function
trigonometry
real analysis
real analysis unsolved

Hi posters!
Does anybody saw before this problem (or some similar one) before. Any comment, hint or idea is welcome.

Let $f: R^{n}\to R$, $g: R^{n}\to R$ ($n=1,2$ is ok) be smooth periodic functions (as differentiable as you want).

For $m$ natural, find ${a_{1},...,a_{m}}$, each $a_{i}\in R^{n}$ such that \[\sum_{i=1}^{m}f(x-a_{i})==g(x)\] Whith "==" I mean identical, i.e. they are the same function.
Typically $g$ is assumed to be a constant function.

Well, ussualy this is not possible so I am trying to minimize: \[\|\sum_{i=1}^{m}f(x-a_{i})-g(x)\|_{N}.\] :?:  :maybe:

With Fourier expansions $f(x)=\sum_{n\in\mathbb Z}b_{n}e^{inx}$ and $g(x)=\sum_{n\in\mathbb Z}c_{n}e^{inx}$, the equation \[\sum_{j=1}^{m}f(x-a_{j})\equiv g(x)\] becomes \[\sum_{j=1}^{m}\sum_{n\in\mathbb Z}b_{n}e^{in(x-a_{j})}\equiv \sum_{n\in\mathbb Z}c_{n}e^{inx}\] which by uniqueness of Fourier coefficients gives \[\sum_{j=1}^{m}e^{-ina_{j}}=c_{n}/b_{n}, \ \forall n\in\mathbb Z\] This gives a criterion to determine when you have an exact solution: compute $c_{n}$ and $b_{n}$, and then check if $c_{n}/b_{n}$ happens to be a trigonometric polynomial.

I don't know what to do about approximation; you should probably take $N=2$ and read about least squares. Maybe you can approximate $c_{n}/b_{n}$ by a trig polynomial.

Thanks.  :)
This sounds like the logical approach. Let's see what can I found.


Let $ \mathbb R^*$ denote the set of non-zero real numbers. Find all functions $ f: \mathbb R^*\rightarrow \mathbb R^*$ such that
\[ f(x^2\plus{}y)\equal{}(f(x))^2\plus{}\frac{f(xy)}{f(x)}\]
for every pair of non-zero real numbers $ x$ and $y$ with $ x^2\plus{}y\not\equal{} 0$.

[quote="math10"]Let $ R^*$ denote the set of nonzero real number.Find all functions $ f: R*\rightarrow R*$ such that:
 $ f(x^2 \plus{} y) \equal{} (f(x))^2 \plus{} \frac {f(xy)}{f(x)}$ for every pair of nonzero real numbers $ x,y$ with $ x^2 \plus{} y\not \equal{} 0$[/quote]

Let $ P(x,y)$ the assertion $ f(x^2\plus{}y)\equal{}f(x)^2\plus{}\frac{f(xy)}{f(x)}$

1) $ f(1)\equal{}1$ and $ f(\minus{}x)\equal{}\minus{}f(x)$ $ \forall x\neq 0$
Let $ f(1)\equal{}a$
Let $ f(\minus{}1)\equal{}b$

$ P(1,1)$ $ \implies$ $ f(2)\equal{}a^2\plus{}1$
$ P(\minus{}1,1)$ $ \implies$ $ f(2)\equal{}b^2\plus{}1$ and so $ b\equal{}\epsilon a$ with $ \epsilon\in\{\minus{}1,\plus{}1\}$

$ P(1,x)$ $ \implies$ $ f(x\plus{}1)\equal{}a^2\plus{}\frac{f(x)}{a}$
$ P(\minus{}1,x)$ $ \implies$ $ f(x\plus{}1)\equal{}a^2\plus{}\epsilon\frac{f(\minus{}x)}{a}$ and so $ f(\minus{}x)\equal{}\epsilon f(x)$

Then :
$ P(x,1)$ $ \implies$ $ f(x^2\plus{}1)\equal{}f(x)^2\plus{}1$
$ P(1,x^2)$ $ \implies$ $ f(x^2\plus{}1)\equal{}a^2\plus{}\frac{f(x^2)}{a}$ and so $ f(x^2)\equal{}af(x)^2\plus{}a\minus{}a^3$

$ P(x,\minus{}1)$ $ \implies$ $ f(x^2\minus{}1)\equal{}f(x)^2\plus{}\epsilon$
$ P(1,x^2\minus{}1)$ $ \implies$ $ f(x^2)\equal{}a^2\plus{}\frac{f(x^2\minus{}1)}{a}$ and so $ f(x^2)\equal{}\frac{f(x)^2}{a}\plus{}a^2\plus{}\frac{\epsilon}{a}$

Comparing these two values for $ f(x^2)$, we get $ f(x)^2(a\minus{}\frac 1a)\equal{}a^3\plus{}a^2\minus{}a\plus{}\frac{\epsilon}{a}$
So, either $ a\equal{}1$ (and then $ \epsilon\equal{}\minus{}1$), either $ f(x)^2$ is a constant, and so $ f(x^2)$ is a constant and so $ f(x)$ is a constant $ c$ for $ x>0$. But this would imply $ c\equal{}c^2\plus{}1$, which is impossible.
Hence the result.

2) $ f(x)\equal{}x$ $ \forall x\in \mathbb R^*$
$ P(1,x)$ $ \implies$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$ $ \forall x\neq \minus{}1$ And so $ f(x\plus{}n)\equal{}f(x)\plus{}n$ $ \forall n\mathbb Z$, $ \forall x\neq n$
An immediate consequence is $ f(n)\equal{}n$ $ \forall n\in\mathbb Z^*$

From $ f(x^2)\equal{}af(x)^2\plus{}a\minus{}a^3$ above, we get $ f(x^2)\equal{}f(x)^2$

Then $ P(x,n)$ $ \implies$ $ f(x^2\plus{}n)\equal{}f(x)^2\plus{}\frac{f(nx)}{n}$ But $ f(x^2\plus{}n)\equal{}f(x^2)\plus{}n\equal{}f(x)^2\plus{}n$ and so $ f(nx)\equal{}nf(x)$
This give us immediately $ f(x)\equal{}x$ $ \forall x\in\mathbb Q^*$

But $ f(x^2)\equal{}f(x)^2$ shows that $ f(x)>0$ $ \forall x>0$. So $ f(x^2\plus{}y)>f(x^2)$ $ \forall x,y>0$ and so $ f(x)$ is increasing for $ x>0$

So $ f(x)\equal{}x$ $ \forall x>0$ and, since $ f(\minus{}x)\equal{}\minus{}f(x)$, we get $ f(x)\equal{}x$ $ \forall x\in \mathbb R^*$


My math teacher asked us if we though math was invented or discovered.... at firsdt i thought the question was clear and didnt need much arguing cuz i felt math had to be invented,like math did not exist before mathematicians actually started laying down the basics nd fundamentals of math ..  xept then my teacher asked me about fermat's last theorem and "pi" n stuff ... n that got me thinking again into how 1+1=2 cant really be invented, it just does =2 ... ya right now thinking about this i'm like  :huh:

Anyone else have some insight on this?

There is an old long thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45408]here[/url] devoted to this question. You may want to read it :)

So where does the Erdös notion of "the book" with all of the best proofs fit into all of this?

From a philosophical point of view, you might try looking up "Platonism" and the various responses and oppositions to that.

[quote="modeler"]God created the integers - Steven Hawking[/quote]

No i didn't   :roll:

so when u think of it that way.. God created integers(or lets just say math here) ,so God [b]invented[/b] and now we are merely discovering what God invented  ... i.e just saying that math is discovered...

from the thread in that link given by fedja (thanks for that btw  :)  ) it seems to come out that math was and is being invented .. i'm still trying to get all my thoughts together about this

[quote="G0D"] n that got me thinking again into how 1+1=2 cant really be invented, it just does =2 ... ya right now thinking about this i'm like  :huh:  [/quote]

Okay, let's put it this way.  God invented this physical world.  We made observations in this physical world, and created a system with symbols and notation to help study and explain it.  So when you ask about who invented 1+1=2, do you mean the symbols and notation?  Or do you mean that fundamental truth?  I don't say anyone can [b]create[/b] a truth.  So nobody invented 1+1=2: based on our notation, that's a definition, an obvious fact which cannot be invented or proven.

Hmm well we invented all the symbols and definitions and axioms (such as 1+1=2 mentioned above). But then after inventing the foundation we have to discover the rest of math (such as Fermat's Last Theorem; we couldn't just INVENT it and claim it was true, we had to discover it and prove it)

The [i]concept[/i] of numbers was discovered. The actual numbers were invented. For example, if something went wrong our number system could be:

Purple, Blue, Green, Red, Yellow, Purple-Red, etc.

The concept of math is a little bit more tricky. Calculus was supposedly invented by Sir Issac Newton. I really don't know  :P .

The question basially is this: if there is another intelligent civilization, will they also come up with the same mathematics?
It is hard to believe that integers were stumbled upon just by accident, it is so mixed up with our lives now. Then on the development of the number system has been fairly methodical (zero, negatives, rationals, surds all evolved naturally). As for things like topology, they are just tools, so I should call it an invention.
As Kent said, read platonism.

The basics are pretty much a priori true, because if they weren't it would be ridiculous. But most of the derived ideas are really just extensions from useful things which we have defined, and in that sense math is invented. Newton and Leibniz invented the idea of the derivative as $ \lim_{h\to 0}\frac{f(x\plus{}h)\minus{}f(x)}{h}$. There are many consequences of the existence of derivatives which we can study, but ultimately the derivative is purely invented. What about infinity? Infinity doesn't exist, anywhere. There are 10^80 electrons in the universe. I used to think time was continuous and therefore a notion of infinity would be there in describing it, but there is a shortest unit of time, the femtosecond, the length of time it takes for an electron to jump orbitals. So much of our mathematics is contingent on the notion of infinity, and yet it's a made-up construct. If we were only confined to discovering things that were "real" rather than inventing useful mathematics, even if it lies purely in the realm of imagination, we would be up the creek without a paddle.


[quote="iwolf"]What is an eclipse with an eccentricity of 0?

Please post below[/quote]

Chuck Norris.

In Soviet Russia, we call that a circle!

Eccentricity is found by $ \frac{c}{a}$, where $ c \equal{} \sqrt{a^{2} \minus{} b^{2}}$, and $ \frac{c}{a}$ can only be $ 0$ when $ a \equal{} b$, therefore it is a circle.

General form of an ellipse - $ \frac{(x \minus{} h)^{2}}{a^{2}} \plus{} \frac{(y \minus{} k)^{2}}{b^{2}} \equal{} 1$

If $ a \equal{} b$, we can multiply by $ a^{2}$ to get the general form of a circle - $ (x \minus{} h)^{2} \plus{} (y \minus{} k)^{2} \equal{} a^{2}$, where a is the radius.

[quote="SnipedYou"]Eccentricity is found by $ \frac {c}{a}$, where $ c \equal{} \sqrt {a^{2} \minus{} b^{2}}$, and $ \frac {c}{a}$ can only be $ 0$ when $ a \equal{} b$, therefore it is a circle.

General form of an ellipse - $ \frac {(x \minus{} h)^{2}}{a^{2}} \plus{} \frac {(y \minus{} k)^{2}}{b^{2}} \equal{} 1$

If $ a \equal{} b$, we can multiply by $ a^{2}$ to get the general form of a circle - $ (x \minus{} h)^{2} \plus{} (y \minus{} k)^{2} \equal{} a^{2}$, where a is the radius.[/quote]

Could've just explained the meaning of eccentricity...

cameron, this is the mathcounts forum. i doubt there are questions about ellipses on it, but i guess this one is easy.

have they ever had a question like this on a test btw?


Bob got thirsty and drank half the water in his canteen.  Later he drank 1/3 of what remained.  A little bit later he drank 1/4 of what remaind, etc.  Joe with a canteen of the same size, drank half the contents on the first drink, half of what remained on the second drink, etc.  After each took 10 drinks, the water Bob had left was how many times greater than Joe had left.

[hide]Is it 93 1/11 times greater? Bob would have (if you start with 1000) 500; 1000/3; 1000/4; 1000/5.... pattern... 1000/11. He has 1000/11 of the 1000 units of water left.

Joe has 500; 250; 125; 125/2; 125/4; 125/8.... 125/128. He has 125/128 of one unit of water left.

To find how many times more water Bob has then Joe, you take x * 125/128 = 1000/11.  You can solve this by working backwards and having 1000/11 * 128/125. This is 8/11 * 128 = 1024/11, which comes out to 93 1/11. [/hide]

I got 2500/66. Dont take my word for it. I did this in a hurry

New, longer explanation.
[hide]Bob: Start with 2520 units of water (fractions are amounts left)
1.    1/2= 1260
2.    2/3= 840
3.    3/4= 630
4.    4/5= 504
5.    5/6= 420
6.    6/7= 360
7.    7/8= 315
8.    8/9= 280
9.    9/10= 252
10.   10/11= 2520/11

Joe: start with 2520 units of water (each number is amount left after drink)
1. 1260
2. 630
3. 315
4. 315/2
5. 315/4
6. 315/8
7. 315/16
8. 315/32
9. 315/64
10. 315/128

To find how many times greater Bob's amount of water left is than Joe's, take

x * 315/128 = 2520/11, where x is the number of times Joe's water is multiplied.

(2520/11)/(315/128)= x

2520/11 * 128/315= 1024/11

Which is the same as 93 1/11.

Am I right, or if you don't think so, what do you think I did wrong? [/hide]
:)  :)  :)


does $ x^2 \plus{}y^3\equal{}z^4$have prime solutions ,$ x,y,z$

no solutions. do mod chasing using 3 and 4.

[hide="Slightly overkill"]If $ z$ is even, we have $ z \equal{} 2$, so $ x^2 \plus{} y^3 \equal{} 16$. We can easily check this has no solutions in primes. Thus, $ z$ is odd, so exactly one of $ x$ and $ y$ must be even.

If $ y$ were even, we need $ x^2 \plus{} 8 \equal{} z^4$. We have $ 8 \equal{} (z^2 \minus{} x)(z^2 \plus{} x)$. We can easily verify that this yields no solutions.

If $ x$ were even, we would have $ 4 \plus{} y^3 \equal{} z^4$. Write this as $ y^3 \plus{} 2^3 \equal{} (z^2)^2 \plus{} 2^2$. We have $ y^3 \plus{} 2^3 \equal{} (y \plus{} 2)(y^2 \minus{} 2y \plus{} 4)$. We have that $ (y\plus{}2, y^2 \minus{} 2y \plus{} 4) \equal{} (y \plus{} 2, y^2 \minus{} 2y \plus{} 4 \minus{} y^2 \minus{} 2y) \equal{} (y\plus{}2, \minus{}4y \plus{} 4) \equal{} (y\plus{}2, y \minus{} 1)$ since $ 4$ and $ y \plus{} 2$ are relatively prime, since $ y$ is odd. But $ (y\plus{}2, y \minus{} 1) \equal{} (y \minus{} 1, 3) \equal{} \mbox{ 1 or 3}$. But if $ 3 | y \plus{} 2$, then $ 3 | z^4 \plus{} 1$, so $ z^4 \equiv \minus{}1 \mbox{ mod 3}$, a contradiction. Thus, we get that $ y \plus{} 2$ and $ y^2 \minus{} 2y \plus{} 4$ must be relatively prime. But we also have that their product is a sum of squares. Thus, $ y \plus{} 2$ cannot be 3 mod 4; otherwise, it has a prime factor that is 3 mod 4 whose prime factor is odd; since $ y \plus{} 2$ and $ y^2 \minus{} 2y \plus{} 4$ are relatively prime, it follows that $ z^4 \plus{} 4$ cannot be a sum of two squares, a contradiction. Also, $ y$ is odd, so it follows that $ y$ must be 3 mod 4. But then, we have $ y^2 \minus{} 2y \plus{} 4 \equiv 3 \mbox{ mod 4}$, yielding a contradiction as before, so this equation has no solutions. [/hide]


geometry
cyclic quadrilateral
perpendicular bisector
geometry proposed

Consider a cyclic quadrilateral $ABCD$, and let $S$ be the intersection of $AC$ and $BD$.
Let $E$ and $F$ the orthogonal projections of $S$ on $AB$ and $CD$ respectively.
Prove that the perpendicular bisector of segment $EF$ meets the segments $AD$ and $BC$ at their midpoints.

This is problem 3 of the BWM 2003, 2nd round. See http://www.mathlinks.ro/Forum/viewtopic.php?t=16761 .

  darij

Call $H, I, J$ the midpoint of $AD, SA, SD$. We need to show that $EH = FH$. This follows from the fact that the two triangles $EIH$ and $HJF$ are congruent.


Okay so we multiply both sides by $abcd$, then we AM-GM it!

Whats' the need even to multiply!!!!  :lol:  :lol:  :lol:

Re-arrangement inequality also works well.

$\frac ab + \frac bc + \frac cd + \frac da \geq  4\sqrt{1} = 4$

we have $\frac ab + \frac bc \geq 2\sqrt{\frac ac}$ and $\frac cd + \frac da \geq 2\sqrt{\frac ca}$
and$\sqrt{\frac ac}+\sqrt{\frac ca}\geq 2$ :lol:

[quote="t0rajir0u"]In case anyone doesn't get it yet:

$\frac{ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} }{4} \ge \sqrt[4]{ \left( \frac{a}{b} \right) \left( \frac{b}{c} \right) \left( \frac{c}{d} \right) \left( \frac{d}{a} \right) } = 1$

(Really, an MO problem shouldn't have a one-line solution.   :huh: )[/quote]the same of $darkdevil$ solution :?

Lol what an easy problem. Just apply AM-GM inequality and get the answer.

 We get that $((a/b) + (b/c) + (c/d) + (d/a))/4 \ge \sqrt{(a/b)(b/c)(c/d)(d/a)}$
The right hand side of the inequality equals to 1 and by taking the denominator i.e $4$ to the right hand side we get the desired inequality.

seriously???
using t2's lemma gives the solution( just to add variety..)

[quote=Evan-Chen123]Yup...We can also use LM, Muirheads etc..[/quote]
For what?

[quote=i_atmmaths]yes we can surely use but in contests like rmo simple inequaities are preferred :)[/quote]

Yeah surely

Indeed its known lol
By AM-GM:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge  4 \cdot \sqrt[4]{\frac{a \cdot b \cdot c \cdot d}{b \cdot c \cdot d \cdot a}}=4$
And equality holds when $a=b=c=d$

[hide=storage]
We have by amgm

$$\frac{\sum_{cyc} \frac{a}{b}}{4} \geq \sqrt[4]{\prod_{cyc} \frac{a}{b}}$$

$\sqrt[4]{\prod_{cyc} \frac{a}{b}} = 1$ so multiply both sides by 4 and we are done, QED $\square$[/hide]

Lol its AM-GM. Let the expression be $K$
Then,
\[\frac{K}{4}\geq 1\implies K\geq 4\]

[quote=Rushil]If $a,b,c,d$ be any four positive real numbers, then prove that \[ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4. \][/quote]
this problem must be a joke. Everyone knows what to use immediately

[hide=?][quote=wardtnt1234][quote=sqing]Let $a,b,c,d$ be  positive real numbers. Prove that $$ \frac{a}{b} +\frac{c}{d}  +\sqrt{\frac{b}{c} + \frac{d}{a}}\geq 3$$
$$\frac{a}{b} + \frac{b}{c} +\sqrt{ \frac{c}{d} + \frac{d}{a}}\geq 3$$[/quote]
also this is not hard[/quote]
[/hide]

[quote=jasperE3][quote=wardtnt1234][quote=sqing]Let $a,b,c,d$ be  positive real numbers. Prove that $$ \frac{a}{b} +\frac{c}{d}  +\sqrt{\frac{b}{c} + \frac{d}{a}}\geq 3$$
$$\frac{a}{b} + \frac{b}{c} +\sqrt{ \frac{c}{d} + \frac{d}{a}}\geq 3$$[/quote]
also this is not hard[/quote]

Unhelpful.

My solution for the first one:
$\frac ab+\frac cd\ge2\sqrt{\frac{ac}{bd}}$
$\sqrt{\frac bc+\frac da}\ge\sqrt2\sqrt[4]{\frac{bd}{ac}}$
$2\sqrt{\frac{ac}{bd}}+\frac{\sqrt2}2\sqrt[4]{\frac{bd}{ac}}+\frac{\sqrt2}2\sqrt[4]{\frac{bd}{ac}}\ge3$[/quote]

same solution

[hide=Very nice. Thanks.][quote=jasperE3][quote=wardtnt1234][quote=sqing]Let $a,b,c,d$ be  positive real numbers. Prove that $$ \frac{a}{b} +\frac{c}{d}  +\sqrt{\frac{b}{c} + \frac{d}{a}}\geq 3$$
$$\frac{a}{b} + \frac{b}{c} +\sqrt{ \frac{c}{d} + \frac{d}{a}}\geq 3$$[/quote]
also this is not hard[/quote]

Unhelpful.

My solution for the first one:
$\frac ab+\frac cd\ge2\sqrt{\frac{ac}{bd}}$
$\sqrt{\frac bc+\frac da}\ge\sqrt2\sqrt[4]{\frac{bd}{ac}}$
$2\sqrt{\frac{ac}{bd}}+\frac{\sqrt2}2\sqrt[4]{\frac{bd}{ac}}+\frac{\sqrt2}2\sqrt[4]{\frac{bd}{ac}}\ge3$[/quote]

[/hide]
$$\frac{a}{b} + \frac{b}{c} +\sqrt{ \frac{c}{d} + \frac{d}{a}}\geq 2\sqrt{\frac{a}{c}}+\sqrt2\sqrt[4]{\frac{c}{a}}
$$$$=2\sqrt{\frac{a}{c}}+\frac{\sqrt2}2\sqrt[4]{\frac{c}{a}}+\frac{\sqrt2}2\sqrt[4]{\frac{c}{a}}\ge3$$

I cannot believe people actually take the effort to write solutions to this. [hide = But to ensure my post isn't pointless, I will give a solution]
We use AM-GM to win. Done.
[/hide]

Let $a,b,c,d$ be positive real numbers. Prove that$$\frac{ a}{b} +\sqrt{ \frac{b}{c} +\frac{c}{d} + \frac{d}{a}}>\frac{12}{5} $$$$\sqrt{\frac{ a}{b}} +\sqrt{ \frac{b}{c} +\frac{c}{d} + \frac{d}{a}}>\frac{5}{2} $$

Let $a,b,c,d$ be  positive real numbers. Prove that $$ \frac{a}{b} +\frac{c}{d}  +\sqrt{\frac{b}{c} + \frac{d}{a}}\geq 3$$
[b]Equality holds when[/b] $b=d=2a=2c.$
$$\frac{a}{b} + \frac{b}{c} +\sqrt{ \frac{c}{d} + \frac{d}{a}}\geq 3$$
[b]Equality holds when[/b] $c=2b=2d=4a.$

[hide]Applying the AM-GM inequality, $$\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}}{4} \ge (\frac{a}{b} * \frac{b}{c} * \frac{c}{d} * \frac{d}{a})^{1/4}$$  or $$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \ge 4$$ Q. E. D.[/hide]

[quote=sqing]Let $a,b,c,d$ be positive real numbers. Prove that$$\frac{a}{b} +\sqrt{ \frac{b}{c} +\frac{c}{d} + \frac{d}{a}}>\frac{12}{5} $$[/quote]

$$\frac ab +\sqrt{ \frac bc +\frac cd + \frac da}\geq\frac ab+\sqrt3\sqrt[4]{\frac ba}=\frac ab+\frac{\sqrt3}4\sqrt[4]{\frac ba}+\frac{\sqrt3}4\sqrt[4]{\frac ba}+\frac{\sqrt3}4\sqrt[4]{\frac ba}+\frac{\sqrt3}4\sqrt[4]{\frac ba}$$$$\geq5\sqrt[5]{\left(\frac{\sqrt3}4\right)^4}>\frac{12}5$$

$\square$.

[quote=sqing]Let $a,b,c,d$ be positive real numbers. Prove that$$\sqrt{\frac{ a}{b}} +\sqrt{ \frac{b}{c} +\frac{c}{d} + \frac{d}{a}}>\frac{5}{2} $$[/quote]

$$\sqrt{\frac ab} +\sqrt{ \frac bc +\frac cd + \frac da}\geq\sqrt{\frac ab}+\sqrt3\sqrt[4]{\frac ba}=\sqrt{\frac ab}+\frac{\sqrt3}2\sqrt[4]{\frac ba}+\frac{\sqrt3}2\sqrt[4]{\frac ba}$$$$\geq3\sqrt[3]{\left(\frac{\sqrt3}2\right)^2}>\frac52$$

$\square$.

Thanks sqing :)


function
abstract algebra
group theory
superior algebra
superior algebra unsolved

1)
let $H,K\leq G$ and H,K,G finite.
show that
$[G: H\cap K]\leq[G: H][G: K]$
2)
if $H,K,L\leq G$ and H,K,G,L finite.
so that $H\subseteq K$ then
$[K: H]\leq[L\cap K][L\cap H]$

1) Consider the obvious "diagonal" map $G/(H \cap K) \to G/H \times G/K$ and show that it is an monomorphism.

ok... let us consider the map you say :)
in my book they say that i have to apply an isomorphism theorem followed
by an application of the lagrange theorem.

considering $f(x)=0,\forall x \in (H \cap K)$ *
and $Im (f)=G/H \times G/K=\left\{(xH,yK)|\forall x,y \in G\right\}$
so the function $f(x)=(xH,xK)$ would satisfy our needs.
we see that indeed * is satisfied because now $Ker(f)=H\cap K$ but i think not all
$Im(f)$ is covered(how can this be solved ?)

now if f is monomorphism as you say we can apply first theorem of isomorphism
for groups and say that $G/kerf \cong imf$
thank you for the idea :)

It seems to me that you consider the map
\[ f : G \to G/H \times G/K , x \mapsto (xH,xK)  \]
That's right, it's obviously a homomorphism with kernel $H \cap K$ (but neither a monomorphism nor an epimorphism! This is another map, which is constructed next). So you get an isomorphism
\[ G/(H \cap K) \cong Im (f)  \]
Now, the image is a subgroup ... I think you can fill in the rest.

should f have been a monomorphism ? is f a good function for our purpose?
isnt $im(f)=G/H \times G/K$?

i'm not sure i can handle the rest.
but i'll write what i can figgure out by myself

i think $[G: (H\cap K)]=|G/(H\cap K)|$ and $|G/H \times G/K|=|G/H|\cdot |G/K|=[G: H]\cdot [G: K]$
now if f is homomorphism(isomorphism) then  $[G: (H\cap K)]=[G: H]\cdot [G: K]$
is this right ?

As I already said, f is neither mono nor epi. The first isomorphism theorem can be applied to every homomorphism.

if f is not mono nor epi then must it be homo ?
why is it important that f is not mono or epi ?

[quote="spx2"]if f is not mono nor epi then must it be homo ?[/quote]
I would not say that. Please check the definitions for homomorphism, monomorphism and epimorphism.
[quote="spx2"]why is it important that f is not mono or epi ?[/quote]
On the one hand, it is not correct! You've already seen that the kernel is $K \cap H$, which is not always $0$, and the images of $f$ obviously don't cover $G/H \times G/K$ in general. On the other hand, it is a unnecessary argument. You just have to know the kernel of $f$, the isomorphism theorem implies directly the assertion.


find all prime p,q and even n>2 such that
\[ p^{n}+p^{n-1}+....+1 = q^{2}+q+1 \]

This topic is similary for http://www.mathlinks.ro/Forum/viewtopic.php?t=149570 :D . But maybe your problem is solved , i hope  :)


function
algebra
functional equation
algebra open

If $f(x)=\frac{ax+b}{cx+d}, g(x)=\frac{px+q}{rx+t}$ then
$f(g(x))=\frac{(ap+br)x+aq+bt}{(cp+dr)x+cq+dt}$.
Therefore you can find square root of matrice.

[quote="lordWings"]I think I read some years ago that the functional equation
$f(f(x))=g(x)$,
where $g(x)$ is a known differentiable bijective equation,
has only one differentiable bijective solution $f(x)$ (the book called it the "functional square root" of $g(x)$),
and infinite many continuous but non-differentiable bijective solutions.
Do I remember correctly? Has anyone any information about this?[/quote]
It is not true. For example $f(f(x))=x$ had solution $f(x)=x$ and $f(x)=-x$.

[quote="Rust"]It is not true.[/quote]
You're right. I can't remember the exact statement. I hope I could remember which book I was reading.
I've been googling "functional square root" for an hour and still can't find a good source (at least one that I can understand).

[quote="mathisfun1"]The original question asks to find a function $f: \mathbb R^{+}\rightarrow \mathbb R^{+}$ that satisfies the equation
$f(f(x))=\frac{3x+1}{x+3}$.

By letting $f(x)=\frac{ax+b}{cx+d}$ and substituting into the original equation, we get
$f(x)=\frac{(2+\sqrt{2})x+2-\sqrt{2}}{(2-\sqrt{2})x+2+\sqrt{2}}$.

Can we say anything about the uniqueness of the solution?[/quote]

On R+, $g(x) = (3x+1)/(x+3)$  is increasing and contracting (toward 1).  Thus you can define $f(x)$ in some arbitrary (but increasing and solving the equation) manner on one interval $(a,g(a))$ to the left of 1, and one interval $(g(b),b)$  to the right of 1, and determine the other values uniquely and consistently from that.


geometry
3D geometry
algebra
factorization
sum of cubes

I didn't think this was olympic material, so...

Find $ \sum_{n\equal{}1}^{2008}\minus{}3n^{3}\plus{}6030n^{2}\minus{}6028n\plus{}2009$.

[quote="Differ"]I didn't think this was olympic material, so...

Find $ \sum_{n \equal{} 1}^{2008}\minus{}3n^{3}\plus{}6030n^{2}\minus{}6028n\plus{}2009$.[/quote]

Its stairtforward. You need to know:
1)$ \sum_{k\equal{}1}^{n}k \equal{}\frac{k(k\minus{}1)}{2}$

2)$ \sum_{k\equal{}1}^{n}k^{2}\equal{}\frac{k(k\plus{}1)(2k\plus{}1)}{6}$.

3)$ \sum_{k\equal{}1}^{n}k^{3}\equal{}\frac{k^{2}(k\plus{}1)^{2}}{4}$.

Now write the summation as,
$ \minus{}3\sum_{n\equal{}1}^{2008}n^{3}\plus{}6030\sum_{n\equal{}1}^{2008}n^{2}\minus{}6028\sum_{n\equal{}1}^{2008}n\plus{}\sum_{n\equal{}1}^{2008}2009$.
And use formulas above.

[hide="Just be careful"]
$ \sum_{n\equal{}1}^{2008}2009 \equal{} 2009\plus{}2009\plus{}...\plus{}2009 \equal{} 2009(2008)$
[/hide]

While I have nothing against the previous solution, my guess is that the problem was intended to be factored or something, so here goes:

[hide="quasi-telescoping-ocity!"]
\begin{eqnarray*}\sum & = &\sum_{n=1}^{2008}(2009-n)(3n^{2}-3n+1)\\ & = &\sum_{n=1}^{2008}(2009-n)(n^{3}-(n-1)^{3})\\ & = & 2008(1^{3}-0^{3})+2007(2^{3}-1^{3})+\cdots+1(2008^{3}-2007^{3})\\ & = & 1^{3}(2008-2007)+2^{3}(2007-2006)+\cdots+2008^{3}(1)\\ & = &\sum_{n=1}^{2008}n^{3}\\ & = & (1004\cdot 2009)^{2},\end{eqnarray*}
by the sum of cubes formula.

Dunno if it's faster than applying all the sum of powers formulas to the original thingy, but it's a way to check answers.

[/hide]


function
number theory unsolved
number theory

Hi!
I have been reading about (Siegel) E-functions and its connection to transcendental number theory. E functions are entire functions which Taylor series have coefficients in some number field and they satisfy a certain growth condition (see for example Wikipedia for exact definition). My problem is the claim: If an E-function is algebraic (that is satisfies an algebraic equation) then it must be a polynomial. At the moment I have not figured out why is it so. Does someone have any idea? Thanks for answers!!

-Michel-

If function f(z) is entire and algebraic then $f(z)\in\mathbb{C}[z]$


Could anyone please check my reasoning for these questions?

1. Predict the overall effects on the amount of C present at equilibrium and the rate of the forward reaction when product is removed from the following equilibrium system: $A+B \rightleftharpoons C$

[hide]The amount of C decreases in the system compared to the original amount at equilibrium. I'm not sure about the rate though.[/hide]

2. A sample of $MgCl_{2}$ is placed in the flame of a Bunsen burner. The component of the substance which is responsible for the production of a flame colour is $Mg^{2-}$. True or False

[hide]False. It is the electrons returning from an excited state to a ground state in the magnesium ion that cause the colour.[/hide]

3. GLC (Gas-Liquid Chromatography) is restricted to compounds that can be easily vapourised, ie. substances that possess molecular weights below 300. Is this statement true or false.

[hide]Not sure.[/hide]

4. The greater the percentage ionisation of an acid in water, the stronger the acid. Is this statement correct or incorrect.

[hide]I think its false but I'm not sure why  :blush:[/hide]

My opinion:
1. At the new equilibrium, the amounts of $A,B$ are less than the original equilibrium. So from $R=K[A]^{m}[B]^{n}$, we have that $R$ decreases. (Unless $m=n=0$?  :maybe: )

4. The strength of an acid depends on its $K_{a}$. But ionization % $= 100\sqrt{\frac{K_{a}}{N}}$, so it depends not only on $K_{a}$.

Hmm we were never taught these formulae at school =(. I was wondering what the $N$ is in the second equation. And also for number 2, I just remembered that magnesium ions don't cause any visible colour... so I'm not sure about that too either now...  :(

You meant $Mg^{2+}$ right?

@ quizzie: You are quite right, $Mg^{2+}$ does not give out a flame of visible colour, the emission spectrum lies outside of the visible spectrum.

4. If we compare solutions of different acids but with the same concentration C (and at the same temperature), then the greater the $\alpha$ (the percentage ionization) then greater will be the $K_{a}$ of the acid and so stronger will be the acid. This because, considering a monoprotic acid HA,

$HA+H_{2}O \rightleftharpoons A^{-}+H_{3}O^{+}$

and, by definition, $\alpha = \frac{[A^{-}]}{C}\times 100$, then

$K_{a}= \frac{\alpha^{2}C}{1-\alpha}$

(to simplify, I have not used the 100 factor in the definition of $\alpha$), which is, for equal concentrations C of acids, a crescent function in $\alpha$.

Well, my $N=$ concentration. So if the concentration is equal, then it depends only on $K_{a}$

For question number 3, instead of "substances that possess molecular weights below 300" we should have "substances that possess boiling points below 300ºC".

Thank you for your responses.

Our textbooks and study guides seem to have "molecular weights below 300" in the explanations, but I think you are correct because it says boiling point in wikipedia too.

So with $I=\sqrt{\frac{K_{a}}{C}}$, does that mean a smaller concentration of the same acid will result in a higher ionization percentage? How does that work though...

Edit: I think what I'm still unclear about is the difference between $K_{a}$, percentage ionization and the strength of an acid (which I thought was a measure of how readily an acid disassociates in water).


it's very easy!
$ \sqrt{a}\plus{}\sqrt{b}\geq \sqrt{a\plus{}b}$
$ \Leftrightarrow 2\sqrt{ab}\geq 0$ wich is obiviously true.

[quote="vazk"]prove:
\[ \begin{gathered} \sqrt {a \plus{} b} \leqslant \sqrt a \plus{} \sqrt b \hfill \\
\forall a,b \in \mathbb{R} \hfill \\
\end{gathered}
\]
Noob here.[/quote]
$ \sqrt a \plus{} \sqrt b \minus{}  \sqrt{a\plus{}b} \equal{} \frac{(\sqrt a \plus{} \sqrt b \minus{}  \sqrt{a\plus{}b})(\sqrt a \plus{} \sqrt b \plus{}  \sqrt{a\plus{}b})}{\sqrt a \plus{} \sqrt b \plus{}  \sqrt{a\plus{}b}} \equal{}$ $ \frac{(\sqrt a \plus{} \sqrt b )^2 \minus{} (\sqrt{a \plus{} b})^2}{\sqrt a \plus{} \sqrt b \plus{}  \sqrt{a\plus{}b}} \equal{} \frac{2\sqrt{ab}}{\sqrt a \plus{} \sqrt b \plus{}  \sqrt{a\plus{}b}} \ge 0$. And thus $ \sqrt a \plus{} \sqrt b \ge \sqrt{a\plus{}b}$.

[quote="vazk"]prove:
\[ \begin{gathered} \sqrt {a \plus{} b} \leqslant \sqrt a \plus{} \sqrt b \hfill \\
\forall a,b \in \mathbb{R} \hfill \\
\end{gathered}
\]
Noob here.[/quote]
It's wrong. Try $ a\equal{}\minus{}1.$

inequality si true  but  not for all a and b.  its only  for postive  reals  a and b.

The condition is true for $ a, b\ge 0$.......
just as babyloverain said.

hello, we have $ \sqrt{a\plus{}b}\le \sqrt{a}\plus{}\sqrt{b}$ $ \forall a,b \in \mathbb{R}$ with $ a,b\geq0$
Sonnhard.


calculus
integration
calculus computations

can someone please Derive L{t sin at}=2as/(s^2+a^2)^2 by differentiating
each side of the interegal

between o to infinity on e^-st*sin at dt=a/(s^2+a^2)

I think that will be easier if you try to solve LHS integral to RHS... :)


A fuller array of greek characters would be nice, especially characters like delta (to denote a triangle) and capital pi (to denote products)

I was reading on Mathlinks that Valentin is setting up LaTeX to work with phpbb somehow. I don't know the details, and I believe it mentioned storing graphics on the server somehow (automatically generated?)... if you're interested, though, Valentin or somebody over there would probably provide more details. It has the potential to be really cool.

The SoSMath has LaTeX implemented - it's somewhere on our ToDo list, but admittedly pretty far down, unless someone can point us to a resource that shows us how to do it pretty quickly.

As an unfortunate non-Latex user, might I suggest that if you do implement it, that there be some help file with explanations for how to do things?  It took me long enough to learn the html for sub and superscripts.

If LaTeX were implemented on this forum, I would be happy to write a basic help file on how to do the most useful things.

i think this would be good... it'd push us to all learn the code, which is something i've wanted to do but never taken the time.

If you are using linux w/apache, i believe you should be able to configure apache to run tex or latex as you would configure php for cgi.  Or, if you prefer, just have php save the latex commands into a text file, run latex using backticks, then bring the output (either dvi or use dvips to convert to ps) into the html.

Can you direct me to a resource regarding how to integrate this with a phpbb board?

well ... without too much pain, there is this approach:
go here and get the tex2im shell script: [url]http://www.nought.de/tex2im.php[/url]

Assuming you have latex installed and in the path, as well as convert (part of Image Magic), you should be able to run tex2im in the command-line using a tex file (just type stuff like \frac{2}{2}, no \begin or \end commands)

Then, in order to integrate into phpbb, in the submit / posting scripts (or in the preview script), add a [LaTeX] set like the url and b set.  I haven't worked with phpbb before, but i am guessing there is some script which detects the commands and does its own processing.  Wherever that is, insert code for the [LaTeX] tag so that it will run tex2im (using backticks `tex2im test.tex` works for php).  It would be recommended to extract the text into a tex file, run tex2im, then replace with a simple img tag that points to the png produced by tex2im.

Let there be light:

[tex]\begin{eqnarray}

\epsilon_{0}\oint{\bf E} \cdot{d}{\bf S} &=&q\nonumber\\

\oint{\bf B} \cdot{d}{\bf S} &=&0\nonumber\\

\oint{\bf B} \cdot{d}{\bf l} &=&\mu_{0}

\left(\epsilon_{0}\frac {d\Phi_{E}}{dt} + i \right)\\

\oint{\bf E} \cdot{d}{\bf l} &=&-\frac{d\Phi_{B}}{dt}\nonumber

\end{eqnarray}[/tex]

[tex]
\sum_{n=1}^{\infty} 1/n^2 = \pi^2 / 6.

\TeX \mbox{ and} \LaTeX!
[/tex]
Thank you!

[quote="tetrahedr0n"]Can someone explain Mr. Rusczyk's equations?
[/quote]

They are Maxwell's Equations (in integral form) - the equations that govern the behaviour of Electromagnetic Fields (and hence light). In high school you learn them separately, and in a simplified form, as Gauss's Law for electricity, Gauss's Law for magnetism, Faraday's Law/Lenz's Law, and Ampere's Law.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq2.html


calculus
integration
function
real analysis
real analysis unsolved

$ \int_{\minus{}\infty}^{\plus{}\infty} \exp( a x \minus{} a b \exp(x) ) dx$
$ 0<a<1 ,0<b<1.$

A book says the answer is   Gamma(a)/(ab)^a.

Gamma is Gamma function.
why ?

Thanks in advance!

you have $ \int_{\minus{}\infty}^{\infty}\;e^{ax}\;e^{\minus{}abe^x}\;dx$

use the substitution

$ y\equal{}ab\,e^x\qquad \Rightarrow \qquad dy\equal{}ab\,e^x\;dx\equal{}y\;dx$

to arrive at

$ \int_0^\infty \;\left(\frac{y}{ab}\right)^a\;e^{\minus{}y}\;\frac{dy}{y}\;\;\;\equal{}\;\;\;\left(\frac{1}{ab}\right)^a\int_0^\infty \;y^{a\minus{}1}\;e^{\minus{}y}\;dy$

do you know the definition of the gamma function ?  :maybe: it says

$ \Gamma(z) \equal{} \int_0^\infty  y^{z\minus{}1} e^{\minus{}y}\,dy\qquad (\text{for}\quad \mathbb{R}e\;z\;>\;0)$

hence, your integral equals $ \boxed{\left(\frac{1}{ab}\right)^a\;\Gamma(a)}$ as you have $ 0\;<\;a\;<\;1$


For all those 6th and 7th graders (and 5th graders too) who are going to join Mathcounts next year, what do you want to do next year in Mathcounts? What level do you want to get to? What scores and ranks do you want at each level? Your goals should be higher than what you had this year, hopefully  :)

My goal: Get to state competition level, 35 at chapters and 1st or second in countdown round.
25 at states
No nationals in sight. my team will kill my scores anyway.

My goal next year is just to get to chapter.  Must try to study... :(

(Jim, don't mock me.)

my goal is to get to nationals because I already got to state this year

That was my goal for a while, too, but after I saw the questions I figured I wouldn't be too disappointed if I only made it to AIME again.


what do you think about illegals in the U.S. let them stay? amnesty? try to send them back?

It would be great if the U.S. government could bring some order into the situation of illegal immigrants.
Or at least of the legal immigrants that apply for an H-1B visa or GC.
Or at least of its own citizens applying for a passport.
Oh well.

Interesting discussion, I'd rather not see it bleed to death.
So in order to keep it going :

- Just how many illegal immigrants are we talking about? (estimates?)
- What are their nationalities (Mexican, Colombian, Brazilian,Cuban,...)
- How well integrated into society are most of them ? (do they speak the language, do they have a job)
- What kind of education do their children get?
-...

Since I don't live there, it's pretty hard for me to get a clear view.  Can anyone provide some insight? :maybe:

[quote="JBL"]There are an estimated 12 million illegal immigrants in the United States.  I assume that most are Mexican, Central & South American, or Caribbean.  Public education in the United States is available to everyone.  Probably many first-generation immigrants (legal and illegal) speak not-so-much English; I don't know if there's a difference between legal and illegal immigrants (controling for the obvious factors).  The children of immigrants generally are bilingual (i.e., there are no prolonged difficulties in integration).

For my money, the biggest problem with illegal immigration is that illegals are (reasonably) afraid of approaching authorities (government or otherwise) for risk of being deported.  Thus, they make excellent targets for crime, can be easily taken advantage of by employers, do not generally get adequate preventative medical care (although this is a problem for the much larger group of uninsured Americans), etc.[/quote]
How many of those 12 million are actually "known" by the authorities? I mean, how many of them are officially not even there?
I can imagine that's pretty much impossible, you can't even send your kids to school?

[quote="sapphyre571"]"Jose" is now the most popular name for baby boys in Texas[/quote]
If you'd rather not give your child a name that was brought into this country by immigrants, please refer to [url=http://www.cedarseed.com/air/nativenames.html]this list[/url].

But one of the factors driving the popularity of "Jose" as a name in the United States is that among common Spanish names, it's one of the easiest for [i]English[/i] speakers to recognize and pronounce. Other names are more subject to being mangled. I've known a "Luis" or two who tired of hearing his name pronounced like "Louise" and told everyone to call him "Lewis." Plenty of Albertos and Robertos drop the "o" off of the name. Second or third generation immigrants from Latin America are just as "typically American" in culture, for better or worse, as were second or third generation immigrants from Italy or Poland 60 years ago.

As a partial answer to fredbel6, most children of undocumented immigrants do attend public schools in the U.S. - they only need proof of actual residency, which can be established by such items as rent receipts or utility bills, to get into school. Some of these children were born in the U.S. and are thus themselves U.S. citizens. However, they then face the unevenness of the U.S.educational system, and are disproportionately grouped into the worst schools. And the neighborhoods in which they disproportionately live have some of the biggest problems with violence and gangs. Some of the gangs are transnational, spanning, for instance, Southern California and El Salvador.

School signs posted in two languages? Small stuff. Every such sign around here is in at least three languages - typically English, Spanish, and Khmer, sometimes with Vietnamese or Tagalog as a fourth or fifth language.

[quote="sapphyre571"]I'm from Texas, so I've seen how bad the immigration problem has gotten. Educators are no longer forced to teach in English until 4th grade, so the native Spanish-speakers have more time to adjust to the language. Public school signs now are Spanish on the front and English on the back. "Jose" is now the most popular name for baby boys in Texas [url=http://www.ssa.gov/cgi-bin/namesbystate.cgi][see here][/url]. I'm sorry, but I though this was an English-speaking country. I'm all for a wall now, because I'm tired of people coming to my country who can't communicate in our national language.[/quote]
Where I come from, activists do defacing campaigns to protect their language. (removing bilingual signs) However, my language is not the strongest language on earth.
I would like to inform you that California, Texas,... were once part of Mexico, in a sense, you're the one who's not willing to adapt.  Or perhaps you should learn a native indian language. :!:
And I think the United States don't really have an official language, it's just "de facto English".

[quote]As a partial answer to fredbel6, most children of undocumented immigrants do attend public schools in the U.S. - they only need proof of actual residency, which can be established by such items as rent receipts or utility bills, to get into school. [/quote]
So the authorities don't really have a system to track illegal undocumented parents, using the children's information?

[quote]Some of these children were born in the U.S. and are thus themselves U.S. citizens.[/quote]
Is that all it takes? Being born in the USA? :maybe:

[quote="fredbel6"]I would like to inform you that California, Texas,... were once part of Mexico, in a sense, you're the one who's not willing to adapt. Or perhaps you should learn a native indian language.
And I think the United States don't really have an official language, it's just "de facto English"[/quote]
Good point.  Also,  the US doesn't have an official language, but individual states have declared official languages.
[quote="fredbel6"]So the authorities don't really have a system to track illegal undocumented parents, using the children's information?[/quote]
I guess they could, but right now I don't think there's a system in place.
[quote="fredbel6"]Is that all it takes? Being born in the USA?[/quote] To have citizenship, you can be born in the US, have a spouse who is a citizen, have a parent or guardian who is a citizen (only works for minors), or be naturalized through a process that takes several years. There may be others, but these are all I know.[/quote]

[quote="vamathletes"][quote="fredbel6"]I would like to inform you that California, Texas,... were once part of Mexico, in a sense, you're the one who's not willing to adapt. Or perhaps you should learn a native indian language.
And I think the United States don't really have an official language, it's just "de facto English"[/quote]
Good point.  Also,  the US doesn't have an official language, but individual states have declared official languages.
[/quote]
Thanks, but apparently Texas doesn't have an official language either?

[quote][quote="fredbel6"]Is that all it takes? Being born in the USA?[/quote] To have citizenship, you can be born in the US, have a spouse who is a citizen, have a parent or guardian who is a citizen (only works for minors), or be naturalized through a process that takes several years. There may be others, but these are all I know.[/quote][/quote]
I find that amazingly mild. :ninja:  This is not the case in many European countries.  :|  In West-European countries, immigration from muslim countries like Morocco, Turkey, Algeria, Tunisia ... is the main issue, and since it is part of the culture of these immigrants, they often have way more children than the average "local" family.

[quote]Educators are no longer forced to teach in English until 4th grade, so the native Spanish-speakers have more time to adjust to the language. [/quote]
Where I come from, where language is a sensitive political issue, everything is neatly written down in the constitution : policemen have to speak these languages in these parts of the country, government funded education must and can only be provided in these regions, road signs must and can only be in these languages here and there,...
However, one can still get education in a language that is "wrong" in that region, but then it won't be funded.

I think France has a similar system : there are some schools where kids are also taught in Breton, but they are not funded (and I think they have to do some final exam eventually to make it official, which probably has to be done in French)

So how many children go to schools where they get lessons in Spanish? :maybe:  In which states? (Texas, New Mexico,...)
And how "mixed" are these schools? Does the average hispanic go to school with kids who only speak English?
I can imagine parents wouldn't send their monolingual anglophone child to a school where spanishspeaking children from immgirants are initially taught in Spanish?

[quote]I can imagine parents wouldn't send their monolingual anglophone child to a school where spanishspeaking children from immgirants are initially taught in Spanish[/quote]
They don't - and that's one factor in the school-quality disparities I spoke of earlier, the isolation of some schools from supportive and politically well-connected parents. But many schools are very large, and what happens in some classrooms is not what happens elsewhere in the same school.

Language of instruction is a hot-button political issue. In California, a public referendum a few years ago outlawed the practice known as "bilingual education" in favor of "English immersion." The referendum had support from at least some Hispanic voters, who felt that children were being kept in the bilingual programs for too long and thus being kept away from the educational benefits present in English-speaking rooms. I should add that the exact definitions of the words used here may not be obvious to the casual observer, and the definitions themselves may be subject to further controversy.

Note also that a school could very well be making good progress in teaching all children English, and yet at the same time, printing up all of its notices and communications in two, three, or four languages - because many of those notices are intended not for the students but for their parents.

I find it interesting to compare the problems with immigration in the USA with ours.  Here, not only the language is a problem, but also the religion : most people are moderate christians or simply atheist, while many immigrants are muslims.
About the language, a common problem is that parents, even if they speak the local language (Dutch in my case), they don't speak it with their children.  They just drop them off, as if they say "isn't learning them the language what school is for?"
The differences are huge : these immigrants are disproportionally absent in the toughest forms of secondary education and higher education (in fact, I have never had an instructor at the university, or a teacher, who was an immigrant).

[quote]
Note also that a school could very well be making good progress in teaching all children English, and yet at the same time, printing up all of its notices and communications in two, three, or four languages - because many of those notices are intended not for the students but for their parents.[/quote]
What you say is true, here we have the problem that the teachers are simply not able to do that.  French can serve as a bridge sometimes, but practically nobody here speaks Arabic or Turkish.   So this often means that the parents are simply not involved in what their kid is doing (when they go to school to discuss the progress of their child, they sometimes bring an older brother or sister with them to serve as a translator).

If I may go into another uber-capitalist rant:

I wouldn't care about illegal immigration at all if the tax system wasn't structured so in a few years I will have to pay for their education and other things. When you start paying people to be poor, you end up with an awful lot of poor people.

That doesn't make any sense: the children of illegal immigrants are (for the most part) citizens.  You're paying for their education in the same way that all taxpayers everywhere pay for the education of all children in public schools.

More generally, the problems caused by illegal immigrants not being hooked into the system (e.g. the fact that they don't pay taxes) are probably best solved by letting them hook themselves into the system without threatening to deport them (i.e. legalizing them).

Okay. I'm a LEGAL immigrant. As my parents went through tons of work to bring me to the US, I think it is completely WRONG that Mexcians are running over the border. For many reasons:

-Mexican constitution says that you get thrown in the slammer if you run over their border
-These illegals are in this for personal gain: they don't pay taxes or serve the United States in any way.
-Legal immigrants work very hard to become Americans. They westernize themselves, learn how to speak English, pay taxes, and serve the country. To see people who just ran over the border to the nearest cheap labor place get the same treatment as the hard-working legals is horrible.

And I'm sick of those Democratic Communists saying we should fell sorry for them. WE SHOULDN'T! THEY ARE COMING INTO THIS COUNTRY DOING NO WORK THAT NATIVES OR LEGAL IMMIGRANTS DO, SO THEY DON'T DESERVE THE SAME TREATMENT!

[color=red]The USA should be a country of natives and legal immigrants; people who WANT to be Americans. [/color]

I say that we send back as many as possible, and bulk up our border patrol.

[quote="SoxFan34"][color=red]The USA should be a country of natives and legal immigrants; people who WANT to be Americans. [/color][/quote]
How do you know the illegals don't want to be Americans? Maybe they are too poor to afford the ~900 USD application fee? Maybe they need a job now in order to feed their families, so they can't wait for their application to be processed? Make it easier for them to become citizens, and most of them will!

What was that about illegal immigrants [url=http://www.nytimes.com/2007/04/16/nyregion/16immig.html?ex=1334376000&en=9b21a57a590afb53&ei=5090&partner=rssuserland]not paying taxes[/url]? This news link wasn't hard to find, and there are plenty of others with a similar message.

My position? I think it should be significantly easier to become a legal resident or citizen.

[quote="JBL"]That doesn't make any sense: the children of illegal immigrants are (for the most part) citizens.  You're paying for their education in the same way that all taxpayers everywhere pay for the education of all children in public schools.

[/quote]
Practicality issues aside, would it not be a great world if all had complete control over our assets and were able to live lives without needing to tax (forcibly remove money from) others to do so?

So yeah, this is going to go the way the other thread went, but I'll try not to be as loony this time, honest!

[quote="PenguinIntegral"]Practicality issues aside, would it not be a great world if all had complete control over our assets and were able to live lives without needing to tax (forcibly remove money from) others to do so?[/quote]
Cynical though it may sound, people are too stupid* for this to work. A lot of the things that the government spends its money on basically involve keeping people from doing particularly stupid stuff. The only way you can have an organization to do this is to tax people. Maybe you could revamp it so that private companies provide that kind of protection, but you'll still have to pay them and I don't see any reason to expect this to be a better system.

*This is perhaps too crude a word. I mean to say uneducated, willfully ignorant, lacking foresight, and a variety of other problems that result in people making poor decisions.

[quote="jmerry"]My position? I think it should be significantly easier to become a legal resident or citizen.[/quote]

Only for foreign-born PhD's.  Why would we want to make it easier for more illiterate Mexicans, er, illegal immigrants to get in and/or stay?  And their children, who are guaranteed to be undereducated as well?  What the U.S. government should do is like, put up this gigantic electric fence across the border with moats all around full of alligators and land mines and boiling acid.  They should also just tell the foreign-born PhD's, "Please stay.  We want you here.  We'll even buy you a house."

[quote="Xevarion"][quote="PenguinIntegral"]Practicality issues aside, would it not be a great world if all had complete control over our assets and were able to live lives without needing to tax (forcibly remove money from) others to do so?[/quote]
Cynical though it may sound, people are too stupid* for this to work. A lot of the things that the government spends its money on basically involve keeping people from doing particularly stupid stuff. The only way you can have an organization to do this is to tax people. Maybe you could revamp it so that private companies provide that kind of protection, but you'll still have to pay them and I don't see any reason to expect this to be a better system.
[/quote]
While I'm not an economist, I'd think it's safe to say that both extremes don't work.  The world has seen the poverty communism can inflict, but we also know that absolute economic freedom can create poverty.
And yes, some things need to be paid for by the community? Who's gonna police the streets? Will each family manage its own foreign affairs? :ninja:

Ok, I can deal with practicality arguments later, but I need to make sure that such a situation is desirable.

PiggyPi: Why? To incorporate such blatant xenophobia into our law system is a mistake. Not just an "I majored in Philosophy" mistake, that's an "invest my life savings in Enron" mistake.

[quote="Piggypi314"]Only for foreign-born PhD's.  Why would we want to make it easier for more illiterate Mexicans, er, illegal immigrants to get in and/or stay?  And their children, who are guaranteed to be undereducated as well?  What the U.S. government should do is like, put up this gigantic electric fence across the border with moats all around full of alligators and land mines and boiling acid.  They should also just tell the foreign-born PhD's, "Please stay.  We want you here.  We'll even buy you a house."[/quote]
Why do the illiterate Mexicans damage our society? They protect American industry by offering cheap labor, thus discouraging outsourcing. They do stuff that people like you wouldn't ever want to do, but that need to be done.
Also, your claim about the children being guaranteed to be undereducated is not a fault of illegal immigration. It's a fault of our education system. Stopping illegal immigration won't fix our education system.
Do you want to create an elite nation of highly educated people? Such a thing is impossible geographically. But I think nationality in terms of belonging to a country is already dying. It stands against the force of globalization. If you want your elite community, stop searching. You already found one, right here on this forum.

Pigpi: Ph.D.s are some of the least productive (in the traditional sense) members of society.  Why would you want to keep [i]them[/i] around?

[quote="SoxFan34"]-These illegals are in this for personal gain: they don't pay taxes or serve the United States in any way.
-Legal immigrants work very hard to become Americans. They westernize themselves, learn how to speak English, pay taxes, and serve the country. To see people who just ran over the border to the nearest cheap labor place get the same treatment as the hard-working legals is horrible.
-[color=red]The USA should be a country of natives and legal immigrants; people who WANT to be Americans. [/color][/quote]

Wait, you've really thought so little about this as to make these statements?  Aren't you a little embarrassed?  -[i]Legal[/i] immigrants come here for their own personal gain.  Why else would anyone pick up and immigrate?  Everyone is looking for a better life, and it's not clear to me why you think [i]your[/i] method of finding it is morally superior.  The rest of your points are similarly absurd: illegal immigrants devote plenty of their own efforts to getting here, presumably because they'd like to be Americans.  As jmerry noted, not all illegals don't pay taxes, and as I noted, those who don't mostly do it out of fear (since anything that attracts attention is risky for illegals, especially when crazed nativists are on the prowl in local & federal government).  (As an aside, it's worth noting that those who do pay taxes are paying for the social security benefits of everyone else, since that's a service they won't be able to collect on.)  The obvious solution, of course, is to make it easier for them to legalize themselves: once the fear of deportation is removed, they'll be in a position where they can integrate into society (pay taxes, seek non-abusive jobs, etc.), thus removing in one swoop all the major problems specific to illegal immigration.

By the way, your entire post could have been summarized in the line, "Now that I'm in, shut the door!"  It would have required much less typing on your part.

[quote="PenguinIntegral"]Practicality issues aside, would it not be a great world if all had complete control over our assets and were able to live lives without needing to tax (forcibly remove money from) others to do so?[/quote]
You missed my point: you complained about paying for their schooling.  There is no way your complaint can be an immigration-specific, let alone illegal immigration-specific, complaint, since schools are provided by taxpayers for all children, and nearly all children in this country are legal.  (I'm accusing you of a non-sequitur.)

Is anyone able to give some relevant statistics :
- what's the average income of an illegal immigrant
- how many of them pay taxes
- are they disproportionally represented in prisons, social welfare programs,...

I guess that those numbers will be quite different, depending on the state. :|

1. http://www.cis.org/articles/2001/mexico/poverty.html
Look at figure 11.
2. I think that most illegal immigrants pay taxes. Look at http://news.minnesota.publicradio.org/features/2005/04/11_tonessb_illegal/
3. I'm not sure.

[quote="JBL"]Ph.D.s are some of the least productive (in the traditional sense) members of society. Why would you want to keep [i]them[/i] around?[/quote]
Umm.. because they are fun to hang out with? :maybe: Okay, I can't think of a good reason either. Considering the job market, more PhDs in math (foreign-born or otherwise) is about the last thing that this country needs.

[b]fredbel6[/b]: Generally, illegal immigrants have less access to US welfare programs than legal white trash. They can get emergency treatment under Medicaid (health care for the poor), but probably not routine care.

I don't have cable television, but I can get about 18 or 19 different over-the-air broadcast TV channels (some of them with poor reception.) Of those, 5 or 6 broadcast exclusively in Spanish, and another 2 in a mixture of other languages. You can get Bollywood music videos, news and talk shows in Farsi, Japanese cooking shows (which can be a real hoot - there's one, that is subtitled, that I've seen fairly often), period dramas in Chinese or Korean (aka "funny hat shows.") Of the large number of Spanish language stations, it seems that most of the content is clearly inspired by U.S. popular culture. You have crude, noisy talk shows that look a lot like American crude, noisy talk shows (like Jerry Springer); you have silly game shows. Scantily-clad women are plentiful, and they tend to be light-skinned and blonde. One channel shows mainstream Hollywood movies with the original English dialog dubbed into Spanish. Another channel shows a lot of "Western" movies - gunslingers riding through rocky countryside menacing or defending the residents of isolated, dusty villages. Such movies were once a staple of Hollywood; their popularity may have faded in the U.S. but it seems they're still being made in Mexico. And then there's sports. If you want to see [i]futbol[/i], even international matches involving the U.S. team, you'll probably be watching on a Spanish-language channel. (One announcer, Andres Cantor, was legendary for just how long he could sustain the vowel in "Goooo...oooool!")

I've been the coach of the Southern California ARML team for a while. If I had to guess, I'd guess that over the years, 75% on the team members have been the children of immigrants, and maybe as many as 50% were immigrants themselves. As to the legal status of those immigrants - naturalized citizen, "green card", student visa, refugee status, dependent of adult on H1b visa, other categories that I don't even know about because I'm not an immigration lawyer - I can't even begin to guess. And I'm not going to ask.

I must say, Bush's guest worker program was a great opportunity to handle this problem, but those idiots in capitol hill made it fail. Also, if they were from another country, say a European country, was there going to be such a problem? No matter how many people say that it's not about race, they are lying, a paart of it is definetely about race.


Trig substitution.

Edit--actually... I don't know if that works.. but it seems sensible.

[quote="chien than"]Find $x;y$ such that:
$\{\begin{array}{l}y+4\sqrt{x-x^{2}}=1\\ \sqrt{x}-\sqrt{1-x}=4y^{3}-3y\end{array}$[/quote]
It is easy to chek, that $-1\le y\le 1$, therefore exist z: y=cosz. Then $4y^{3}-3y=cos3z$ and $1+cos6z=2cos^{2}3z=2(\sqrt x-\sqrt{1-x})^{2}=2-4\sqrt{x-x^{2}}=2-1+cosz$. It give $cos6z=cosz$.
$0=cosz-cos6z=2sin\frac{7z}{2}sin\frac{5z}{2}$.
If $x=sin^{2}t$, then $sin2t=sin^{2}\frac{z}{2},\sqrt 2 sin(t-\frac{\pi }{4})=cos3z.$
Therefore solution is $y=cosz,x=sin^{2}t$ were $z=\frac{2\pi k}{7},k=0,1,2,3,4,5,6,z=\frac{2\pi m}{5},m=1,2,3,4.$


algebra
polynomial
function
algebra unsolved

Find all roots with multiplicity biggest than 1 of polynomial $ \prod_{i \equal{} 1}^{n} (x^i \minus{} i)$ ( $ n \in \mathbb{N}$ ).

Look at all factors and the modulii of their roots. Study the function $ x^{\frac{1}{x}}$.
The answer is only $ \pm \sqrt{2}$

I try to solve but I don't do this.

Can you explain your idea ?

Tnx


You mean $ \sum_{k\equal{}1}^{n}\frac{1}{(2k\plus{}1)\plus{}0.7^{2k\minus{}1}}$? I don't believe there's a good closed form for that.

Aha, but I think you meant $ \sum_{k\equal{}1}^{n}\frac{1}{(2k\plus{}1)\cdot 7^{2k\minus{}1}}$, and you just chose to use the most ambiguous notation for multiplication in the world in order to confuse us.

If it's the second one, Mathematica says your closed form is $ 7 \left(\minus{}1\plus{}7 \text{ArcTanh}\left[\frac{1}{7}\right]\minus{}\frac{49^{\minus{}1\minus{}n} \text{Hypergeometric2F1}\left[1,\frac{3}{2}\plus{}n,\frac{5}{2}\plus{}n,\frac{1}{49}\right]}{3\plus{}2 n}\right)$

:|

[quote="grn_trtle"]You mean $ \sum_{k \equal{} 1}^{n}\frac {1}{(2k \plus{} 1) \plus{} 0.7^{2k \minus{} 1}}$? I don't believe there's a good closed form for that.

Aha, but I think you meant $ \sum_{k \equal{} 1}^{n}\frac {1}{(2k \plus{} 1)\cdot 7^{2k \minus{} 1}}$, and you just chose to use the most ambiguous notation for multiplication in the world in order to confuse us.

If it's the second one, Mathematica says your closed form is $ 7 \left( \minus{} 1 \plus{} 7 \text{ArcTanh}\left[\frac {1}{7}\right] \minus{} \frac {49^{ \minus{} 1 \minus{} n} \text{Hypergeometric2F1}\left[1,\frac {3}{2} \plus{} n,\frac {5}{2} \plus{} n,\frac {1}{49}\right]}{3 \plus{} 2 n}\right)$

:|[/quote]
yes second notation is right but your answer is very difficult

$ \mathrm{arctanh}(a x)\equal{}\sum_{n\equal{}1}^\infty \frac{a^{2n\minus{}1}}{2n\minus{}1}x^{2n\minus{}1}$.
So, $ a^2 \ \mathrm{arctanh}\left(\frac{1}{a}\right)\equal{}\sum_{n\equal{}1}^\infty \frac{1}{(2n\minus{}1)a^{2n\minus{}3}}\equal{}\sum_{n\equal{}0}^\infty \frac{1}{(2n\plus{}1)a^{2n\minus{}1}}$.
Thus,
$ a^2 \ \mathrm{arctanh}\left(\frac{1}{a}\right)\minus{}a\equal{}\sum_{n\equal{}1}^\infty \frac{1}{(2n\plus{}1)a^{2n\minus{}1}}$
$ \iff \sum_{n\equal{}1}^\infty \frac{1}{(2n\plus{}1)a^{2n\minus{}1}}\equal{}a\left(a \ \mathrm{arctanh}\left(\frac{1}{a}\right)\minus{}1\right)$

If, $ a\equal{}7$,
$ \boxed{\sum_{n\equal{}1}^\infty \frac{1}{(2n\plus{}1)7^{2n\minus{}1}}\equal{}7\left(7 \ \mathrm{arctanh}\left(\frac{1}{7}\right)\minus{}1\right)}$

He wants the sum up to $ n$ terms, not the infinite sum.

Sorry.
I overlook "up to $ n$ terms".
So, I misunderstood it when it was "up to $ \infty$".


algebra
polynomial
modular arithmetic
number theory
relatively prime
number theory unsolved

Let $ W(x)$ be a polynomial  with integer coefficients such that there are two distinct integer at which $ W$ takes coprime values. Prove that there exists an infinite set of integers such that the values $ W$ takes at them are pairwise coprime.

[quote="shoki"]let these two integers be $ a,b$ then by the chinese remainder system we can find $ X$ such that
$ X \equiv a \pmod {W(b)}, X \equiv b \pmod {W(a)}$
then obviously $ W(X) \equiv W(a) \pmod {W(b)}$ and it's the same for $ W(a)$ which implies that $ W(X)$ is relatively prime to both  $ W(a)$ and $ W(b)$.
now we can also choose $ Y$ such that
$ Y \equiv a \pmod {W(X)},Y \equiv b \pmod{W(a)},Y \equiv X \pmod {W(b)}$
now the sequence $ Y,X,a,b$ will also have the property mentioned.continue in the same way to get the infinite sequence.
note that (obviously) $ X,Y$ and all integers obtained in the same way can be chosen in such a way that none of them appears twice in the sequence.[/quote]
your solution similar to my solution  :)  (I had using the inductive method)
But I think that this problem has other solutions  :D

Take $x_0=a\wedge x_1=b\wedge \forall_{n\in Z_+} x_{n+1}=b+(a-b)kW(x_0)+W(x_0)\cdot W(x_1)\cdot...\cdot W(x_n)z$ where $z\in Z_+$ is such a big number that $x_{n+1}$ is different from previous elements of string and $k,m\in Z$ are such numbers that $kW(x_0)+mW(x_1)\cdot...\cdot W(x_n)=1$. They exist because $W(x_0)$ and $W(x_1)\cdot...\cdot W(x_n)$ are coprime.
Our infinite set is then $\lbrace x_0,x_1,x_2,...\rbrace$


[quote="Arne"]Find all real numbers $x$, $y$, $z$ and $w$ such that $\sqrt{x-y} + \sqrt{y-z} + \sqrt{z-w} + \sqrt{w+x} = x + 2$.[/quote]
\[ \sqrt{x-y} + \sqrt{y-z} + \sqrt{z-w} + \sqrt{w+x} = x + 2 \]
\[ \iff (\sqrt{x-y}-1)^2+(\sqrt{y-z}-1)^2+(\sqrt{z-w}-1)^2+(\sqrt{w+x}-1)^2=0 \]

or we can Cauchy it

$(1+1+1+1)((x-y)+(y-z)+(z-w)+(w+x))\geq (\sqrt{x-y}+\sqrt{y-z}+\sqrt{z-w}+\sqrt{w+x})^2=(x+2)^2$

$\Longleftrightarrow (x-2)^2\leq 0$ $\implies x=2$ . Equality holds when $x-y=y-z=z-w=w+x=1$ .Hence

$(x,y,z,w)=(2,1,0,-1)$

 :)


induction
combinatorics unsolved
combinatorics

Let $ A,B,C$ be subsets of positive integers. The sets $ A,B,C$ are called a "partitioning" of $ mathbb{N}$ if they satisfy the following conditions:
(i) $ A,B,C\neq$empty set
(ii) $ A \cap B \equal{} B \cap C \equal{} C \cap A \equal{}$ empty set
(iii) $ A \cup B \cup C \equal{} \mathbb{N}$

Prove that there does not exist a partitioning of $ \mathbb{N}$ such that the following conditions are
satisfied:
(1) For all $ a \in A, b \in B, a \plus{} b \plus{} 2008 \in C$,
(2) For all $ b \in B, c \in C, b \plus{} c \plus{} 2008 \in A$,
(3) For all $ c \in C, a \in A, c \plus{} a \plus{} 2008 \in B$.

Here is an outline of my approach, and I can fill in the details if needed. It's a bit long but if proves a stronger statement: such sets don't exist for any number instead of "2008" and without the condition (iii).

1. Assume that such $ A,B,C$ exist, and replace them with $ A \plus{} 2008$, $ B \plus{} 2008$, $ C \plus{} 2008$. The new sets are still nonempty, still disjoint, and they now satisfy

(*) $ A \plus{} B \subseteq C$, $ A \plus{} C \subseteq B$ and $ B \plus{} C \subseteq A$.

So we got rid of the pesky 2008 and got a more natural question. We will ignore the fact that the union of the original $ A,B,C$ is $ \mathbb{N}$ and prove is a stronger statement: If three sets exist satisfying (*), then they cannot be pairwise disjoint.

2. We first reduce to the case where $ A \cup B \cup C$ includes at least one odd number. Indeed, if $ 2^k$ divides every number in $ A \cup B \cup C$, then dividing all the numbers by $ 2^k$ creates three sets which satisfy (*) and are disjoint only if the original ones are.

3. Take any three elements $ a \in A$, $ b \in B$ and $ c \in C$. Consider linear combinations $ xa \plus{} yb \plus{} zc$ where $ x,y,z$ are non-negative integers. Use induction to show that $ xa \plus{} yb \plus{} zc \in A$ whenever the following conditions are both satisfied:

(a) $ y$ and $ z$ have the same parity, but $ x$ has a different parity, AND
(b) If $ y$ and $ z$ are both zero, then $ x \equal{} 1$.

Similar rules apply for determining when $ xa \plus{} yb \plus{} zc \in B$ or $ C$ (just switch the role of $ x$ with $ y$ and $ z$, respectively). The induction works when you use all three rules simultaneously, and it's very easy. To demonstrate this, we know that $ 2a \plus{} 7b \plus{} 5c \in A$, but we don't know anything about $ 3a \plus{} 7b \plus{} 5c$, and also nothing about $ 5a$.

4. Now, find $ a \in A$, $ b \in B$ and $ c \in C$ such that $ b \minus{} c$, $ c \minus{} a$ and $ a \minus{} b$ are not all even. This can be done due to step 2. Note that these numbers sum to 0 so they cannot all be odd, either. Now pick some arbitrary large coefficients $ x,y,z$, also not all even and not all odd, such that $ x > |b \minus{} c|$, $ y > |a \minus{} c|$ and $ z > |a \minus{} b|$, and such that the parity of $ (x,y,z)$ differs from the parity of $ (b \minus{} c,c \minus{} a,a \minus{} b)$. Then $ N \equal{} xa \plus{} yb \plus{} zc \equal{} (x \plus{} (b \minus{} c))a \plus{} (y \plus{} (c \minus{} a))b \plus{} (z \plus{} (a \minus{} b))c$ are two representations of the same number with coefficients that satisfy the conditions of step 3 but with different consequences.

For example if $ b \minus{} c$ is even while $ c \minus{} a$ and $ a \minus{} b$ are odd, we choose $ x \equal{} 1000$, $ y \equal{} 2000$ and $ z \equal{} 3001$ (assuming these are large enough). We get $ 1000a \plus{} 2000b \plus{} 3001c \in C$ and $ (1000 \plus{} (b \minus{} c))a \plus{} (2000 \plus{} (c \minus{} a))b \plus{} (3001 \plus{} (a \minus{} b))c \in B$, but these two are the same number so $ B \cap C \neq \emptyset$. QED

----

BTW there [b]do[/b] exist pairwise disjoint sets satisfying $ A \plus{} B \subseteq C$, $ A \plus{} C \plus{} 1 \subseteq B$ and $ B \plus{} C \plus{} 2 \subseteq A$ (easy exercise).


im tying to solve this equation for a riddle (?... yea i know) but im not a maths person.

If anyone could solve it for me it would be greatly appreciated. :)

[img]http://www.simplecodeworks.com/mazeto-3/prognostication/IMAGE.GIF[/img]

Thanks in advance.

The sum of the first $n$ squares can be found by using the formula $\frac {n(n+1)(2n+1)}{6}$. You can prove this by mathematical induction. Since you are not a "maths person", that may not interest you. At least verify that the formula works for the first few integers to satisfy yourself. In your case, $n=25$, so the sum is 25(26)(51)/6 = 5525.


Here's a question that's been nagging me for a while.  I have no idea whether a solution is attainable.

We take a series of numbers, where the nth term is the sum of the first n fibbonaci numbers.  What is the pattern behind the factors of the numbers?
I found that if n=3k or 3k-1, then it is divisible by 2, and if n=8k, 8k-1, or 8k-3, it is divisible by 3.  Does anyone have an idea about whether you can find a rule for what values of n the number is divisible by some number r, and can anyone provide a proof for any specific values of r (such as the ones that I have already provided)?  The second part should be fairly easy with induction, but I don't know where to go with the first part.

use of computers is fine, because this is fairly tough.  It would be impressive to see it entirely without computers, though.

Before you go any further, try and find the value F1 + F2 + ... + Fn, where Fi is the ith Fibonacci number.

The generalized formula for the ith Fibonacci number may also help:

F(n) = 1/ :rt5: (((1+ :rt5:)/2)^n - ((1 - :rt5:)/2)^n)

I tried using both of these things, to no avail.  Any more help?

To evaluate the sum F(1)+F(2)+...+F(n), try subbing F(k)=F(k+2)-F(k+1) and see what cancells. I think you'll like what you get.  :D

(I don't know how to write subscripts... anyone want to enlighten me?)

[sub] [/sub] with the [] replaced by <>, or now with latex

a_i = [tex] a_i [/tex]

I think that I understand the basic reasoning behind it.  Thanks! :D


If fence posts are set 10 meters apart how many posts are needed to fence a rectangular field 400 meters by 200 meters?

116 posts are required
Each side has two posts at the corners that are common with another side

$ 2\left(\frac{400}{10}\plus{}\frac{200}{10}\right)\equal{}120.$
Therefore, the answer is $ 120.$

[hide="this is the correct answer"]they are sold in sets of 10 so we get '

400/10=40

200/10=20

and we add these 2 together and get 60 and multiply by 2 and get 120

so the answe is 120 [/hide]

[b][i][u]please rate my post thanks!![/u][/i][/b]

[hide]Answer is 120  2(400/10+200/10)=120[/hide]
Solution and answer above.

Here is a different solution:

40010 is 40 so there are 41 posts on each length, 200/10 is 20 so there are 21 posts on each width so 41*2+21*2=124 but we counted each post on the corners twice and since there are 4 corners which means 4 posts so 124-4 is 120. Our answer is [b]120.[/b]


In the Island M, there was a murder last night. Witnesses saw one person shooting the guy, but no one could see him well enough to identify him. In addition, the five men A, B, C, D, and E were all seen near the place of the murder shortly after and before the murder. The police brought them in for questioning and their responses were as follows:

A: I didn't do it if and only if D didn't do it.

B: A did do it if I didn't do it.

C: If I did it and A didn't do it, D didn't do it.

D: I didn't do it or B did do it, or C did it.

E: I did it.

It is known for a fact that ONLY one of them did do it.

If Only One person told the truth, who did it?

And who told the truth?

I'm not really sure
could u say if I'm right?
[hide] B did it and told the truth[/hide]

[hide]
D did it and C is telling the truth[/hide]


inequalities
geometry
complex numbers
3D geometry
geometry solved

Prove than if A,B,C,D are non-coplanars, AB*CD, AC*BD and AD*BC can be lenghts of a triangle non-degenerate

I don't know if this is a solution, but it is sure an idea: have you tried a  solution using complex numbers in this problem?  :?

I know that Ptolemey's inequality is true in space (use a proof with inversion). So u could try it. :?

I don't know what I was thinking, I thought it was a plane geometry problem...  Ptolemy inequality is one of the solutions. In 3D geometry it can be proved like this:

we know that in space Ptolemy inequality is:
AD*BC+AB*DC>AC*BD which can be proved like so: you can show that you can "move" one of the points so that it can be coplanary with the other 3 and then apply Ptolemy inequality in plane: AD*BC+AB*DC>=AC*BD and you do the same with the other two inequalities to show that you can have a triangle with the lenghts: AD*BC;AB*DC;AC*BD.

Hope this time I am right!  :D  :D
tschuess!


[quote="Anavel_Gato"]
$ \equal{} \left(1 \minus{} \frac {1}{2005!}\right)\frac {2004!}{2005! \minus{} 1}$
[/quote]

why is that : How did you win that :o

$ \left(\sum_{k \equal{} 1}^{2004}\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}\right)...$
Hi, Thanks a lot alex,I understand now and I think Anavel Gato should had write : $ \left(\sum_{k \equal{} 1}^{2004}\left(\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}\right)\right)...$

[quote="enndb0x"]Anavel Gato should had write : $ \left(\sum_{k \equal{} 1}^{2004}\left(\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}\right)\right)...$[/quote]

Yes,that's better I think.
Sorry for confusing you,enndb0x :P


I saw this somewhere else and thought it would be appropriate for these forums. :) Not sure if it has been posted already.

http://movies.collegehumor.com/media/movies/finite.wmv

Edit: Remember to right-click -> save link as...

Haha
I think it's great how it ends with QED

funny...

haha that's great, all the reason more for me to look forward to college... :D


[b]Note:[/b] [i]The solution below is highly non-rigorous because it assumes $ f$ to be defined at all integer arguments, whereas it obviously does not have to be. A correct solution would just use instances of the recursive definition, along with the given value, to strategically compute f at particular arguments. Still, it can be useful to make such assumptions on a short answer contest.[/i]

[hide]View x as a parameter. Then, the difference quotient $ \frac {f(y \plus{} f(x)) \minus{} f(y \plus{} 7)}{f(x) \minus{} 7} \equal{} \frac {x}{f(x) \minus{} 7}$ is invariant of $ y$, so $ f$ takes the form $ f(x) \equal{} ax \plus{} b$

Plugging that into our equation, we get $ (a^2 \minus{} 1) x \plus{} a(b \minus{} 7)$ is identically zero for all $ x$.

From the first part, we find that $ a \equal{} \pm 1$, and thus, $ a(b \minus{} 7) \equal{} 0 \implies b \equal{} 7$

Using the fact that $ f(2) \equal{} 5$, we conclude that $ f(x) \equal{} 7 \minus{} x$, and thus $ f(11) \equal{} \minus{} 4$[/hide]


I want to write [code](2x^2-x+4)/(x^3+4x)

as a sum of fractions so that i can integrate it.    This particular one is difficult because of the irreducible quadratic we get.  Plz donot try and integrate it. Just write it as a sum of fractions.

And what is the idea behind doing this. consider this.

1/(x^2+5x+6) writing it as sum of fractions i factorze the denominator.

so 1/(x+2)(x+3)= A/(x+2) + B/(x+3) and

1/(x+2)(x+3)= {A(x+3)+B(x+2)}/(x+2)(x+3)}

1 = A(x+3)+B(x+2)

My teacher said this will be like an identity. so subsitute some x-values get the equations and ur done. But then i find that it acts as identity only when i chose x solutions which i would get by setting the quadratic to zero. which in this case are -3 and -2. But had i substituted say 6 and 7 the equation wouldn't be an identity right!  So the point i'm making is how do u deal with ones like above?[/code]

You should look for the identity in the form:

[tex]\frac{2x^2-x+4}{x(x^2+4)}=\frac{A}{x}+\frac{Bx+C}{x^2+4}[/tex]

When there is an irreductible quadratic in the denominator then the numerator has to be linear not constant.

You get:

[tex]2x^2-x+4=Ax^2+4A+Bx^2+Cx=(A+B)x^2+Cx+4A[/tex] so

[tex]A+B=2,\ C=-1,\ 4A=4[/tex]

Finally it is:

[tex]\frac{2x^2-x+4}{x(x^2+4)}=\frac{1}{x}+\frac{x-1}{x^2+4}[/tex]


We simply use the triangular numbers formula $ \frac{n(n \plus{} 1)}{2}$ and plug in $ 20$ for $ n$:

$ \frac{20 \cdot 21}{2} \equal{} 10 \cdot 21 \equal{} \boxed{210}$


inequalities
logarithms
inequalities proposed

[quote="The soul of rock"]Given $ 0 < a\le b\le c\le d$. Prove that
\[ \sqrt{\frac{a^{2}}{a^{2}\minus{}ab\plus{}b^{2}}}\plus{}\sqrt{\frac{b^{2}}{b^{2}\minus{}bc\plus{}c^{2}}}\plus{}\sqrt{\frac{c^{2}}{c^{2}\minus{}cd\plus{}d^{2}}}\plus{}\sqrt{\frac{d^{2}}{d^{2}\minus{}da\plus{}a^{2}}}\le 4\]
[/quote]

[b]Sorry , i repair ...[/b]

$ \rightarrow\sum\sqrt\frac{1}{1\minus{}\frac{b}{a}\plus{}\frac{b^{2}}{a^{2}}}\leq 4$

Have $ \frac{b}{a}\equal{} x;\frac{c}{b}\equal{} y;\frac{d}{c}\equal{} z;\frac{a}{d}\equal{} t$

$ \rightarrow\sum\sqrt{\frac{1}{1\minus{}x\plus{}x^{2}}}\leq 4$

*$ f(x) \equal{}\frac{1}{1\minus{}x\plus{}x^{2}}$

$ f'(x) \equal{}\frac{x^{2}\minus{}3x\plus{}2}{(1\minus{}x\plus{}x^{2})^{2}}$

f(x) have max when x=1

$ \rightarrow$ prove easy  :lol:

@Phạm Thành Quang , please check your solution before you post a reply  :( .OK ?

I have found the following inequality (my problem comes from your idea)
$ \bigstar$ [b]Problem[/b]
Let $ a_{1},a_{2},...,a_{n}$ be positive real numbers and $ a_{1},a_{2},...,a_{n}\in\left[0,\frac{5\minus{}\sqrt{5}}{2}\right]$. They are such that $ a_{1}a_{2}...a_{n}\equal{} 1$. Prove that:
\[ \sqrt{\frac{1}{1\minus{}a_{1}\plus{}a_{1}^{2}}}\plus{}\sqrt{\frac{1}{1\minus{}a_{2}\plus{}a_{2}^{2}}}\plus{}...\plus{}\sqrt{\frac{1}{1\minus{}a_{n}\plus{}a_{n}^{2}}}\le n\]
The inequality is also true with $ a_{1},a_{2},...,a_{n}\ge\frac{5\sqrt{5}}{2}$. The Soul Of Rock's problem is equivalent to the following inequality:
$ \bigstar$ [b]Problem[/b]
Let $ a,b,c,d$ be positive real numbers and  such that $ abcd \equal{} 1$. Prove that:
\[ \sqrt{\frac{1}{1\minus{}a\plus{}a^{2}}}\plus{}\sqrt{\frac{1}{1\minus{}b\plus{}b^{2}}}\plus{}\plus{}\sqrt{\frac{1}{1\minus{}c\plus{}c^{2}}}\plus{}\sqrt{\frac{1}{1\minus{}d\plus{}d^{2}}}\le 4\]

Your problem is totally easy and have nothing to do with tangent line plus logarithm but my condition is more interesting. :wink:

Using Karamata's Majorization Inequality, we have

\[ \sqrt{\frac{a^{2}}{a^{2}\minus{}ab\plus{}b^{2}}}\plus{}\sqrt{\frac{b^{2}}{b^{2}\minus{}bc\plus{}c^{2}}}\plus{}\sqrt{\frac{c^{2}}{c^{2}\minus{}cd\plus{}d^{2}}}\le 2\plus{}\sqrt{\frac{a^{2}}{d^{2}\minus{}da\plus{}a^{2}}}.\]

yes, my problems is not hard to solve. It spent 1 minutes to find out.
But your problem is also not hard  :) I will try ... :arrow:

[quote="The soul of rock"]Given $ 0 < a\le b\le c\le d$. Prove that
\[ \sqrt{\frac{a^{2}}{a^{2}\minus{}ab\plus{}b^{2}}}\plus{}\sqrt{\frac{b^{2}}{b^{2}\minus{}bc\plus{}c^{2}}}\plus{}\sqrt{\frac{c^{2}}{c^{2}\minus{}cd\plus{}d^{2}}}\plus{}\sqrt{\frac{d^{2}}{d^{2}\minus{}da\plus{}a^{2}}}\le 4\]
[/quote]
Actually, we have

$ \sqrt{\frac{a^{2}}{a^{2}\minus{}ab\plus{}b^{2}}}\le\frac{2a}{a\plus{}b}$.

Thus, we have to show that

$ \frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}d}\plus{}\frac{d}{d\plus{}a}\le 2$,

which is not hard. In addition, the initial inequality holds for $ n$ numbers.

[quote="Vasc"]

Thus, we have to show that

$ \frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}d}\plus{}\frac{d}{d\plus{}a}\le 2$,

which is not hard. In addition, the initial inequality holds for $ n$ numbers.[/quote]

How to do this, please?

[quote="Vasc"]
Thus, we have to show that

$ \frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}d}\plus{}\frac{d}{d\plus{}a}\le 2$,
[/quote]

It is equivalent to $ \frac{(a\minus{}c)(b\minus{}d)(ac\minus{}bd)}{(a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)}\leq 0$ obviously true  :)

Nice, Silouan. I am waiting now a solution for n numbers. This means is to prove that

$ \frac{a_{1}}{a_{1}\plus{}a_{2}}\plus{}\frac{a_{2}}{a_{2}\plus{}a_{3}}\plus{}\frac{a_{n}}{a_{n}\plus{}a_{1}}\le\frac{n}{2}$,

for $ 0 < a_{1}\le a_{2}\le ...\le a_{n}$.

[quote="Vasc"]This means is to prove that

$ \frac{a_{1}}{a_{1}\plus{}a_{2}}\plus{}\frac{a_{2}}{a_{2}\plus{}a_{3}}\plus{}\frac{a_{n}}{a_{n}\plus{}a_{1}}\le\frac{n}{2}$,

for $ 0 < a_{1}\le a_{2}\le ...\le a_{n}$.[/quote]
Actually, the following nice statement holds:

[i]If $ 1\le k\le n\minus{}1$ is an integer number, and $ x_{1},x_{2},...,x_{n}$ are nonnegative numbers such that $ x_{1}\le x_{2}\le ...\le x_{n}$, then

$ \frac{x_{1}}{kx_{1}\plus{}x_{k\plus{}1}}\plus{}\frac{x_{2}}{kx_{2}\plus{}x_{k\plus{}2}}\plus{}...\plus{}\frac{x_{n}}{kx_{n}\plus{}x_{k}}\le\frac{n}{k\plus{}1}$.[/i]


Να αποδειχθεί ότι το 3-πλάσιο των εμβαδών των εδρών ενός τετραέδρου είναι μικρότερο από το άθροισμα των τετραγώνων των ακμών του.

Για κάθε τρίγωνο ισχύει ότι $ 3(ABC)\equal{}\frac{1}{2}(absinC\plus{}bcsinA\plus{}casinB)<\frac{1}{2}(ab\plus{}bc\plus{}ca)\le\frac{1}{2}(a^2\plus{}b^2\plus{}c^2)$.
Για να αποδείξουμε το προβλημα εφαρμόζουμε το ίδιο και για τις 4 έδρες του τετραέδρου.

Ωραίος Τάσο, και εγώ έτσι το πήγα. Μόνο μην ξεχάσεις να αποδείκεις ότι δεν ισχύει το $ \equal{}$, δηλαδή ότι αν το τετράεδρο είναι κανονικό δεν ισχύει η ισότητα.

Ορφέας

Στο προηγούμενο post απέδειξα $ 3(ABC)<\frac{1}{2}(a^2\plus{}b^2\plus{}c^2)$ (αν δεν το πρόσεξες) άρα δεν χρειάζεται να ελένξω το $ \equal{}$.

Σωστός. Έχεις δίκιο


probability
trigonometry
LaTeX
floor function

[quote="cancer"]Let $ a,b$ be roots of $ z^{224}=1$. Find the probability that $ |a-b|\geq 1$. (Assume that all roots are chosen with equal probability.)[/quote]
If $ \zeta$ is a 244-th root of unity then $ a=\zeta^{n}$ and $ b=\zeta^{m}$ for (WLOG) some $ 1\leq m\leq n \leq 224$. So $ |a-b| = \left| \zeta^{n}-\zeta^{m}\right| = \left| \zeta^{n-m}-1\right|$. Simplify with some trig to get $ 2\sin \frac{\pi (n-m)}{244}$

EDIT: LaTeX dissappeared and I got to leave now. I try to continue but I think my approach works out.

[quote="ZetaX"]The result is (as any probability that is equaly distributed on a finite set) a rational number.
Your result isn't.[/quote]
I didnt post a complete solution.  :wink:

All I shown was that $ |a-b| = 2\sin \frac{\pi (n-m)}{244}$.

Since we want that $ |a-b|\geq 1$ it means,
$ \sin \frac{\pi (n-m)}{244}\geq \frac{1}{2}$

To answer that question consider a similar problem,
$ \sin \frac{\pi k}{244}\geq \frac{1}{2}$ for $ 1\leq k\leq 244$.

It seems to me for $ k=1,2,...,40$ we have $ \sin \frac{\pi k}{244}< \frac{1}{2}$. And for $ k=41,42,...,122,123,...203$ we have $ \sin \frac{\pi k}{244}\geq \frac{1}{2}$. But then the graph starts going down so we have $ \sin \frac{\pi k}{244}< \frac{1}{2}$ for $ 204,...,244$.

So you can think of your problem being formulated as follows:

Given integers $ 1\leq m\leq n \leq 244$ what is the probability that $ 41 \leq n-m \leq 203$?

Which is, $ \frac{203-41+1}{243}= \frac{163}{243}$

Wow, that seem so complicated... either I'm missing something [b]huge[/b]... or ...

[hide="Solution"]This can be rephrased as, given 224 equally spaced points around a unit circle, what is the probability that two randomly selected points have a distance greater than or equal to 1. WLOG fix the first point. The arc between the first and second points must then be at least 60 degrees, or one-sixth of a circle.

$ \frac{224}{6}= 37. \overline{3}$, so there are $ \lfloor 37. \overline{3}\rfloor = 37$ invalid points on either side.

This makes the probability $ \frac{223-37 \times 2}{223}= \frac{149}{223}$

This would make sense, because from any given point on a unit circle, the probability that another point on the unit circle is a distance of at least 1 away is equal to $ \frac{360^\circ-2 \times 60^\circ }{360^\circ}= \frac{2}{3}\approx \frac{149}{223}$
[/hide]

[quote="A M Y"]Wow, that seem so complicated... either I'm missing something [b]huge[/b]... or ...
[/quote]
I got the answer wrong cuz I did if for $ 244$ instead of $ 224$, but anyways it seems our answers match. :)

Its really not [b]that[/b] complicated. I thought about doing what you said, but I was afraid geometric approach is going to be complicated.


abstract algebra
linear algebra
matrix
superior algebra
superior algebra solved

How would I prove that any group with 2 or 3 elements must be abelian?

Two possibilities:

1) If you know Lagranges' theorem it's an easy corallary.

2) You can write up all possible group-tables (using some system, otherwise you would get stuck)

[quote]2) You can write up all possible group-tables (using some system, otherwise you would get stuck)[/quote]
I'm going to assume our friend here doesn't know Lagrange's Theorem, so this is the way to go. This is [i]very easy[/i] for the 2 and 3 element groups.

To start with, we know that $e*a=a*e=a$ for all $a$ so the top row and left-hand column must simply be lists (in the same order) of the members of the group.

The next big issue is this: if $ax=ay$, then $x=y$. This is easy in a group, becuase you can just multiply through on the left by $a^{-1}.$ A direct consequence of this: you can't repeat the same element twice on the same row of the multiplication table. Each row must be a permutation of the list of elements. The same can be said for each column.

So for the 2-element case we start out with the table

\[\begin{matrix}e & a \\ a & ? \end{matrix}\]
But the fact that you can't repeat elements on a row leaves you no choice: that "?" has to be $e$. That's the only possible multiplication table.

For the 3-element case, you start out with

\[\begin{matrix}e & a & b \\ a & ?& ? \\ b & ? & ? \end{matrix}\]
So now, try out the possible things that you can put in place of the question marks. You'll see that there's little if any choice here as well.

Oh - how can you tell if what you get is abelian? There's a very easy way to spot that in the table, but you can figure out that condition as well.

I think Herstein's [i]Topics in Algebra[/i] has a fiendish early exercise in which he want you to prove that all groups of order $\le 5$ are abelian by this kind of table-chasing. Of course, as soon as you learn Lagrange's Theorem ...

[quote="Kent Merryfield"][quote]
I think Herstein's [i]Topics in Algebra[/i] has a fiendish early exercise in which he want you to prove that all groups of order $\le 5$ are abelian by this kind of table-chasing. Of course, as soon as you learn Lagrange's Theorem ...[/quote][/quote]
Yup I remeber the problem very well.Gave me a headache... :(


probability
modular arithmetic
Euler
limit
number theory open
number theory

Hello, when studying class groups etc this surprising weird problem popped up

How 'much' of the squarefree numbers are 7 mod 8 ?
My guess is $\frac{1}{6}$ of all squarefree numbers

I at least need that it is less than $\frac{1}{2}$
with how much i mean :

$lim_{n\rightarrow\infty} \frac{\textrm{number of squarefree numbers that are}\equiv 7 mod 8,\leq n}{\textrm{number of squarefree numbers below n}}$

So, numbers $\equiv 0, 4 \bmod 8$ cannot be squarefree, I think that is where your guess comes from.  However, numbers $\equiv 2 \bmod 8$, when divided by two, are $\equiv 1 \bmod 4$, which can be square, and numbers $\equiv 6 \bmod 8$, when divided by two, are $\equiv 3 \bmod 4$, which cannot (even though they can be the product of an odd square and, say, three), which would affect the probability distribution, so... I cannot help much.  It is a complicated problem, I'm afraid  :|

it is indeed, for a moment i thought of multiplying with seven , to prove that 1 mod 8 has just as many , but that does not work when the original 7 mod 8 number is itself divisible by seven, and if that would work , you could prove that the even numbers have a density of one, which shows that i have confused cardinality with density

Showing that the density is less than $\displaystyle \frac{1}{2}$ is actually pretty easy.

The probability that an integer is square-free is known to be $\displaystyle \frac{6}{\pi^2}$.  (Here by probability I really mean density.)  The probability that an integer is $7 \bmod 8$ is of course $\frac{1}{8}$; in particular, the probability that an integer is both square-free and $7 \bmod 8$ is at most $\frac{1}{8}$.

Thus, given that an integer is square-free, the probability that it is $7 \bmod 8$ is at most $\frac{1/8}{6/\pi^2} = \frac{\pi^2}{48}$, which is less than $0.21$.

I think I see how to push the argument to show that the density is actually $\displaystyle \frac{1}{6}$, but since you don't need it, I will stop here.

[quote="Ravi B"] in particular, the probability that an integer is both square-free and $7 \bmod 8$ is at most $\frac{1}{8}$.
[/quote]

How can you conclude this?  If the two probabilities aren't independent events then this might be a lower bound rather than an upper bound  :huh:

The set of integers that are both square-free and $7 \bmod 8$ is a subset of the set of integers that are $7 \bmod 8$.  Thus the density of the first set is at most the density of the second set.

But i am asking for the density of the squarefree 7 mod 8 numbers within the squarefree numbers, not the entire set of integers?

Right, but I already answered that in the second-to-last paragraph of my first post.

Let me explain again.  Let $n$ be a large integer.  Let $A$ be the set of integers from 1 to $n$ that are square-free.  Let $B$ be the set of integers from 1 to $n$ that are $7 \bmod 8$.  Let $C$ be the set of integers from 1 to $n$ that are both square-free and $7 \bmod 8$.

It is a known result that $|A| \sim \frac{6}{\pi^2} n$.  It is trivial that $|B| \sim \frac{1}{8} n$.  Because $C$ is a subset of $B$, we have $|C| \le (\frac{1}{8} + o(1)) n$.  Thus
\[ \frac{|C|}{|A|} \le \frac{1/8 + o(1)}{6/\pi^2} = \frac{\pi^2}{48} + o(1) \le 0.21.  \]
That's what you wanted to know.

And yes, we can tighten the argument to show that the density is actually $1/6$.

Let $a$, $b$ be positive integers such that $(a, b)=1$. Consider the sum
\[ S_{a, b}(x) = \sum_{ \substack{ n \le x \\ n \equiv a \pmod{b} }} \mu^2(n) ,  \]
where $\mu(n)$ is the M\"obius function. We use the identity $\mu^2(n) = \sum_{ d^2 \mid n} \mu(d)$ and find
\[ S_{a, b}(x) = \sum_{ \substack{ n \le x \\ n \equiv a \pmod{b} }} \sum_{ d^2 \mid n } \mu(d) = \sum_{ d \le \sqrt{x} } \mu(d) \sum_{ \substack{ n \le x \\ n \equiv a \pmod{b} \\ n \equiv 0 \pmod{d^2} }} 1 .  \]
Since $(a, b)=1$, the inner sum vanishes if $(b, d)\not=1$ and equals $\frac{x}{b d^2} + O(1)$ if $(b, d)=1$. Therefore
\[ S_{a, b}(x) = \frac{x}{b} \sum_{ \substack{ d \le \sqrt{x} \\ (d, b) = 1} } \frac{\mu(d)} {d^2} + O(\sqrt{x}) = \frac{x}{b} \sum_{ \substack{ d = 1 \\ (d, b)= 1 }}^{\infty} \frac{\mu(d)}{d^2} + O (\sqrt{x}) .  \]
Usinng the Euler identity we find
\[ S_{a, b}(x) = \frac{x}{b} \prod_{ p \nmid b } \Big( 1 - \frac{1}{p^2} \Big) + O (\sqrt{x})  \]
(the product is taken over all primes $p \nmid b$). Hence
\[ \lim_{x \to \infty} \frac{S_{a, b}(x)}{S_{1, 1}(x)} = \frac{1}{b} \prod_{ p \nmid b } \Big( 1 - \frac{1}{p^2} \Big) \prod_{ p } \Big( 1 - \frac{1}{p^2} \Big)^{-1} = \frac{1}{b} \prod_{ p \mid b } \Big( 1 - \frac{1}{p^2} \Big)^{-1} .  \]
If $b = 8$ then the later expression equals $1/6$.


Find x, y, z:
$\left\{ \begin{array}{l} xz + xy - yz = \frac{1}{{a^2 }} \\ xy + yz - zx = \frac{1}{{b^2 }} \\ zy + xz - yx = \frac{1}{{c^2 }} \\ \end{array} \right.$

Row_1 plus Row_2 gives u xz in terms of a,b,c, etc
multiply all and find xyz in terms of a,b,c and than find the values for each variable.
You can approach this problem as an augmented matrix but the calculations are larger i think!


What do you think the cutoff will be this year?

73, like last year. This one seemed to be about the same difficulty as last year's, and yet I still managed to screw it up because I spent an hour working on question B4 the wrong way.

Well, at least I'll be taking the repechage...

what do you guys think the repechage cutoff will be?

Do you get an email if you make CMOR?

[quote="Bacteria"]Well, the moment of truth will come this week: I got my CMO e-mail on the 17th of December last year...[/quote]
Dude are you seriously nervous?

Yep, I'm fairly sure that you get an e-mail for CMOR.

Nope I'm not nervous; 77 gets me CMO and APMO basically 100% of the time.

Anyone received an email for CMO/Repechage invite yet??
If no, when is it expected to be sent? Cuz i got my repechage thing on the 15th last year

I haven't got mine yet but they should be sent out in the next couple days.

The average of 39.3 has been released (43.9 last year), so assuming the same standard deviation the CMO cutoff should be around 43.9-39.3=4.6 less. It was 73 last year so that gives a cutoff of around 68 or 69 for an educated guess...

ARE YOU SERIOUS!!! That means I either made it by the skin of my teeth or missed it by a nanometer...

I will EXTREMELY be upset if its the latter...

Are all the CMOR invitations sent out at the same time/day?

No... i was wondering how long it will take them to post the result... :|

Yep, I was bored so I took like 45 minutes to calculate a standard deviation of 13.521 for last year.

@athunder
Which one did you qualify for (CMO?)

he got the cmor, which means that everyone who scored a 67+ is going to make something.

I hear the cmo cutoff is 71...any confirmation? And why do i still not have the email?

LOL are all the invitations sent on the same day? i have a 66 and i didnt get one yet >_______>

Hmmm, I don't think so but its a possbility. Nobody that I know of has received a CMO invitation yet so perhaps they send CMO after CMOR invitations...

[quote="athunder"]66 is good enough to make it, just wait, or you could have misprinted your email address.[/quote]
Lets say i did that, the results will be out before the actual CMOR questions are posted, so i can always check with CEMC?

I got into CMOR!!  :) , so im guessing $ \ge66$?

I just got my CMO email! =]
I got a 69...I think I'm really close to the cut off. Just for reference, I got it at 3:27 Eastern Time.

Wow; the cutoff is less than seventy :o
I got my CMO email at 3:28 EST today

So evidently the CMO cutoff is exactly 69, since Nike got a 69 and made it, and I got a 68 and didnt make it.

 :(

I really dunno but

when I am looking at my mark

(because Im invited to CMO)

I think the cutline is 73

because I solved everything except last question.

For last question, I got a)

and got only answer for c)

I haven't gotten any emails about CMO yet, though I think I qualified for it(I got 73). Is this just because they have not been sending to all of people yet?


I dont know to solve it, please help me with a solution:
For $ c$ a real positive number, prove that:
$ (\frac {\frac {1}{c} \plus{} c}{2})^{(2n \plus{} 1)^2}\geq\frac {c^{2n \plus{} 1} \plus{} \frac {1}{c^{2n \plus{} 1}}}{2}$

I frankly think the inequality you gave is not true. It must be changed the side like this
$ \frac{c^{2n\plus{}1}\plus{}\frac{1}{c^{2n\plus{}1}}}{2} \geq (\frac{c\plus{}\frac{1}{c}}{2})^{2n\plus{}1}$
On a regular basis of math solving, I think it's not hard to prove this inequality.
Apparently, we consult the following inequality as an example for the above inequality:
For every n>1, we have $ \frac{a^n\plus{}b^n}{2} \geq ( \frac{a\plus{}b}{2})^n$

But it seems that his exponent on both sides is not the same. He has an extra exponent, a squaring in
$ (\frac {\frac {1}{c} \plus{} c}{2})^{(2n \plus{} 1)^2}$

So it amounts to showing (for your case)
$ \frac{a^n\plus{}b^n}{2} \equal{}?\equal{} ( \frac{a\plus{}b}{2})^{2n}$ which is quite different.