω. Εδώ συζητάμε το δεύτερο. Αν δηλαδή τα 2 κριτήρια που ανέφερα ισχύουν.

[quote="gpapargi"]Επειδή το ολοκλήρωμα του αθροίσματος είναι ίσο με το άθροισμα των ολοκληρωμάτων μπορώ να εξετάσω κάθε όρο χωριστά. Με ενδιαφέρει ο χειρότερος. Αν κάποιος από τους 2 απειρίζει το διπλό ολοκλήρωμα τότε απειρίζεται και το άθροισμα. Αν και τα 2 δίνουν πεπερασμένο ολοκλήρωμα τότε το ίδιο ισχύει και για το άθροισμα.[/quote]

Η ενέργεια $ E[f] \equal{} \iint f(x)f(y) \log|x \minus{} y| dx dy$ δεν είναι γραμμική, είναι ένα διπλό ολοκλήρωμα που δίνει και cross-terms στην ενέργεια $ E[f \plus{} g]$ ενός αθροίσματος. Για να ισχύει αυτό που λες, θα πρέπει να έχουμε $ \iint f(x) g(y) \log|x \minus{} y| dx dy$ οποτεδήποτε τα διπλά ολοκληρώματα των $ f,g$ είναι πεπερασμένα. Είναι τόσο προφανές αυτό όταν π.χ. και η $ f$ και η $ g$ μπορούν να έχουν ουσιώδεις ασυνέχειες σε [i]κάθε[/i] ρητό του $ [0,1]$?

Π.χ. τι γίνεται όταν $ f(x) \equal{} \sum_{n \equal{} 1}^\infty \frac {2^{ \minus{} n}}{\sqrt {|x \minus{} r_n|}}$ όπου $ \{r_n\}$ μία απαρίθμηση των ρητών του $ [0,1]$?

Cheerio,

Durandal 1707

PS: Για να μην παρεξηγηθώ, το παραπάνω επ'ουδενεί δε μειώνει τη διαισθητική αξία της ιδέας σου... I mean, that's what's really going on ;) Απλώς, το μόνο που λέω είναι ότι υπάρχουν μερικές τεχνικές δυσκολίες που πρέπει να ξεπεραστούν όταν επιτραπεί στην f να έχει άπειρες ουσιώδεις ασυνέχειες. Let's not dwell there :)


If $ \left|BC\right| \equal{} a$, $ \left|AC\right| \equal{} b$, $ \left|AB\right| \equal{} c$, $ 3\angle A \plus{} \angle B \equal{} 180{}^\circ$ and $ 3a \equal{} 2c$, then find $ b$ in terms of $ a$.

$\textbf{(A)}\  \frac {3a}{2} \qquad\textbf{(B)}\  \frac {5a}{4} \qquad\textbf{(C)}\  a\sqrt {2} \qquad\textbf{(D)}\  a\sqrt {3} \qquad\textbf{(E)}\ \frac {2a\sqrt {3} }{3}$

My answer differs from all the listed ones, yet I think I made no error according to the statement such as it is. Please verify both the solution and the statement.

[hide]$ 3\alpha \plus{} 2\beta \equal{} 180^\circ\implies 2\alpha \plus{} \beta \equal{} \gamma$

$ \beta \equal{} \gamma \minus{} 2\alpha\implies 3\alpha \plus{} 2(\gamma \minus{} 2\alpha) \equal{} 180^\circ\implies\boxed{\alpha \equal{} 2\gamma \minus{} 180^\circ}$

$ \beta \equal{} \gamma \minus{} 2(2\gamma \minus{} 180^\circ)\implies\boxed{\beta \equal{} 360^\circ \minus{} 3\gamma}$

$ {\sin\alpha\over a} \equal{} {\sin\gamma\over c}\implies{\sin(2\gamma \minus{} 180^\circ)\over a} \equal{} {\sin\gamma\over{3a\over 2}}\implies \minus{} \sin 2\gamma \equal{} {2\over 3}\sin\gamma\implies \minus{} 2\sin\gamma\cos\gamma \equal{} {2\over 3}\sin\gamma\implies\boxed{\cos\gamma \equal{} \minus{} {1\over 3}}$

$ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos\gamma\implies{9a^2\over 4} \equal{} a^2 \plus{} b^2 \plus{} {2a\over 3}b\implies b^2 \plus{} {2a\over 3}b \minus{} {5a^2\over 4} \equal{} 0$

Solving the quadratic and taking only the positive solution, we get

$ b \equal{} \frac { \minus{} {2a\over 3} \plus{} \sqrt {{4a^2\over 9} \plus{} 5a^2}}{2} \equal{} \frac { \minus{} {2a\over 3} \plus{} {7a\over 3}}{2}\implies\boxed{b \equal{} {5a\over 6}}$[/hide]

I have corrected the problem.
$ 3\angle A\plus{}2\angle B \equal{} 180{}^\circ$  will be $ 3\angle A\plus{}\angle B \equal{} 180{}^\circ$.

[hide="Solution to the corrected problem"]$ 3\alpha\plus{}\beta\equal{}180^\circ\implies\gamma\equal{}2\alpha\implies\beta\equal{}180^\circ\minus{}3\alpha$

$ {\sin\alpha\over a}\equal{}{\sin 2\alpha\over{3a\over 2}}\iff {3\over 2}\sin\alpha\equal{}2\sin\alpha\cos\alpha\iff\cos\alpha\equal{}{3\over 4}$

$ b^2\equal{}{9a^2\over 4}\plus{}a^2\plus{}3a^2\cos 3\alpha$

$ \cos 3\alpha\equal{}4\cos^3\alpha\minus{}3\cos\alpha\equal{}4\cdot{27\over 64}\minus{}{9\over 4}\equal{}\minus{}{9\over 16}$

$ b^2\equal{}{13a^2\over 4}\minus{}{27a^2\over 16}\equal{}{25a^2\over 16}\iff\boxed{b\equal{}{5a\over 4}}$[/hide]

Let $Q$ on $AB$ such that $CQ$ bisect $\angle ACB$. Let $\angle A=\alpha$ and $\angle B=\beta$. Observe that $\angle ACB=2\alpha$ and $\triangle ABC$ similiar with $\triangle CBQ$. We have
\[\frac{QC}{AC} = \frac{CB}{AB} = \frac{2}{3} \Longleftrightarrow QC= \frac{2b}{3} \]
\[\frac{QB}{CB} = \frac{CB}{AB} = \frac{2}{3} \Longleftrightarrow QB = \frac{2a}{3}\]
Because $AQ=QC$, we have
\[\frac{2b}{3} = AB-BQ = \frac{3}{2}a - \frac{2}{3} a = \frac{5}{6}a\]
And then $\displaystyle \boxed{b=\frac{5a}{4}}$.

3A+B=180°
A+B+C=180°
SO,
2A=C
sin 2A=sin C
2cos A=sin C/ sin A
 sin C/ sin A=c/a( Sine rule)
2cos A=c/a
3a=2c, c/a=3/2
cos A=3/4
b²+c²-a²/ 2bc=3/4
b²+(9a²)/4 +a²=6bc/4
{4b²+(5a²)}/4=9ab / 4
5a²-9ab+4b²=0
(a-b)(5a-4b)=0
If a-b=0
a=b
angle A=45°
cosA=1/√2
But cos A=3/4
So,
5a-4b=0
b=5a / 4


analytic geometry
vector
geometry
parallelogram

1.) Given points $ u \equal{} (3, \minus{}1, 2)$ and $ v \equal{} (5, 5, 5)$, give the coordinates of a point $ w$ which is on the line segment connecting the two points but nearer to $ u$ than $ v$.

2. Give an example of point t which is not collinear to $ uv$.

1. Compute the midpoint of U and V to get a point on the line, W(4, 2, 3.5), which is equally close to both. Now find the midpoint of W and U to get another point on the line closer to U than V, P(4.5, 3.5, 3.75).

2. Clearly UV is not the locus of (3,-1,2)+a(0,1,0), so setting a=1 gives a point off the line: (3,0,2).

[b]Notes:[/b]
A line is a locus of points of the form $ \vec{u} \plus{} a\vec{v}$, where $ \vec{u}, \; \vec{v}$ are n-dimensional vectors.

For part 1, this means that the mean of any number of points on the line is also on the line:
$ \frac {\sum_{j \equal{} 1}^k (\vec{u} \plus{} a_j \vec{v})}{k} \equal{} \frac {k \vec{u} \plus{} \vec{v} \sum_{j \equal{} 1}^k a_j }{k} \equal{} \vec{u} \plus{} \left( \sum_{j \equal{} 1}^k \frac {a_j}{k} \right) \vec{v}$, and is thus part of the locus.

For part 2, this means that if we keep the same $ \vec{v}$ but change our $ \vec{u}$, we will get a new line parallel to the original line, which means that any point on this new line is not on the original line. in terms of vectors U and V given, the original line was $ \vec{U} \plus{} a \vec{(V \minus{} U)}$. We notice that $ (0,1,0)$ is not of the form $ \vec{V \minus{} U}$, and thus adding this vector to the line will add to the constant vector, and not the scaled vector. It thus makes a parallel line.

Wow, that's a good explanation. But im totally new to this vector thing. I know that addition of vectors would produce a point that would form a parallelogram (in 2d space) which we all know. But how do I know that the two lines are parallel? pls give example in 2d space.


Prove that $ n^2 \minus{} n \plus{} 41$ is always prime for $ n \in [ \minus{} 39,40]$ and n is an integer

Correct me if I am wrong, but can't you use induction for -39 <--> 40?

No you cannot because this set is finite.

It's a finite set, so you can always just test each value or consider it mod every prime up to $ 41$.

Of course, there is more going on than that.  See [url=http://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions]the Wikipedia article[/url].

[quote="FOURRIER"]Prove that $ n^2 \minus{} n \plus{} 41$ is always prime for $ n \in [ \minus{} 39,40]$[/quote]
I think, you are wrong. Try $ n\equal{}\frac{\sqrt5\plus{}1}{2}.$ :wink:

It was meant to be $ n\in \{\minus{}39,\minus{}38,...,40,41\}$ :wink:

[quote="polskimisiek"]It was meant to be $ n\in \{ \minus{} 39, \minus{} 38,...,40,41\}$ :wink:[/quote] Without 41  :wink:


function
limit
calculus
integration
real analysis
real analysis unsolved

Given a continuous function $ f : \mathbb{R} -> \mathbb{R}$, let
\[ f^*(x) = \lim_{h->0^+} \frac{f(x+2h)-f(x-h)}{3h} \]
Assume that $ f^*(x)$ is defined for all $ x \in \mathbb {R}$ and $ f^* \geq 0$. Prove that $ f$ is increasing on $ \mathbb {R}$.

I would try to prove that the indefinite integral of $ f$ is convex instead. For integrals we have a lot of convergence theorems, so it shouldn't be hard to emulate some proofs from Calc I-class.

That maybe true, but I haven't found a way to make use of the convergence theorems in this problem.  :maybe:
Can you enlighten me ?

I can't solve it using my suggestion  :D . No matter what I try I end up needing to somehow bound $ \frac {f(x \plus{} 2h) \minus{} f(x \minus{} h)}{3h}$ from below by something that is integrable over compact intervals (or otherwise deal with the negative part of the fraction). If we could do that, then Fatou's lemma would say that

$ \int_a^b f^*(x) \, dx \leq \varliminf_{h\to 0^ \plus{} } \int_a^b \frac {f(x \plus{} 2h) \minus{} f(x \minus{} h)}{3h} \, dx \equal{} f(b) \minus{} f(a)$, the last equality by continuity.

But, I think it is of no use.

This is a standard "first crossing time trick" problem. The first step is to add $ \delta x$ to $ f$ to ensure that the limit is strictly positive (if we can show that $ f(x) \plus{} \delta x$ is increasing for all $ \delta > 0$, we can conclude that $ f$ itself is non-decreasing). Now we know that for every $ x$, we have $ f(x \plus{} 2h)\ge f(x \minus{} h)$ for all sufficiently small $ h$. Let $ a\in\mathbb R$, $ f(a) \equal{} c$, $ \varepsilon > 0$. Our aim is to show that $ f(x) \ge c \minus{} \varepsilon$ for all $ x > a$. Let $ s \equal{} \sup\{x: f\ge c \minus{} \varepsilon\text{ on }[a,x]\}$. The continuity of $ f$ implies that $ s > a$ and $ f\ge c \minus{} \varepsilon$ on $ [a,s]$. But then, if $ s$ is finite, $ f(s \plus{} 2h)\ge f(s \minus{} h)\ge c \minus{} \varepsilon$ for all small $ h$ and we see that there is $ s' > s$ such that $ f\ge c \minus{} \varepsilon$ on $ [a,s']$, which contradicts the definition of $ s$. Thus, $ s \equal{} \plus{} \infty$, and we are done.

Let's try to generalize now. Fix some numbers $ a_1,\dots a_n; c_1,\dots,c_n$ such that $ \sum_j c_j\equal{}0$, $ \sum_j c_j a_j > 0$ and assume that for every $ x$, we have $ \sum_{j}c_jf(x \plus{} a_jh) > 0$ if $ h$ is small enough. Must $ f$ be non-decreasing?


group theory
superior algebra
superior algebra solved

The statement must be wrong- I can show that $a^{5}=e$, but certainly not $a=e$ or $b=e$. It's possible for $a$ to be an element of order 5 and for $b$ to be an element of order 2 in the dihedral group with ten elements.

[quote="LiMa"]Hello rustam,

maybe you meant the problem mentioned in this thread:
http://www.mathlinks.ro/Forum/viewtopic.php?t=90501

LiMa[/quote]

Yes, it was my fault , I did typo.
Thanx.


hey! what are hardest geometry problem you have?
please give me five of them as soon as you can?
thank!

Construct the Trisection of an angle using a compass and a strait edge  :P.

[quote="eagerkill"]no one have more hardest geometry problem :(
i want five hardest geometry problem with solution
who can give me? :lol:[/quote]
You think you are wery well in geometry? :D
If yes,geometry have more unsloved problems,if you can solve any of it. :rotfl:

there are so many hard problems in geometry section...

dont u want to search them?

Eagerkill, I give you a test : solve [b]("kill")[/b] at least five problems from the [u]unsolved geometry problems - section[/u] and after that we say again !

i not good at geometry but i like it
so i want many hard problem
please don't lauch me

It is O.K. It isn't the your case. There are many dificult problems in this section. Success !


calculus
integration
calculus computations

"Show that if f(x) has a period p, the average value of f is the same over any interval of length p".

This is more of a physics problem using the general average of a function. I was thinking about breaking the integral into the sum of 2 integrals:

INT(f(x),x,a,p)+INT(f(x),x,p,a+p)=INT(f(x),x,a, a+p). Sub in x=t+p in the last integral.

This is where I am hung up. Can someone lend a hand?.

Cody

You have it - just keep going! So far, you've said this:

$\int_a^{a+p}f(x)\,dx=\int_a^pf(x)\,dx+\int_p^{a+p}f(x)\,dx$

Now you substitute $x=t+p$ or $t=x-p$ into the last integral:

$\int_p^{a+p}f(x)\,dx=\int_0^af(t+p)\,dt=\int_0^af(t)\,dt.$

We have just explicitly used that $f$ is periodic. But now, we can just let $t=x$ again:

$\int_0^af(t)\,dt=\int_0^af(x)\,dx$ so

$\int_a^{a+p}f(x)\,dx=\int_a^pf(x)\,dx+\int_0^af(x)\,dx= \int_0^pf(x)\,dx.$

There's an implicit hypothesis in this, which is that $f$ is integrable over any bounded interval under whatever theory of integration you are assuming.

You are undoubtedly drawing pictures which assume that $a\in(0,p).$ The proof actually works for any $a$, but if you feel more comfortable you can also WLOG first shift the whole thing by an integer multiple of $p$.

Thank you very much. I see I was on the right track. One other question though. What is a WLOG first shift?.

Thanks again,
Cody

"What is a WLOG first shift?."

Just leave the word "first" out and understand "shift" as the verb in that sentence. If necessary, make the substitution $u=x+np$ for some $n\in\mathbb{Z}$, which you can do without changing anything.

Also, "WLOG" stands for "without loss of generality".


i kinda have a problem. I have no new mail in my Private Messages icon up in the menu, and the icon is yellow and green instead of blue and black. Can anyone help me? :?

don't worry, same thing's happening with me. You get used to it after a while. It's nothing big.

[quote="G-UNIT"]Each and every time I look at my highlighted Private Messages icon, my heart fills with joy.  The fact that someone would  bother to write to me, gives me more happiness and inspirational imaginable.  Its what keeps me breathing each day.  And then, when I foolishly click on it, and I find no one new has messaged me.....my heart breaks.  I feel no one cares for me.  No one appreciates me.  I am alone and unwanted. :(

  How could you daresay its "nothing big"?[/quote]

Yep, but I correspond with *a few* more people then probably most users here. My current completely-full inbox has PM's from 21 unique users. None of whom is a relative of mine. :)  How about ya'll? How many unique users have PMed you in the last 100 PMs?

[quote="G-UNIT"]Each and every time I look at my highlighted Private Messages icon, my heart fills with joy.  The fact that someone would  bother to write to me, gives me more happiness and inspirational imaginable.  Its what keeps me breathing each day.  And then, when I foolishly click on it, and I find no one new has messaged me.....my heart breaks.  I feel no one cares for me.  No one appreciates me.  I am alone and unwanted.

  How could you daresay its "nothing big"?[/quote]

You get even more happiness when the PM pop-up comes up. Yay for more happiness!

[quote="solafidefarms"]Yep, but I correspond with *a few* more people then probably most users here. My current completely-full inbox has PM's from 21 unique users. None of whom is a relative of mine. :)  How about ya'll? How many unique users have PMed you in the last 100 PMs?[/quote]

ten, technically. but one was r.. rus... er... richard, about a class. a BUNCH were mcrawford. and then half of the rest were "please dont bump old threads" or "please edit this post" so I don't think they really count.

I feel the same way as G-UNIT. :D :D :D Good thing that never happened to me! :rotfl:

I have the same problem.  I think it comes from deleting messages without reading them.  Does anybody know how to fix it?

I have 22 different people in my message box, but that goes way back to december so that doesn't really count.

Mine only goes back to Aug. 29, but I never delete anything.

I've got 40 different people in my inbox. But I often delete PMs so the actual number of people I have been corresponding with using PM might be more than 40.


Hi!

How can I make delimiters like a $ \bigg\{$ of all equations of eg an align-environment? E.g. something like
\[ \begin{cases}
  a\equal{}b&\qquad\qquad\qquad (1)\\
  c\equal{}d&\qquad\qquad\qquad (2)\\
  e\equal{}f&\qquad\qquad\qquad (3)
\end{cases}
\]

In particular, I can't use the cases-environment because I need to tag each equation automatically.

Thanks
robs

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=cases]cases package[/url]. The following code then gives exactly the example you give
[code]\begin{numcases}{}
 a=b \\
 c=d \\
 e=f
\end{numcases}[/code]


Here is a recurence that seems very hard to me, if you don't know how it appeared!!!
So, $a_1=0,a_2=0,a_3=1$ and as the title suggests $\displaystyle a_n=a_{[\frac{n}{2}]}+a_{[\frac{n+1}{2}]}$.
It would be nice to find a direct formula for $a_n$, for example if we consider the binary representation of $n={\overline{x_1\ldots x_k}}_2$, what would the answer be ;) ?
If hard, I can offer a hint! Also, it has a combinatorics solution, so moderators, please don't move it!!!

You forgot to say what we are suppossed to do.


Over the past few months, my ratings kept going up and down by .1 very often. (they've been staying around the same number, it's just sometimes it's .1 higher sometimes it's .1 lower)

However, whenever I check my past posts, I see that none of them are rated.

What is causing the fluctuation in rating?

Rating on a post is only shown when multiple people rate it. So you won't always see a post as rated, even if it is.


I'm not sure if people on this forum are fimiliar with MATLAB and flops, but here goes.

find the flop count for the following sequence of commands:
a=0;
..for p=1:n
....for q=p:n
......for r=q:n
......a=a+1;
....end
..end
end

I am not looking for a number, but a summation formula interms of n.

I know this is close:

sum(i,n) i = n(1-n)/2
and
sum(i,n) i^2 = n(1-n)(2n-a)/6

and i am tempted to find

sum(i,n) i^3

Does anybody have any idea how to go abouts solving this?

[quote="Kent Merryfield"]Minor technical quibble: is $a: =a+1$ really a "flop"? It looks like an integer. Of course, if you just ran this program for a given value of $n,$ and output $a$ at the end of the program, that output value of $a$ would simply be the count you're looking for.[/quote]

So what your saying is that the output a for some n is the number of flops?  :o

I'm not sure if I am following this  :maybe:

Infact, could you please explain to me how matlab will calculate a for some given n?  I can't seem to figure this out.


If u had 52 cards in a deck(clubs, spades, hearts, diamonds); Aces-kings;and u put them in 4 equal stacks. stack 1 is in order from Ace-King. stack 2 is in order from King-Ace.stack 3 is in order from Ace-King. stack 4 is in order from King-Ace. then , u take 1 card each round from each stack, until all the cards r gone. u lay all of the cards face up, what is the one card that show up in the middle with 4 of the same right next to each other?

[hide]The middle cards of the A-K and K-A stacks will be the same. They will be (1+13)/2=7. Since you will be taking all of the middle cards in the same round, it will be the 4 "7"s that appear all next to each other.[/hide]

is this problem too stupid; no point? or is it ok? y isn't anyone else posting any replies??

The question isn't stupid. It is actually very fun. :)

Only thing is that sometimes, people don't reply to all the questions. And especially for some people in this forum, they mostly post on the Intermediate rather than Getting Started so that's why it was not getting a lot of points.

Posting problems are always good. Keep up your good work!

 :)

I think the problem was just a big confusing to understand, I'll try to rewrite it.

A standard deck of 52 cards is divided into four stacks. Two of them go from 1-K, and two go from K-1.

You then merge these four mini-stacks into one big stack by taking one card from each mini-stack and putting that set of 4 cards onto the top of the merged big stack until all the cards have been merged into the big stack.

In this big stack, a chain of 4 of the same rank cards (suits are irrelevant) can be found in the middle. What is the rank of those cards?

thx. that does make it more easier to read.  :)


First of all, thank you for the new pages on [tex]\LaTeX[/tex] - they look quite nice.

Now, a (minor) note:
I don't know why, but on this page: http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
The \displaystyle versions of \int and \oint are missing the upper right hand portion of the symbols.

edit: Just noticed that you have a note on there about that already

edit2: <pedantic>On [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#fonts]this page[/url] "though" should be "thought"</pedantic>

That's a result of the parsing latexrender does on very tiny images (it LaTeXes the source, then runs it through imagemagick, which I think is responsible for the snipping).  The arrows come out more nicely in LaTeX than you see on that page, too.


Denote by $ s(n)$ and $ k(n)$ the sum and the number of decimal digits of $ n$. Find all the integers $ n$ such that $ k(n).s(n^2) \equal{} n \plus{} 1$.

If $ k(n)\equal{}k$, then $ 10^{k\minus{}1}\le n <10^k, s(n)\le 9k$ and from $ 10^{k\minus{}1}<n\plus{}1\le 9k^2$ we get $ k\le 2$.
If k=1, $ s(n^2)\equal{}n\plus{}1$ had not solution.
If k=2 $ s(n^2)\equal{}\frac{n\plus{}1}{2}<36\to n\plus{}1<72\to n\plus{}1<62\to n\plus{}1<60.$
Therefore we can chek only $ n\equal{}11,13,15,...,57$ (24 numbers) and find solution.

You have a mistake: my problem is $ s(n^2)$ not $ s(n)$! :)


vector
geometry
3D geometry
sphere
induction
Functional Analysis
real analysis

Let $ y_k$ be a sequence of vectors in the infinite-dimensional Hilbert space $ H$ such that $ \sup_k(x,y_k)\ge 1$ for all unit vectors $ x\in H$. Does it follow that there exists a unit vector $ x\in H$ such that
a)$ \limsup_{k\to\infty}(x,y_k) > 0$
b)$ \limsup_{k\to\infty}(x,y_k)\ge 0.9$
c)$ \limsup_{k\to\infty}(x,y_k)\ge 1$?

I decided to try a version between b) and c):
(d) Does $ \forall x \sup_k (x,y_k) > 1$ imply $ \exists x\limsup (x,y_k) > 1$?

Let's call a spherical cap [i]proper[/i] if it is contained in some hemisphere. Statement d) has a nice geometric formulation:
(*) Any cover of the Hilbert space sphere by open proper spherical caps covers some point infinitely many times.
(The caps in question are $ \{z\colon |z| \equal{} 1, (z,y_k) > 1\}$)

A finite-dimensional version of (*) is that some point of $ S^n$ is covered by at least $ n \plus{} 1$ caps.
[b]Proof:[/b] by induction. Trivial when n=1. If the total $ n$-volume of caps is infinite, also trivial. Otherwise there is a cap $ C$ of maximal radius. Its boundary is an $ n \minus{} 1$ dimensional sphere, also covered by proper spherical caps. (By maximality of $ C$ and because a ball cannot contain more than half of a sphere unless it has greater radius.) By induction hypothesis, some point on $ \partial C$ is covered by at least $ n$ caps. So is a neighborhood of that point. But a part of that neighborhood lies in $ C$, which gives $ n \plus{} 1$.

Returning to the Hilbert space, we see that there are points covered by arbitrarily many sets, which of course isn't what we want. In the special case $ |y_k|\searrow 1$ the argument for $ S^n$ seems to work: restrict to the boundary of the largest cap, then to the boundary of the largest cap there, etc. - the sets are closed and nested, and their diameters tend to $ 0$, so there is a common point $ x$. This common point is covered by infinitely many caps (their closures, but this does not really matter). So, (d) is true when $ |y_k|\searrow 1$.

Ther is a classical theorem of Corson (Corson, Harry H. Collections of convex sets which cover a Banach space. (English) [J] Fundam. Math. 49, 143-145 (1961)) which says that if an infinite dimensional reflexive Banach space $ X$ is covered by a family of convex closed and bounded sets, then this covering is not locally finite, i.e. there is a point $ x$, such that every neighborhood of $ x$ meets infinitely many  members of the covering. Some recent improvements see in Fonf, Vladimir P.; Zanco, Clemente Finitely locally finite coverings of Banach spaces. (English) [J] J. Math. Anal. Appl. 350, No. 2, 640-650 (2009)

From the Corson's theorem one can deduce version b) of Fedja's question. Let us demonstrate this. Denote as usual $ B_H$ and $ S_H$ the unit ball and the unit sphere of $ H$ respectively. Fix an $ \varepsilon > 0$ and assume contrary that $ \limsup_{k\to\infty}\langle h ,y_{k}\rangle < 1 \minus{} \varepsilon$ for every $ h \in S_H$. Then in particular the sequence of functionals $ \langle .,y_{k}\rangle$ is pointwise bounded, so by the uniform boundness principle $ \sup_n \|y_n\| \equal{} M < \infty$. Define
\[ A_n \equal{} \{x \in (1 \plus{} \varepsilon)B_H: \langle x,y_{k}\rangle \ge 1\}\]
Then all $ A_n$ are bounded convex and closed,  $ \bigcup_n A_n
\supset (1 \plus{} \varepsilon)B_H \setminus B_H$ and $ A_n \bigcap
\frac{1}{M} B_H \equal{} \emptyset$. Consider the covering of $ H$ by $ B_H$ and $ A_{n,m}: \equal{} (1 \plus{} \varepsilon)^{m\minus{}1}A_n$, $ n,m \in \mathbb N$. By the Corson's theorem there is a point $ x \in H$ such that every neighborhood of $ x$ meets infinitely many of $ A_{n,m}$. Since
\[ A_{n,m} \bigcap \frac{(1 \plus{} \varepsilon)^{m\minus{}1}}{M} B_H \equal{}
\emptyset,\]
 $ A_{n,m}$ with big $ m$ cannot be close to $ x$. So there is an $ m_0$ such that every neighborhood of $ x$ meets infinitely many of $ A_{n,m_0}$. This means that $ x \in (1 \plus{} \varepsilon)^{m_0} B_H$ and $ \limsup_n \langle x,y_{n}\rangle \ge (1 \plus{} \varepsilon)^{m_0\minus{}1}$. Then $ h \equal{} \frac{x}{\|x\|}$ is the unit vector that we need to get the contradiction:
\[ \limsup_n \langle h ,y_{n}\rangle \ge (1 \plus{} \varepsilon)^{\minus{}1}.\]


MIT college Harvard Yale Princeton Stanford Columbia

To all the current juniors (or anybody else, I guess): where are you guys planning on applying so far?
I, myself, have no idea...

Two reaches.
Two regular schools.
One saftey.

[quote="mirebeka"]I'm trying for some of the ivies and MIT, and i'm still trying to come up with a good safety.  Anyone know of any?  Especially ones in the northeast.[/quote]
Michigan Technological University (mtu.edu) is very good for engineering (and other stuff). I believe it's the #2 public school in the nation for engineering (second to Georgia Tech).

[quote="kidnappedturtle"]To all the current juniors ... : where are you guys planning on applying so far? [/quote]

I'm looking at applying to either Harvard or Yale early.  If I get deferred, I'll send out applications to MIT, Yale or Harvard (the one I didn't apply to early), Princeton, Stanford, UC Berkeley, and maybe Columbia or Brown.  If I get accepted early, I'll be sending out regular apps to only MIT and Yale/Harvard (and maybe Princeton).

I'm looking for a place where I can pursue mathematics at a graduate level and be surrounded by fellow students who excel at math.  But I'm not certain that I want to go into math, so I desire a school with challenging classes in a rich variety of areas (physics, linguistics, biology, history, languages, literature, etc.)  FInally, I play the violin and would like a school that fosters a healthy music environment, complete with a top-notch orchestra, various student-run ensembles, and preferably string faculty.  I also would like to stay relatively close to home (CT).

In my mind, my preferences narrow my search to Harvard and Yale, and possibly to MIT.  But of course I wouldn't mind to attend any of the places on my complete list.

[quote="cn2_71828182846"]In my mind, my preferences narrow my search to Harvard and Yale, and possibly to MIT.[/quote]

By request, I'll reply to this. I think based on the criteria you set up, I'd prefer Princeton to Yale (even though Princeton is not in Connecticut, and Yale is). That's because earlier threads on this forum suggest that even though Yale has undeniably strong math PROFESSORS, it doesn't have the same kind of coherent undergraduate program that, say, Princeton has. I think helpful forum participant Gauss202 may be persuaded to tell you what he liked about Princeton, and I'm not sure that we have a Yalie here.

I largely agree with your list of top math schools, and what programs you look for other than math. Indeed, your selection criteria (other than preferring to be near CT) sound much like the criteria that I THINK my son will be applying to his college search several years from now. My concern, thinking about the idea of "safety school," is where you can apply that you can be reasonably SURE to get into, because all the schools you aspire to are hard schools to get into. Because we live in Minnesota, and my son is already in an accelerated math program at the U of Minnesota, the U of Minnesota (my alma mater, for a Chinese language major) is already in place as a "safety school" for my son. I guess JBL and I constitute a choir of two voices saying that one can NEVER be sure of getting into any particular "top" school, so I'm just wondering what other school you would apply to just to be reasonably sure you get in somewhere.

Hope this helps!

[quote="cn2_71828182846"]Well, what do you think would be a good safety?  I have thought about this myself, and have had trouble coming up with a school.  The issue is that I do not truly desire to attend any school that might fit that category.[/quote]
State schools are often a good choice for safety schools, and are frequently also a good choice for an education.

[quote="JS1527"][quote="cn2_71828182846"]The issue is that I do not truly desire to attend any school that might fit that category.[/quote]

Listen, it's great if you get into one of those schools, but believe it or not, you can still be happy in a school that is not ranked in the top 25.[/quote]

With all due respect, how can you be so sure?

[quote="JS1527"]I go to University of Rhode Island right now and I am happy.  If you look it up in the US news rankings it is in the third level of schools (there are four levels, I think).[/quote]

Again with all due respect, you do not convince me that I would be happy at a school not in the "top 25."  I see that you are happy at University of Rhode Island; that is great.  But what makes you believe I would be happy at a place like that?  This is not to say that there is anything wrong with your school, but that we might have different interests.

It's just that there are some elements of a school I believe to be necessary for me to be happy there.  (I listed these characteristics above.)  It seems to me that the only schools that meet these criteria are in the "top 25," so to speak.

[quote="cn2_71828182846"]I'm looking for a place where I can pursue mathematics at a graduate level and be surrounded by fellow students who accel at math.[/quote]

btw, it's excel, not accel. I'm not trying to be picky, but it's good if you know this, especially on a college app.

Thanks, Yasha.  It has been edited.

It occured to me, however, that to write "students who accel in math" could be considered short-hand for "students who accelerate through mathematics courses," i.e. those who skip grades in math.  But, after all, I wrote "students who accel 'at' math," not "students who accel 'in' math."

By the way, I read earlier that you used to live in New Haven.  When did you live there, and what school(s) did you attend?

Summer 2000 through Summer 2003. Went to a public school in Hamden in 6th grade, Hopkins for 7th and 8th.

Hamden? Awesome!  Which elementary school?

Well let me ask you this: people on this forum are always warning how hard it is to get into schools in the top 25, even with lots of credentials.  What are you going to do if you don't get into a top 25 school?  If you don't apply to anywhere else, you are in trouble.  If you did, then will you just resign yourself to unhappiness?

Yeah, PM me when you get your acceptance letters.  (Also, on your list of academic credentials, be sure to mention your unique ability to create new words, as well as innovative grammatical structures to go along with them.)


induction
function
algebra proposed
algebra

Suppose that you are given some subsets of the set of real numbers $X_{1,1},X_{1,2},...,X_{2,1},X_{2,2},...$ such that $X_{n,j+1}$ is a subset of $X_{n,j}$ for all $n,j$, $X_{j,n}$ is a subset of $X_{j+1,n}$ for all $j,n$ and also for all $n$, the union of all $X_{j,n}$ is the whole set of real numbers. Is it true that we can choose $n_{1}<n_{2}<...$ such that the intersection of all $X_{n_{i},i}$ is infinite?

I think the answer is "YES"
[b]All of the following is wrong:!!!!!!![/b]
I think this problem refers to the well-known problem [b]" the number of the real numbers is more than the number of the integer"[/b](Is it been expressed like that?)
But, I have little knowledge about this ,so I can't give a regorous proof ,though I have some idea of it.
I hope my idea is correct:
We call a set$M$   "[b]Very large[/b]"   iff there isn't such a bijection $f: N-\to M$ .
Firstly,If all of $A_{i},(i=1,.....)$  is [b]Not Very Large[/b],
then $\cap_{i=1}^{\infty}A_{i}$ is [b]Not Very Large[/b].

Secondly $R$ is[b]Very Large[/b],
Thirdly there must exist $X_{i,j}$ such that $X_{i,j}$  is [b]Very Large[/b] and then We take $n_{1}=i$ and $X_{n_{1},1}$is [b]Very Large[/b]
Forthly,If we have found $n_{1},.....n_{k}$ and $\cap_{i=1}^{k}X_{n_{i},i}$ is  [b]Very Large[/b],then We can found $n_{k+1}$ such that
$\cap_{i=1}^{k+1}X_{n_{i},i}$ is  [b]Very Large[/b].
I think the third and forth is based on the first one.So if the first is true,then all is true.
I hope I am correct. :D

Your very large means uncountable. And the problem is the following: if I understood correctly, you prove by induction the existence of $n_{1}<n_{2}<...<n_{k}<...$ such that each of the intersections of the sets $X_{n_{1},1}$, then $X_{n_{1},1}$ and $X_{n_{2},2}$, then the next three and so on are uncountable. However, this does not imply that the intersection of all $X_{n_{j},j}$ is uncountable (just imagine that $X_{n_{j},j}$ you construct are $[j,\infty)$).

Oh, you understood me correctly.But I am not correct :D

But I have a question :
If $A_{i}=[i,\infty)$ then $\cup_{i=1}^{\infty}A_{i}$ is infinte or finite or $\emptyset$? I can't understand such a complex problem with two many $\infty$ :D
Thanks.

Hawk Tiger, here we speak about intersections and not unions. The intersection of all $A_{i}$ is clearly void, but their union is infinite. I agree however that the problem is very strange...

Very sorry.All of the $\cup$ should be replaced by $\cap$ .It's a typo.
I have understand why it is void now( so stupid question :blush: )
I think I am unable to disccuss the problem you proposed...... :blush:

Harazi, this problem is essentially equivalent to your last question in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=continuum&t=79703]this[/url] thread, namely:

If $(f_{n})_{n}$ is a sequence of real functions defined on the real line which converges pointwise to $0$, must there exist an infinite set on which $(f_{n})$ converges uniformly?

The answer is 'yes', and the proof in that topic should be easy to adapt.


geometry
analytic geometry
quadratics
trapezoid
complex numbers
algebra
quadratic formula

Let $r$ be a nonzero real number. The values of $z$ which satisfy the equation \[ r^4z^4 + (10r^6-2r^2)z^2-16r^5z+(9r^8+10r^4+1) = 0  \] are plotted on the complex plane (i.e. using the real part of each root as the x-coordinate
and the imaginary part as the y-coordinate). Show that the area of the convex quadrilateral with these points as vertices is independent of $r$, and find this area.

[hide="Answer"]We begin by factoring the quartic into two quadratics. The final term factors to $9r^8+10r^4+1=(9r^4+1)(r^4+1)$. If we assume that the first term factors to $(r^2z^2)^2$, then we factor the quartic to $(r^2z^2+2rz+(9r^4+1))(r^2z^2-2rz+(r^4+1))$. Using the quadratic formula on each of these, we find the roots to be $\frac{-1}{r}\pm 3ri$ and $\frac{-1}{r}\pm ri$.

Plotting these points on the imaginary plane, we have an isosceles trapezoid with height $\frac{2}{r}$ and bases $2ri$ and $6ri$. Therefore, the area of the trapezoid is $\frac{2ri+6ri}{2}\cdot \frac{2}{r}=\boxed{8i}$.[/hide]

I forgot to divide the roots by $2$ in the solution that I sent in. >_< Also, are you supposed to include the $i$ in the area?

no, how can you have an imaginary area?
when you state the bases you are giving their coordinates, not the distance
and how did you get bases to be 2r and 6r?
you have the roots correct, 3ri and ri, why did you multiply by 2?

so, [hide="my solution"]
label your roots $x_1+y_1*i, x_2+y_2*i$ and their conjugates
because coefficient of z^3 term is 0 sum of the roots is 0 and $x_1=-x_2$
so height is $(x_1- -x_1)=2x_1$ and bases are $2y_1$ and $2y_2$
Area = $2x_1(y_1+y_2)$
Now get the roots as jesusfreak said, and so you have
$x_1=1/r , y_1=r , y_2=3r$ so plug this in for the area formula above and you get 8.[/hide]

Do you think it matters if we said $6r$ and $2r$ like you did, rather than $|6r|$ and $|2r|$ because distance has to be positive?

[quote="JesusFreak197"]Because you have $3r+|-3r|=6r$ and $r+|-r|=2r$...[/quote]

Oh, I see what you did.


Aratati ca pentru orice $n\in{\bf N}$ exista o infinitate de numere $a_1,a_2,b_1,b_2\in{\bf Q}^*$ si $\theta\in{\bf R}$ astfel incat \[ \sin\theta=a_1+b_1\sqrt{n} \] si \[ \cos\theta=a_2+b_2\sqrt{n} \].

Fie $n\in{\bf N}$ fixat arbitrar.
Pentru fiecare $p\in{\bf N}$, fie
$a_1=\frac{1}{p^{2}+2}$ si $b_1=\frac{p^{2}+1}{(p^{2}+2)([\sqrt{n}]+1)}$.
Avem $a_1+b_1\sqrt{n}=\frac{1}{p^{2}+2}+\frac{(p^{2}+1)\sqrt{n}}{(p^{2}+2)([\sqrt{n}]+1)}=y_p\in{(0;1)}$,
de unde deducem că există o infinitate de valori $\theta=\theta_p\in{\bf R}$ astfel încât $\sin{\theta}=y_p$.
Analog se tratează cosinusul.

Sunteti sigur ca exista $a_2,b_2\in{\bf Q}$ astfel ca \[ \cos\theta=\sqrt{ 1-\left(\frac{1}{p^2+2}+\frac{(p^2+1)\sqrt{n}}{(p^2+2)([\sqrt{n}]+1)\right)^2}=a_2+b_2\sqrt{n}\; ? }\]

Prin ridicare la patrat si identificarea termenilor rationali si a celor irationali in egalitatea anterioara apare un sistem in $a_2$ si $b_2$ care nu stiu daca are solutie rationala (nu vad o solutie rapida a acestui fapt).

De aceea cred ca solutia dumneavoastra este incompleta. Ea asigura doar pentru $\sin\theta$ reprezentarea ceruta, dar nu si pentru $\cos\theta$.

Mai incercati!

Ţinând cont că $\sin^2{\theta}+\cos^2{\theta}=1$, ar trebui să avem
$(a_1+b_1\sqrt n)^2+(a_2+b_2\sqrt n)^2=1\Leftrightarrow (a_1^{2}+a_2^{2}+b_1^{2}n+b_2^{2}n)+2(a_1b_1+a_2b_2)\sqrt{n}=1$ (1).
Facem $a_1b_1+a_2b_2=0$ şi fie $\frac{a_1}{b_2}=-\frac{a_2}{b_1}=k\in Q^{*}\Rightarrow a_1=kb_2, a_2=-kb_1.$
Ţinând cont de (1), mai trebuie să avem  $a_1^{2}+a_2^{2}+b_1^{2}n+b_2^{2}n=1\Leftrightarrow (k^2+n)(b_1^2+b_2^2)=1.$ (2)
Alegem $k=\frac{n-4}{4}$ ($n$ diferit de $4$; cazul $n=4$ este banal) şi relaţia (2) se scrie $(\frac{n+4}{4})^2 (b_1^2+b_2^2)=1.$ (3)
Pentru $p\in N, p>1$ arbitrar, luăm $b_1=\frac{4}{n+4}\cdot\frac{p^2-1}{p^2+1}$ şi $b_2=\frac{4}{n+4}\cdot\frac{2p}{p^2+1}$ care asigură valabilitatea relaţiei (3)
Este evident că există o infinitate de valori $\theta=\theta _p$ care verifică ipotezele.


Let $b>a\geq 2$ be positive integers. Prove that if number $a+k$ is coprime to number $b+k$, for all $k=1,2,...,b-a$, then $a,b$ are consecutive numbers

Among the $b-a$ consecutive integers $a+1,a+2,\ldots,a+(b-a)$ one is divisible by b-a. But if $b-a \mid a+k$, then $b-a \mid a+k+(b-a) = b+k$, then $b-a \mid (a+k,b+k)$. From the hypotesis of coprimality, $(a+k,b+k)=1$, then $|b-a| = 1$.


Consider four distinct digits.

Construct the greatest four digit number out of those digits.

Now construct the smallest number, again from those four digits.

If the difference of the two numbers consists of the same four digits as chosen originally, can you find the four digits?

:?:

If the mods feel that its not suitable here, please move it to the appropriate folder.

[quote="abhishek.nanda"]i got the number 7641 though i'm not sure whether its a unique solution.[/quote]

Thanks, how to prove this? or just direct calculation over the set of 4 digit numbers?

[quote="Gyan"]Hint: [hide]There aren't too many numbers  to try ... ( about 10)  such that
 the digits are distinct ..
 Their sum is  18 or 27
 ( becuase The diff  would be divisible by 9) so even a bruteforce method is  will take a minute or two ...[/hide][/quote]

of course, this is there. But I was wondering if this number can be arrived logically.

Anyways, thanks for the help.

though i didnt find any totally logical sol u can put a few restictions and then try out diff n.os


function
limit
induction
calculus
calculus computations

For a function $ f(x) \equal{} 4x \minus{} x^2$, a sequence $ \{a_n\}$ is defined by $ a_1 \equal{} c\ (0 < c < 2),\ a_{n \plus{} 1} \equal{} \sqrt {f(a_n)}\ (n \equal{} 1,\ 2,\ 3,\ \cdots)$.

(1) Prove that $ a_n < 2,\ a_n < a_{n \plus{} 1}\ (n \equal{} 1,\ 2,\ 3,\ \cdots)$.

(2) $ 2 \minus{} a_{n \plus{} 1} < \frac {2 \minus{} c}{2}(2 \minus{} a_n)\ (n \equal{} 1,\ 2,\ 3,\ \cdots)$.

(3) Find $ \lim_{n\to\infty} a_n$.

a)by induction for $ n \equal{} 1$ it's true,next suppose that for $ n$ we have $ 0 < a_n < 2$ then since $ (a_n \minus{} 2)^2 > 0$ we have  $ 0 < 4a_n \minus{} a_n^2 < 4$ so  $ 0 < a_{n \plus{} 1} < 2$.

we have $ a_{n \plus{} 1}^2 \minus{} a_n^2 \equal{} 2a_n(2 \minus{} a_n) > 0$ and since $ a_n$ and $ a_{n \plus{} 1}$ are positive we have $ 0 < a_n < a_{n \plus{} 1}$.

b)$ 2 \minus{} a_{n \plus{} 1} \equal{} 2 \minus{} \sqrt {4a_n \minus{} a_n^2} \equal{} \frac {4 \minus{} 4a_n \plus{} a_n^2}{2 \plus{} \sqrt {4a_n \minus{} a_n^2}} \equal{} (2 \minus{} a_n)\frac {2 \minus{} a_n}{2 \plus{} \sqrt {4a_n \minus{} a_n^2}}$.

in other hand we have $ 2 \minus{} a_n < 2 \minus{} c$ and $ 2 \plus{} \sqrt {4a_n \minus{} a_n^2} > 2$. so $ 2 \minus{} a_{n \plus{} 1} < \frac {2 \minus{} c}{2}(2 \minus{} a_n)$.

c)by induction we prove that for every integer $ n$ we have $ 2 \minus{} a_n < (\frac {2 \minus{} c}{2})^{n \minus{} 1}(2 \minus{} a_1)$ and since $ \frac {2 \minus{} c} {2} < 1$ then $ (\frac {2 \minus{} c}{2})^{n \minus{} 1}(2 \minus{} a_1)$ go to $ 0$ ,and from a) $ a_n$ converge to a real $ x$ therefore $ 2 \minus{} x \le 0$ so $ x \equal{} 2$.

Your solution is a typical solution for Japanese Highschool students. But I have two questions.

[quote="tchebytchev"]
in other hand we have $ 2 \minus{} a_n < 2 \minus{} c$[/quote]

Why can you say that?

[quote="tchebytchev"]

c)by induction we prove that for every integer $ n$ we have $ 2 \minus{} a_n < (\frac {2 \minus{} c}{2})^{n \minus{} 1}(2 \minus{} a_1)$ and since $ \frac {2 \minus{} c} {2} < 1$ then $ (\frac {2 \minus{} c}{2})^{n \minus{} 1}(2 \minus{} a_1)$ go to $ 0$ ,[/quote]

You should say that $ 0<\frac{2\minus{}c}{2}<1$ or $ \left|\frac{2\minus{}c}{2}\right|<1$.

[quote="kunny"]
in other hand we have $ 2 \minus{} a_n < 2 \minus{} c$[/quote]

because $ a_n> a_{n\minus{}1}>.....>a_1 \equal{}c$  then $ \minus{}a_n<\minus{}c$ so $ 2\minus{}a_n <2\minus{}c$.

[quote]You should say that $ 0 < \frac {2 \minus{} c}{2} < 1$ or $ \left|\frac {2 \minus{} c}{2}\right| < 1$[/quote]

yes of course since  $ 0<c<2$ then $ \minus{}2<\minus{}c<0$ so $ 0<2\minus{}c<2$ . :)

O.K. :)

Here is my solution to the former part of (1).

Suppose that $ 0 < a_n < 2$, we have $ a_{n \plus{} 1} \equal{} \sqrt {4a_n \minus{} a_n^2} \equal{} \sqrt {a_n(4 \minus{} a_n)}\leq \frac {a_n \plus{} (4 \minus{} a_n)}{2} \equal{} 2$, but the equality doesn't hold, which completes the proof by induction.


Suppose $A=\{1,2,\dots,2002\}$ and $M=\{1001,2003,3005\}$. $B$ is an non-empty subset of $A$. $B$ is called a $M$-free set if the sum of any two numbers in $B$ does not belong to $M$. If $A=A_1\cup A_2$, $A_1\cap A_2=\emptyset$ and $A_1,A_2$ are $M$-free sets, we call the ordered pair $(A_1,A_2)$ a $M$-partition of $A$. Find the number of $M$-partitions of $A$.

SWEEEEEEEEET  :D

Let's show a more general result, i.e. let $n \geq 3$ be an odd positive integer and consider $A = \{1,2,\dots , 2n\}$ and $M = \{n, 2n+1, 3n+2\}$.  We show that there are exactly $2^{(n+1)/2}$ $M$-partitions.  Define

$X = \{1,2,\dots , n-1\}$
$Y = \{n, n+1\}$
$Z = \{n+2, n+3, \dots , 2n\}$

Now, note that (this is the crucial observation?) for any element $x \in X$, there are exactly two other elements $y, y' \in A$ such that $x+y$ and $x+y'$ are in $M$.  Furthermore, one of $y$ and $y'$ is in $X$ and the other is in $Z$.    (Note that clearly we have $x \neq y, y'$ as the three elements in $M$ are odd.)  Similarly, for any element $z \in Z$, the same holds.

Consider the $4$-tuples

$(k, n-k, n+k+1, 2n-k+1)$ for $1\leq k \leq \frac{n-1}{2}$

and note that the sum of any two adjacent terms (and the first and last) is in $M$.  Moreover, the union of all such $4$-tuples for $k = 1,2,\dots , \frac{n-1}{2}$ will be the union of $X$ and $Z$.  Note that these adjacent sums are the ONLY possible sums of elements in $A\Y$ which in $M$, by our previous observations.  Thus, we either have

$k, n+k+1 \in A_1$ and $n-k, 2n-k+1 \in A_2$
or $n-k, 2n-k+1 \in A_1$ and $k, n+k+1 \in A_2$ for $k = 1,2,\dots , \frac{n-1}{2}$

Now, we may account for $Y$ by noting that either

$n \in A_1$ or $n+1 \in A_1$ and only one element in $Y$ can be in each $A_i$.

Thus, we have a total of

$2^{(n-1)/2} \cdot 2 = 2^{(n+1)/2}$ possible $M$-partitions!

well done!!!! have u solved it urself.
  the proof is elementry one but to check its validity u need to think a lot.
  but how come u created X,Y and Z? how much time did it take to solve this problem?
 regards
         amir siddique

Thanks!

Yeah, I admit I left out some details, but I guess I started out trying to characterize all pairs of elements which sum to an element in $M$.  So, just looking at a random element of $A$ (let's say $1$ for now)... if $1\in A_1$ we have $1000, 2002 \in A_2$, so $1002, 1003 \in A_1$, etc... until it "circles" around and repeats again.  This happens with all elements except those in $Y$, so it seems natural to create $X$ and $Z$ to characterize this.  Now, taking another element besides $1$, you can basically use the same idea.  Also, I suppose generalizing it actually makes the problem easier to think about...

(I don't really remember how long I spent on it--I think I worked on it before without much progress and gave up, but returned to it recently and got it in an hour or so.)


number theory
prime factorization
number theory proposed

Solve in the set of positive integers the equation \[ \[x^3 - y^3 = 2005 \left(x^2 - y^2 \right)\]

Obviously, $ x\equal{}y$ is a solution. So, from now on assume $ x\not\equal{}y$ and set $ \varrho\equal{}e^{\frac{2 \pi i}{3}}$. This implies $ x^2\plus{}xy\plus{}y^2\equal{}2005(x\plus{}y)$ and $ x\plus{}y|xy$, i.e. $ x\equal{}ra(a\plus{}b),y\equal{}rb(a\plus{}b)$ with $ r,a,b$ positive integers and $ (a,b)\equal{}1$. Substitution yields $ a^2\plus{}ab\plus{}b^2\equal{}\frac{2005}{r}$ and $ r\in\{1,5,401,2005\}$ due to the prime factorization of $ 2005$. In the Euclidean ring $ \mathbb{Z}[\varrho]$ we identify the LHS as $ N(a\minus{}\varrho b)\equal{}(a\minus{}\varrho b)(a\minus{}\varrho^2b)$, the norm of $ a\minus{}\varrho b$. Since $ 5$ and $ 401$ remain primes in $ \mathbb{Z}[\varrho]$ (alternatively, note that $ \forall z\in\mathbb{Z}[\varrho]: \,N(z)\equiv 0,1\bmod{3}$ holds), we can only have a solution, if $ r\equal{}2005$. However, $ a^2\plus{}ab\plus{}b^2\equal{}1$ has no solution in the positive integers.


topology
advanced fields
advanced fields unsolved

Let $X$ be a topological space. $f, g: X \to \mathbb{S}^{n}$  continuous. Show that if for every $x \in X$  , $f(x) \ne-g(x)$ then $f$ and $g$ are homotopic.

[quote="mlok"]Let $\gamma_{x}\colon [0,1]\to S^{n}$ be the geodesic arc between $f(x)$ and $g(x)$. Define $F_{t}(x)=\gamma_{x}(t)$ -- this is a homotopy.

[It's not really necessary to re-post a question if you've made a typo in the title.][/quote]

How can you show that  $F_{t}(x)=\gamma_{x}(t)$ is continuous ?

Quoting the entire preceding post is not necessary either.

Let $Y=\{(a,b)\in S^{n}\times S^{n}\colon a\ne-b\}$. For any $(a,b)\in Y$ there is a unique geodesic $\gamma_{a,b}\colon [0,1]\to S^{n}$ with $\gamma_{a,b}(0)=a$ and $\gamma_{a,b}(1)=b$, parametrized according to the arclength, that is, $d(\gamma_{a,b}(t),\gamma_{a,b}(t'))=|t-t'|d(a,b)$. To prove that the map $\Phi\colon (a,b,t)\mapsto \gamma_{a,b}(t)$ is continuous from $Y\times [0,1]$ to $S^{n}$, it suffices to prove that it is locally Lipschitz in each variable separately. For $t$ this is obvious. To prove that $\Phi$ is locally Lipschitz in $b$, assume for convenience that $a$ is the North Pole, and write down an explicit formula for $\gamma$.

mlok, how about writing down the homotopy?

if you see $S^{n}\subset \mathbb{R}^{n}$, and set $p = (x,y,z)$, the homotopy should be \[H(p,t) = \frac{tf(p)+(1-t)g(p)}{|tf(p)+(1-t)g(p)|}\] :p


linear algebra
matrix
linear algebra unsolved

HI

is the first problem in which, you have to prove that for every integer X value the determinant of the matrix not is cero.

[hide]Use the formula $ det\equal{} \sum_{J \in S_n} \epsilon (J) a_{j_11}a_{j_22}...a_{j_nn}$ mod 2. Only one is not even yielding an odd determinanat, hence not 0...[/hide]


Hi guys
I play online casino games for many years now
Usually I play the same games..blackjack, the roulette, and if im really in a crazy mood I play some dice
I saw the game [url=http://www.casino-on-net.com]baccarat[/url] in one site and I wanna learn more about this game..:)
Where can I read more about it?

Wikipedia is your friend. :)
http://en.wikipedia.org/wiki/Baccarat


For a quadrilateral ABCD with perpendicular diagonals intersecting at E. EF, EG, EH and EM are perpendicular to AB, BC, CD, and AD, respectively. Prove that lines FG, AC, and MH are concurrent.

It is enough to see that $ FGHM$ is cyclic due to

$ \angle FGH \equal{} \angle ABE \plus{} \angle ECD \ , \ \angle FMH \equal{} \angle EAB \plus{} \angle EDC \ \Longrightarrow$

$ \angle FGH \plus{} \angle FMH \equal{} \angle ABE \plus{} \angle ECD \plus{} \angle EAB \plus{} \angle EDC \equal{} 90^{\circ} \plus{} 90^{\circ} \equal{} 180^{\circ}$

$ FGCA$ is also cyclic due to $ \angle BFG \equal{} \angle BEG \equal{} \angle BCA$ $ \Longrightarrow$ $ FG$ and $ AC$ are antiparallel WRT rays $ BA,BC.$ Similarly $ MHCA$ is cyclic $\Longrightarrow$ $ AC,FG,MH$ concur at the radical center of $ \odot(FGHM) ,\odot(FGCA)$ and $\odot(MHCA)$