If anyone of you has used the online books and/or online tests for SAT II Physics, Chemistry and Math IIC, how do you compare them with the Princton Review and Barron's prep books? Do the online notes cover EVERYTHING that is required. Also, are the test right on track as far as question difficulty is concerned?
The online 'bad' thing I found with my Barron's book is that since I have an indian reprint on poor , old paper, it's not fun to read although the subject matter is concerned.
Also, what about the experiments in Physics and Chemistry. Please recommend online sources for knowing abpout lab  equipment and standard experiments done in US high schools.

Sparknotes' test prep books are horrible.  They have multitudes of wrong information/answers.  It is probably due from the fact that they are fairly new in the test prep market and only have first editions out for sale.  They will most likely be better in a few years.

Princeton Reivew is usually right on the money.  You can't really go wrong with them.  Barron's however is mostly harder than the material on the actual test/exam or whatever you're preparing for.  Take that how you wish, but I choose Barron's because its difficulty level is slightly higher than that of the actual material, which in most cases makes the testing easier because I have a better grasp.  This isn't the case with everyone though, I just find that it helps me because I am somewhat of an overachiever.

Having spent  nearly 50$\$$ over a period of 2 weeks using the SAT II Physics, Math IIC, Chemistry Practice tests offered by SparkNotes test prep , I have one advice, heed to it ----
 [color=red][color=cyan][size=200]NEVER EVER USE SPAKNOTES!!!!!![/size][/color][/color]

Their typesetting is pitiful . In maths, no formula or equation is legible. The question says use the graph/diagram shown and there is no diagram. You mark an option and the computer says you didn't. You mark the correct option given in the explanation as well but the computer marks it wrong. The question solved in the explanation is totally different from the one actually in the test. Wrong questions (confirmed by other books) have been put in with wrong answers. It's a total mayhem and waste of money and more importantly time!!! You don't get your accurate assessment!!!!!!
Again, I'd advice you to look for something else. I really regret wasting my time and money!!!!!!

Having spent $\$$0 over a period of 1 day using just one practice Chem SAT II, I would actually say it's pretty accurate to the actual test, and I ran into no issues in terms of typesetting or answers/solutions. Then again, my chem is really rusty since I took it last school year. I dunno, I found it to be an okay indicator of how I'm doing.

Personally, I thought their assessment was really cool. They show you exactly how many of what types of problems you got wrong/right, and tell you what concepts to work on more than others. It's pretty nifty, in my opinion.


calculus
integration
logarithms
inequalities
real analysis
real analysis unsolved

$ P \equal{}\int_{0}^{1}\int_{0}^{1}\left(\frac{\log^{2}(x y)}{x y\plus{}1}\right)^{2}dxdy$

a substituition $ x \equal{} e^{\minus{}\phi }$ , $ y \equal{} e^{\phi }\theta$

using mister Jacob

$ \left(\frac{\log^{2}(x y)}{x y\plus{}1}\right)^{2}\left|\left(\begin{array}{ll}\frac{\partial x}{\partial\theta }&\frac{\partial x}{\partial\phi }\\ \frac{\partial y}{\partial\theta }&\frac{\partial y}{\partial\phi }\end{array}\right)\right| d\theta d\phi \equal{}\frac{\log^{4}(\theta )}{(\theta\plus{}1)^{2}}(d\theta d\phi )$

Solving the inequalities we get $ 0 <\theta < 1\land 0 <\phi <\minus{}\log (\theta )$

$ P \equal{}\int_{0}^{1}\int_{0}^{\minus{}\log (\theta )}\frac{\log^{4}(\theta )}{(\theta\plus{}1)^{2}}d\phi d\theta \equal{}\minus{}\int_{0}^{1}\frac{\log^{5}(\theta )}{(\theta\plus{}1)^{2}}\, d\theta$

$ \int_{0}^{1}\frac{\log^{5}(\theta )}{(\theta\plus{}1)^{2}}\, d\theta \equal{}\int_{0}^{1}\left(\sum_{i \equal{} 0}^{\infty }\log^{5}(\theta )\theta^{i}(i\plus{}1) (\minus{}1)^{i}\right)\, d\theta \equal{}\sum_{i \equal{} 0}^{\infty }(i\plus{}1) (\minus{}1)^{i}\int_{0}^{1}\log^{5}(\theta )\theta^{i}\, d\theta$

it's very trivial to demonstrate that $ \int_{0}^{1}\log^{c_{1}}(\theta )\theta^{c_{2}}\, d\theta \equal{}\minus{}\left(\minus{}\frac{1}{c_{2}\plus{}1}\right)^{c_{1}\plus{}1}\Gamma\left(c_{1}\plus{}1\right)$

$ \int_{0}^{1}\log^{5}(\theta )\theta^{i}\, d\theta \equal{}\minus{}\frac{120}{(i\plus{}1)^{6}}$

$ P \equal{}\sum_{i \equal{} 0}^{\infty }\frac{120 (\minus{}1)^{i}}{(i\plus{}1)^{5}}\equal{} 120\sum_{i \equal{} 1}^{\infty }\frac{(\minus{}1)^{i\minus{}1}}{i^{5}}$

$ \zeta (s)\equal{}\frac{\sum_{i\equal{}1}^{\infty }\frac{(\minus{}1)^{i\minus{}1}}{i^{s}}}{1\minus{}2^{1\minus{}s}}$

$ P \equal{}\frac{225\zeta (5)}{2}$

and we are donne

wow!  :) very nicely done !  :coolspeak:


integration
calculus
function
algebra unsolved
algebra

Do there exist method withoud integrals or rather not...?

Intuitively, (without Calculus), we need an approximation of form

$ \frac{1}{x^3} \approx \frac{\alpha}{x(x\minus{}1)(x\plus{}1)} \plus{} \frac{\beta}{x(x\minus{}1)(x\plus{}1)(x\plus{}2)} \plus{} ...$

The problem is: Finding such an approximation for any function is not easy.


is this inequality true? can someone generalize and prove it ?what is the name of this inequality ?i just guessed it my self.

$ ab\geq\frac {b}{a}$

If $ b$ is a positive real, then we can cancel and it reduces to $ a \geq \frac1a$. This is true only if $ a \in [\minus{}1,0)$ or $ a \in [1,\infty )$ but never otherwise. (If $ b$ is a negative real, then the inequality is flipped: $ a \leq \frac1a$)

So that inequality isn't true in general.

what does $ a \in [ \minus{} 1,0)$ or $ a \in [1,\infty )$ mean ?

the symbol $ \in$ means "is an element of."  The notation $ (a,b)$ means "on the interval $ a,b$" or "between $ a$ and $ b$."  A "$ ($" means that the endpoint is included in the interval, while a $ [$ means the endpoint is not included in the interval.'

So, for example, the expression $ a \in [\minus{}1,0)$ means $ a$ can have the value of any real number between $ \minus{}1$ and $ 0$, including $ \minus{}1$ but NOT including $ 0$.  The second expression, $ a \in [1,\infty)$ means that $ a$ can be any real number greater than or equal to $ 1$.

I think you made a little mistake while typing that:

[quote="jrshoch"]A "$ ($" means that the endpoint is included in the interval, while a $ [$ means the endpoint is not included in the interval.'[/quote]
That should say, "A '$ ($' means that the endpoint is not included in the interval, while a $ [$ means the endpoint is included in the interval."

Everything else is correct. :)

Another notation, which I know they use in part of Europe, is $ [ a ; b[$.  Where the the first $ [$ means inclusive and the second $ [$ means exclusive.

[b]The Law of Small Cases[/b] (my own designation :) ):  Before positing a hypothesis, test small cases.

In this case, let (for example) $ a \equal{} 2, b \equal{} \minus{}2$ to get the contradiction $ \minus{}4 \ge \minus{}1$.

Or: if it's true for $ (a,b)$, it isn't true for $ (\minus{}a,b)$ nor $ (\minus{}b,a)$.

$ ab\geq \frac {b}{a}\Longleftrightarrow b\left(a \minus{} \frac {1}{a}\right)\geq 0$

$ \Longleftrightarrow \frac {b}{a}(a \plus{} 1)(a \minus{} 1)\geq 0$

$ \Longleftrightarrow ab(a \plus{} 1)(a \minus{} 1)\geq 0\ \wedge a\neq 0$

$ \Longleftrightarrow ab\geq 0\ \wedge (a \plus{} 1)(a \minus{} 1)\geq 0 \vee ab\leq 0\ \wedge (a \plus{} 1)(a \minus{} 1)\leq 0\ \wedge a\neq 0$

$ \Longleftrightarrow ab\geq 0\ \wedge a\leq \minus{} 1,\ 1\leq a \vee ab\leq 0\ \wedge \minus{} 1\leq a\leq 1\ \wedge a\neq 0$

thank you everyone. how about this

$ a^b\geq ab$
is this always true ?

$ (\minus{}1)^{\minus{}1} \equal{} \minus{}1 \not\geq (\minus{}1)*(\minus{}1) \equal{} 1$

So, no.

An inequality along those lines that [i]is[/i] true:

$ (1 \plus{} x)^r \ge 1 \plus{} rx$

for $ r \le 0, r \ge 1$ and $ x > \minus{}1$.

its true for $ a,b\in\mathbb{N}$ isnt it ?

Nope, let $ a\equal{}1$ and $ b>1$, you have $ 1\equal{}1^b<b$  :huh:

In the positive integers, it [i]is[/i] true for $ a > 1$.  Can you prove it?  (Try induction.)

i cant do it , can you show us please in details

[quote="Elio (n)"]Hi !

$ a^{b}\ge ab$

Let be $ b \equal{} 1$ we get  $ a^{1} \equal{} a*1$

Let be  $ a^{b}\ge ab$ true

$ a^{b}*a\ge ab*a$    and         $ a^{b \plus{} 1}\ge a^{2}b$

$ a^{2}b > ab$  so $ a > 1$

Is this what you mean ?  :)[/quote]I'm a little confused about your work... Which variable are you using induction on? Is it $ a$? Or is it $ b$?  :what?:


linear algebra
matrix
vector
linear algebra unsolved

Let $ M$ is a matrix. Maximize $ S \equal{} \sum_{i}P_{i}^{T}MP_{i}$ with constrains $ P^{T}P \equal{} I$.

I know, that if matrix $ M$ is symmetric, all eigenvalues are real and maximizers are eigenvectors of $ M$ (by Lagrange method).
But is my case $ M$ is not symmetric and eigenvalues can be complex. Is any way to maximize $ S$ only with $ P$ with real components? Thanks for attention.

I assume $ S\equal{}P^TMP$ and $ P$ a real vector with norm$ 1$.
With Lagrange method, there exists a real $ \lambda$ s.t. for all vector $ H$:
$ H^TMP\plus{}P^TMH\plus{}\lambda{P^T}H\equal{}0$. Then $ P^TM^T\plus{}P^TM\plus{}\lambda{P^T}\equal{}0$ or $ (M\plus{}M^T)P\plus{}\lambda{P}$.
Conclusion: the max (it exists!) is obtained with $ P$ an eigenvector of $ M\plus{}M^T$.


topology
advanced fields
advanced fields theorems

Hello.

Trying to learn topology.
Does anyone have som tips on topology books?
Any advices? problem books?
perquisites ?

I understand that topology is a huge subject, but I'm initially just
interested in the basic (undergrad.) concepts.

I already know what everyone is going to say: Munkres, Munkres, Munkres, ...

(P.S.-I'm a personal fan as well it isn't just everyone else)


For how many positive integer values of $ x$ is the sum $ x^2\plus{}4x\plus{}4$ less than 20?

$ x^2 \plus{} 4x \plus{} 4$ factors as $ (x \plus{} 2)(x \plus{} 2)$. So we're trying to find for how many values $ (x \plus{} 2)^2 < 20$.

This simplifies as:

$ x \plus{} 2 < 2\sqrt{5}$

$ 2\sqrt{5} \approx 4$, so only $ \boxed{2}$ values work for this inequality.

$x^{2}+4x+4<20$

$x^{2}+4x-16<0$

$-2-2\sqrt{5}<x<-2+2\sqrt{5}$

However the problem asks for all the positive integer values of $x$, thus:

$0<x<-2+2\sqrt{5}$

$-2+2\sqrt{5} \approx 2$

Hence there are only two positive integer values that work.


can you help me with this ineq.i need some proofs:
JBMO,2003.   prove that
\[ \frac{{1 \plus{} {x^2}}}{{1 \plus{} y \plus{} {z^2}}} \plus{} \frac{{1 \plus{} {y^2}}}{{1 \plus{} z \plus{} {x^2}}} \plus{} \frac{{1 \plus{} {z^2}}}{{1 \plus{} x \plus{} {y^2}}} \ge 2\]
for any real numbers $ x,y,z\ge \minus{}1$


Let $a,b\in \mathbb N$ such that $\forall n\in \mathbb N$ we have $a^{n}+n|b^{n}+n$. Prove that $a=b$.


Given a triangle $ABC$, its internal angle bisectors $BE$ and $CF$, and let $M$ be any point on the line segment $EF$.
 Denote by$S_A, S_B$ and $S_C$ the areas of triangles $MBC, MCA$ and $MAB$, respectively.
Prove that
$\frac{\sqrt{S_B} + \sqrt{S_C} }{\sqrt{S_A}} \leq  \sqrt{\frac{AC + AB}{BC}}$
and determine when equality holds.


function
algebra
polynomial
algebra proposed

Let $S\in R$ satisfy the following condition:

For $a,b,c,d$ not being $0,1$, if $a+b+c+d=S, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=S$, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=S$. What is $S$?

We can choose $c=-a,d=-b, a^{2}+b^{2}=2$ for $S=0$, and $c={1\over a}, d={1\over b}, a+b+{1\over a}+{1\over b}=2$ for $S=2$, but it still has to be investigated whether those are only values.

They're not the only values; there's an interval containing zero that works.

Write $f(x)=(x-a)(x-b)(x-c)(x-d)=x^{4}-s_{1}x^{3}+s_{2}x^{2}-s_{3}x+s_{4}$; the $s_{i}$ are the elementary symmetric functions of $a,b,c,d$.
Clearing denominators, our relations are
$s_{1}=s$, $s_{3}=s_{4}s$, and $s(1-s_{1}+s_{2}-s_{3}+s_{4})=4-3s_{1}+2s_{2}-s_{3}$ ($sf(1)=f'(1)$).
Using the first two to eliminate $s_{1}$ and $s_{4}$ from the last, we get $(2-s)(2-s+s_{2}-s_{3})=0$. Either $s=2$ (already noted) or  $s_{2}=s_{1}+s_{3}-2=s+ss_{4}-2$. Letting $s_{4}=t$, $f(x)$ becomes
$x^{4}-sx^{3}+(s+st-2)x^{2}-stx+t$.
We want to know when this $f$ has only real roots. For $s=0$ and $0\le t\le 1$, it has four (with multiplicity) real roots, as previously noted. The roots of a polynomial are continuous in the coefficients, and we can only switch from real to pairs of complex roots at a point with a double root. We have two paths along which the polynomial has double roots, and the region between them has only real roots. Where these paths intersect, we have a triple root.


On hypotenuse $ AB$ of a right triangle $ ABC$ a second right triangle $ ABD$ is constructed with hypotenuse $ AB$. If $ \overline{BC} \equal{} 1, \overline{AC} \equal{} b$, and $ \overline{AD} \equal{} 2$, then $ \overline{BD}$ equals:

$ \textbf{(A)}\ \sqrt {b^2 \plus{} 1} \qquad\textbf{(B)}\ \sqrt {b^2 \minus{} 3} \qquad\textbf{(C)}\ \sqrt {b^2 \plus{} 1} \plus{} 2$
$ \textbf{(D)}\ b^2 \plus{} 5 \qquad\textbf{(E)}\ \sqrt {b^2 \plus{} 3}$

[hide="Click for solution"]
Let $ DB\equal{}x$.  Then $ x^2\plus{}4\equal{}b^2\plus{}1 \implies x\equal{}\sqrt{b^2\minus{}3}$, or $ \boxed{\textbf{(B)}}$.
[/hide]


[b]My friend and I are still stuck on:[/b]

For each $ n \in \omega$, let $ X_n$ be the set $ \{0, 1\}$, and let $ \tau_n$ be the discrete topology on $ X_n$. For each of the following subsets of $ \prod_{n \in \omega} X_n$, say whether it is open or closed (or neither or both) in the product topology.

(a) $ \{f \in \prod_{n \in \omega} X_n | f(10)\equal{}0 \}$
(b) $ \{f \in \prod_{n \in \omega} X_n | \text{ }\exists n \in \omega \text{ }f(n)\equal{}0 \}$
(c) $ \{f \in \prod_{n \in \omega} X_n | \text{ }\forall n \in \omega \text{ }f(n)\equal{}0 \Rightarrow f(n\plus{}1)\equal{}1 \}$
(d) $ \{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)\equal{}0 \}|\equal{}5 \}$
(e) $ \{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)\equal{}0 \}|\leq5 \}$

Recall that $ \omega\equal{}\mathbb{N} \cup \{ 0 \}$

[b]Background:[/b]

If it helps you can think of $ \prod_{n\in \omega} X_n$ as $ \prod_{n\equal{}0}^{\infty} X_n$. We define $ \prod_{n\equal{}0}^{\infty}X_n$ to be the set of all functions $ f: \mathbb{N} \to \{ 0 , 1\}$ that satisfies $ f(n) \in \{ 0 , 1\}$.

There's a nice graphical representation of the product topology on $ Y^X$ (i.e. the product of the space $ Y$ $ |X|$ times). Namely, if we draw $ X$ as an "$ x\minus{}$axis" and $ Y$ as a "$ y\minus{}$axis", then elements in $ X^Y$ are "graphs of functions" in the $ X\minus{}Y$ "plane". An open nbhd of an element $ f$ is the set of all functions $ g$ whose graphs are close to the graph of $ f$ at finitely points. We get different nbhds by varying the closeness to $ f$ and/or the set of finite points.

In our case the product space is $ 2^\omega\equal{}2^\mathbb{N}$, whose "plane" looks like two copies of the naturals $ \mathbb{N}$. In other words, if you were to imagine this as a 'subset' of $ \mathbb{R}^2$, it's just the set $ \{(n,i) : n \in \mathbb{N}, i \in \{0,1\}\}$.

[b]Progress:[/b]

So remember that open sets in the infinite product topology is really just having all but finitely many the whole space and the rest are open. Since the individual factors are discrete, you only need to check that all but finitely many are the whole space.

e.g. in (a) the 10th coordinate has a specific value, but all other coordinates can be whatever, so this is certain open.

[b]Progress:[/b]

Take set (b). Let $ B \equal{} \{f \in \prod_{n \in \omega} X_n | \;\exists n \in \omega \; f(n)\equal{}0 \}$. If $ f\in B$ then there exists m such that f(m)=0. Then the set $ \{g \in \prod_{n \in \omega} X_n |\; g(m)\equal{}0\}$ is an open neighbourhood of f contained in B. [b]Therefore B is open.
[/b]
It's usually more difficult to check when a set is closed. You have to look at its complement and decide whether that is open. Sometimes this is straightforward. For example, the complement of set (a) is the set of all f such that f(10)=1. [b]That is open, so set (a) is closed as well as open.[/b]

For a slightly less easy example, look at set (c). Let $ C \equal{} \{f \in \prod_{n \in \omega} X_n | \text{ }\forall n \in \omega \text{ }f(n)\equal{}0 \Rightarrow f(n\plus{}1)\equal{}1 \}$. If $ f\notin C$ then there exists m such that f(m)=f(m+1)=0. Then $ \{g \in \prod_{n \in \omega} X_n |\; g(m)\equal{}g(m\plus{}1)\equal{}0\}$ is an open neighbourhood of f containing no points of C. [b]Therefore the complement of C is open and so C is closed.[/b]
[b]
[color=red]I still do not know how to do parts (d.) and (e.)[/b][/color]


a)Let $t_n=$ the number of involutions of $S_n${permutations $\sigma$ of $[1,..,n]$ which satisfy $\sigma\cdot\sigma=Id$} .Show that $t_n$ verifies the following relation:$t_n=t_{n-1}+(n-1)t_{n-2}$.
b)How do you write $(1,2,...,n)$ as the product of $2$ involutions?

Isn't it very easy? :? All you need to do is look at the cycle decomposition of these permutations: they can be involutions iff they're decomposed as the product of some disjoint transpositions, and some fixed points (i.e. cycles of length $1$). This means that $t_n$ is equal to the number of coverings of $1,2,\ldots,n$ with "bricks" of length $1$ and $2$ (and each such covering yields a different involution, since the disjoint cycle decomposition is unique up to the order of the cycles, which we're not interested in anyway).

There are two possibilities: $n$ is covered by a brick of length $1$, and, in this case, the number of coverings is just $t_{n-1}$, or $n$ is part of a brick of length $2$. In each such case, the number of coverings is $t_{n-2}$, and there are $n-1$ posibilities for this situation, according to which brick $n$ is part of: $(1,n),(2,n),\ldots,(n-1,n)$.

The idea of the bricks can be used in another form :

Any involution satisfies exactly one of the two following cases :
- If $f(n)=n$ then the involution is determined by its restriction to $\{1, \cdots, n-1 \}$ which has to be an involution on this set. Conversely, any such restriction leads to exactly one involution, so that there are $t_{n-1}$ involutions in that case.
- If $f(n)=p$ where $p<n$ then $f(p)=n$ and the involution is determined by its restriction on $\{1, \cdots, n-1 \} \setminus \{p \}$, which has to be an involution too on this set with $n-2$ elements. Since there are $n-1$ choices for $p$, this leads as above to exactly $(n-1)t_{n-2}$ involutions in that case and we are done.

Pierre.

I forgot to add the second question.And one more question....Can we write any permutation as a product of involutions?


Hello,

Can someone solve this question by Murihead's Inequality. Please add as much details as possible  :oops:

For any positive $ x,y,z$ prove that: \[ \frac{\left(x\plus{}y\plus{}z\right)^{2}}{3} \ge x\sqrt{yz}\plus{}y\sqrt{zx}\plus{}z\sqrt{xy}\]

[hide="Solution"]We may transform the given inequality into $ \frac {(x \plus{} y \plus{} z)^2}{9}\ge\frac {x\sqrt {yz} \plus{} y\sqrt {zx} \plus{} z\sqrt {xy}}{3}$.

Applying AM-GM to RHS, we have $ \left(\frac {x \plus{} y \plus{} z}{3}\right)^2\ge (\sqrt [3]{xyz})^2$.

Now, it suffices to prove that $ \sqrt [3]{x^2y^2z^2}\ge \frac {x\sqrt {yz} \plus{} y\sqrt {zx} \plus{} z\sqrt {xy}}{3}$.

Substitute $ x \equal{} a^6, y \equal{} b^6, z \equal{} c^6$ for convenience.

Then, we are to prove that $ 3a^4b^4c^4\ge a^6b^3c^3 \plus{} a^3b^6c^3 \plus{} a^3b^3c^6$ which is a direct consequence of Muirhead Inequality $ \sum_{\text{sym}}a^4b^4c^4\ge \sum_{\text{sym}}a^3b^3c^6$. $ \blacksquare$[/hide]

EDIT: Darn, so close.

[hide]$ (4,4,4)$ does not majorize $ (6,3,3)$.[/hide]

[hide="Solution"]Note that $ (1,1,0)$ majorizes $ \left(1,\frac{1}{2},\frac{1}{2}\right)$, therefore
\[ \frac{2(xy\plus{}yz\plus{}xz)}{3} \geq \frac{2(x\sqrt{yz}\plus{}y\sqrt{zx}\plus{}z\sqrt{xy})}{3}\] by Muirhead's Inequality.

Also, $ (2, 0,0)$ majorizes $ \left(1,\frac{1}{2},\frac{1}{2}\right)$, therefore
\[ \frac{x^2\plus{}y^2\plus{}z^2}{3} \geq \frac{x\sqrt{yz}\plus{}y\sqrt{zx}\plus{}z\sqrt{xy}}{3}\] by Muirhead's Inequality.

Adding these two together, \[ \frac{\left(x\plus{}y\plus{}z\right)^{2}}{3}\ge x\sqrt{yz}\plus{}y\sqrt{zx}\plus{}z\sqrt{xy}\] as desired. $ \blacksquare$[/hide]

[quote="FantasyLover"][hide="Solution"]We may transform the given inequality into $ \frac {(x \plus{} y \plus{} z)^2}{9}\ge\frac {x\sqrt {yz} \plus{} y\sqrt {zx} \plus{} z\sqrt {xy}}{3}$.

Applying AM-GM to RHS, we have $ \left(\frac {x \plus{} y \plus{} z}{3}\right)^2\ge (\sqrt [3]{xyz})^2$.

Now, it suffices to prove that $ \sqrt [3]{x^2y^2z^2}\ge \frac {x\sqrt {yz} \plus{} y\sqrt {zx} \plus{} z\sqrt {xy}}{3}$.

Substitute $ x \equal{} a^6, y \equal{} b^6, z \equal{} c^6$ for convenience.

Then, we are to prove that $ 3a^4b^4c^4\ge a^6b^3c^3 \plus{} a^3b^6c^3 \plus{} a^3b^3c^6$ which is a direct consequence of Muirhead Inequality $ \sum_{\text{sym}}a^4b^4c^4\ge \sum_{\text{sym}}a^3b^3c^6$. $ \blacksquare$[/hide][/quote]
sorry, I don't think that $ \sum_{sym}a^4b^4c^4 \ge \sum_{sym}a^3b^3c^6$
In fact $ \sum_{sym}a^4b^4c^4 \le \sum_{sym}a^3b^3c^6$
[hide="my Solution"]
since$ \sum_{cyc}(a\minus{}b)^2 \ge0$
$ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$.........(1)
$ a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca)\ge 3(ab\plus{}bc\plus{}ca)$
$ (a\plus{}b\plus{}c)^2\ge 3(ab\plus{}bc\plus{}ca)$........(2)
By (2), we have ,
$ \frac{(x\plus{}y\plus{}z)^{2}}{3} \ge (xy\plus{}yz\plus{}zx)$
put $ a\equal{}\sqrt{yz},b\equal{}\sqrt{xz},c\equal{}\sqrt{xy}$ into (1),we have
$ xy\plus{}yz\plus{}zx \ge x\sqrt{yz}\plus{}y\sqrt{zx}\plus{}z\sqrt{xy}$
$ \therefore \frac{(x\plus{}y\plus{}z)^{2}}{3}\ge x\sqrt{yz}\plus{}y\sqrt{zx}\plus{}z\sqrt{xy}$
[/hide]

Thanks a lot
But am I right that "∫FaiLurE∮" didn't use Murihead's inequality?
And could you guys please till me why "FantasyLover" mijorizing is wrong?

Thanks in advance  :)

Fantasy's majorizing is wrong because (6,3,3) $ \ge$ (4,4,4), not the reverse

Solution:

Expanding, multiplying by 2, we have $ (2,0,0)\plus{}2(1,1,0) \ge 3(1,\frac{1}{2},\frac{1}{2})$, which is clearly true by Muirhead: $ (2,0,0)\ge (1,\frac12,\frac12), 2(1,1,0)\ge 2(1,\frac12,\frac12)$

By the way, $ (a,b,c)\equal{}\sum_{sym} x^ay^bz^c$

Remark: In this notation, Schur's inequality looks like this:

$ (a\plus{}2b,0,0)\plus{}(a,b,b)\ge 2(a\plus{}b,b,0)$

$ \frac{(x\plus{}y\plus{}z)^2}{3} \geq xy \plus{}yz \plus{}zx$

Now we use majorization

$ (1,1,0) \succ \left(1,\frac{1}{2} .\frac{1}{2} \right)$ , and we're done  :)


Hi there, first time poster here. This question has recently popped up and I was wondering about possible methods of solutions. It honestly has me stumped. Well, here goes.

The voltage of a certain thermocouple as a function of temperature is given by E=6.2T + 0.0002T^3. What is the approximate change in voltage if the temperature changes from 100 to 101 degrees celsius.


Let a,b,c>0 Prove that
$ \sum\frac{a}{b\plus{}c} \geq \frac{3}{2}$

Let a,b,c>0 Prove that
${ \sum\frac{a^2}{a+b} \geq \frac{\sqrt{3(a^2+b^2+c^2)}}{2}}$

[quote="tran300491"]Let a,b,c>0 Prove that
${ \sum\frac {a^2}{a + b} \geq \frac {\sqrt {3(a^2 + b^2 + c^2)}}{2}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124434

[quote="tran300491"]Let a,b,c>0 Prove that
${ \sum\frac {a^2}{a + b} \geq \frac {\sqrt {3(a^2 + b^2 + c^2)}}{2}}$[/quote]
By Tchebychev we have:
\[ \sum\frac {a^2}{a+b} \geq \frac{1}{3}(a^2+b^2+c^2)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}) ,\]
then we have to prove that:
$ {\sqrt {(a^2 + b^2 + c^2)}(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}) \geq \frac{3\sqrt{3}}{2}}$
but by Cauchy-Shwarz we have that:
$ \sqrt {(a^2 + b^2 + c^2)}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}) \geq \frac{1}{\sqrt{3}}(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}) \geq  \frac{3\sqrt {3}}{2}.$

[quote="rachid"][quote="tran300491"]Let a,b,c>0 Prove that
${ \sum\frac {a^2}{a + b} \geq \frac {\sqrt {3(a^2 + b^2 + c^2)}}{2}}$[/quote]
By Tchebychev we have:
\[ \sum\frac {a^2}{a + b} \geq \frac {1}{3}(a^2 + b^2 + c^2)(\frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{a + c}) ,
\]
[/quote]
Are you sure?

Let a,b,c>0 Prove that
$ \sum\frac{a^2}{b\plus{}c} \geq \frac{3\sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}}{2}$

[quote="tran300491"]Let a,b,c>0 Prove that
$ \sum\frac {a^2}{b \plus{} c} \geq \frac {3\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}}{2}$[/quote]
$ \sum_{cyc}\frac {a^2}{b \plus{} c} \geq \frac {3\sqrt [5]{\frac {a^5 \plus{} b^5 \plus{} c^5}{3}}}{2}$ is true too .  :wink:

[quote="tran300491"]Let a,b,c>0 Prove that
$ \sum\frac {a}{b \plus{} c} \geq \frac {3}{2}$[/quote]
It's really is Nesbitt  :maybe:

OK. It was the famous ineq. I found 25 proofs for it but I would like to know your proofs.

here
http://www.mathlinks.ro/viewtopic.php?p=1140269#1140269  ;)

If 0<a,b,c<1 such that$ abc\equal{}(1\minus{}a)(1\minus{}b)(1\minus{}c)$ then
$ a^2\plus{}b^2\plus{}c^2\plus{}2abc \geq1$

[quote="tran300491"]If 0<a,b,c<1 such that$ abc \equal{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$ then
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \geq1$[/quote]

and
$ a\plus{}b\plus{}c\plus{}2(a^2\plus{}b^2\plus{}c^2)\geq6$

[quote="tran300491"]If 0<a,b,c<1 such that$ abc \equal{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$ then
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \geq1$[/quote]
Let $ a \equal{} \frac {1}{1 \plus{} x},$ $ b \equal{} \frac {1}{1 \plus{} y}$ and $ c \equal{} \frac {1}{1 \plus{} z},$ where $ x > 0,$ $ y > 0$ and $ z > 0.$
Then $ xyz \equal{} 1$ and we need to prove that
$ \frac1{(1 \plus{} x)^2} \plus{} \frac1{(1 \plus{} y)^2} \plus{} \frac1{(1 \plus{} z)^2} \plus{} \frac2{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)}\geq1,$
which is equivalent to $ x^2 \plus{} y^2 \plus{} z^2\geq3.$

[quote="tran300491"][quote="tran300491"]If 0<a,b,c<1 such that$ abc \equal{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$ then
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \geq1$[/quote]

and
$ a \plus{} b \plus{} c \plus{} 2(a^2 \plus{} b^2 \plus{} c^2)\geq6$[/quote]
Are you sure? :wink:

By the way, the following inequality is true:
$ a \plus{} b \plus{} c \plus{} 2(a^2 \plus{} b^2 \plus{} c^2)\geq3$

If a,b,c>0 such that abc=1 then
$ \sum\frac{1}{(1\plus{}a)^2}\geq\frac{3}{4}\plus{}\frac{a\plus{}b\plus{}c\minus{}3}{(1\plus{}a)(1\plus{}b)(1\plus{}c)}$

If a=1 then  LHS=RHS

-> We have$ LHS\minus{}RHS\equal{}(\frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\minus{}\frac{3}{2})^2  \geq0$

Done :rotfl:

And this is new inequality

if a,b,c>= such that abc=1  then

$ \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c} \geq  1\plus{}sqrt{\frac{2}{(1\plus{}a)(1\plus{}b)(1\plus{}c)}}$

I'm sorry for my typing error  , my problem is

 $ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \geq 1 \plus{} \sqrt{\frac {2}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}}$

( with a,b,c>0  such that abc=1)

[quote="tran300491"]I'm sorry for my typing error  , my problem is

 $ \frac {1}{1 + a} + \frac {1}{1 + b} + \frac {1}{1 + c} \geq 1 + \sqrt\frac {{2}{(1 + a)(1 + b)(1 + c)}}$[/quote]
I think it have mistake too. you should check more carefully

Where is the denominator under the root??

[quote="bunhiacovski"][quote="tran300491"]I'm sorry for my typing error  , my problem is

 ${ \frac {1}{1 + a} + \frac {1}{1 + b} + \frac {1}{1 + c} \geq 1 + \sqrt\frac {2}{(1 + a)(1 + b)(1 + c)}}$[/quote]
I think it have mistake too. you should check more carefully[/quote]

[quote="tran300491"]I'm sorry for my typing error  , my problem is

 $ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \geq 1 \plus{} \sqrt {\frac {2}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}}$

( with a,b,c>0  such that abc=1)[/quote]
It's equivalent to $ a^2\plus{}b^2\plus{}c^2\geq3.$ :wink:

[quote="tran300491"]If a,b,c>0  such that abc=1 then
$ \sqrt {\frac {1}{1 \plus{} a}} \plus{} \sqrt {\frac {1}{1 \plus{} b}} \plus{} \sqrt {\frac {1}{1 \plus{} c}} \leq2\sqrt {1 \plus{} \frac {1}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}}$

(Trieu Bon)[/quote]
setting $ x^2 \equal{} \frac {1}{1 \plus{} a}, y^2 \equal{} \frac {1}{1 \plus{} b}$ and $ z^2 \equal{} \frac {1}{1 \plus{} c}$
$ abc\equal{}1\Leftrightarrow (1 \minus{} x^2)(1 \minus{} y^2)(1 \minus{} z^2) \equal{} x^2y^2x^2$
we need prove that $ x \plus{} y \plus{} z\le 2\sqrt {1 \plus{} x^2y^2z^2}$  :wink:

[quote="tran300491"]If a,b,c>0  such that abc=1 then
$ \sqrt {\frac {1}{1 \plus{} a}} \plus{} \sqrt {\frac {1}{1 \plus{} b}} \plus{} \sqrt {\frac {1}{1 \plus{} c}} \leq2\sqrt {1 \plus{} \frac {1}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}}$

(Trieu Bon)[/quote]
Let $ a \equal{} \frac {y}{x},$ $ b \equal{} \frac {z}{y}$ and $ c \equal{} \frac {x}{z},$ where $ x,$ $ y$ and $ z$ are positive numbers such that $ x \plus{} y \plus{} z \equal{} 1.$ Then
$ \sum_{cyc}\sqrt {\frac {1}{1 \plus{} a}}\leq2\sqrt {1 \plus{} \frac {1}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt {\frac {x}{x \plus{} y}}\leq2\sqrt {1 \plus{} \frac {xyz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)}}.$
But $ f(x) \equal{} \sqrt x$ is a concave function.
Hence, $ \sum_{cyc}\sqrt {\frac {x}{x \plus{} y}} \equal{} \sum_{cyc}\frac {x \plus{} z}{2}\sqrt {\frac {4x}{(x \plus{} y)(x \plus{} z)^2}}\leq\sqrt {\sum_{cyc}\frac {2x}{(x \plus{} y)(x \plus{} z)}} \equal{}$
$ \equal{} \sqrt {\sum_{cyc}\frac {2x(x \plus{} y \plus{} z)}{(x \plus{} y)(x \plus{} z)}} \equal{} 2\sqrt {1 \plus{} \frac {xyz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)}}.$
Jensen forever! :)

(from  PVT.LV_SUYLUAN.KHAMPHA)
IF a,b,c>o such that abc=1  then
$ \frac{a}{(1\plus{}a)^2}\plus{}\frac{b}{(1\plus{}b)^2}\plus{}\frac{c}{(1\plus{}c)^2} \leq \frac{1}{4}\plus{}\frac{4}{(1\plus{}a)(1\plus{}b)(1\plus{}c)}$

[quote="lenhamquy"][quote="tran300491"](from  PVT.LV_SUYLUAN.KHAMPHA)
IF a,b,c>o such that abc=1  then
$ \frac {a}{(1 \plus{} a)^2} \plus{} \frac {b}{(1 \plus{} b)^2} \plus{} \frac {c}{(1 \plus{} c)^2} \leq \frac {1}{4} \plus{} \frac {4}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}$[/quote]
setting $ 1 \plus{} a \equal{} x,1 \plus{} b \equal{} y$ and $ 1 \plus{} c \equal{} z$
$ \Leftrightarrow abc(2(ab \plus{} bc \plus{} ca) \minus{} 3abc)^2\ge 0$[/quote]
Are you sure?
NOTE:'='<->a=b=c=1
I think you write
<->$ xyz[2(xy\plus{}yz\plus{}zx)\minus{}3xyz]^2 \geq0$

[quote="arqady"][quote="tran300491"]If a,b,c>0  such that abc=1 then
$ \sqrt {\frac {1}{1 + a}} + \sqrt {\frac {1}{1 + b}} + \sqrt {\frac {1}{1 + c}} \leq2\sqrt {1 + \frac {1}{(1 + a)(1 + b)(1 + c)}}$

(Trieu Bon)[/quote]
Let $ a = \frac {y}{x},$ $ b = \frac {z}{y}$ and $ c = \frac {x}{z},$ where $ x,$ $ y$ and $ z$ are positive numbers such that $ x + y + z = 1.$ Then
$ \sum_{cyc}\sqrt {\frac {1}{1 + a}}\leq2\sqrt {1 + \frac {1}{(1 + a)(1 + b)(1 + c)}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt {\frac {x}{x + y}}\leq2\sqrt {1 + \frac {xyz}{(x + y)(x + z)(y + z)}}.$
But $ f(x) = \sqrt x$ is a concave function.
Hence, $ \sum_{cyc}\sqrt {\frac {x}{x + y}} = \sum_{cyc}\frac {x + z}{2}\sqrt {\frac {4x}{(x + y)(x + z)^2}}\leq\sqrt {\sum_{cyc}\frac {2x}{(x + y)(x + z)}} =$
$ = \sqrt {\sum_{cyc}\frac {2x(x + y + z)}{(x + y)(x + z)}} = 2\sqrt{1 + \frac {xyz}{(x + y)(x + z)(y + z)}}.$
Jensen forever! :)[/quote]

If a,b,c>o then
$ \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}$

If a,b,c>0 then
$ \sum(\frac{a}{b\plus{}c})^{\frac{2}{3}} \geq 1$

[quote="tran300491"]If $ a,b,c$ be  positive numbers such that $ abc \equal{} 1,$ then

$ \sqrt {\frac {1}{1 \plus{} a}} \plus{} \sqrt {\frac {1}{1 \plus{} b}} \plus{} \sqrt {\frac {1}{1 \plus{} c}} \leq2\sqrt {1 \plus{} \frac {1}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}}.$

(Trieu Bon)[/quote]Let $ a \equal{} x^3 ,b \equal{} y^3,c \equal{} z^3,$ then

$ \sum_{sym}\sqrt {\frac {1}{1 \plus{} a}}\leq2\sqrt {1 \plus{} \frac {1}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}}\Longleftrightarrow$

$ \sum_{sym}\sqrt {\frac {yz}{yz \plus{} x^2}}\leq2\sqrt {1 \plus{} \frac {x^2y^2z^2}{(yz \plus{} x^2)(zx \plus{} y^2)(xy \plus{} z^2)}}$

$ \equal{} 2\sqrt {\frac {(y^2z \plus{} z^2x \plus{} x^2y)(yz^2 \plus{} zx^2 \plus{} xy^2)}{(yz \plus{} x^2)(zx \plus{} y^2)(xy \plus{} z^2)}}.$

But $ \sum_{sym}\sqrt {yz(y^2 \plus{} zx)(z^2 \plus{} xy)} \equal{} \sum_{cyc}\sqrt {(y^2z \plus{} z^2x)(yz^2 \plus{} xy^2)}$

$ \leq\sqrt {\sum_{cyc}{(y^2z \plus{} z^2x)}\sum_{cyc}{(yz^2 \plus{} xy^2)}} \equal{} 2\sqrt {(y^2z \plus{} z^2x \plus{} x^2y)(yz^2 \plus{} zx^2 \plus{} xy^2)}.$

Cauchy enough. :)


integration
calculus
calculus computations

Couldnt solve the following:

$\int \frac{ x*e^\arctan(x)}{ (1+x^2)^\frac{3}{2} }$

I tried $t=arctan(x)$

but what do I do with the power $3/2$ in the denominator...

Any help would be great, thx.


Έχουμε από τη μια μεριά Παοκτσήδες και από την άλλη Αριανούς. Αν από τους Παοκτσήδες γίνει ένας Αριανός, τότε οι Αριανοί είναι ίσοι με τους Παοκτσήδες. Επίσης, αν ένας Αριανός γίνει Παοκτσής, τότε οι Παοκτσήδες είναι διπλάσιοι από τους Αριανούς. Να βρεθεί το πλήθος των οπαδών της κάθε ομάδας.

ωραια ασκηση ...


group theory
abstract algebra
real analysis
real analysis unsolved

We know that the measurable subgroups of $\mathbb{R}$ have measure $0$. But does $\mathbb{R}$ have any non-measurable subgroups ?

Sure. For example, let $\{\alpha_{i}\}_{i\in I}$ be a Hamel basis of $\mathbb R$ over $\mathbb Q$, fix some $i_{0}\in I$, and look at the subgroup $G$ of $\mathbb R$ consisting of those elements whose $i_{0}$'th coefficient in the basis $\{\alpha_{i}\}$ is zero. $G$ cannot have measure zero, because $\mathbb R$ is a countable union of translates of $G$, and it cannot be measurable with positive measure because $G+G=G$ which contains no non-degenerate intervals, and it's well-known that if $S$ is a measurable set of positive measure, then $S+S$ contains an interval.


Ring Theory
superior algebra
superior algebra unsolved

(a) Let $\displaystyle \mathcal K$ be a commutative field and $n$ be an integer, $n \geq 2$. Determine
\[ Z \left( \mathcal M_n \left( \mathcal K \right) \right) = \left\{ A \in \mathcal M_n \left( \mathcal K \right) | AX = XA, \, \forall X \in \mathcal M_n \left( \mathcal K \right) \right\} ,  \]
and prove that the subring $Z \left( \mathcal M_n \left( \mathcal K \right) \right)$ and $\mathcal K$ are isomorphic.

(b) Prove that the rings $\mathcal M_n \left( \mathbb R \right)$ and $\mathcal M_n \left( \mathbb C \right)$ aren't isomorphic.

[i]Marian Andronache, Ion Savu[/i]

hi,
Let $A$ be $\in \mathbb{M}_{n}(\mathbb{K})$,multiplying by matrices of type D say all entries except one daigonal entry to be  zero and with that diagonal entry different from $1,0$,and then comparing entries on both sides of the equation $AD=DA$ u will get that all nondiagonal entries are zero,similarly by multiplying by matrices of type having all entries except one entry(say in position(1,2)) zero u can get all diagonall entries should  be equal,so the centre is  the set of scalar matrices,which is isomorphic to $K$.

For the second part if there existed an isomorphism,the centre on both sides should be isomorphic which in turn implies $\mathbb{R} \cong \mathbb{C}$ which is not true as one is algebraically closed while the other is not.


You can look on the AoPS book store, there are a lot of good books on there.  You might as well get yourself acquainted with the IMO, because this is your last chance of getting into it.  Good luck!  Also, the Intermediate books are coming out in early 2007, you can also get those when they come out.  Volume 2 is a better idea than Volume one for high schoolers.

[quote="illmaticman"]I am very late to get into this, someone reccomended this forum to me, and i decided to give it a try. I am going into grade 12, and i live in Canada. My problem solving skills aren't great, and i haven't participated in alot of contests. I am interested in doing well in grade 12, and doing well on the Euclid contest. Most of all I'd like to gain problem solving skills. For a guy like me, who's skills aren't very good, would you reccomend vol 1 and 2, or please suggest any books, I really need them. Thanks[/quote]
you sound a lot like me, except i started late in grade 10. Now i'm going into 11. I think, since you live in canada, know most of vol.1 and some of vol.2. YOu should buy both though, and also get Acops (art adn craft of problem solving).
And like Mgao said, go to
http://www.artofproblemsolving.com/Books/AoPS_B_Item.php
hope that helps. :)

And make sure you hang out on this site a lot too.  You learn a ton just from all the discussions in the forum...

Btw, Inspired by Nature, you're just about exactly like me, except that I got in February of sophomore year rather than April. :wink:


calculus
parameterization
algebra
function
domain
analytic geometry

Can someone explain how to find parametric equations for a circle or semicircle?

use y=sin t; x=cos t for a circle. Specify the domain of t to make arcs.[/list]

"The standard parametrization is (cos(theta),sin(theta)), and you can pick your favorite interval if you want a circle or semicircle or something."

Since it's 3 diminsional, would one of the equations be equal to 0?

Those equations assume there is no t-axis.

[quote="Mathhelp"]"use y=sin t; x=cos t for a circle. Specify the domain of t to make arcs."

What if it's a semicircle?  How does it work if you have the point at which it's centered at and its radius?[/quote]

A semicircle is just an arc of radius  :pi:

I have the center point and the radius of a semicircle.  How do I find parametric equations for  x=  y= and z=?

Okay --

do you understand what the equations x = sin :theta:, y = cos :theta: give you?  It'll be a circle in the x-y plane of radius 1, right?

Now, if you want a circle in 3 dimensions, it can be more complicated.  It depends on the orientation -- what is the orientation of the circle you want?  If it happens to lie in one of the x-y, y-z, or x-z planes, then it's easy.  If it's tilted, that's problematic.

If you want to make the radius different from 1, you have to multiply the parametric equations by r.

To move the center, you just add the center coordinate to x and y.  (If it used to be centered at (0, 0), and you add 2 to x and 1 to y, it will just move it so that it's the same circle centered at (2, 1).)

If you want an arc, you have to figure out limitations on :theta: in the parametric equations so that it only goes part of the way around.  In your case, if you want a semi-cricle, it would have to be something like 0 :le: :theta: :le: :pi:


If the integers are $ a,a\plus{}1,a\plus{}2,...,a\plus{}5$ the difference is $ \boxed{5}$


Hello,

find all positive integers $ n$ such that

$ \frac{2010}{n^{2}\minus{}1}$ is also a positive integer.

We have to let $ n^2 \minus{} 1$ be a factor of $ 2010$.  So $ n^2$ has to be a factor of $ 2011$.  The only two factors of $ 2011$ are $ \{1,2011\}$.  However, $ n^2 \neq 1$, because then $ \frac{2010}{0}$ would be undefined.  Thus, there are no answers.

[quote]We have to let $ n^2 \minus{} 1$ be a factor of $ 2010$. So $ n^2$ has to be a factor of $ 2011$.[/quote]

We can all do without such a piece of advice!

Since $ n^2\minus{}1$ is a factor of $ 2010\equal{} 2\cdot 3\cdot 5\cdot 67$, one will just try all (finitely not that many) cases.

[quote="ernie"]So $ n^2$ has to be a factor of $ 2011$.[/quote]

That's like saying because $ \frac{4}{2}$ is an integer, then $ \frac53$ has to be one too.

[quote="ernie"]$ n^2$ has to be a factor of $ 2011$.[/quote]

 :lol: I think you realize this now, but that's not correct.

[hide="Solution"]
$ (n\plus{}1)(n\minus{}1)|2\cdot 3\cdot5\cdot 67$

Adding one to all of these will give the potential answers $ 3,4,6,68$.
Subtracting one gives $ 1,2,4,66$

Only $ \boxed{2}$ works, I think.
[/hide]

2010=2*3*5*67. in order to 2010/(n^2-1) is a positive integer, n^2-1 must be estimated number of 2010 with n is a integer so that n=4,2(n^2-1=15=3*5;n^2-1=3) is the right answer.

I think $ 2010/(n^2\minus{}1)$ can be factored as $ 1005/(n\minus{}1) \minus{} 1005/(n\plus{}1)$ and the problem turns to sequentive odd factor hunt, at least we have less to worry about ^^

The thing is, if you factor out the bottom to $ (n \plus{} 1)(n \minus{} 1)$ you get n cannot be 1.
Because 0 is neither negative nor positive, it cannot be counted. The statement, [quote]find all positive integers  such that [/quote] restricts all the negative numbers.

Now that we have our limits, we know that n has to be a number that, after going through the $ n^{2} \minus{} 1$, is a factor of 2010 because of the statement, [quote]is also a positive integer.[/quote]

Just list out the factors of 2010 and get n using the $ n^{2} \minus{} 1$

Please correct me if I made a mistake in any way.

I am not sure if I understood what you said, but even if you dont factor $ n^2 \minus{}1$ you still get that $ n$ cannot be 1.

In this problem, $ n$ chasing isnt hard at all, since you dont have much options anyway. Solutions are $ n\equal{}4$ and $ n\equal{}2$.


search
number theory unsolved
number theory

Show that the sum of all the primitive roots modulo $ p$ is congruent to $ \mu (p\minus{}1)$ modulo $ p$ and what can we say for the sum of all the primitive roots modulo $ p^{\alpha}$, $ (\textrm{mod} \; p^{\alpha}\ )$ ?


analytic geometry
graphing lines
slope
function

Find the angle of inclination, to the nearest tenth of a degree, of each of the following lines.

y=-3x+8       y=4     y=2/3x+1

how do you do this?

please and thanks!!

Use the fact that $m=tan(\alpha)$, where $m$ is the slope of the line, and $\alpha$ is the angle of inclination.

I'm assuming you mean the angle that those lines form with the X-axis.

I'll post a picture in a second.

Clearly, for Y = 4, there is no angle of inclination ($\angle = 0^{\circ}$).

Looking at the first graph ($y=\frac{2}{3}X+1$), we see that the graph intercepts the X axis at $(-1.5, 0)$ and the Y axis at $(0,1)$. We can make a right triangle with the Y axis, and use $arctan(\frac{1}{1.5})$ to come up with the (rough) angle $\boxed{\measuredangle = 33.69^{\circ}}$

Repeat this process of finding a right triangle using intercepts, then using the arctan function to find the angle.


AMC
AIME
geometry
geometric transformation
rotation

similar to the aops solution, i put the particle in the complex plane and then multiplied by cis pi/4 and added 10 for every move. but, i used the fact than cis pi/4 is $ \sqrt{i}$ to make thing much easier:
starting position: 5
5$ \sqrt{i}$+10
10$ \sqrt{i}$+5
5$ \sqrt{i}$
5
so we see, like the aops solution, that it repeats after every 4 moves. except with less algebra.
so we take 10$ \sqrt{i}$+5 and replace $ \sqrt{i}$ with cis pi/4 and compute |p|+|q|

[b]Theorem:[/b]  Every orientation-preserving isometry of $ \mathbb{R}^2$ that fixes a point is a rotation about that point.

Letting $ \omega \equal{} e^{ i \frac{\pi}{4} }$ denote a primitive eighth root of unity, the transformation $ z \mapsto \omega z \plus{} 10$ is orientation-preserving and clearly fixes some point, hence is rotation about some point, which we can see by writing

$ z \minus{} \frac{10}{1 \minus{} \omega} \mapsto \omega \left( z \minus{} \frac{10}{1 \minus{} \omega} \right)$.

And we know why rotations are periodic.  :)

Please try to continue discussion about problems in the contests section in their linked to topics: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088469#1088469


inequalities
triangle inequality
geometry unsolved
geometry

Suppose there are $4n$ line segments of unit length inside a circle of radius $n$. Furthermore, a straight line $L$ is given. Prove that there exists a straight line $L'$ that is either parallel or perpendicular to $L$ and that $L'$ cuts at least two of the given line segments.

We will show that that line perpendicular(line $x$) or parallel(line $y$) to $L$, passing through the center $O$ of the circle, satysfies our conditions. Let us enumerate our segments with numbers $1,2,...,4n$.For each $i$ let $x_i$ and $y_i$ denote perpendicular projections of the i-th segment to lines $x$ and $y$ respectively. Assume, that our lines $x$ anf $y$ are "bad". That means that no two of segments $x_1,x_2,...,x_{4n}$ have a point in common (and the same for segments $y_1,y_2,...y_{4n}$) Thus we have \[ x_1+x_2+...+x_{4n}<2n \] and \[ y_1+y_2+...+y_{4n}<2n \] as $2n$ is the diameter of our circle. But for each $i$ we obtain \[ (x_i+y_i)^2=(x_i^2+y_i^2)+2x_i y_i\geq 1 \] so $x_i+y_i\geq 1$. Which means that  \[ (x_1+x_2+...+x_{4n})+( y_1+y_2+...+y_{4n})\geq 4n \] so we've already obtained a contradiction  :lol:  which means that either $x$ or $y$ are 'good' !!!

[quote="jastrzab"]But for each $i$ we obtain \[ (x_i+y_i)^2=(x_i^2+y_i^2)+2x_i y_i\geq 1  \] so $x_i+y_i\geq 1$. [/quote]
You can actually use the triangle inequality to show this. ;)

hy

i wonder what were the results of the selection test?

how many people solved 3 or more problems?

[quote="jastrzab"] That means that no two of segments $x_{1},x_{2},...,x_{4n}$ have a point in common (and the same for segments $y_{1},y_{2},...y_{4n}$) Thus we have
\[x_{1}+x_{2}+...+x_{4n}<2n \]
and
\[
y_1+y_2+...+y_{4n}<2n
[/quote]

Hey, this is not necessary. The projections can be the same but the segments may not be cut by the lines at all.


In the forums under "online classes," I see the four moderator buttons (delete, move, lock, split) but as far as I know, I am not a mod (yet? ;)). What's goin on?

The same goes for me...I seem to have the same "mod power" buttons in all of the forums for my online classes.

[quote="rrusczyk"]Fixed now?[/quote]

Nope. I'm still a fake mod :)

I'm also a fake mod in the Mathcounts problem series board.

Can you actually do anything with your moderator powers?  In other words, if you click on move this topic, will it let you try?  (Please don't actually move it!)


calculus
integration
calculus computations

Why is this an hard integral?

[hide]$ \frac {x^3 \minus{} x^2 \plus{} 1}{x^4 \plus{} x^2 \plus{} 1} \equal{} \frac {x^3 \minus{} x^2 \plus{} 1}{(x^2 \plus{} 1)^2 \minus{} x^2} \equal{} \frac {x^3 \minus{} x^2 \plus{} 1}{(x^2 \minus{} x \plus{} 1)(x^2 \plus{} x \plus{} 1)}$

$ \equal{} \frac {\frac {3}{2}x \plus{} 1}{x^2 \plus{} x \plus{} 1} \minus{} \frac {1}{2} \frac {x}{x^2 \minus{} x \plus{} 1}$

and now its straightforward.[/hide]


geometry
3D geometry
number theory
relatively prime
number theory solved

Determine all paire (x,y) of positive integers such that x^3-y^2=2000000
thanks.
_________________

I get (300,5000) and (129, 383) as the only solutions in positive integers.  I wont post my entire solution, since the number 2000000 makes things long.

case 1: gcd(y,2000000)=1, then x^3 = (y+1000 \sqrt -2)*(y-1000 \sqrt -2) and each factor on the right is coprime (proof similar to the one used in one of the mathlinks contest) so each factor must be a cube, and we have (y+1000 \sqrt -2) = (a+b \sqrt -2)^3 with a,b integers, or equivalently 1000 = b*(3a 2 - 2b 2) and y = a*(a 2 -6b 2 ) so b|1000, and we can try all possible values for b, and solve for a, and if a is an integer, plug in to get y, and then solve for x.  And this method yields the solutions (129,383) and (300,5000) and even though 5000 is not relatively prime to 2000000, we accept it but we need to look for more.

case 2:  gcd(y,2000000)!=1.  here, either 2|y or 5|y, this is a bit hard to explain, but ill just consider one case for now, and the other case is similar.  So, suppose 5|y, then 5|x, so 5a=x, 5b=y for integers a,b and we get
125a^3-25b^2=2000000 => 5a^3-b^2=80000 => 5|b, so 5c=b
5a^3-25c^2=80000 => a^3-5c^2=16000 => 5|a, so 5d=a
125d^3-5c^2=16000 => 25d^3-c^2=3200 => 5|c, so 5e=c
25d^3-25e^2=3200 => d^3-e^2=128.

Again, we have two cases to consider, but if we consider e to be relatively prime to 128, and we use the Z[ \sqrt -2] method from case 1, we dont get any solutions in which e is relatively prime to 128, so 2|e, then 2|d, so 2f=d, 2g=e and
8f^3-4g^2=128 => 2f^3-g^2=32 => 2|g, 2h=g
2f^3-4h^2=32 => f^3-2h^2 = 16 => 2|f, 2i=f
8i^3-2h^2=16 => 4i^3-h^2=8 => 2|h, so 2j = h
4i^3-4j^2 = 8 => i^3-j^2=2 and we have hit rock bottom!
if we had started with 2|y instead of 5|y, we would have come to the same conclusion using the same descent method and seeing no relatively prime solutions.  So now we just have to solve i^3-j^2=2, which is just i^3=(j + \sqrt -2)(j - \sqrt -2), and we use a similar method as in case 1 on top, to see that j = 5, and if we work our way back up, y = 5000, and x = 300

yielding the two solutions above.


inequalities
function
inequalities unsolved

[quote="ashrafmod"]$4^{1/3}-3^{1/3}< 3^{1/3}-2^{1/3}$[/quote]
It is equivalent to $\frac{4^{1/3}+2^{1/3}}{2}< 3^{1/3}$ and this inequality is jus Jensen for concave function $f(t)=t^{1/3}$.

[quote="ashrafmod"]$9^{1/3}-8^{1/3}<3^{1/3}-2^{1/3}$[/quote]
Transform it in the form $9^{\frac13}+2^{\frac13}<8^{\frac13}+3^{\frac13}$. Then since $(9,\; 2)\succ (8,\; 3)$ and the function $f(t)=t^{\frac13}$ is concave, it is straightforward by Karamata.

[quote="Rzeszut"][quote="Hawk Tiger"]just use calculator :P[/quote]
Try to prove $\sqrt[3]{200!+1}+\sqrt[3]{100!-1}< \sqrt[3]{200!}+\sqrt[3]{100!}$ with a calculator.  :D[/quote]
Haha,Just a joke.


geometry
incenter
inradius
trigonometry
geometry proposed

For any triangle $ABC$ with inradius $r$, angle-bisectors $w_{a},w_{b},w_{c}$ and incentre $I$, prove the  identity
\[\frac{1}{w_{a}IA}+\frac{1}{w_{b}IB}+\frac{1}{w_{c}IC}=\frac{1}{2r^{2}}.\]

We have
1)$w_{a}=\frac{2bc}{b+c}\cdot \cos{\frac{A}{2}}$, etc.
2)$\sin{\frac{A}{2}}=\frac{r}{IA}$, etc.
3)$\cot{\frac{A}{2}}=\frac{p-a}{r}$, etc.
4)$r=\frac{S}{p}$.
5)$S=\sqrt{p(p-a)(p-b)(p-c)}$.

Now, by assertions above problem is easy, but maybe my solution is long  :oops:


calculus
integration
function
geometry
calculus computations

Hi everyone,

Can someone please suggest me some calculus problem books that are challenging????

Thanks

1,Boris Demidovich ,  Problems in Mathematical Analysis

2,  Problems in Mathematical Analysis 1: Real Numbers, Sequences and Series (Student Mathematical Library, V. 4) by W. J. Kaczor

3,  Problems in Mathematical Analysis II (Student Mathematical Library, Vol. 12) by W. J. Kaczor

4,  Problems in Mathematical Analysis III (Student Mathematical Library,) by W. J. Kaczor

5) Problems and Theorems in Analysis I: Series, Integral Calculus, Theory of Functions (Classics in Mathematics) (Paperback)
by George Polya, Gabor Szegö, D. Aeppli (Translator), C.E. Billigheimer

6) Problems and Theorems in Analysis. Volume II: Theory of Functions. Zeros. Polynomials. Determinants. Number Theory. Geometry (Classics in Mathematics) (Paperback)
by George Polya, Gabor Szegö, C.E. Billigheimer

I think  :wink:


if\[ x \plus{} y \plus{} z \plus{} w \equal{} 0
\]

prove that\[ xyz \plus{} yzw \plus{} zwx \plus{} wxy \equal{} \sqrt {(yz \minus{} wx)(zx \minus{} wy)(xy \minus{} zw)}
\]

[hide="Solution"]
Notice that the Left Hand Side is equal to $ xy(z \plus{} w) \plus{} wz(x \plus{} y) \equal{} (xy \minus{} zw)(z \plus{} w)$ since $ x \plus{} y \plus{} z \plus{} w \equal{} 0$.  Similarly, the the Left Hand Side is equal to $ (xz \minus{} wy)(y \plus{} w)$ and $ (yz \minus{} wx)(x \plus{} w)$.  Now, the given inequality is true if and only if\[ (xy \minus{} wz)(z \plus{} w) \equal{} \sqrt {(yz \minus{} wx)(zx \minus{} yw)(xy \minus{} zw)}\iff\]\[ (xy \minus{} wz)^2(z \plus{} w)^2 \equal{} (yz \minus{} wx)(xz \minus{} yw)(xy \minus{} zw)\iff\]\[ (xy \minus{} yz)(z \plus{} w)^2 \equal{} (xz \minus{} wx)(xz \minus{} yw)\iff\]\[ (z \plus{} w)(xz \minus{} yw)(y \plus{} w) \equal{} (xz \minus{} wx)(wz \minus{} yw)\iff\]\[ zy \plus{} wz \plus{} wy \plus{} w^2 \equal{} yz \minus{} wx\iff w(x \plus{} y \plus{} z \plus{} w) \equal{} 0\]which is true and we are done. [/hide]


The probability of snow for each of the next three days is $ \frac{3}{4}$. What is the probability that it will not snow any of those three days? Express your answer as a common fraction.

[quote=Ryan2010]this isn't even complementary counting[/quote]

well it just wants you to find the complement. probably a lvl 1 problem tho

[quote=Ryan2010]no its level 11 lol[/quote]

whoaaaaaa. weird


Let the isosceles triangle ABC with angle A = 100 degs be given and the points D, E and F respectively on its sides as follows:
D on (BC in order B-C-D and angle CAD = 10 degs
E on (BA, in order B-A-E and angle ACE = 20 degs
F on (AC) and angle CBF = 10 degs
Prove that |AE| = |CD| = |CF|.

Best regards,
sunken rock

[b][u]Proof[/u].[/b] Suppose w.l.o.g. $ BC \equal{} 2$ . Observe that $ AB \equal{} AC \equal{} \frac {1}{\cos 40}$ .

Apply the [b]SINUS[/b]' [u]theorem[/u] in the triangles $ ACE$ , $ ACD$ , $ BCF$ respectively :

$ \begin{array}{ccccccc} \frac {AE}{\sin \widehat {ACE}} \equal{} \frac {AC}{\sin\widehat {AEC}} & \Longleftrightarrow & \frac {AE}{\sin 20} \equal{} \frac {AC}{\sin 80} & \Longleftrightarrow & AE & \equal{} & \frac {\sin 20}{\cos 40\cos 10}\ . \\
 \\
\frac {CD}{\sin \widehat {CAD}} \equal{} \frac {AC}{\sin\widehat {ADC}} & \Longleftrightarrow & \frac {CD}{\sin 10} \equal{} \frac {AC}{\sin 30} & \Longleftrightarrow & CD & \equal{} & \frac {\sin 10}{\cos 40\sin 30}\ . \\
 \\
\frac {CF}{\sin \widehat {CBF}} \equal{} \frac {BC}{\sin\widehat {BFC}} & \Longleftrightarrow & \frac {CF}{\sin 10} \equal{} \frac {BC}{\sin 130} & \Longleftrightarrow & CF & \equal{} &  \frac {2\sin 10}{\sin 50}\ .\end{array}$

In conclusion, $ AE \equal{} CD \equal{} CF \equal{} \frac {2\sin 10}{\cos 40}$ .

I was expecting this proof from you, Virgil, however, my solution is whole synthetic.

Best regards,
sunken rock

Circle (A) with center A and radius AB = AC cuts BF again at G.
Isosceles $ \triangle CAG \cong \triangle ACE$ by SAS or ASA => AE = CG.
$ \angle CFG \equal{} \angle CBF \plus{} \angle BCF \equal{} \frac {_1}{^2} \angle CAB \equal{} \angle CGF$ => CF = CG.
AD bisects $ \angle CAG$ => ACDG is a kite => DA bisects $ \angle CDG.$ $ \angle CDA \equal{} 30^\circ$ => $ \triangle CDG$ is equilateral => CD = CG.


How many different sums can be made by adding two or more distinct numbers from the set $ \{3, 4,5 \}$?

We can do $ 3\plus{}4$=$ 7$, $ 3\plus{}5$=$ 8$, $ 4\plus{}5$=$ 9$, and finally $ 3\plus{}4\plus{}5$=$ 12$.
This gives us a total of $ \boxed {4}$ different sums.

Alternatively, we could have noted that each number can be paired with one of two other numbers, giving us $ 3 \times 2 \equal{} 6$ total ways. We would then add 1, for 3+4+5, giving us a total of $ \boxed{7}$.

Because there were only a few numbers, my way wouldn't be much faster than jxl's, but as the number of elements in the set increase, so does the time it takes to count all the cases out.

But if you just do $ 3 \times 2$, aren't you over counting?  Or am I just missing something?

I think isabella meant you have to divide by 2 after getting 6, and then add 1.

Yes she's counting everything except the $ 3\plus{}4\plus{}5$ twice. You have to divide by $ 2$, otherwise $ 3\plus{}4$ would be counted as a different sum than $ 4\plus{}3$. jxl28's solution is correct.


Let $a=(13-x)^{\frac{1}{3}}, \ b=(13+x)^{\frac{1}{3}}$

Then we have the system of two equations:

$a^{3}+b^{3}=26, \ \ \ a^{2}+4b^{2}=5ab$

The last one can be rewrited as $(a-b)(a-4b)=0$

This evidently leads to 2 solutions $x_{1}=0, \ x_{2}=-\frac{63}{5}$ without any extra condition:
[quote]where 5x belongs to Z[/quote]


function
real analysis
real analysis unsolved

Let $ f,g: R \rightarrow R$ be two continuous functions. Suppose $ g$ is Lipschitz continuous and define $ F: R^2 \rightarrow R$  by $ F(x,y) \equal{} f(x)g(y)$.    Find minimal conditions for $ f$ so that $ F$ is locally Lipschitz.

Here is what I have thought about it

- even $ f$ being continuous, $ F$ is not necessary globally Lipschitz (e.g. $ f(x) \equal{} g(x) \equal{} x$)
- My guess is that $ f$ must be bounded or something like that, but I don't know how to prove the statement.

Danken!

[quote="fedja"]I assume the question means "Find all $ f$ such that for every Lipschitz $ g$, the function $ F$ is locally Lipschitz". Then the answer is that $ f$ must be locally Lipschitz. The necessity is trivial (consider $ g\equiv 1$ and move in $ x$-direction only). See if you can prove sufficiency. Note the word "locally", it is important.[/quote]

I needed it to edit my comment here since I realize that what I want was what you just write  :P


See the attachment.

Best regards,
sunken rock

To: jrrbc

What do you mean by "no"?
Both sketches are similar, using the same algorithm.

Best regards,
sunken rock


geometry
3D geometry
prism
rectangle
algebra
polynomial
search

Given a rectangular prism, how can you make four cuts that divide it into 16 identical pieces?
This problem is a brain teaser, the given solution involves stacking the pieces after three cuts.
This brings up two questions of mine:
Is it possible to dissect a rectangle into 16 equal pieces with 4 cuts? (the implied answer is no, but I'm looking for a proof)
Is it possible to dissect a rectangular prism into 16 equal pieces with 4 cuts? Again, I'm pretty sure the answer is no, but I want a proof.

EDIT: The pieces have to be equal, i failed to specify that earlier :oops:. Also, no stacking/moving pieces.

[quote="JBL"]For your two questions, I assume that you mean without folding or moving pieces.  In that case, the questions are equivalent to the question, "What is the maximal number of pieces into which a fixed number of lines (or planes) can divide a plane (or space)?"  Both answers are polynomial sequences.  The first is easier to prove, and I'm sure it's appeared on the forum before.  The last time something similar popped up ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=969692530&t=151093]here[/url]), my advice was to "think inductively."

I think it also helps to consider first the 1-dimensional version, "What is the maximal number of pieces into which a fixed number of points can divide a line?" and build up from there.[/quote]
Oops, I meant for the pieces the thing is partitioned into to be equal :oops:
I'll edit the post

And, in dimension $ n \equal{} 4$ or higher, it is a simple task to divide an $ n$-cube into 16 pieces with four hyperplane cuts, just as it's easy to divide a square or cube into four pieces with two cuts, or to divide a cube into eight pieces with 3 cuts.  In a coordinate system with coordinates $ (x_{1}, x_{2},\ldots, x_{n})$, we take the vertices to be the elements of $ \{\minus{}1, 1\}^{n}$ and take the hyperplanes to be the planes $ x_{i}\equal{} 0$.  Choosing $ k$ of these for $ 0\leq k\leq n$ will divide the hypercube into $ 2^{k}$ congruent rectangular boxes.

Here's a fundamentally different (and harder!) question: is it possible to divide a square (actually, it should be nicer for an equilateral triangle or regular hexagon, I think) into 7 pieces of equal area with three cuts?  (I believe that this question in the case of a circle has been posted on the forum, and the answer is "no."  But I don't know where that thread is.)

[quote="t0rajir0u"][quote="diophantient"]no stacking[/quote]

It's not possible.

[hide="Here are the polynomial sequences to which JBL refers..."] The maximum number of pieces into which the plane can be cut by $ n$ lines is given by

$ {n\choose 0}\plus{}{n\choose 1}\plus{}{n\choose 2}$

For $ n \equal{} 4$ this gives $ 11$, not $ 16$.

The maximum number of pieces into which space can be cut by $ n$ planes is given by

$ {n\choose 0}\plus{}{n\choose 1}\plus{}{n\choose 2}\plus{}{n\choose 3}$

For $ n \equal{} 4$ this gives $ 15$, again not $ 16$. [/hide]
Edit:  Very interesting proofs for these formulae are given in Engel, and I state them in this form because they are suggestive of both the proofs and of the generalization.  Although the proofs require some fairly creative thinking, you want to use what is known as the [i]extremal principle[/i], which is the principle that given some collection of objects you look at the subset that maximizes something.  (It's a fairly hard proof to rediscover.  I may post it later.)

Edit #2:  I readily concede that JBL's suggestion is [b]much[/b] easier to follow.[/quote]
But with the rectangle you have four extra lines (the boundaries) and with a prism you have 6 extra faces (the boundaries).

It doesn't matter: the edges of the rectangle may divide one of the regions of the plane formed by our lines into two pieces, but only one of those two pieces is inside the rectangle.  This is more generally true of any convex shape: you can't cut it into more pieces than you can cut the plane (with straight-line cuts).  Suppose the contrary: then by pigeonhole there must exist two regions, $ A$ and $ B$, of our shape which lie in the same region of the plane.  (The regions I refer to are those formed by the lines.)  Choose any point in $ A$ and any point in $ B$, and draw the line segment between them.  This line segment is entirely in our shape (because it is convex) and it is entirely in the region of the plane which contains $ A$ and $ B$ (because the regions of the plane so-formed are convex), so $ A$ and $ B$ are in fact connected to each other and so are in fact the same region, contradiction.

Note that this is not true for concave shapes: for any concave shape in the plane, there exists a line which divides it into $ 3 > 2$ regions.

Shorter me: I'll give you 10 dollars if you can show me how to cut a rectangle into 8 pieces with 3 lines and no cheating.

[quote="t0rajir0u"]

Edit:  Very interesting proofs for these formulae are given in Engel, and I state them in this form because they are suggestive of both the proofs and of the generalization.  Although the proofs require some fairly creative thinking, you want to use what is known as the [i]extremal principle[/i], which is the principle that given some collection of objects you look at the subset that maximizes something.  (It's a fairly hard proof to rediscover.  I may post it later.)

[/quote]

hmm i tried to prove it but i wasnt very successful. could you post a proof of that?

(sorry, i know i shouldnt revive old threads but it was related to my original question)


This was an AMC problem at some point, and it also showed up on a local math competition. It is not in the contests section.

Let $ x$, $ y$, and $ z$ be three positive real numbers whose sum is $ 1$. If no one of these numbers is more than twice any other, then the minimum possible value of the product $ xyz$ is:
A. $ \frac{1}{32}$
B. $ \frac{1}{36}$
C. $ \frac{4}{125}$
D. $ \frac{1}{127}$
E. None of these

Intuitively, the answer should be $ (x,y,z)\equal{}(\frac{1}{2},\frac{1}{4},\frac{1}{4})$, thus $ \frac{1}{32}$, but I am looking for a proof.

[quote="ghjk"]How about the case when x=y=z=1/3 and gives the xyz=1/27<1/32.This triple (x,y,z) also satisfies the restriction since they are all equal.I think E is correct.[/quote]

1/27 is greater than 1/32  :wink:

Imagine x, y, z as the sides of a triangle?  I dunno.

$ \frac{3}{7}$, $ \frac{3}{7}$, and $ \frac{1}{7}$ [among other values] can be sidelengths of a triangle but are not possible values of $ x$, $ y$, and $ z$.

So far, I have by AM-GM that xyz >or= $ 3/(1/x^3 \plus{} 1/y^3 \plus{} 1/z^3)$ so we want to maximize $ 1/x^3 \plus{} 1/y^3 \plus{} 1/z^3$. Does this lead anywhere?

[hide="Slightly Less Intuitively than OP"]Making the numbers as close together as possible will give a larger value by AM-GM, so we want the values to be as extremal as possible.

Thus, let $ (a,b,c) \equal{} (t,xt,2t)$ for $ 1 \leq x \leq 2$, where we have made the smallest and largest values as distant as possible.

Then, $ a \plus{} b \plus{} c \equal{} 1 \implies t \equal{} \frac {1}{x \plus{} 3} \implies abc \equal{} \frac {2x}{(x \plus{} 3)^3}$

This achieves its minimum over that interval with $ x \equal{} 1$, for which we get $ \frac {1}{32}$, thus this is the answer[/hide]


[quote="April"]For a positive integer $ n$, let $ S(n)$ denote the sum of its digits. Find the largest possible value of the expression $ \frac {S(n)}{S(16n)}$.[/quote]
First notice that $ S(10^kn) \equal{} S(n)$ so we see: $ \frac {S(n)}{S(16n)} \equal{} \frac {S(10000n)}{S(16n)}$ So we can assume $ 625 \mid n$. Let $ n \equal{} 625k$. Then $ \frac{S(n)}{S(16n)} \equal{} \frac{S(625 \cdot k)}{S(10000k)} \equal{} \frac{S(625 \cdot k)}{S(k)}$.
For $ k \equal{} 1$ we see that $ \frac{S(625)}{S(1)} \equal{} 13$. And we just have to prove the obvious $ S(a) \cdot S(b) \ge S(ab)$. Then $ \frac{S(625 \cdot k)}{S(k)} \le S(625) \equal{} 13$. So the max is $ \boxed{13}$

[quote="erbed"]$S(a)S(b) \geq S(ab) $ doesnt seem so obvious. Anyone can give a proof?[/quote]
Let $a$ resp. $b$ have decimal representation $a_0 \ldots a_n$ resp. $b_0 \ldots b_m$, i.e.
\[ a = \sum_{i = 0}^n 10^i a_i, b = \sum_{i = 0}^m 10^i b_i, a_i,b_i \in \{0,1,\ldots,9\} \]
Then $S(a)S(b) = \sum_{i = 0}^n \sum_{j = 0}^m a_ib_j$, on the other hand:
\[ ab = \sum_{i = 0}^n \sum_{j = 0}^m 10^{i+j} a_i b_j \]
So:
\[ S(a)S(b) = \sum_{i = 0}^n \sum_{j = 0}^m S(10^{i+j} a_i b_j) \ge S \left ( \sum_{i = 0}^n \sum_{j = 0}^m 10^{i+j} a_i b_j  \right ) = S(ab) \]
At the last step I use $S(x) + S(y) \ge S(x+y)$ which you can easily prove yourself.

We claim that the answer is 13, achieved by 625.

We wrote $16n$ as a sum of $S(16n)$ powers of 10. When this is divided by 16, it becomes a sum of a bunch of "625"s that occupy consecutive place values (some of them possibly going into decimals). Therefore, the new sum of digits can be at most 13 times the original, since each term $10^a$ becomes $6\cdot 10^{a-2}+2\cdot 10^{a-3}+5\cdot 10^{a-4}$ which contributes 13, but there could possibly be "carries" that further decrease the sum of digits, so we are done


inequalities
topology
real analysis
real analysis unsolved

I am puzzled by a proof in Evans book"PDE" : section 5.6, thm6, "general sobolev inequality".
In the proof, step3,

$ U$ is a bounded open set in $ R^n$ with $ C^1$ boundary, $ u \in W^{k,p}(U)$. And if $ k>\frac{n}{p}$ and $ \frac{n}{p}$ is an integer. Set $ l\equal{}\frac{n}{p}\minus{}1$.

(By repeatedly using Gagliado-Nirenberg-Sobolev inequality) we have $ u \in W^{k\minus{}l,r}(U)$, for $ r\equal{}\frac{pn}{n\minus{}pl}\equal{}n$.

then he said "Hence Gagliado-Nirenberg-Sobolev inequality shows $ D^{\alpha}u \in L^q(U)$ for all
$ n<\equal{}q<\infty$ and all $ \alpha <\equal{}k\minus{}l\minus{}1$.

I think Gagliado-Nirenberg-Sobolev inequality only genetee some special $ q$, i.e. $ q$ has to be conjugate of some number.
But how do we get conclution for all $ n<\equal{}q<\infty$ ?

[quote="mlok"]Since $ U$ is bounded, the spaces $ L^q(U)$ are nested (Holder's inequality).[/quote]

If $ p<q$, then $ L^q \subset L^p$. So if $ f \in L^q$ for some $ q$, then $ f \in L^p$ for all $ p<q$,
but how do we know $ f \in  L^p$ for all large $ p$ upto $ \infty$? which is my problem.

You have $ D^{\alpha}u\in W^{1,n}\subset L^q$ for any $ q<\infty$.

[quote="mlok"]You have $ D^{\alpha}u\in W^{1,n}\subset L^q$ for any $ q < \infty$.[/quote]
I see: let me prove why this is true by Gagliado-Nirenberg-Sobolev inequality.
 Since $ D^{\alpha}u\in W^{1,n}(U)$. take a number $ \beta < n$ but arbitrarily near $ n$.
So $ D^{\alpha}u\in W^{1,\beta}(U)$ since $ L^p(U)$ is nested. Now $ \beta$'s conjugate = $ \frac{\beta n}{n\minus{}\beta}$ is arbitrarily large, so  by Gagliado-Nirenberg-Sobolev inequality:

$ D^{\alpha}u\in L^q(U)$ for any $ q < \infty$ since $ L^p(U)$ is nested.


Just consider mod 4... $n \text{(mod 2)}+(-1)^{n}-1 \in{0,1}\text{ (4)}\implies n \equiv 0 \text{(2)}$ which is fancy for saying $n$ is even...
Now the equation's solutions must satisfy $16n^{4}+49^{n}+47$ is a perfect square. ${(7^{n})}^{2}+(16n^{4}+47)=(7^{n}+a)^{2}\implies a^{2}+2a*7^{n}=16n^{4}+47$
more junk then modding out by seven... in the end the best I get is that $n\equiv 0,2 (14)$

Necessary, not sufficient.  $n = 0, 2, 14$ don't give squares.

The first step is to prove that n must be even.

Another hint : There are finitely many values of $n$ that work.

Okay if n is even ...
[hide="hint"]try a few values (n=4 may even work)
[hide="more hint"]You don't have to try too many before you realize that there aren't any
[hide="because"](Cause  there is no  perfect square between $(7^{n/2})^{2}$ and $(7^{n/2}+1)^{2}$) [/hide][/hide][/hide]

sorry, when i said 0,2, (14) I meant if there was a solution it had to be congruent to 0 or 2 mod 14...

I know what you meant.  $0, 2, 14$ are the first three numbers that are congruent to $0, 2 \bmod 14$ and none of them work, so you only gave necessary, not sufficient, conditions.  And if Gyan is right, they may not even have been necessary  :o

BTW, n=4 works, and it is neither 2 nor 0 mod 14.
(Hint given before in my post,  almost gives away the whole solution)

Wow, this is an easy problem... I guess in general when you have something to the nth power and other terms with n to the something the nth power grows way too quickly... that's why the expression must be between the two squares you listed. I was looking for an all-mods solution but I don't think there is one

[quote="moogra"][quote="jclarkemathL314159"]Find, with proof, all natural numbers $n$ such that $n^{4}+7^{n}+47$ is a perfect square.[/quote]

no not true. 1 doesn't wrok[/quote]

You really really misread the question  :D He asked "Find all natural number s.t. ... Prove your answer." and not "Proof that for all natural numbers ..."

thats a big mix up
i think if you thought that and checked case n=1 you could be pretty sure you read the problem wrong...

So basically, we have to find the values for $m$ (with $n=2m$) for which
$(7^{m}+1)^{2}>7^{m}+16m^{4}+47$ (and afterwards check manually for smaller $m$)
That is
$7^{m}>8m^{4}+23$
OK, from an analytical point of view this will happen sooner or later, because an exponential grows in the end faster than a polynomial. Just use an appropriate computer program to approximate the least possible m...
But how can one actually solve such an equality in a [i]nice[/i] fashion, i.e. without forcing the use of analysis or approximations? Are there any techniques? Similar inequalities like Bernoulli $(1+x)^{m}>1+mx, x>-1$ that transform powers to factors? The latter doesn't seem to be very fruitful here...

BTW, the above equation holds for $m>3.9$. So only $n=0$ and $n=2$ to be checked. No solutions, I fear.

[hide]
Mod 4 tells us that $n$ is even: $n= 2k$.

So we want to find when $16k^{4}+49^{k}+47$ is a perfect square.  First we know that $16k^{4}+49^{k}+47 > (7^{k})^{2}$.

Now we can show that $16k^{4}+49^{k}+47 < (7^{k}+1)^{2}$ when $k > 3$.  Expand, cancel the $49^{k}$, subtract 1, and divide by two to get $7^{k}> 8 k^{4}+23$.  We can easily show that it holds for $k=4$: $2401 > 2071$.  Now suppose it works for $k = m \ge 4$: $7^{m}> 8 m^{4}+23$.  Multiply both sides by $7$ to get $7^{m+1}> 56 m^{4}+23*7 > 56 m^{4}+23$.  Now we can show that $56 m^{4}> 8 (m+1)^{4}$.  It rearranges to $7 < \left(\frac{m+1}{m}\right)^{4}$.  The inequality is true when $m=2$, and the right side is decreasing, so, by induction, $(7^{k}+1)^{2}> 16 k^{4}+49^{k}+47$ wen $k > 3$.

So now we just check $16 k^{4}+49^{k}+47$ for $k=1,2,3$.  They give $112, 2704,$ and $118992$.  The only square is $2704 = 52^{2}$.  So the only solution is $\boxed{n = 4}$.[/hide]

yeah, i'd just like to point out that normally you'd use mods in a problem where you got to disprove something, not in one where you have to find something... can't really explain it without using my hands to draw a picture in the air

:lol: Lol treething

Where'd you get that idea from?

[quote="B23"]
BTW, the above equation holds for $m>3.9$. So only $n=0$ and $n=2$ to be checked. No solutions, I fear.[/quote]
$n$ is twice your $m$ and $n=4$ (or m=2), as pointed out before, is  indeed a solution of the original equation.

FWIW, ( knowing, $3^{4}< 2^{4}\cdot 7$ or $(3/2)^{4}< 7$)
one can  show  as long as $m>1$

$(m+1)^{4}\le ((3/2)m)^{4}< 7 m^{4}$
One can use induction

Thus if   the relationship ($7^{m}>m^{4}$)  is true for  $m$, it is true for $m+1$  too, and you can check that  it is true for $m=4$ thus it is true for all $m>4$

(BTW  knowing max of $(x^{(1/x)}$  is when $x=e$ (IOW since $4>e$ then $7^{m}>4^{m}>m^{4}$ if $m>4$ so it is not too hard to guess (or you can even try, m=1,2,3,4  till you find that 4 is already too big) the number 4)

Added later: Did not see, someone has already pointed out essentioally what has been pointed out here ..


algebra
Integer sequence
Integer Polynomial

Suppose $f$ is a polynomial in $\mathbb{Z}[X]$ and m is integer .Consider the sequence $a_i$ like this $a_1=m$ and $a_{i+1}=f(a_i)$ find all polynomials $f$ and alll integers $m$ that for each $i$:
\[ a_i | a_{i+1}\]

This is really weird. Are you sure all of them can be found and presented in a nice way?

$a_i|a_{i+1}=f(a_i)$ is equivalent to $a_i|f(0)$. First of all, all polynomials for which $f(0)=0$ and all numbers $m\in\mathbb Z$ are Ok. Now assume $f(0)\ne 0$. This means that $\{a_i\}$ has only finitely many values, so from some point on it becomes periodic. Another thing we can observe is that $a_i\ne 0$ because then nothing would make sense. If the period the sequence eventually gets is $>2$, then we get a contradiction, because there must be some $i$ for which $|a_i|>|a_{i+1}|$. Now there are two cases left:

1) For $n\ge n_0$ we have $f(a_n)=-a_n$;
2) For $n\ge n_0$ we have $f(a_n)=a_n$.

I have no idea if there's a nice way to classify all these polynomials (and $a_1$s). We know one thing: in the first case $f(x)+x$ must have both $k$ and $-k$ as roots for some integer $k$, and in the second case $f(x)-x$ must have at least an integral root.

I can't find any question of this sort in Iran NMO (Round 1/2/3)..

Is it Iran TST 2004? or is this question just wrong?

As grobber mentioned, I doubt there could possibly be a clear formula for the solution..

[Edit] I think this is the solution:

(i) For all $ f$ with integer coefficients such that $ f(0)\equal{}0$: m can be any integer.
(ii) If $ f(0) \neq 0$, then $ f(f(x)) \equal{} x$ must yield an integer solution.
$ m$ must be an integer solution of the equation $ f(f(x))\equal{}x$.

(The proof for the (ii) part involves the following lemma - if a polynomial $ f$ satisfies $ f^k (a) \equal{} a$ for some integer $ a$ and some natural number $ k$, then $ f$ must satisfy $ f(f(a)) \equal{} a$ - which is easy (something similar to this was on the USAMO))

after 12 year and It have't a full solution  : :)


symmetry
geometry
LaTeX
Pythagorean Theorem

Square ABCD has side lengths of 13 units.  Point E lies in the interior of the square such that AE=5 units and BE= 12 units.  What is the distance from E to side AD? Express your answer as a mixed number.

[hide="SOLUTION"]$ \triangle{ADE}$, like $ \triangle{ABE}$, is a $ 5$-$ 12$-$ 13$ right triangle. To find the altitude to the hypotenuse, we take the product of the length of the legs, in this case $ 60$, and divide it by the length of the hypotenuse. This yeilds $ \boxed{\frac {60}{13}}$, thus our answer.[/hide]

@Kingsmasher, the answer should be expressed as a mixed number so:

$ \frac{60}{13}$ = $ \boxed{4\frac{8}{13}}$

I am afraid neither of these answers is correct...lol

we are looking for the distance to AD not AB.....

the correct answer is 1 12/13....use pythagorean theorem....

Umm...the Distance from E to AB and AD are the same. AEB is a right triangle with side lengths 5,12,13. The distance from E to AD is the same as the altitude of triangle AED. AE=5, and AD=13, therefore, by symmetry we can say that ED=12. Thus, it is a right triangle, with area 12(5)/2 = 30. We also know that 13(x)/2 = 30, for the altitude from E to AD being x, solving gives x=60/13.

[quote="limac"]Umm...the Distance from E to AB and AD are the same. [/quote]

Can you explain this?

I actually intended for the rest of my explanation for expand on my initial claim...but it is flawed. In order for 60/13 to be the answer, E must have been the midpoint of the diagonal DB, which it is clearly not on as $ 5 \plus{} 5 \not \equal{} 13\sqrt2.$ Sorry about that...

[hide="Take Two"]By the previous solution, the distance between E to AB is 60/13. Thus, we have a right triangle with hypotenuse 5 and legs of x and 60/13. (This right triangle is the one where x is the distance between E and AD, and the other leg is on AD; however, it is congruent to the one with one leg on AB and the other as the projection to AB). By the Pythagorean Theorem, the second leg is $ \sqrt{5^2\minus{}\frac{60}{13}^2}\equal{}4\frac{8}{13}$[/hide]

[quote="randomguy64"][hide="Take Two"]By the previous solution, the distance between E to AB is 60/13. Thus, we have a right triangle with hypotenuse 5 and legs of x and 60/13. (This right triangle is the one where x is the distance between E and AD, and the other leg is on AD; however, it is congruent to the one with one leg on AB and the other as the projection to AB). By the Pythagorean Theorem, the second leg is $ \sqrt {5^2 \minus{} \frac {60}{13}^2} \equal{} 4\frac {8}{13}$[/hide][/quote]

and incidentally, $ 4\frac{8}{13}$ is the same as $ \frac{60}{13}$, the same answer we have been getting all this time

[quote="randomguy64"]By the Pythagorean Theorem, the second leg is $ \sqrt {5^2 \minus{} \frac {60}{13}^2} \equal{} 4\frac {8}{13}$[/quote]

Actually, I did not get $ 4\frac{8}{13}$ when I computed the square root. In fact, I got $ 1\frac{12}{13}$, as follows:

$ \sqrt {5^2 \minus{} \left(\frac {60}{13}\right)^2} \equal{} \sqrt{25 \minus{} \frac{3600}{169}} \equal{} \sqrt{\frac{169 * 25 \minus{} 3600}{169}} \equal{} \sqrt{\frac{169(25) \minus{} 144(25)}{169}} \equal{}  \sqrt{\frac{25^2}{13^2}} \equal{} \frac{25}{13} \equal{} 1\frac{12}{13}$

Sorry for the messy latex; I am new here.

[hide="::first:: For a more comprehensive answer, CLICK HERE :first: "]
First draw a diagram. Draw ABCD then put E in the appropriate position. The lengths 5, 12, and 13 are lengths of a right triangle. Draw a line parallel to AB with E as one of its points. We will call  it MN.

$ AM \equal{} BN\equal{} y$
$ ME \equal{} x$
$ EN \equal{} 13\minus{}x$

Lets get the equations down.

$ (13\minus{}x)^2 \plus{} y^2\equal{}12^2$
$ x^2\plus{}y^2 \equal{} 5^2$

$ 169\minus{}26x\plus{}x^2\plus{}y^2 \equal{} 144$
$ x^2\plus{}y^2 \equal{} 25$

$ 169\minus{}26x \equal{} 119$
$ 50\equal{}26x$
$ x\equal{}\frac{50}{26} \equal{} \boxed{1 \frac{12}{13}}$

Hope this explains it for you. :)
[/hide]


geometry
parallelogram
geometry proposed

ABCD is a parallelogram. B is the point on AB so that 2AF=FB and K is the point on AD so that 3DK=DA. Intersection point of CF and BK is the point M and area of triangle MFB is 4. Find area of ABCD.

Let $ { M } = BK \cap CD$ and, after some easy calculation we get $ CM = 2 BF$, that is, $ CM = 2 MF$ and, consequently, we have $ [BCF] = 3 [BMF]$. If we now take G on CD with $ FG \parallel BC$, the area of the parallelogram BCGF will be  $ [BCGF] = 6 [BMF]$, and, with area of  AFGD being half of area of BFCG, we shall finally get $ [ABCD] = 9 [BFM] = 36$.

Best regards,
sunken rock


function
integration
limit
topology
absolute value
real analysis
real analysis unsolved

Suppose $(X,\mu)$ is a finite measure space, $f_{n}\colon X \to \mathbf R$ is a sequence of measurable functions such that $f_{n}\to 0$ in measure. Let $\varphi\colon\mathbf R_{+}\to\mathbf R_{+}$ be a continuous function such that
\[\sup_{n\in\mathbf N}\int \varphi(|f_{n}|) d\mu < \infty \]
and
\[\lim_{s\to\infty}\frac{\varphi(s)}s = \infty. \]
Show that $f_{n}\to 0$ strongly in $L^{1}(\mu)$.

// $\mathbf R_{+}: = [0,+\infty)$.

[b]P.S.[/b] Note that this problem generalizes [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152057]this one[/url].

This can't possibly be true. We can just choose $\phi$ to be identically zero on $[0,1]$ and then we have no control of any sequence $f_{n}$ with values smaller than 1 in absolute value. Should the limit condition be as $s\to 0$? I guess so.

Plus, I think you mean $f_{n}\to 0$ in norm. Strongly means something different for me...

I have no doubt that volodja meant convergence in $L^{1}$. As to functions less than $1$ in absolute value, don't forget that they are assumed to converge to $0$ in measure ;). When I was a student, they used to teach this criterion in regular measure theory course on par with dominated convergence theorem. I do not really know what is the right word in English for such families of functions (equiintegrable? :?) but the idea is that if you have a measure space $(X,\mu)$ with finite measure and a family $f_{n}$ of integrable functions such that for every $\varepsilon>0$ there exists $\delta>0$ such that $\int_{E}|f_{n}|\,d\mu<\varepsilon$ for all $n$ whenever $\mu(E)<\delta$, then convergence of $f_{n}$ to $f$ in measure implies that $f\in L^{1}$ and  $\|f_{n}-f\|_{L^{1}}\to 0$. The condition in volodja's problem is a simple sufficient condition for "equiintegrability".

Sentences "$(f_{n})$ converges strongly" and "$(f_{n})$ converges in the norm topology" have the same meaning.

Ok, I believe this problem is very simple to do directly. I wonder why I didn't solve the simple special case in my exam. Time pressure I suppose. Oh well. However, I don't see why the continuity condition is necessary - is it only to ensure that $\varphi(|f_{n}|)$ is measurable?

Let $\varepsilon > 0$. Choose $M > 0$ s.t. $x \geq M \longrightarrow \epsilon \varphi(x) \geq x$.  Put $F_{n}= \{x \, ;\, |f_{n}(x)| \geq M\}$. Then
$\int |f_{n}| \; d\mu = \int_{X \setminus F_{n}}|f_{n}| \; d\mu \;+\; \int_{F_{n}}|f_{n}| \; d\mu \leq \int_{X \setminus F_{n}}|f_{n}| \; d\mu \;+\; \varepsilon \cdot \sup_{n\in\mathbf N}\int \varphi(|f_{n}|) d\mu$
So by bounded convergence
$\varlimsup_{n}\int |f_{n}| \; d\mu \leq \varepsilon \cdot \sup_{n\in\mathbf N}\int \varphi(|f_{n}|) d\mu$
Since $\varepsilon$ is arbitrary
$\lim_{n}\int |f_{n}| \; d\mu = 0$

Are you sure that Bounded Convergence Theorem may be applied to
\[\int_{X \setminus F_{n}}|f_{n}| d\mu?\]

One more step is needed.  :wink:

What step? $\chi_{X\setminus F_{n}}\cdot |f_{n}|$ is a sequence of measurable functions converging in measure to $0$, $\chi_{X\setminus F_{n}}\cdot|f_{n}| \leq M$ and $M$ is an integrable function.  So bounded convergence gives $\lim_{n}\int \chi_{X\setminus F_{n}}\cdot|f_{n}| \; d\mu = 0$ which is exactly what we want.

Oh, I missed this. I'm sorry. Good proof, indeed.


I didn't know the police phone number was a prime  :)

[quote="Dnas"]19, 109, 1009, and 10009 are all primes. 100009 isn't though. :([/quote]
that's weird

Does anyone here know a formula for calculating primes, if there is one?

Here are some cool primes that you did not mention:
Fibonacci sequence segments: 1123, 235813, 3581321 are prime.
701, 7001, 70001, and 700001 are primes.
1888081808881 is a prime number that reads the same upside down or in a mirror.

[quote="chess64"][quote="math92"]i got a  cool formula...take a number, square it, and if you mulyiply, or divide, or subtact, or add 1 you will get a multiple of five.
example: 6^2=36
                36-1=35 which is a multiple of 5[/quote]

that's amazing! I mean, it's [b]so[/b] useful!  :lol: I don't memorize these kinds of formulas because they don't help me at all.[/quote]
i didn't think you would be so amazed

[quote="bob123"][quote="chess64"][quote="math92"]i got a  cool formula...take a number, square it, and if you mulyiply, or divide, or subtact, or add 1 you will get a multiple of five.
example: 6^2=36
                36-1=35 which is a multiple of 5[/quote]

that's amazing! I mean, it's [b]so[/b] useful!  :lol: I don't memorize these kinds of formulas because they don't help me at all.[/quote]

it's just using the fact that the square of any number has to end in 0,1,4,5,,6, or 9...[/quote]
really, it's pretty obvious

Umm.. easy way $1993 \equiv 1\mod4$ units digit is $2$

[quote="nat mc"]Is it fast?[/quote]

Really fast. It calculated $2^{2004}$ in 3 seconds.

[quote="nat mc"]WOW!!!  Thats amazing, wouldn't that overflow or something?[/quote]

It's an application (Cabamap). You have to learn how to use it though. It works on a principle called stacking. It's simple, you'll get it quickly if you don't already know how it works.

well i know that there's a prime thing for $2^{2^x}+1=prime$ but it doesn't work for 5 so 2^32+1=prime.

[quote="chess64"][quote="mathgeek2006"]I have this calculator app called Cabamap that calculates primes, but it doesn't work all the time...not for the larger numbers at least, but of course those are too hard to do.  It also can perform very large calculations.  Once in a state team practice, we were doing a past nationals test.  I believe it said to do something like 2^1993 and find the last digit.  I did it, then arrowed over to the last digit.  Of course that was just practice.[/quote]

Lol, I have that too! Except it takes a [i]long[/i] time to get to the last digit when there are lots of digits.[/quote]
That's why I started it, held the over button and did the other problems.  When we had like 5 minutes left, I checked it, and sure enough, there was the last digit....

[quote="Boy Soprano"]There is no formula for finding the $x$th prime—yet.  People are still trying to find one, though.  There are, however, certain formulae for producing [i]probable[/i] primes.  Perhaps the most famous one generates what are called Mersenne primes, or primes of the form
$2^{P}-1$,
where $P$ is a prime.  This does not always work, but it produces a number with a high probability of being prime.  It was used, I believe, to produce the largest known prime, namely
$2^{13,466,917}-1$
at least as of 2002.  It was discovered in 2001.
Sources :  Interview with Dr. Edwards of SPSU math department; John dePillis's [i]777 Mathematical Conversation Starters[/i].[/quote]

The search of Mersenne primes is an international search where users from around the world download a program that tests these numbers. The program is free and you can be a helper in this seach to! Its like a screen saver. Incidently, IBM or Intel (not sure which lol) uses this program to test their chips.

Thats pretty neat! Why are Mersenne primes special, though?  :?

[quote="236factorial"]I think that if a three digit number is divisible by 15 and has its digits add up to 15, when it is divided by 15, it will produce a prime number.[/quote]
Don't fall for it. Try the number 960= 64*15 :)
(In fact any such number which ends in zero,  when divided by 15 will be even)


geometry
perimeter
rectangle
3D geometry
Pythagorean Theorem

new leaderboard:
Tare & Theone-6
leeyuhang-3
notfunny-1

1. If the perimeter of a rectangle is 100, what is the shortest diagonal the rectangle could have?

I have to go my dad is gonna kill me...

squares are rectangles... so it can be 25 on each side so its 625+625 or 1250 but then u haf to sqrt it so sqrt1250...

pythagorean theorem

Since the shortest would be a square 100/4=25 so a square with a side of 25. Multiply :rt2: and get 25:rt2:

Arg, stupid web connection, couldn't see this for like 2 minutes :evil: :o

me? no... i just kept on refreshin da page to keep on seein if there wuz another round and as soon as there wuz i just answered it

no forgot to divide again theone got it
notfunny forgot to simpilfy

[quote="theone853"]How'd u get that so fast? are you cheating????[/quote]
Now now accusation is not good for it doesn't solve anything and it makes you look like a sore loser. :)

Pshaw, I'd love it if we set times like before. This random posting problem thing really isn't working for me (I'll have to refresh every few seconds but in order to do that I have to go to a page that doesn't use phpBB then come back here which takes some time)

[quote="theone853"]How'd u get that so fast? are you cheating????[/quote]
me or not funny?

To not funny: I bow down to you for you to forgive me for calling you a cheater, i was kidding anyways.

[quote="notfunny"]it doesnt matter u can write it either way  :sqrt:1250 is also simplest form[/quote]
no i don't think so
1250/25=50 so 5 :sqrt: 50
50/25=2 so
5*5 :sqrt:2
25 :sqrt:2

haha in your faces(except tare,)
i showed work:p

OK I'm going, I'll be back in an hour. Don't start without me! (and if you guys do please diversify the points so it'd be easier to catch up)

actually i showed work too... lyk the 4th post

ok to settle fights theone and i both get da point (we posted at da same time)

ok gotta go eat lunch dont start without me

To topper: Just wondering if I could answer problems from old UNL math day contests like the vwery first one, to break the tie? I just got the answer.

wait, u mean da first edition? thats cheap... but if u mean da 1st installment of da fourth edition... den i think dats ok

You misunderstood me. I meant to break the tie for the first edition. Its a three-way tie between me, tare, and confuted.

Actually I'm one point ahead of you two so even if you do solve the problem it'll still be a tie between you and me. Oh well.

Not me, I'll be still logged on but I'll be fully back in 30 mins.

If at least two users that feel like answering questions are logged in at 5:30pm, cst, I'll post the other 4 questions.

You don't have to post solutions because of the importance of posting first.

Answers must be in simplest form.

Any other issues?

I don't see any problems.

I even changed my time on AoPS to CST so I know when you are talking about.

2.A car gets 20 miles per gallon of gas.  By modifying the engine at a cost of $400, the car will get 25 miles per gallon.  If gasoline costs $1.25 per gallon, what is the least number of miles the car would have to be driven after the change in order to make back the cost of the modification?

3. If f(x) = x^3 + 3, then f(a/b), b not equal to 0 equals
a.) (a^2 + 3b^2)/b^2
b.) (a^2 + 3)/(b^2 + 3)
c.) (a^2 + 3)/b^2
d.) (a^2 + 3)/b
e.) a^2/b + 3

4. If the product (4^5)(5^13) is written as an integer, then what is the number of digits in the integer product? (no calculators)

5. Unit cubes are assembled into a larger cube and then two nonopposite faces of the larger cube are painted.  The large cube is dissassembled after the paint is dry and it is discovered that between two hundred and three hundred of the unit cubes have no paint on any of their faces.  How large was the larger cube?

The wheels on the bus goes round and round, round and round, round and round...

Oh boy. Another tie because I give the point to notfunny for 1, cuz he posted it first and it was the right answer.

the tie, continues, I'll get the next 5 on right now...


I need some help on this problem.

(A U C) intersect [(A  intersect B) U (C' intersect B)] = A intersect B

I've been trying to work this out.  The basic set of identities doesn't cover (A U C) or
(C' intersect B).  What am I not seeing here?  Is this more complex than I think?[/list]

I have two suggestions:

(1) Draw a Venn diagram.  Plot what each set looks like.

(2) Let $x$ be an element in the set on the right-hand side and show that it is in the set on the left-hands side, and vice-versa.

Could you write my original problem down and go step by step?  I'm kinda confused from the 1st line you wrote down after your substituted symbols.

[quote="10000th User"]This one easy!
Do like multiplication. My teacher tell me I write $\cap=\times$ and $\cup=+$
$(A+C)\times[(A\times B)+(C'\times B)]=(A+C)\times[(A+C')\times B]$

But I read notes and say:

Sometimes they are different from ordinary arithmetic:

We know that $A\times(B+C)=(A\times B)+(A\times C)$

But, $A+(B\times C)=(A+B)\times(A+C)$

[Edit]Oh sorry I forget to finish

Put $[(A+C)\times(A+C')]\times B=[A+(C\times C')]\times B$. Now you do last step and finish!!![/quote]

Hello, OK, I'm sorry, my notes is very confusing too :P

$(A\cup C)\cap[(A\cap B)\cup(C'\cap B)]=(A\cup C)\cap[(A\cup C')\cap B]$ Distributive law

$[(A\cup C)\cap(A\cup C')]\cap B=[A\cup(C\cap C')]\cap B$ Distributive law

$[A\cup(C\cap C')]\cap B=(A\cup\not\hspace{-0.15cm}\bigcirc)\cap B$ Empty set

$(A\cup\not\hspace{-0.15cm}\bigcirc)\cap B=A\cap B$


algorithm
combinatorics proposed
combinatorics

I thought my post had not gotten through because I didn't see it there when I tried, sorry.


linear algebra
matrix
number theory
greatest common divisor
function
vector
linear algebra unsolved

First idea? We can make the gcd into an inner product. If $ f_n(x)\equal{}\begin{cases}1&x\equal{}\frac{k}{n}, 0\le k<n\\0&\text{otherwise}\end{cases}$, $ \langle f_n,f_m\rangle\equal{}\sum_x f_n(x)f_m(x)\equal{}\text{gcd}(n,m)$.
That makes this matrix a Gramian, and makes it positive definite (since these $ n$ functions are linearly independent). The determinant is the square of a volume, but this description has far too many dimensions to work well.

OK, can we write it as a Gramian in a better way? We can sort the fractions $ \frac{k}{n}$ by their lowest denominator; for a given denominator $ d$, we either have all or none of those fractions represented. There are $ \phi(d)$ of these, so they contribute $ \phi(d)$ to the inner products if present. Since only $ n$ denominators are possible, we can put everything in a matrix $ M$. The entries are $ m_{ij}\equal{}\begin{cases}\sqrt{\phi(i)}&i|j\\0&\text{otherwise}\end{cases}$, with the $ j$th column representing $ f_j$. The $ jk$ entry of $ M^TM$ is $ \sum_d m_{dj}m_{dk}\equal{}\sum_{d|j\text{ and }d|k}\phi(d)\equal{}\langle f_j,f_k\rangle\equal{}\text{gcd}(j,k)$.

Now we know $ M^TM\equal{}A$. $ M$ is an upper triangular matrix with diagonal entries $ \sqrt{\phi(j)}$, so $ \det M\equal{}\prod_{j\equal{}1}^n \sqrt{\phi(j)}$ and $ \det A\equal{}\prod_{j\equal{}1}^n \phi(j)$.

A teaching solution:
Step 1: with a pocket calculator , you can see that, for small values of $ n$, $ det(A)>0$ and therefore $ A$ is surely a positive definite symmetric (PDS) matrix (why ?).
Fact: If you want to calculate numerically the det of a PDS matrix then the best method is the Cholesky's decomposition (see Wikipedia): $ A\equal{}CC^T$ where $ C$ is a lower triangular matrix with positive diagonal.
Step 2: with hand (if you can spend a long time) or with Maple you can conjecture that:
$ c_{ij}\equal{}\sqrt{\phi(j)}$ if $ j|i$ else $ 0$.
-(Now when a searcher want to write a paper then he uses intensively Maple, Mathematica or Matlab ; why ?; because it is the best manner of manufacturing conjectures or of checking if a conjecture is correct / When I see a student that doesn't want to do math with a computer then I am :mad:)-
Step 3: Use the recursive Cholesky's algorithm. Let $ B$ be the $ C$-value for the dimension $ n\minus{}1$. You seek $ l$ a row of dimension $ n\minus{}1$ and a positive real $ \alpha$ s.t. $ C\equal{}\begin{pmatrix}B&0\\l&\alpha\end{pmatrix}$. We obtain $ Bl^T\equal{}[gcd(1,n)\cdots,gcd(n\minus{}1,n)]^T$ and $ \alpha^2\equal{}n\minus{}\parallel{}l\parallel{}^2$.
The vector  $ l\equal{}[l_{j}]$, with $ l_j\equal{}\sqrt{\phi(j)}$ if $ j|n$ else $ 0$, satisfies the first equation.
$ \parallel{}l\parallel{}^2$ is the cardinal of the integers $ \leq{n}$ that are not prime with $ n$. Then $ \alpha^2\equal{}\phi(n)$.

[quote="loup blanc"]Now when a searcher want to write a paper then he uses intensively Maple, Mathematica or Matlab[/quote]As it happens, I don't have any of those tools installed; I'll occasionally do some numerical stuff in a spreadsheet, but that's it. The other path is to be clever and lucky; this problem didn't require any computation at all. There was, however, no guarantee that what I tried would work.

See also
http://www.mathlinks.ro/viewtopic.php?t=45
http://www.mathlinks.ro/viewtopic.php?t=37587

By the way, this problem looks like an innocent olympiad question - but it's an auxiliary lemma for the so-called Kóvasz lemma:

Let $ \phi$ and $ \psi$ be two permutations of the set $ \left\{1,2,...,n\right\}$, and let $ k$ be a field (not necessarily of zero characteristic). Let $ P\in k^{n\times n}$ and $ Q\in k^{n\times n}$ be the permutation matrices of the permutations $ \phi$ and $ \psi$ over $ k$. If the matrices $ P$ and $ Q$ are similar, then the permutations $ \phi$ and $ \psi$ are conjugate in $ S_n$.

This lemma, in turn, is used in modular representation theory: See Lemma 7.2 in
R. Gow, B. Huppert, R. Knörr, O. Manz, W. Willems, [i]Representation Theory in Arbitrary Characteristic[/i], CEDAM 1993.


parameterization
algebra unsolved
algebra

hi every body,
the equation x^4+ax^3+bx^2+cx+d=0 can write in form
(x^2+px+q)(x^2+rx+s)=0 and I try to solve p,q,r,s with a,b,c,d are parameter but I'm unsuccessfull can you help me.
sorry for my bad english :blush:

Well, no wonder you didn't make it: it is quite hard, actually.

See this for methods of solving quartic equations:

[url]http://mathworld.wolfram.com/QuarticEquation.html[/url]
[url]http://www.sosmath.com/algebra/factor/fac12/fac12.html[/url]

little bit advanced:

[url]http://planetmath.org/encyclopedia/GaloisTheoreticDerivationOfTheQuarticFormula.html[/url]

oh.... now I know why I didn't make it
thanks you hsiljak


Hi everyone.
At the Junior Balkan I heard our team leader saying that the next Balkan Olympiad will be held in Romania. ;) Is it true???

[quote="ikap"]Yes. That is true. And more than that, it will be held in my hometown  :([/quote]Yes, but nevertheless, this means that you will be present at the BMO next year, if you are in the team or not ! :D:D
It will probably be my first balkan MO since 2001 :P this time hopefully as coordinator.

That's the most sadistic thing that I heard this week.