function
algebra
functional equation
algebra unsolved

find all functions $f,g,h: \mathbb R \to \mathbb R$ such that
\[f(x+y)+g(x-y)=2h(x)+2h(y), \quad \forall x,y \in \mathbb R.\]

[quote="KMK00009"]find all real functions f,g,h such that
f(x+y)+g(x-y)=2h(x)+2h(y)[/quote]

Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\plus{}g(x\minus{}y)\equal{}2h(x)\plus{}2h(y)$

$ P(x,0)$ $ \implies$ $ f(x)\plus{}g(x)\equal{}2h(x)\plus{}2h(0)$ and so $ g(x)\equal{}2h(x)\minus{}f(x)\plus{}2h(0)$

$ P(x,x)$ $ \implies$ $ f(2x)\plus{}g(0)\equal{}4h(x)$ and so $ f(x)\equal{}4h(\frac x2)\minus{}g(0)\equal{}4h(\frac x2)\minus{}4h(0)\plus{}f(0)$

So we have a mandatory expression for $ f(x)$ and $ g(x)$ depending from $ h(x)$ :

$ f(x)\equal{}4h(\frac x2)\minus{}4h(0)\plus{}a$
$ g(x)\equal{}2h(x)\minus{}4h(\frac x2)\plus{}6h(0)\minus{}a$

Plugging this in the original equation, we get a functional equation on $ h(x)$ :

$ 4h(\frac {x\plus{}y}2)\minus{}4h(0)\plus{}a\plus{}2h(x\minus{}y)\minus{}4h(\frac {x\minus{}y}2)\plus{}6h(0)\minus{}a\equal{}2h(x)\plus{}2h(y)$

$ \iff$ $ 2(h(\frac {x\plus{}y}2)\minus{}h(0))\plus{}$ $ h(x\minus{}y)\minus{}h(0)\minus{}2(h(\frac {x\minus{}y}2)\minus{}h(0))$ $ \equal{}(h(x)\minus{}h(0)\plus{}(h(y)\minus{}h(0)$

And so, using $ u(x)\equal{}h(x)\minus{}h(0)$ :
New assertion $ Q(x,y)$ : $ 2u(\frac {x\plus{}y}2)\plus{}u(x\minus{}y)\minus{}2u(\frac {x\minus{}y}2)\equal{}u(x)\plus{}u(y)$ with $ u(0)\equal{}0$

adding $ Q(x,y)$ and $ Q(x,\minus{}y)$, we get new assertion $ R(x,y)$ : $ u(x\minus{}y)\plus{}u(x\plus{}y)\equal{}2u(x)\plus{}u(y)\plus{}u(\minus{}y)$

subtracting $ R(y,x)$ from $ R(x,y)$, we get $ u(x\minus{}y)\minus{}u(y\minus{}x)\equal{}u(x)\minus{}u(\minus{}x)\minus{}u(y)\plus{}u(\minus{}y)$

And so $ v(x\minus{}y)\equal{}v(x)\minus{}v(y)$ where $ v(x)\equal{}u(x)\minus{}u(\minus{}x)$ and so $ v(x)$ is a solution of Cauchy's equation.

So $ u(x)\equal{}p(x)\plus{}c(x)$ where $ p(x)$ is an even function and $ c(x)\equal{}\frac{v(x)}2$ is a solution of Cauchy's equation and $ p(0)\equal{}c(0)\equal{}0$

Plugging this is $ R(x,y)$, we get $ p(x\minus{}y)\plus{}p(x\plus{}y)\equal{}2p(x)\plus{}2p(y)$ which implies $ p(2x)\equal{}2p(x)$ and so $ u(2x)\equal{}2u(x)$

Using this property, $ Q(x,y)$ may be written : $ u(x\plus{}y)\equal{}u(x)\plus{}u(y)$ and $ u(x)$ is also a solution of Cauchy's equation.

And it's immediate to check back that $ Q(x,y)$ and $ u(0)\equal{}0$ $ \iff$ $ u(x)$ is a solution of Cauchy's equation.

And this gives the general solution of the requested equation :

Let $ u(x)$ any solution of Cauchy's equation :

$ f(x)\equal{}2u(x)\plus{}a$
$ g(x)\equal{}4b\minus{}a$
$ h(x)\equal{}u(x)\plus{}b$

[quote="pco"][quote="pco"][quote="KMK00009"]find all real functions f,g,h such that
f(x+y)+g(x-y)=2h(x)+2h(y)[/quote]

... Plugging this is $ R(x,y)$, we get $ p(x \minus{} y) \plus{} p(x \plus{} y) \equal{} 2p(x) \plus{} 2p(y)$ which implies $ p(2x) \equal{} 2p(x)$ ...[/quote]

Sorry, but I made a mistake here : This implies $ p(2x) \equal{} 4p(x)$ and not $ p(2x) \equal{} 2p(x)$ :(

So we have to solve $ p(x \minus{} y) \plus{} p(x \plus{} y) \equal{} 2p(x) \plus{} 2p(y)$ with $ p(0) \equal{} 0$

It's very easy to show that $ p(ax) \equal{} a^2p(x)$ $ \forall a\in\mathbb Q$, $ \forall x$

Which gives infinitely many solutions :
Using an hamel basis $ {b_i}$ and $ u_i\in\mathbb R$, you get solution $ f(\sum x_ib_i) \equal{} \sum u_ix_i^2b_i$

And so at least the family of solutions :

Let $ c(x)$ any solution of Cauchy's equation
Let $ p(x)$ any function in the form above

$ f(x) \equal{} p(x) \plus{} 2c(x) \plus{} a$
$ g(x) \equal{} p(x) \plus{} 4b \minus{} a$
$ h(x) \equal{} p(x) \plus{} c(x) \plus{} b$

[u][b]And maybe some others [/b][/u]... (I dont know) .

[u][b]If continuity constraint is added[/b][/u], we have then $ p(x) \equal{} cx^2$ and $ c(x) \equal{} dx$ and the solutions :
$ f(x) \equal{} cx^2 \plus{} 2dx \plus{} a$
$ g(x) \equal{} cx^2 \plus{} 4b \minus{} a$
$ h(x) \equal{} cx^2 \plus{} dx \plus{} b$[/quote]
c(x) be solution of Cauchy's equation but c is function from R to R so c is accord by 3 condition,monotone or continuous or bound function


superior algebra
superior algebra unsolved

In which extensions over $ \mathbb {Q}$ is a rational prime $ q$ totally ramified? All I can think of are the fields contained in some $ \mathbb{Q}(\zeta_{q^i},\sqrt[q^i]{q})$. Are there many more?

you can write down the absolute Galois group of a $ p$-adic local field, $ p > 2$ (or $ p\equal{}2$ and $ i \in K$).


Find all the integers $x,y,z,t$ with $z>1$ and :  $(1+x^t)^z=1+x^y$

As far as I remember, it is from Moscow 1999.
Have a look here for a generalization :
http://www.mathlinks.ro/Forum/viewtopic.php?t=28571

Pierre.

You may also have a look here for the original problem :

http://groups.google.fr/group/sci.math.num-analysis/browse_thread/thread/9e48c97522375b1e/a1919fd1f1e0f9ad?q=pierre+bornsztein+and+moscow&rnum=1&hl=fr#a1919fd1f1e0f9ad

Pierre.


(a) is similar to (e) in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110792.


vector
function
calculus
derivative
inequalities proposed
inequalities

\[ \forall x,y \in \mathbb R^+: f(x,y)= \min\left(x,\frac{y}{x^2+y^2}\right) \]

Find all $x_0,\ y_0$ so that $\forall x,y\in \mathbb R^+ : f(x_0,y_0) \ge f(x,y)$. what is the value of $f(x_0,y_0)$?

We are trying to make $f(x,y)$ as big as we can. so let's fix $x$ in $f(x,y)$. in order to maximize $f(x,y)$, $\frac{y}{x^2+y^2}$ will have to be as big as possible.

Now we see that $\displaystyle \frac{d}{dy} \frac{y}{y^2+x^2}=\frac{x^2-y^2}{(x^2+y^2)^2}$, so $x$ is a maximum point for $f(x,y)$ viewed as a function of $y$: $f(x,x) = \textrm{min} \left( x , \frac{1}{2x} \right)$. (probably a direct argument works, but i'm lazy now)

But $x \mapsto x - \frac{1}{2x}$ is an increasing function. (let's not use again derivatives: take $x>y$ and then it is easy to prove that their images satisfy a similar relation) For $x \in \left( 0,\frac{\sqrt{2}}{2} \right]$, $\frac{1}{2x}$ is the bigger, so $f(x)=x$. in the other interval we have $f(x)=\frac{1}{2x}$. therefore the maximum value of $f$ is $\displaystyle \frac{\sqrt{2}}{2}$, and is achieved in $\left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$. (i don't think it can be achieved in other points)

Hopefully, I did not mess this up  :D

Excellent solution. I would've never thought to use $\min \leq GM$
 :P


sabet konid ham neheshtie $6x^{2}+5x+1\equiv0\pmod{m}$ baraye har modav'el tabie $m$ daraye javab ast ba vojoodi ke moadeleye $6x^{2}+5x+1=0$ daraye javabe sahih nist  :?:

6x^2+5x+1=0 (mod m) pas (2x+1)(3x+1)=0 (mod m)
 farz konid adade m, r ta amele 2 dashteh bashad pas m=s.2^r ke s fard ast.
x ra tori tayin mikonim ke 2x+1=0 (mod s) , 3x+1=0 (mod 2^r)
ebteda x ra az roye rabeteye dovom bedast miavarim:
(dar payin manzoor az (a)* varoone zarbiye a ast.)

3x+1=0 (mod 2^r) pas x=-(3)* (mod 2^r) yani x=k.2^r-(3)* ke k adady tabeyi ast.(chon (3,2^r)=1 pas varoone zarbi vojood darad.)

pas kafist az beine x haye bala x ra toori biyabim ke 2x+1=0 (mod s)
 yani x=-(2)* (mod s) yani adade k ra tori biyabim ke                          k.2^r-(3)*=-(2)* (mod s) ke chon (2^r,s)=1 chonin k-yi vojood darad. pas be eazaye har m adady manande x vojood darad ke (2x+1)(3x+1)=0 (mod m) yani haman chizi ke mikhastim. :)


calculus
integration
logarithms
trigonometry
calculus computations

$ I\equal{}\int_0^{\infty } \frac{\sqrt[3]{x}\text{Log}[x]}{x^2\plus{}1} \, dx\equal{}\int_0^{\infty } \frac{9y^3\text{Log}[y]}{y^6\plus{}1} \, dy\equal{}9\left[\int_0^1 \frac{y^3\text{Log}[y]}{y^6\plus{}1} \, dy\plus{}\int_1^{\infty } \frac{y^3\text{Log}[y]}{y^6\plus{}1} \, dy\right]$
$ \equal{}9\left[\int_0^1 \frac{y^3\text{Log}[y]}{y^6\plus{}1} \, dy\minus{}\int_0^1 \frac{\text{yLog}[y]}{y^6\plus{}1} \, dy\right]\equal{}\frac{9}{4}\left[\int_0^1 \frac{2y\left(y^2\minus{}1\right)\text{Log}\left[y^2\right]}{\left(y^2\right)^3\plus{}1} \, dy\right]\equal{}\frac{9}{4}\left[\int_0^1 \frac{(z\minus{}1)\text{Log}[z]}{z^3\plus{}1} \, dz\right]\equal{}\frac{9}{4}\left[\int _0^1(z\minus{}1)\text{Log}[z]\left(1\minus{}z^3\plus{}z^6\plus{}\text{...}\text{...}.\right)dz\right]\text{.Using} \text{the} \text{fact} \text{that} \int _0^1x^n\text{Log}[x]dx\equal{} \minus{}\frac{1}{(n\plus{}1)^2} \text{for} n\equal{}0,1,2,3\text{...}\text{we} \text{get}I\equal{}\frac{ 9}{4}\left[\left\{\frac{1}{1^2}\minus{}\frac{1}{2^2}\plus{}\frac{1}{3^2}\text{...}\text{...}\text{...}\right\}\minus{}\left\{\frac{1}{3^2}\minus{}\frac{1}{6^2}\text{...}\text{...}.\right\}\right]\equal{}\frac{9}{4}\left(1\minus{}\frac{1}{9}\right)\left(\frac{\pi ^2}{12}\right)\equal{}\frac{\pi ^2}{6}$

Right,  :lol:

My solution,
\[ I \equal{} \int_0^{ \plus{} \infty } {\frac {{\sqrt [3]{x}\ln x}} {{1 \plus{} x^2 }}} dx\]
Put,
\[ x^2 \equal{} \phi\]

\[ I \equal{} \frac{1}
{4}\int_0^{ \plus{} \infty } {\frac{{\phi ^{\frac{{ \minus{} 1}}
{3}} \ln \phi }}
{{1 \plus{} \phi }}} d\phi\]

Now consider the integral:
\[ J(\lambda ) \equal{} \int_0^{ \plus{} \infty } {\frac {{\phi ^{\lambda \minus{} 1} }} {{1 \plus{} \phi }}} d\phi \equal{} \Gamma \left( \lambda \right)\Gamma \left( {1 \minus{} \lambda } \right) \equal{} \pi \csc \left( {\pi \lambda } \right)\]

\[ \underbrace \Rightarrow _{Leibinz}J'(\lambda ) \equal{} \int_0^{ \plus{} \infty } {\frac {{\phi ^{\lambda \minus{} 1} \ln \phi }} {{1 \plus{} \phi }}} d\phi \equal{} \frac {d} {{d\lambda }}\left\{ {\pi \csc \left( {\pi \lambda } \right)} \right\} \equal{} \frac {{ \minus{} \pi ^2 \cos \left( {\pi \lambda } \right)}} {{sen^2 \left( {\pi \lambda } \right)}}\]
Now, note that:
\[ I \equal{} \frac {1} {4}J'\left( {\frac {2} {3}} \right) \equal{} \frac {{ \minus{} \pi ^2 \cos \left( {\frac {{2\pi }} {3}} \right)}} {{4sen^2 \left( {\frac {{2\pi }} {3}} \right)}} \equal{} \frac {{\pi ^2 }} {6}\]
:o  Bye!  :)


There was some question raised about the answer to problem 5.

If four coins are tossed, what is the probability that exactly three of them show heads or exactly three of them show tails?  Express you answer as a common fraction.

Here are all the possibilities with the ones we need to answer the problem.
H	H	H	H
T	H	H	H	X
H	T	H	H	X
H	H	T	H	X
H	H	H	T	X
T	T	H	H
H	T	T	H
H	H	T	T
H	T	H	T
T	H	T	H
T	H	H	T
T	T	T	H	X
T	T	H	T	X
T	H	T	T	X
H	T	T	T	X
T	T	T	T

This show that 8 out of 16 or $ \frac{1}{2}$ is the probability.

Any one is welcome to give other (easier) solutions other than working it all out.  I did it this way to show that the method I used in class gives the correct answer.

pick 3 out of 4 coins to be heads so 4c3=4 and since it can be tails also, multiply by 2 to get 8 and the total number of combs is 2^4=16 so 8/16=1/2.


quadratics
inequalities
calculus
derivative
algebra unsolved
algebra

[b](1) If (x-a)*(x-b)+(x-b)*(x-c)+(x-c)*(x-a) = 0,
then the roots of the equation are
ans=(a) real
       (b)equal
       (c)not real.

is there is any other method other then algebra (means Inequality, or using calculus)[/b]

[quote="jagdish"](1) If $ (x \minus{} a)(x \minus{} b) \plus{} (x \minus{} b)(x \minus{} c) \plus{} (x \minus{} c)(x \minus{} a) \equal{} 0$,
then the roots of the equation are:

(a)real
(b)equal
(c)not real

is there is any other method other then algebra (means Inequality, or using calculus)[/quote]

$ (x \minus{} a)(x \minus{} b) \plus{} (x \minus{} b)(x \minus{} c) \plus{} (x \minus{} c)(x \minus{} a) \equal{} 3 x^2 \minus{} 2(a \plus{} b \plus{} c)x \plus{} (ab \plus{} bc \plus{} ca) \equal{} 0$

note that by AM-GM, $ 4(a \plus{} b \plus{} c)^2 \ge 12(ab \plus{} bc \plus{} ca)$.

therefore, the answer is [b](b)[/b] if $ a \equal{} b \equal{} c$; otherwise, the answer is [b](a)[/b].

The condition that $ a>b>c$ must have been given,if so then the question is very easy.
let $ f(x)\equal{} (x\minus{}a)(x\minus{}b)\plus{}(x\minus{}b)(x\minus{}c)\plus{}(x\minus{}c)(x\minus{}a)$
$ f(a)\equal{}(a\minus{}b)(a\minus{}c)$ and $ f(b)\equal{}(b\minus{}c)(b\minus{}a)$ so $ f(a)f(b)<0$ means exactly one root of the equation lies between $ a$ and $ b$.So option $ a$ is correct.

I presume that a,b,c are given to be real and distinct.

Since $ g(x) \equal{} (x\minus{}a)(x\minus{}b) \plus{} (x\minus{}b)(x\minus{}c) \plus{} (x\minus{}c)(x\minus{}a)$ is the derivative of $ f(x) \equal{} (x\minus{}a)(x\minus{}b)(x\minus{}c)$

which has three real roots, namely a,b and c, by Rolle's theorem, g(x) has two real roots lying between these roots.

hello, we get
$ x_1\equal{}\frac{1}{3}(a\plus{}b\plus{}c)\plus{}\frac{1}{3}\sqrt{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ac}$
$ x_2\equal{}\frac{1}{3}(a\plus{}b\plus{}c)\minus{}\frac{1}{3}\sqrt{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ac}$
because we have $ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}ac\plus{}bc$ we get real solutions and in the case $ a\equal{}b\equal{}c$
one real solution.
Sonnhard.


$ ABCDEF$ is a convex 6-gon such that $ AB\equal{}DE,BC\equal{}EF,CD\equal{}FA$. Let $ A_1, B_1,C_1,D_1,E_1,F_1$ are middle points of segments $ AC,BD,CE,DF,EA,FB$. Prove that segments $ A_1D_1,B_1E_1,C_1F_1$ are collinear (passing single point).

Ok this is a very nice problem but it's not the first time it appears in an olympiad.
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1837873716&t=82897


The Fibonacci sequence $ F_{1}, F_{2}, F_{3}, ...$ is defined by $ F_{1} \equal{} 1, F_{2}\equal{} 1$ and

$ F_{n\plus{}1} \equal{} F_{n} \plus{} F_{n\minus{}1}\,\, (n \ge 2)$.

Write down the values of $ F_{3}, F_{4}, ..., F_{10}$.

Let $ S \equal{} \sum_{i\equal{}1}^{\infty} \frac{1}{F_{i}}.$

(i) Show that $ \frac{1}{F_{i}} > \frac{1}{2F_{i\minus{}1}}$ for $ i \ge 4$ and deduce that $ S > 3$.

Show also that $ S < 3\frac{2}{3}$.

(ii) Show further that $ 3.2 < S < 3.5$.

[b]I.a)[/b] We will prove, by strong induction, that $ \frac {F_{n \minus{} 1}}{F_n} > \frac {1}{2}$.

Verified thath $ \frac {3}{2} \equal{} 1.5 > 0.5 \equal{} \frac {1}{2}$, assume that

$ \frac {F_{n \minus{} 1}}{F_n} > \frac {1}{2} \quad \text{and} \quad \frac {F_{n \minus{} 2}}{F_{n \minus{} 1}} > \frac {1}{2}$.

By applying the definition on $ \frac {F_{n \minus{} 1}}{F_n} > \frac {1}{2}$:

$ \frac {F_n \minus{} F_{n \minus{} 2}}{F_{n \plus{} 1} \minus{} F_{n \minus{} 1}} > \frac {1}{2}$

$ F_n \minus{} F_{n \minus{} 2} > \frac {1}{2}(F_{n \plus{} 1} \minus{} F_{n \minus{} 1})$

$ F_n > \tfrac{1}{2} F_{n \plus{} 1} \plus{} (F_{n \minus{} 2} \minus{} \tfrac{1}{2}F_{n \minus{} 1})$

$ \frac {F_{n}}{F_{n \plus{} 1}} > \frac {1}{2} \plus{} (F_{n \minus{} 2} \minus{} \tfrac{1}{2}F_{n \minus{} 1})$

We note that $ (F_{n \minus{} 2} \minus{} \tfrac{1}{2}F_{n \minus{} 1}) > 0$ for our assumptions, so we proved by induction that $ \frac {F_{n \minus{} 1}}{F_n} > \frac {1}{2} \quad \forall n\geq4$.

[b]I.b)[/b] We deduce that S > 3

By I.a) we find that

$ 4 \cdot \big( \tfrac{1}{F_3} \plus{} \tfrac{1}{F_4} \plus{} \dots \big) > 2 \cdot \big( \tfrac{1}{F_2} \plus{} \tfrac{1}{F_3} \plus{} \dots \big) > \big( \tfrac{1}{F_1} \plus{} \tfrac{1}{F_2} \plus{} \tfrac{1}{F_3} \plus{} \dots \big)$

Iterating 4 times we get

$ 16 \cdot \big( \tfrac{1}{F_5} \plus{} \dots \big) > \big( \tfrac{1}{F_1} \plus{} \tfrac{1}{F_2} \plus{} \tfrac{1}{F_3} \plus{} \dots \big)$

Thus:

$ 16( S \minus{} (\tfrac{1}{F_1} \plus{} \tfrac{1}{F_2} \plus{} \tfrac{1}{F_3} \plus{} \tfrac{1}{F_4})) > S$

$ 16( S \minus{} \tfrac{17}{6} ) > S$

$ S > \tfrac{136}{45} > 3$

[quote="AndrewTom"]Thanks, Haile. Unfortunately I cannot view hidden content on my computer: when I click on it, it goes to the top of the page.[/quote]

Ok, I edited my solution, now it is not hidden

Would anyone like to finish this? (Thanks.)

While I have not got the $ 3\frac{2}{3}$ in part (i), I have got the stronger result $ S < 3.5$ and, in part (iii), instead of $ 3.2$, I have the stronger result $ S > 3\frac{1}{3}$.

I first found that $ \frac{F_{n\minus{}1}}{F_{n}} < \frac{2}{3}$, for $ n \ge 5$, and proved it by mathematical induction in exactly the same way that Haile did for $ \frac{F_{n\minus{}1}}{F_{n}} > \frac{1}{2}$. This gave me $ 3(\frac{1}{F_{5}} \plus{} \frac{1}{F_{6}} \plus{} ...) < 2(\frac{1}{F_{4}} \plus{} \frac{1}{F_{5}} \plus{} ...)$ so $ 3( S \minus{} \frac{17}{6}) < 2(S\minus{}\frac{5}{2})$ so $ S < 3\frac{1}{2}$.

I then found that $ \frac{F_{n\minus{}1}}{F_{n}} > \frac {3}{5}$ for $ n \ge 6$ and so on (as above), getting $ 5(S\minus{} \frac{91}{30}) > 3(S\minus{}\frac{17}{6})$ and so $ S > 3\frac{1}{3}$.

I'm now wondering how to get the $ 3\frac{2}{3}$ and $ 3.2$ in the question. Any ideas?

By playing around with some numbers, $ \frac{F_{n\minus{}1}}{F_{n}} < \frac{3}{4}, n \ge 5$ gives $ S < \frac{23}{6}$, not quite $ 3\frac{2}{3}$ while $ \frac{F_{n\minus{}1}}{F_{n}} > \frac{11}{20}, n \ge 3$ gives $ S > 3\frac{2}{9} > 3.2$.  Any improvements on these results would be appreciated.


Let $ x,y,z$ be real numbers with $ 0\le x,y,z \le 1$ and $ x \plus{} y \plus{} z \equal{} 2$.  Prove:
\[ 1\le \frac {x}{x \plus{} 1} \plus{} \frac {y}{y \plus{} 1} \plus{} \frac {z}{z \plus{} 1}\le 2
\]

$ \frac {x}{x \plus{} 1} \plus{} \frac {y}{y \plus{} 1} \plus{} \frac {z}{z \plus{} 1} \ge \frac {x}{1 \plus{} 1} \plus{} \frac {y}{1 \plus{} 1} \plus{} \frac {z}{1 \plus{} 1}\equal{}1$. As for the right inequality, $ f(x)\equal{}\frac{x}{x\plus{}1}\equal{}1\minus{}\frac{1}{x\plus{}1}$ is concave, so we have, by Jensen inequality, that $ \frac {x}{x \plus{} 1} \plus{} \frac {y}{y \plus{} 1} \plus{} \frac {z}{z \plus{} 1}\le \frac{3\frac{x\plus{}y\plus{}z}{3}}{\frac{x\plus{}y\plus{}z}{3}\plus{}1}\equal{}\frac{6}{5}<2$.

For the right inequality, another approach would be noting that, $ \displaystyle\sum_{cyc} \frac{x}{x\plus{}1} \le \sum_{cyc} x \equal{} 2$ which follows from trivial inequality.


Find the roots of the equation $ \sqrt{2x\minus{}\sqrt{8x\minus{}4}}\plus{}x^{\log_{x}2(2x\minus{}1)}\equal{} 5$.

[hide="Solution"]
Notice that $ x^{\log_{x}2(2x\minus{}1)}\equal{} 2(2x\minus{}1)$ and we must have $ x >\frac{1}{2}$

Let $ y \equal{} 2x\minus{}1$, we have: $ \sqrt{y\plus{}1\minus{}2\sqrt{y}}\plus{}2y \equal{}\sqrt{\left(\sqrt{y}\minus{}1\right)^{2}}\plus{}2y \equal{} 5\Longrightarrow\left|\sqrt{y}\minus{}1\right|\plus{}2y \equal{} 5$

Now its obvious that $ y\neq1$

If $ y > 1$ or $ x > 1$ we have: $ 2y\plus{}\sqrt{y}\minus{}6 \equal{} 0$ and $ \sqrt{y}\equal{}\frac{\minus{}1\pm7}{4}$ so $ \sqrt{y}\equal{}\frac{3}{2},y \equal{}\frac{9}{4}$ in other words $ x \equal{}\frac{13}{8}$

If $ y < 1$ or $ x < 1$ we have: $ 2y\minus{}\sqrt{y}\minus{}4 \equal{} 0$ and $ \sqrt{y}\equal{}\frac{1\pm\sqrt{33}}{4}$ so $ \sqrt{y}\equal{}\frac{1\plus{}\sqrt{33}}{4}$ but this is impossible because $ y < 1$.

So the only solution is $ \boxed{x \equal{}\frac{13}{8}}$.[/hide]


Μετα το φορτο εργασιας και τα προβληματα συνδεσης  τα ξαναλεμε.....

Δίνεται η ακολουθια $ a_n$ με $ a_1\equal{}2007^{2007}$ και εστω $ S_n$ το αθροισμα των ψηφίων του $ a_n$.
Αν $ a_n\equal{}S_{n\minus{}1}$ για καθε $ n>1$ να βρεθεί το $ \sum_{i\equal{}5}^{2005}{a_i}$.

ΚΑΛΗ ΧΡΟΝΙΑ ΣΕ ΟΛΟΥΣ

[quote="xp"]Μετα το φορτο εργασιας και τα προβληματα συνδεσης  τα ξαναλεμε.....

Δίνεται η ακολουθια $ a_n$ με $ a_1 \equal{} 2007^{2007}$ και εστω $ S_n$ το αθροισμα των ψηφίων του $ a_n$.
Αν $ a_n \equal{} S_{n \minus{} 1}$ για καθε $ n > 1$ να βρεθεί το $ \sum_{i \equal{} 5}^{2005}{a_i}$.

ΚΑΛΗ ΧΡΟΝΙΑ ΣΕ ΟΛΟΥΣ[/quote]

$ a_1 \equal{} 2007^{2007} \leq 10^{4x2007} \equal{} 10^{8028}$ ,
$ a_2 \leq \equal{} 9x8029 \equal{} 72261$ ,
$ a_3 \leq 5x9 \equal{} 45$ ,
$ a_4 \leq 12$    (γιατί ο 39 έχει το πιο μεγάλο άθροισμα ψηφίων μεταξύ των αριθμών μικρότερων ή ίσων του 45).

Επειδή ο 2007 είναι πολλαπλάσιο του 9, έπεται ότι $ a_4$ = (μη μηδενικό) πολλαπλάσιο του 9. Άρα  $ a_4$ = 9. Τα υπόλοιπα είνα ρουτίνα.

Τάκης Π.

                             [color=blue][size=150]ΚΑΛΗ ΧΡΟΝΙΑ [/size][/color]

Καλωσοριζω το νεο μελος Παπαδοπουλο Τακη στην παρεα μας
Ετσι ειναι η λυση και αν προσεχα λιγο περισσότερο στο αθροισμα και το ζητουσα απο 4 εως 2004 θα ηταν 2009......που μολις ηρθε


MATHCOUNTS email AMC AIME USA(J)MO USAMO

anybody know the state scores of the top individuals?

in particular, weina bao and seung-sohn  :lol:

omg i suck at math i just got 23 and 24 on SCHOOL sprints 2 weeks in a row.

[quote="mathcrazed"]Are you serious? I laugh at you... and here I thought that you were getting better, not worse -_-.  good luck next year anyways, virginia's a tough state :D. And when our 8th grade class goes to D.C., Youkow and I are gonna stalk you and find out where you live :)[/quote]

yes i am serious and i suck....

btw what books do you use?

No books lol... I just... kinda... take Mr. Fish's tests lol :) I do like no extra math related to mathcounts =D
We're starting Calc classes you know... on Monday. WHY'D YOU MOVE I HATE YOU :D

calc? lucky!!!

highest here is A2Trig  (combined class)

if i email mr fischer do you think he'll give them to me in a scan or something?

ya. He knows your good :D, but don't you have another coach there?? I think it was Mr. Frost... or someone that Mr. Fish knew. My dad's teaching calc now, cause a lot of random 8th graders who... aren't that good want to learn, like me. I suck too, Yuqing, don't worry... I've already missed one on the workouts and warmups T.T.

Youkow and I are taking pre-cal at syc... cept it's kinda just like us randomly doing problems :D

[quote="mathcrazed"]I suck too, Yuqing, don't worry... I've already missed one on the workouts and warmups T.T. [/quote]
i've already missed like 7 and we've done only half the thing... :(

i'm worried about mathcounts now...

lol... umm... you probably just need to get back in the "groove" lol. You're just having some slight issues cause you just restarted math. I've been doing math, like AIME and USAMO, over the whole summer so ya :D

[quote="infinity4ever"]don't worry yuqing you'll do just fine at the real competition.

CALC?!?!?!?! now i feel really stupid...  :(  :(  :([/quote]
his dad teaches a private class...

@Lyndon: we have A2Trig here in a PUBLIC SCHOOL!!!! something impossible in indiana... :P

[quote="mathcrazed"]ermm... Don't feel too bad... I'm just special :D.  Dude, Justin freaking turned my room temperature down by ten degrees the night before the competition LOL. I HAD A FREAKING COLD THAT DAY.[/quote]

hahaha

[quote="mathcrazed"]HAD A FREAKING COLD THAT DAY.[/quote]

oh yeah i had a 100.5 degree fever... :P

not like that mattered lol

[url=http://kevan.org/brain.cgi?indianamath]Secret of Competition Success[/url]

[quote="mathcrazed"]uhh... no, actually, it's not impossible =)
Cause, if you go to like Eastwood (LOL), then you can take classes at their city high school :D:D[/quote]

uhh... no, actually it's not unimpossible  :rotfl: .


conics parabola AMC AIME calculus integration

Parabola P has vertex at the origin and directrix y=-1. Find all lattice points on P that are less than or equal to 2007 units away from the point (0,1).

This seems like a good possible AIME question (one of the easier ones)


calculus
integration
derivative
trigonometry
real analysis
real analysis unsolved

(The upperlimit= pi, the lowerlimit=0   (1+sin(x))  dx)

Do you just take the anti-derivative, and then solve insert the upper and lower limit? so that x-cos(x) is the antiderivative?

$ \int_0^{\pi} (1\plus{}\sin x)\ dx\equal{}\left[x\minus{}\cos x\right]_0^{\pi}\equal{}(\pi \minus{}\cos \pi)\minus{}(0\minus{}\cos 0)\equal{}\pi \plus{}2$.


geometry
calculus
projective geometry
geometry proposed

Given triangle $ ABC$ whose $ AA_1$ is bisector. Let $ I$ be a point on segment $ AA_1$. Denote $ B_1 \equal{} BI \cap AC, C_1 \equal{} CI \cap AB, M \equal{} A_1C_1 \cap BI, N \equal{} A_1B_1 \cap CI$.
Show that: $ AI$ is the bisector of angle $ MAN$

I have a solution using harmonic division (pretty nice! :D). However, I'm interested in alternate solutions. Could anyone solve it?

[color=darkblue][b]Attention, I"ll use another notations ![/b][/color]

[quote="ll931110"][color=darkred]Given triangle $ ABC$ whose $ AD$ is bisector. Let $ I$ be a point on segment $ AD$. Denote $ E\in BI \cap AC$ ,

$ F\in CI \cap AB$ , $ M \in DF\cap BI$ , $ N \in DE \cap CI$ . Show that $ AI$ is the bisector of angle $ \widehat {MAN}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] [u]Prove easily[/u] that the following "strong" lemma :[/color]

[quote][color=darkred][b][u]LEMMA[/u].[/b] Given are six collinear points $ (Z,B,X,D,Y,C)$ in this order so that the divisions $ (Z,B,D,C)$ ,

$ (B,X,D,C)$ , $ (B,D,Y,C)$ are harmonically. Then the division $ (Z,X,D,Y)$ is also harmonically.[/color]

[hide="Proof."]
[color=darkblue]$ \left\|\begin{array}{c} (z \plus{} d)(b \plus{} c) \equal{} 2(bc \plus{} zd) \\
\ (b \plus{} d)(x \plus{} c) \equal{} 2(bd \plus{} xc) \\
\ (c \plus{} d)(y \plus{} b) \equal{} 2(cd \plus{} yb)\end{array}\right\|\ \implies$ $ \left\|\begin{array}{c} z \equal{} \frac { \minus{} d(b \plus{} c) \plus{} 2bc}{ \minus{} 2d \plus{} (b \plus{} c)} \\
 \\
x \equal{} \frac {(2b \minus{} c)d \minus{} bc}{d \plus{} (b \minus{} 2c)} \\
 \\
y \equal{} \frac {(2c \minus{} b)d \minus{} bc}{d \plus{} (c \minus{} 2b)}\end{array}\right\|\ \implies (z \plus{} d)(x \plus{} y) \equal{} 2(zd \plus{} xy)$ .

[b]Example.[/b] $ \boxed{\frac {ZB}{30}\equal{}\frac {BX}{6}\equal{}\frac {XD}{4}\equal{}\frac {DY}{5}\equal{}\frac {YC}{15}}$ $ \equal{}\frac {ZC}{60}\equal{}\frac {BD}{10}\equal{}\frac {DC}{20}\equal{}\frac {CB}{30}\equal{}\frac {CD}{20}$

and $ \frac{DB}{DC}\equal{}\frac {ZB}{ZC}\ \ ,\ \ \frac {XB}{XD}\equal{}\frac {CB}{CD}\ \ ,\ \ \frac {YD}{YC}\equal{}$ $ \frac {BD}{BC}\ \implies\ \frac {DX}{DY}\equal{}\frac {ZX}{ZY}$.[/color][/hide][/quote]

[color=darkblue]Denote $ X\in AM\cap BC$ , $ Y\in AN\cap BC$ and $ Z\in EF\cap BC$ . [u]Prove easily[/u] that  $ AZ\perp AD$ and the divisions $ (Z,B,D,C)$ ,

$ (B,X,D,C)$ , $ (B,D,Y,C)$ are harmonically. From the upper lemma obtain that the division $ (Z,X,D,Y)$ is also harmonically.

Since $ AZ\perp AD$ obtain $ \widehat {DAX}\equiv\widehat {DAY}$ , i.e. $ AI$ is the bisector of angle $ \widehat {MAN}$ . [/color]

[quote="Virgil Nicula"][color=darkblue][b]Attention, I"ll use another notations ![/b][/color]

[quote="ll931110"][color=darkred]Given triangle $ ABC$ whose $ AD$ is bisector. Let $ I$ be a point on segment $ AD$. Denote $ E\in BI \cap AC$ ,

$ F\in CI \cap AB$ , $ M \in DF\cap BI$ , $ N \in DE \cap CI$ . Show that $ AI$ is the bisector of angle $ \widehat {MAN}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] [u]Prove easily[/u] that the following "strong" lemma :[/color]

[quote][color=darkred][b][u]LEMMA[/u].[/b] Given are six collinear points $ (Z,B,X,D,Y,C)$ in this order so that the divisions $ (Z,B,D,C)$ ,

$ (B,X,D,C)$ , $ (B,D,Y,C)$ are harmonically. Then the division $ (Z,X,D,Y)$ is also harmonically.[/color][/quote]

[color=darkblue]Denote $ X\in AM\cap BC$ , $ Y\in AN\cap BC$ and $ Z\in EF\cap BC$ . [u]Prove easily[/u] that  $ AZ\perp AD$ and the divisions $ (Z,B,D,C)$ ,

$ (B,X,D,C)$ , $ (B,D,Y,C)$ are harmonically. From the upper lemma obtain that the division $ (Z,X,D,Y)$ is also harmonically.

Since $ AZ\perp AD$ obtain $ \widehat {DAX}\equiv\widehat {DAY}$ , i.e. $ AI$ is the bisector of angle $ \widehat {MAN}$ . [/color][/quote]
Coud you post your solution of your lemma. My problem is it :(

It is simply. Is geometry on a line, i.e. algebraic calculus ! You can use an well-known property :

 the division $ (A,B,C.D)$ (in this order) is harmonically $ \Longleftrightarrow (a \plus{} c)(b \plus{} d) \equal{} 2(ac \plus{} bd)$ .

[quote="Virgil Nicula"]It is simply. Is a geometry on a line, i.e. algebraic calculus ![/quote]
And I see Z, M, N are collinear. Can you prove it :D

Sorry, I don't use $ Z\in MN$ ! See above in "hide" my proof of the lemma. You wanted [u]another proof[/u].
I think that you solved this problem in two rows with  harmonical divisions. It is already evidently.
My lemma is very strong and has many apllications. Try solve it with harmonical divisions. It is easy !


so the goal of the game is to say "moo."  however, you can't remember about the game.  as soon as you remember about the game, you lose, and you have to say "i lost."  if you say "moo" without remembering that you're in the game, you win!  but if you lose, nothing happens to you.  you're still playing.  everyone is always playing all the time.

i lost.

has anyone out there ever won?  could anyone who's won please post here? :lol:

I lost
How can you know that someone said moo whitout remembering the game? Maybe someone really good can move the conversation whitout anyone notice what is he doing

yea, I've won several times...

I lost now though, eh

All these poor people knowing about it will get paranoid and crazy and  whatever, help them!
Please delete this topic immediatly! And delete every new thread about this! And delete every user who tries to restart it! And when there is no other way: delet the whole Forum! Or destroy the internet!

Otherwise, in some weeks, MathLinks will need a self-help group of people who cannot forget the game anymore.

I think this game is very meaningful in many aspects, like computer.
[url]http://en.wikipedia.org/wiki/COW_programming_language[/url]

I lost.

I've won a couple of times. I've also played a version where there's no way to win. That may sound incredibly stupid, but it's fun when a bunch of people all randomly shout "I lost!"

u have no idea

have u ever played RPS evolution?

guess what it is

Is it anything like Rock, Paper, Scissors: The Collectable Card Game? (Another stupid but fun game. We math people are easily amused...)

rock paper shoes evolution:

play and each time u win a round, u go up a level

levels:

nothing
egg
chick
chicken
rooster
dinosaur
t rex
power ranger

really stupid, but even more fun

I lost. :(
Now that i think of it I won this a lot of times(if saying moo before knowing about the game counts as a win).
Used to say moo to my friends all the time  :D

say a word in another langauage that they dont knwo

it gets them annoyed :)

i know...
some of my chinese friends talk in chinese to get me annoyed  :lol:

tell her to recite "imagine if ur memory lasted only 3 seconds" -2 times

it will take her 3 seconds to say it :)

[quote="bubka"][quote="akalra1"]i can speak in hindi

nobody knows what it is!

o yea, i lost in the moo game, but my baby brother Sam won

imagine if ur memory lasted only 3 seconds
imagine if ur memory lasted only 3 seconds
imagine if ur memory lasted only 3 seconds
imagine if ur memory lasted only 3 seconds
...[/quote]

aap sach me hindi bol sakte hai? Badia hai! Kya baat bhaiia.

[hide="By the way"][color=blue][size=200][b]MOO![/b][/size][/color][/hide][/quote]

u from india?

cool

(i am)

[quote="MysticTerminator"]wtf is everyone in this forum indian

I lost[/quote]

yes


Per and Kari were walking up the stairs of a tower. Per was constantly 52 steps ahead of Kari. When Per was half-way up the stairs, he said to Kari: "When I've reached the top, you'll be three times as far up as you are now." The number of stairs in the tower is

$ \textbf{(A)}\ 104 \qquad \textbf{(B)}\ 156 \qquad \textbf{(C)}\ 208 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 312$

[hide="Solution"]Let $ p$ be the total number of steps, and $ k$ be the number of the step on which Kari stands when Per reaches the top (step $ p$).

Per was constantly 52 steps ahead of Kari, so $ p \equal{} k \plus{} 52$. The second piece of information gives $ k \equal{} 3\left(k \minus{} \frac {p}{2}\right)$.

The second equation can be simplified to $ 3p \equal{} 4k$, and then rewritten as $ k \equal{} \frac {3}{4}p$.

Substituting $ k$ into the first equation, $ p \equal{} \frac {3}{4}p \plus{} 52$. When simplified, $ p \equal{} 208\ \Rightarrow\ \mathrm{C}$.[/hide]


calculus
derivative
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $p$ be a prime and $|G|=p^{n}$, prove that $|G'|\leq p^{n-2}$. And as usual can you make it tighter?

Hi.

Of course this holds only if $n\geq 2$. There is a normal subgroup of order $p^{n-2}$ (as of every other order) and since groups of order $p^{2}$ are abelian, we get $|[G,G]| | p^{n-2}$.

This upper bound is sharp since $[Q_{8}, Q_{8}]=\{\pm 1\}$ which has index 4.


calculus
integration
function
floor function
calculus computations

Show that
\[ \int_{0}^{n}\sum_{i=0}^{n-1}\left\lfloor x+\frac{i}{n}\right\rfloor dx=\frac{1}{2}n(n+1)(n-1) \]
$ n$ is an positive integer. ($ \lfloor\pi\rfloor=3$)

can we also use this?:
\[ \sum_{i=0}^{n-1}\left\lfloor x+\frac{i}{n}\right\rfloor = \lfloor nx\rfloor \]

please reply

Yes, using Hermite's identity gets you right to the answer.

$ \int_{0}^{n}\sum_{i=0}^{n-1}\lfloor x+\frac{i}{n}\rfloor dx = \int_{0}^{n}\lfloor nx \rfloor dx = \frac{1}{n}\cdot \int_{0}^{n^{2}}\lfloor u \rfloor du$

And the result follows, if you consider what the graph of $ \lfloor u \rfloor$ looks like.


${ f: \mathbb{R} - {\pm1,0,2,\frac {1}{2}}}\longrightarrow\mathbb{R}\;;\;f(x) - \frac {x}{x + 1}\cdot f(\frac {1}{1 - x}) = \frac {1}{1 - x}$

These are probably easy but I'm having a hard time finding which value to replace $ x$ with. How do I proceed to solve this problem?

EDIT: how do I put { } on latex?

You can use \{ and \}.  Example: $ \{1,2\}$.

[hide="Hint"]
Note that $ g(x)\equal{}\frac{1}{1\minus{}x}$ is cyclic, so my guess is that substituting $ g(x)$ in for $ x$ a few times will help.
[/hide]


Let a_1,a_2,a_3,...a_n be n distinct natural numbers,
Prove that 1/(a_1)^2+1/(a_2)^2+....+1/(a_n)^2<2

This is a REALLY good problem.....

[hide]Isn't it less than :pi:2/6 since that's what the sum of the reciprocal squares is equal to?[/hide]

There are proofs of it already on the forum (probably under Advanced Topics). It has been discussed a couple of times before.

In fact, here are [url=http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf]fourteen[/url].

I have to say that I prefer the telescoping sum solution, even though it wasn't what popped into my head.  Mostly because you don't need to know much of anything to do it, which is always nice.  Or, rather, it requires you to know a technique rather than a fact.


[hide="For #1"]
Whatever speed the swimmer swims, and whatever the speed of the current is, after 2 hours she will have gone upstream for an hour, and downstream for an hour. Since the current slows her done by c (the speed of the current) in the first hour, and speeds her up by c in the second, after 2 hours whe will be 2c downstream from the bridge. But the hat will also be 2c downstream from the bridge, so she gets the hat back after 2 hours.

In this case, that means the speed of the current is 1/2 mph

I might have strayed a bit too far into algebra on that one...
[/hide]

[hide="For #3"]
Every 12 hours, the hands meet 11 times.
Since it clearly takes the same amount of time between each of these (rotate the clock slightly...) 12/11 hours have passed, which means that it is 1:05 and about 27 seconds.
[/hide]

[hide] Your answer for #1 is pretty much what I had in mind. However, an easier way to phrase it:

Relative to the [i]hat[/i], the swimmer travels at constant speed. So, from the point of view of the hat, the swimmer swims away for an hour, then swims back, which of course must have taken an hour. In these two hours, the hat drifted 1 mile, so the current must be 1/2 mph

For #3, how do you know that the hands meet 11 times every 12 hours? Seems like you'd have to go through some algebra to find this first. [/hide]

Sunrise was when Pat and Dana started their pointless journey.
 :roll:

That wasn't the most enLIGHTening answer  :roll:

[hide]

The hands cross 11 times in 12 hours because they clearly cross once at 12, once between 1 and 2, once between 2 and 3 ... once between 10 and 11 (and not between 11 and 12, because at this point we're back to 12 again).

[/hide]

[quote="Phelpedo"]All of the below are pretty straightforward, messy algebra problems. But see if you can find a solution with minimum "crunching" by recasting the problem or considering alternate points of view.

1) A person dives from a bridge into a river and swims upstream through the water for $1$ hour at constant speed. She then turns around and swims dwnstream through the water at the same rate of speed. As the swimmer passes under the bridge, a bystander tells her that her hat fell into the river as she originally dived. The swimmer continues dwnstream at the same rate of speed, catching up with the hat at anotoher bridge exactly 1 mile downstream from the first one. What is the speed of the current in miles per hour?

2) Two towns, $A$ and $B$, are connected by a road. At sunrise, Pat begins biking from $A$ to $B$ along this road, while simultaneously Dana begins biking from $B$ to $A$. Each person bikes at a constant speed, and they cross paths at noon. Pat reaches $B$ at $5$pm while Dana reaches $A$ at 11:15 pm. When was sunrise?

3) What is the first time after $12$ o'clock that the hour and minute hands meet? (avoid algebra!)[/quote]
Are these from zeits?

Indeed (although they are classic problems). And a hint: number 2 is a geometry problem.


inequalities
logarithms
calculus
function
rearrangement inequality
inequalities unsolved

[quote="Bojan Basic"]What is the easiest way to prove the following:
\[\left(\frac x y\right)^z\cdot\left(\frac y z\right)^x\cdot\left(\frac z x\right)^y\leqslant 1\]
for positive reals $x\leqslant y\leqslant z$.[/quote]

the inequality is equivalent to $\frac{a-1}{\ln a} \ge \frac{b-1}{\ln b}$, where $a=\frac{z}{y}$ and $b=\frac{x}{y}$. Calculus shows that the function $\frac{x-1}{\ln x}$ is an increasing function, so the result follows.

True for (x,y,z) implies true for (kx,ky,kz), so normalize $x+y+z = 1$.

By weighted AMGM, $LHS \leq xz/y + yx/z + zy/x$.

But $\sum x^2yz \geq \sum (yz)^2$ by Muirhead.

[quote="Singular"]
But $\sum x^2yz \geq \sum (yz)^2$ by Muirhead.[/quote]

sign reversed.

I don't understand? // how is this wrong?
[b]
Muirhead:  Let $(a_1, a_2, \ldots, a_n)$ and $(b_1, b_2, \ldots, b_n)$ be nondecreasing real sequences such that $\sum_{i=1}^k a_i - b_i \geq 0$ for $k = 1\ldots(n-1)$ with equality when $k=n$, and $\sigma$ be a permutation of $\{1,2,\dots,n\}$.  Then $\sum_{\sigma} \sum_{i=1}^n x_i^{a_{\sigma_i}} - x_i^{b_{\sigma_i}} \geq 0$. [/b]

----------
So, in this case, consider (1,1,2) and (0,2,2) as the sequences.  Then $1-0 \geq 0$, $2-2 \geq 0$, and $4 - 4 = 0$, and we have the result that $\sum_{sym} x^2yz \geq \sum_{sym} x^2y^2$, implying $x^2yz + xy^2z + xyz^2 \geq x^2y^2 + x^2z^2 + y^2z^2$ as required.

[quote="Singular"]
But $\sum x^2yz \geq \sum (yz)^2$ by Muirhead.[/quote]

If you let $a = xy$, $b = yz$, and $c = zx$ you have $\sum ab \ge \sum a^2$ which is definitely wrong...

we are talking about non-[b]increasing[/b] sequences in majorization

dang, that means that my notes on muirhead is wrong

Let x=1, take logs and use rearrangement inequality  :P

Edit: I guess this is similar to siuhochung's solution, but mine only uses the fact that log is increasing, as opposed to calculus. (notwithstanding the fact that log is defined using calculus....)


When 10! seconds have elapsed, how many weeks have gone by?

Simply: lets divide by $ 60$ (the number of seconds in a minute) multiplied by $ 60$ (the number of minutes in an hour) multiplied by $ 24$ (the number of hours in a day) multiplied with $ 7$ (the number of day in a week). See it visually here:$ \frac{{10!}}{{60 \cdot 60 \cdot 24 \cdot 7}} \equal{} 6$ so its $ 6$ weeks.


LaTeX
geometry
3D geometry
prism
geometric transformation
rotation
vector

[img]http://www.thiel.edu/mathproject/CNAT/Chapt1/Exercises/wpdoc1%7Bimage32%7D.gif[/img]

Given that ray OQ is congruent to ray PV and ray RW, and that ray VT is congruent to OR, show that ray WT is congruent to ray OP.

[Sorry I don't know how to make the ray sign in latex]

[quote="AdithyaGanesh"]With the information given, can we conclude that the figure is a cube?  And by ray, do you mean segment?[/quote]

No I mean ray.  That's what the problem said. =/

Strange.

How about my second question?  Can we conclude that it's a cube?

It doesn't have to be a cube; it could be a rectangular prism (right?).

Yeah..I believe it doesn't have to be a cube.

[quote="grn_trtle"]Aren't all rays congruent? There exists a composition of a rotation and translation that maps any ray to any other ray..[/quote]

Perhaps vector would be a better word, especially since rays are infinite and the given "rays" are not.

If we can assume that the rays are vectors, then my solution works.

[hide="My Solution"]
Two vectors are congruent if their magnitude and direction are congruent. Thus, we have $ \overline{OQ}\parallel{}\overline{PV}\parallel{}\overline{RW}$, $ OQ \equal{} PV \equal{} RW$, $ \overline{VT}\parallel{}\overline{OR}$, $ VT \equal{} OR$. We will use the fact that a quadrilateral with two opposites sides that are parallel and congruent is a parallelogram. (This can be proven easily using SAS.) We can conclude that $ OQVP$ is a parallelogram so that $ \overline{QV}\parallel{}\overline{OP}$ and $ QV \equal{} OP$. We have can also conclude that $ RWQO$ is a parallelogram from the given so $ \overline{WQ}\parallel{}\overline{RO}$ and $ WQ \equal{} RO$. We have $ WQ \equal{} RO \equal{} VT$ and $ \overline{WQ}\parallel{}\overline{RO}\parallel{}\overline{VT}$ so $ WTVQ$ is also a parallelogram. Therefore, $ \overline{WT}\parallel{}\overline{QV}\parallel{}\overline{OP}$ and $ WT \equal{} QV \equal{} OP$, as desired.
[/hide]


Hint: By AM-GM, $ \sqrt{6}|xy^2|\le 2x^2\plus{}3y^4$


inequalities
function
geometry
geometric transformation
rotation
special factorizations

not much time now, maybe take first ineq. (the given) move all to lhs so geq zero, then multiply thru by a+b+c, then move neg terms to rhs
just an idea

[quote="Franz Joseph"]Assume WLOG that $a \leq b \leq c$. We know that $a+b+c>abc$ and since $abc<a^{2}+ab+ac \leq a^{2}+b^{2}+c^{2}$. We deduce that $a^{2}+b^{2}+c^{2}>abc$.

QED.

Is that right? :maybe:[/quote]
Why do you have $abc<a^{2}+ab+ac=(a+b+c)a$? You do not necescerally have $a\ge 1$

[quote="Phelpedo"]However that is a minor hole because if $a \leq b \leq c$ and $a \geq 1$, then $a^{2}+b^{2}+c^{2}\geq a+b+c$.[/quote]

Yes, so the case $a \not\geq 1$ is still totally unaddressed.

No it's not...look at Franz's proof.

$abc< a+b+c< a(a+b+c) < a^{2}+b^{2}+c^{2}.$

Here is a solution involving boring(but easy) case work:
Assume $a\le b\le c$
1. $a< 1$: Then $c^{2}\ge bc\ge abc$ and so $a^{2}+b^{2}+c^{2}> abc$
2. $a\ge 1$. Because $a,b,c\ge 1$ we have $a^{2}\ge a, b^{2}\ge b, c^{2}\ge c$ so $a^{2}+b^{2}+c^{2}\ge a+b+c>abc$
So in both cases $a^{2}+b^{2}+c^{2}>abc$. QED
I am still trying to find a nice solution(ie using am-gm, schur, jensen, etc)

[quote="Phelpedo"]Hm where are you getting the 3 from? I can get it to be $2\sqrt{2}$ but not 3.[/quote]
what do u mean?
$(a^{2}+b^{2}+c^{2})^{2}\ge 3(a+b+c)abc$ by simple am-gm or muirhead...

[quote="Phelpedo"]Hm I just confused myself. Maybe I multiplied out wrong:

$(a^{2}+b^{2}+c^{2})^{2}=a^{4}+b^{4}+c^{4}+2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})= \\ (a^{2}-\sqrt{2}bc)^{2}+(b^{2}-\sqrt{2}ac)^{2}+(c^{2}-\sqrt{2}ab)^{2}\\ \geq 2\sqrt{2}(a^{2}bc+b^{2}ac+c^{2}ab) = 2\sqrt{2}(a+b+c)(abc)$.

Since there is an equality case, $2\sqrt{2}$ should be the upper bound...unless it can't be achieved?[/quote]
I fail to see why the second equality holds.  :huh: Just filling $(a,b,c)=(1,1,1)$ in disproves the second equality and the first inequality.

[quote="Kurt Gödel"][quote="Phelpedo"]Hm I just confused myself. Maybe I multiplied out wrong:

$(a^{2}+b^{2}+c^{2})^{2}=a^{4}+b^{4}+c^{4}+2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}) \rightarrow \\ (a^{2}-\sqrt{2}bc)^{2}+(b^{2}-\sqrt{2}ac)^{2}+(c^{2}-\sqrt{2}ab)^{2}\\ \geq 2\sqrt{2}(a^{2}bc+b^{2}ac+c^{2}ab) = 2\sqrt{2}(a+b+c)(abc)$.

Since there is an equality case, $2\sqrt{2}$ should be the upper bound...unless it can't be achieved?[/quote]
I fail to see why the second equality holds.  :huh: Just filling $(a,b,c)=(1,1,1)$ in disproves the second equality and the first inequality.[/quote]

I shouldn't have put an equal sign. It's not an equality, it's simply the step before the inequality obtained by completing the square:

$a^{4}+2b^{2}c^{2}= (a^{2}-\sqrt{2}bc)^{2}-2\sqrt{2}a^{2}bc$.

[quote="Phelpedo"]I shouldn't have put an equal sign. It's not an equality, it's simply the step before the inequality obtained by completing the square:

$a^{4}+2b^{2}c^{2}= (a^{2}-\sqrt{2}bc)^{2}-2\sqrt{2}a^{2}bc$.[/quote]
I guess that should be $a^{4}+2b^{2}c^{2}= (a^{2}+\sqrt{2}bc)^{2}-2\sqrt{2}a^{2}bc$ or $a^{4}+2b^{2}c^{2}= (a^{2}-\sqrt{2}bc)^{2}+2\sqrt{2}a^{2}bc$. However, this doesn't really help to solve the problem, does it?  :P

Because proving that [quote="Phelpedo"]$(a^{2}-\sqrt{2}bc)^{2}+(b^{2}-\sqrt{2}ac)^{2}+(c^{2}-\sqrt{2}ab)^{2}\geq 2\sqrt{2}(a+b+c)(abc)$ [/quote] or even $(a^{2}+\sqrt{2}bc)^{2}+(b^{2}+\sqrt{2}ac)^{2}+(c^{2}+\sqrt{2}ab)^{2}\geq 2\sqrt{2}(a+b+c)(abc)$, leads to nothing, I think. Elaborate your idea  :)

I'll write my proof more clearly:

I think we can all agree that \[(a^{2}+b^{2}+c^{2})^{2}= a^{4}+b^{4}+c^{4}+2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}) = (a^{2}-\sqrt{2}bc)^{2}+(b^{2}-\sqrt{2}ac)^{2}+(c^{2}-\sqrt{2}ab)+2\sqrt{2}(a+b+c)(abc).\] So, we can say that \[(a^{2}+b^{2}+c^{2})^{2}> 2\sqrt{2}(a+b+c)(abc).\] Since we are given $a+b+c>abc$, we substitute: \[(a^{2}+b^{2}+c^{2})^{2}> 2\sqrt{2}(abc)^{2}.\] We can now take square roots to get

$a^{2}+b^{2}+c^{2}> \sqrt{2\sqrt{2}}abc$, which is a stronger result than the problem asks for. My question was how mysticterminator and rem got 3, which is even stricter than $2\sqrt{2}$.

Ah, thanks for the proof!  :)

[quote="Phelpedo"]My question was how mysticterminator and rem got 3, which is even stricter than $2\sqrt{2}$.[/quote]
So you basically want to prove $(a^{2}+b^{2}+c^{2})^{2}\geq 3abc(a+b+c)$. Expand both sides to get the equivalent \[\sum_{\text{symm}}a^{4}+2\sum_{\text{symm}}a^{2}b^{2}\geq 3\sum_{\text{symm}}a^{2}bc,\] which is true after Muirheading $(4,0,0)\succ (2,1,1)$ once and $(2,2,0)\succ (2,1,1)$ twice.

Is there also an AM-GM way to get it (like rem said)?

[quote="Kurt Gödel"][quote="me@home"]So what exactly the difference between cyclic sums and symmetric sums? Because according to my understanding, you really should be using cyclic, not symmetric, when you say "$\sum_{\text{symm}}a^{4}+2\sum_{\text{symm}}a^{2}b^{2}\geq 3\sum_{\text{symm}}a^{2}bc,$"
Isn't symmetric the one that gives a ton more terms?[/quote]

I'm gonna explain it for three variables: if you have a function $f(a,b,c)$ we say that \[\sum_{\text{cyc}}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b)\ \ \text{and}\ \ \sum_{\text{symm}}f(a,b,c) = f(a,b,c)+f(a,c,b)+f(b,a,c)+f(b,c,a)+f(c,a,b)+f(c,b,a).\] (Generally: cyclic sums of $n$ variables = $n$ terms (cyclic rotation of variables until the first element is back in place) and symmetric sums of $n$ variables = $n!$ terms (all permutations of the variables)) I hope it's clear :P

Anyways, remark that saying that \[2(a^{4}+b^{4}+c^{4})+4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}) = (a^{4}b^{0}c^{0}+a^{4}c^{0}b^{0}+\cdots+c^{4}b^{0}a^{0})+2(a^{2}b^{2}c^{0}+a^{2}c^{2}b^{0}+\cdots+c^{2}b^{2}a^{0}) = \sum_{\text{symm}}a^{4}+2\sum_{\text{symm}}a^{2}b^{2}\geq 3\sum_{\text{symm}}a^{2}bc = 3(a^{2}bc+a^{2}cb+\cdots c^{2}ba) = 6(a^{2}bc+b^{2}ca+c^{2}ab)\] is completely equivalent to saying that \[(a^{4}+b^{4}+c^{4})+2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}) = \sum_{\text{cyc}}a^{4}+2\sum_{\text{cyc}}a^{2}b^{2}\geq 3\sum_{\text{cyc}}a^{2}bc = 3(a^{2}bc+b^{2}ca+c^{2}ab).\] The only difference is that you can't Muirhead the latter  :D[/quote]
Thank you so much for clearing that up for me... I really need to learn Muirhead :wink:

[quote="MysticTerminator"]I used AM-GM?[/quote]

Rem said [quote="rem"]what do u mean?
(a^{2}+b^{2}+c^{2})^{2}\ge 3(a+b+c)abc by simple am-gm or muirhead...[/quote]

This an ineq from Mediterranean Mo 2001, look in Inequality Theorem and formulas section.

With the small corrections for $a \geq 1$. However, mysticterminator's solution gives a much tighter bound: in fact, $a^{2}+b^{2}+c^{2}> 1.7abc$.

By $RMS-AM$, $\sqrt{\frac{a^{2}+b^{2}+c^{2}}3}\geq \frac{a+b+c}3$.
By $RMS-GM$, $\sqrt{\frac{a^{2}+b^{2}+c^{2}}3}\geq \sqrt[3]{abc}$, i.e. $\frac{\left( a^{2}+b^{2}+c^{2}\right)^{3/2}}{3 \sqrt 3}\geq abc$.

Multiplying the two inequalities we get that \[\left( a^{2}+b^{2}+c^{2}\right)^{2}\geq \frac{\sqrt 3 \cdot 3 \sqrt 3}3 \cdot (a+b+c)abc = 3abc(a+b+c) .\]


function
combinatorics proposed
combinatorics

The set $\{1,2,\dots\>,n\}$ is called $P$. The function $f: P \to \{1,2,\dots\>,m\}$ satisfies \[f(A\cap B)=\min (f(A), f(B)).\] What is the relationship between the number of possible functions $f$ with the sum $\displaystyle \sum_{j=1}^m j^n$?

There is a nice and easy solution to this. Too bad I did not think of it...

this has already been posted. see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24010


$O$ is inside $\triangle ABC$,  $PQR$ is the pedal triangle of $O$.

Prove $OA.OB.OC \ge (OP+OQ)(OQ+OR)(OR+OP)$.

Here is my solution:

We have $OR+OQ = OA (sin{\measuredangle{OAB}} + sin{\measuredangle{OAC}}) \leq 2.OA.sin{\frac{A}{2}}$.

Hence, analogously we obtain $OP+OR \leq 2OB.sin{\frac{B}{2}}$, $OQ+OP \leq 2OC.sin{\frac{C}{2}}$. By multiplying all up we have

$RHS \leq 8.OA.OB.OC. \prod{sin{\frac{A}{2}}} \leq OA.OB.OC$

[hide="To Alex"] How is everything going? [/hide]


Is there any way to clear a console in java? I am using Eclipse. I know console control is not Java's strong point because it wasn't designed to do so.. but I'm still hoping there's some way.

I tried something cheap like:

[code]
for(int i=0; i<100; i++)
{
     System.out.println("");
}
[/code]

But that messes up the cursor placement.

Thanks!

You could check if there's a version of curses/ncurses for java. I haven't checked for one, though. (Curses allows easy manipulation of virtual terminals)

Hm, I searched on Google and there's nothing promising. Thanks for the idea, though.
Do you normally have to download curses? How are they used after download?

It's a library. In C++ you would include a header file which gives you access to a bunch of functions that come with the library. In curses's case, the functions allow you to manipulate the terminal. To use a library with C++, you install it (it should come with an installer) and then you can essentially just use it in your programs (although you do need special compile flags).

Java doesn't have any standard libraries that would do what you want, you will have to use some third party software or write your own implementation. This project looks promising but I must say I haven't actually tried it... :).

http://sourceforge.net/projects/javacurses/

On Linux and most unix variants, the following printf code for a form feed will clear the screen:

System.out.println( "\f" );

It does not appear to work on a windows cmd/command shell.

[quote="kevmo314"]Or you can stay cheap.

[code]for(int i=0; i<100; i++)
{
     System.out.println();
}[/code][/quote]

100 is a magic number :)


cheghad tv mibinin?tv ro bishtar mipasandin ya cinema ro?iranian or american or indian or...?
che bazigari ro doos darin?chera?
man montazere javabam :)

If you are going to post something, can't you do it in ENGLISH???


number theory
prime factorization
number theory unsolved

[hide="Solution"] Is $ p$ a prime? If yes: $ \sigma(p^2) \equal{} p^2 \plus{} p \plus{} 1$ because all of $ p^2$'s divisors are $ p^2$, $ p$ and $ 1$.
So, $ \sigma(\sigma(p^2)) \equal{} \sigma(p^2 \plus{} p \plus{} 1)$.[/hide]

Try $ p \equal{} 2$ for the non-equal part. You have made a mistake!

Then suggest a way to fix it. For instance let $ p$ be an odd prime.

Why don't you fix it? How to fix it?
Maybe like: $ \sigma(p^2\plus{}p\plus{}1)\equal{}2p^2$ may be true?

Well for $ n\equal{}3$, it works.
$ \sigma(3^2\plus{}3\plus{}1) \equal{} \sigma(13) \equal{} 13\plus{}1 \equal{} 14$
and for $ 2p^2, 2(3)^2 \equal{} 18$, so no yours isn't true.

[quote="Fermat's Little Turtle"]Well for $ n \equal{} 3$, it works.
$ \sigma(3^2 \plus{} 3 \plus{} 1) \equal{} \sigma(13) \equal{} 13 \plus{} 1 \equal{} 14$
and for $ 2p^2, 2(3)^2 \equal{} 18$, so no yours isn't true.[/quote]

Don't get that 'no yours isn't true'. What have you meant?

[quote="ssibseya"]prove following...

when $ p$ is odd prime,
$ \sigma(n) \equal{} \text{sum of n's divisor} \\
\sigma(\sigma(p^2)) \equal{} \sigma(p^2 \plus{} p \plus{} 1)\neq 2p^2$

In other words, when p is odd prime prove "$ p^2$ is not super perfect"[/quote]
[hide="Solution"]
Assume the contrary.  We prime factor $ p^2 \plus{} p \plus{} 1 \equal{} p_1^{e_1}p^{e_2}\cdots p_n^{e_n}$.  Then,\[ 2p^2 \equal{} \sigma (p^2 \plus{} p \plus{} 1) \equal{} (1 \plus{} p_1 \plus{} p_1^2 \plus{} \cdots \plus{} p_1^{e_1})(1 \plus{} p_2 \plus{} \cdots \plus{} p_2^{e_2})\cdots (1 \plus{} p_n \plus{} \cdots \plus{} p_n^{e_n})\]Yet, since the sum of the exponents of the prime factorization of $ 2p^2$ is three, we have that there at most three such terms in the factorization of $ \sigma (p^2 \plus{} p \plus{} 1)$ shown above.  This means that there are at most three prime factors of $ p^2 \plus{} p \plus{} 1$.  Yet, notice if there were three factors, then for some prime factor of $ p^2 \plus{} p \plus{} 1$, say $ s$, we would have that $ 1 \plus{} s \plus{} s^2\cdots \plus{} s^k \equal{} 2$, which is impossible since $ s\ge 2$.  Thus, there are at most $ 2$ prime factors of $ p^2 \plus{} p \plus{} 1$.  Assume that there are $ 2$ prime factors of $ p^2 \plus{} p \plus{} 1$, say $ q$ and $ r$.  Now, let $ p^2 \plus{} p \plus{} 1 \equal{} q^xr^y$.  Then, we first notice that $ p^2 \plus{} p\equiv 0\bmod 2$, so $ 2$ is not a factor of $ p^2 \plus{} p \plus{} 1$, which means that $ q$ and $ r$ are odd, and thus, above or equal to $ 3$ since they are primes.  Now, we get that\[ \sigma (p^2 \plus{} p \plus{} 1) \equal{} (q^x \plus{} q^{x \minus{} 1} \plus{} q^{x \minus{} 2}\cdots \plus{} 1)(r^y \plus{} r^{y \minus{} 1}\cdots \plus{} 1)\]which means that\[ \sigma (p^2 \plus{} p \plus{} 1)\le q^xr^y(1 \plus{} \frac {1}{q} \plus{} \frac {1}{q^2}\cdots )(1 \plus{} \frac {1}{r} \plus{} \frac {1}{r^2}\cdots)\]so\[ 2p^2\le (p^2 \plus{} p \plus{} 1)(\frac {q}{q \minus{} 1})(\frac {r}{r \minus{} 1})\]Since $ r\ne q$ and $ r$ and $ q$ are primes, we see that $ \frac {q}{q \minus{} 1}\cdot \frac {r}{r \minus{} 1}\le \frac {3}{2}\cdot \frac {5}{4}$.  Hence,\[ 2p^2\le (p^2 \plus{} p \plus{} 1)\frac {15}{8}\implies 16p^2\le 15p^2 \plus{} 15p \plus{} 15\implies p^2 \minus{} 15p \minus{} 15\le 0\]This means that $ p\le \frac {15 \plus{} \sqrt {285}}{2}$, so $ p\le 15$.  This yields possibilities $ p \equal{} 3, 5, 7, 11, 13$.  We can check these five cases to see that they fail.

Now, if $ p^2 \plus{} p \plus{} 1$ has only one prime factor, say $ q$, then we let $ p^2 \plus{} p \plus{} 1 \equal{} q^x$.  We do the exact same thing as above to get that\[ 2p^2\le (\frac {q}{q \minus{} 1})(p^2 \plus{} p \plus{} 1)\le \frac {3}{2}\cdot (p^2 \plus{} p \plus{} 1)\implies p^2 \minus{} 3p \minus{} 3\le 0\]This means that $ p\le 3$.  Yet, since $ p$ is odd by the problem statement, we see that $ p \equal{} 3$, which is a contradiction after checking.  Hence, $ \sigma (p^2 \plus{} p \plus{} 1)\ne 2p^2$, as desired. [/hide]

[hide="proof"]
Lemma: $ \sigma(n)\not\equal{}2$
Proof:
Clearly, $ \sigma$ is multiplicative. Then we can write:
suppose $ \sigma(n)\equal{}2$. Then:\[ \sigma(p^a)\sigma(q^b)\cdots\equal{}(p^a\plus{}...p\plus{}1)(q^b\plus{}...q\plus{}1)(\cdots)\equal{}2\implies n\equal{}p^a\implies p\equal{}1\] contradiction.

Suppose first that $ p^2\plus{}p\plus{}1\equal{}q^a$. Then if $ a\equal{}1$, $ \sigma(p^2\plus{}p\plus{}1)\equal{}p^2\plus{}p\plus{}2<2p^2$ for odd primes $ p$.
If $ a>1$ then \[ \sigma(p^2\plus{}p\plus{}1)\geq (p^2\plus{}p\plus{}1)^2\plus{}p^2\plus{}p\plus{}2>2p^2\plus{}p\plus{}2>2p^2\]

Thus we can write $ ab\equal{}p^2\plus{}p\plus{}1$ with $ (a,b)\equal{}1$. So $ \sigma(p^2\plus{}p\plus{}1)\equal{}\sigma(a)\sigma(b)\equal{}2p^2$
Since $ 2$ is not in the range of $ \sigma$, we must have:
\[ \sigma(a)\equal{}p,\sigma(b)\equal{}2p\]
Thus $ a,b$ are prime powers.

Suppose there exists one of them that is greater than exponent $ 1$; then \[ 2p^2\geq (a^2\plus{}a\plus{}1)(b\plus{}1)>(p^2\plus{}p\plus{}1)(a\plus{}1)\geq2(p^2\plus{}p\plus{}1)>2p^2\] as $ a\equal{}0$ is impossible (because then the prime would be $ 1$)

Thus both $ a,b$ are prime - thus odd primes, as $ ab$ is odd.

But then $ \sigma(a)\equal{}a\plus{}1\equal{}p$ contradiction.
[/hide]

[quote="me@home"][hide="proof"]
Lemma: $ \sigma(n)\not \equal{} 2$
Proof:
Clearly, $ \sigma$ is multiplicative. Then we can write:
suppose $ \sigma(n) \equal{} 2$. Then:
\[ \sigma(p^a)\sigma(q^b)\cdots \equal{} (p^a \plus{} ...p \plus{} 1)(q^b \plus{} ...q \plus{} 1)(\cdots) \equal{} 2\implies n \equal{} p^a\implies p \equal{} 1
\]
contradiction.

Suppose first that $ p^2 \plus{} p \plus{} 1 \equal{} q^a$. Then if $ a \equal{} 1$, $ \sigma(p^2 \plus{} p \plus{} 1) \equal{} p^2 \plus{} p \plus{} 2 < 2p^2$ for odd primes $ p$.
If $ a > 1$ then
\[ \sigma(p^2 \plus{} p \plus{} 1)\geq (p^2 \plus{} p \plus{} 1)^2 \plus{} p^2 \plus{} p \plus{} 2 > 2p^2 \plus{} p \plus{} 2 > 2p^2
\]
Thus we can write $ ab \equal{} p^2 \plus{} p \plus{} 1$ with $ (a,b) \equal{} 1$. So $ \sigma(p^2 \plus{} p \plus{} 1) \equal{} \sigma(a)\sigma(b) \equal{} 2p^2$
Since $ 2$ is not in the range of $ \sigma$, we must have:
\[ \sigma(a) \equal{} p,\sigma(b) \equal{} 2p
\]
Thus $ a,b$ are prime powers.

Suppose there exists one of them that is greater than exponent $ 1$; then
\[ 2p^2\geq (a^2 \plus{} a \plus{} 1)(b \plus{} 1) > (p^2 \plus{} p \plus{} 1)(a \plus{} 1)\geq2(p^2 \plus{} p \plus{} 1) > 2p^2
\]
as $ a \equal{} 0$ is impossible (because then the prime would be $ 1$)

Thus both $ a,b$ are prime - thus odd primes, as $ ab$ is odd.

But then $ \sigma(a) \equal{} a \plus{} 1 \equal{} p$ contradiction.
[/hide][/quote]

Actually, I get the solution already. Your simple  but nice Idea p^2+p+1=a^~*b^~ is nice. Thanks!!!

[quote="me@home"][hide="proof"]
Lemma: $ \sigma(n)\not \equal{} 2$
Proof:
Clearly, $ \sigma$ is multiplicative. Then we can write:
suppose $ \sigma(n) \equal{} 2$. Then:
\[ \sigma(p^a)\sigma(q^b)\cdots \equal{} (p^a \plus{} ...p \plus{} 1)(q^b \plus{} ...q \plus{} 1)(\cdots) \equal{} 2\implies n \equal{} p^a\implies p \equal{} 1
\]
contradiction.

Suppose first that $ p^2 \plus{} p \plus{} 1 \equal{} q^a$. Then if $ a \equal{} 1$, $ \sigma(p^2 \plus{} p \plus{} 1) \equal{} p^2 \plus{} p \plus{} 2 < 2p^2$ for odd primes $ p$.
If $ a > 1$ then
\[ \sigma(p^2 \plus{} p \plus{} 1)\geq (p^2 \plus{} p \plus{} 1)^2 \plus{} p^2 \plus{} p \plus{} 2 > 2p^2 \plus{} p \plus{} 2 > 2p^2
\]
Thus we can write $ ab \equal{} p^2 \plus{} p \plus{} 1$ with $ (a,b) \equal{} 1$. So $ \sigma(p^2 \plus{} p \plus{} 1) \equal{} \sigma(a)\sigma(b) \equal{} 2p^2$
Since $ 2$ is not in the range of $ \sigma$, we must have:
\[ \sigma(a) \equal{} p,\sigma(b) \equal{} 2p
\]
Thus $ a,b$ are prime powers.

Suppose there exists one of them that is greater than exponent $ 1$; then
\[ 2p^2\geq (a^2 \plus{} a \plus{} 1)(b \plus{} 1) > (p^2 \plus{} p \plus{} 1)(a \plus{} 1)\geq2(p^2 \plus{} p \plus{} 1) > 2p^2
\]
as $ a \equal{} 0$ is impossible (because then the prime would be $ 1$)

Thus both $ a,b$ are prime - thus odd primes, as $ ab$ is odd.

But then $ \sigma(a) \equal{} a \plus{} 1 \equal{} p$ contradiction.
[/hide][/quote]

but $ p^2 \plus{} p \plus{} 1 \equal{} q^2$ --->
\[ \sigma(p^2 \plus{} p \plus{} 1) \equal{} (p^2 \plus{} p \plus{} 1)^2 \plus{} p^2 \plus{} p \plus{} 2
\]
is somewhat wrong thought to me. Why such a so???


[quote="designtheory"]Solutions to problems in this issue should arrive no later than 29 th October, 2009. Solutions received after this date will only be considered if there is time before the publication of the solutions. Include your name, affiliation for each submission.

Each problem is given in English and Vietnamese to serve international readership.

Email: mathvn2008@gmail.com

Read more: http://mathvn.org/viewpage.php?page_id=4[/quote]

The problems are not original,some of them were on CRUX,and most of the highschool part I have seen before.


Compute

$ \sum_{k,n\equal{}1}^\infty \frac{1}{k^2\, n^2\, (k\plus{}n)^2}$

The answer should be $ \frac13 \zeta(6)$ . Can you prove this?

An approach, which I don't know how practical it is:
$ \sum_{k,n \equal{} 1}^\infty \frac {1}{k^2\, n^2\, (k \plus{} n)^2} \equal{} \sum_{n \equal{} 1}\frac {1}{n^2}\sum_{a \plus{} b \equal{} n}\frac {1}{(ab)^2}$
Let $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {x^n}{n^2}, g(x) \equal{} f^2(x) \equal{} \sum_{n \equal{} 1}^{\infty} x^n \sum_{a \plus{} b \equal{} n}\frac {1}{(ab)^2} \equal{} \sum a_ix^i$.
We want to calculate $ h(1)$, where $ h(x) \equal{} \sum \frac {a_ix^i}{i^2}$.
$ x(h'(x)x)' \equal{} g(x)$
$ x(f'(x)x)' \equal{} \frac {1}{1 \minus{} x} \implies f'(x) \equal{} \frac { \minus{} \ln(x \minus{} 1)}{x}$
$ g(x) \equal{} f^2(x)$
So we may be able to find $ h$ explicitly.

Turns out that $ f$ is "famous" and called Polylogarithm: $ f(z) \equal{} Li_2(z)$.
So $ x(h'(x)x)' \equal{} Li_2(x)^2, h(0) \equal{} 0, h'(0) \equal{} 0$.


A glassy-eyed student stared at an "$x_{1}y_{2}$," which was written on the scratch paper. After some doodling the student wrote
$x_{1}y_{2}\\ \\ (1) x_{1}y_{2}y_{3}x_{4}\\ \\ (2) x_{1}y_{2}y_{3}x_{4}y_{5}x_{6}x_{7}y_{8}...$

Where each successive line is formed by copying the line above, then copying it again backwards and continuing the subscripts.

Prove that in line $n$, the sum of the $n$-th powers of the x-subscripts is equal to the sum of the $n$ powers of the y-subscripts.

I honestly do not understand the question, can't even get through it!?

the sum of the n-th powers of the x-subscripts?
What does that means?

oh, sorry. I am still having problem with my English... :(

But the equality for line 2 doesn't hold... :?:

It's more similar to http://www.mathlinks.ro/Forum/viewtopic.php?t=30998 than one may think...


WLOG a+b+c=1, then we can use isolated fudging...

$\frac{a^{2}}{1-a}\ge ma+n$

Determine $m$ and $n$ such that $m+3n=\frac{1}{2}$ and $\frac{a^{2}}{1-a}-ma-n$ has a double root at $a=\frac{1}{3}$

[hide="Try this"]
$\frac{a^{2}}{b+c}+\frac{b+c}{4}\geq a$

(It's Cauchy)

The rest is for you  :) [/hide]


$a,b$ are positive integers and $2^{n}a+b$ is perfect square for all of the natural numbers $n$. prove: $a=0$

Dude, this has been discussed like four or five times already :).


vector
linear algebra
linear algebra unsolved

1. Let Z denote the set of all intergers with addition defined in the usual way and define the scalar multiplication, denoted *, by

alpha*k = [[alpha]] * k    for all k elements in Z

where [[alpha]] denotes the greatest interger less than or equal to alpha.
Show that Z, together with these operations, is not a vector space. Which axioms fail to hold?

Solution? From what I assume the axiom must have to do with scalar multiplication. Since Axioms 6 and 7 (alpha + beta)x = alpha x + beta x  for any scalars... and... (alpha X beta)x = alpha(beta X x)   for any scalars. By rounding these scalars down in the middle of the equations, it causes the axioms to not hold for all values?

Just wondering if I'm anywhere near the right track.

Yes. Just produce a counterexample for each axiom, and show that the others hold.

A pair of [hide="counterexamples"]$a=1.5,b=1,x=1$: $ax+bx=2,(a+b)x=3x=3$
and $a=1.5,b=1.5,x=1$: $(ab)x=2.25x=2$, $a(bx)=a(1.5\cdot 1)=a\cdot 1=1$.[/hide]


I am confused about the meaning of the question.

Q) Which of the following will give always the SN2 mechanism in acidic as well as basic conditions?

A) $ CH_3 \minus{} O \minus{} CH_3$
B) $ C_6H_5 \minus{} O \minus{} CH_3$
C)$ CH_3 \minus{} CH \minus{} CH_2$ ( Sad attempt at making an epoxide)
 ................... \_O _/

I have never wondered about reaction medium i.e. acidic or basic for SN2, I mean I only care about protic/aprotic ,polar/non polar but how does acidity or basicity matter?

The cleavage of open ethers occurs only at acidic conditions. However, epoxides open not only in acidic conditions but also in basic ones. Also,

[quote="Conjurer"]how does acidity or basicity matter?[/quote]

the product obtained from the epoxide depends on the conditions - in other words, the direction of opening is different.

Can you explain the epoxide case. But the answer is all of these.

What do you mean by the answer is all of these?

The eppxode case: in basic conditions the least substituted carbon is attacked, while in acidic conditions it is the most substituted carbon that suffers attack. Now is your turn to explain why.

Sir, the answer says all these 3 substrates will always give SN2 mechanism in acidic as well as basic medium.

Then your source is wrong: dimethyl ether is unreactive towards bases. Even anisole I doubt it will react in alkaline medium. The solutions in a book are not always correct.

[quote="Carcul"]Then your source is wrong: dimethyl ether is unreactive towards bases. Even anisole I doubt it will react in alkaline medium.[b] The solutions in a book are not always correct[/b].[/quote]

Yes, especially Indian books!  :lol:  :D


Wanted to know which rounds (sprint, target, countdown) typically used by schools to determine their MC team?

I can't say really, but my school uses sprint+target+AMC 8. I think the AMC 8 gets a good amount of weight because it shows who can perform under the pressure.

We used to do sprint and targets but this year was different. We did 2 sprints and a set of targets.

It varies alot between teams.

Last year, we used the AMC 8, and half the school sprint and half the target.
Two years ago, we used half the sprint (there weren't many ppl to choose from though)
Three years ago, we used rankings from MOEMS, I think.

We used 5 school tests to pick our four, then Mr. Frost used team rounds (done individually by around 10 people that scored high), and saw how hard they were actually working, to pick the alternate 4. We didn't do a countdown.

Last year the other coach only did one school test (2006) to determine the team, then did a countdown.

It sort of depends.

This year, our school used half the Sprint Round, the Target, and the countdown.

Then our coach had the top 8 do the team round individually and then the top 4 made it.

target+sprint.... simple and more simple

our school use
2 sets(sprint+target) of school
2 sets of chapter
and 1 set of state( to really separate the people that can solve tough problems vs. the only can solve easy ones)
and 2 rounds of cd for fun

then the chosen top 4(the team get extra stuff to do)
the second 4 get less stuff
and the 2 subs get little extra problems to work on

[quote="bpms"][quote="jhredsox"][quote="Walk Around The River"]we use...

the school sprint round
a random chapter sprint round

and avg those.  highest 9 are on team, but the team must have at least 3 7th/6th graders.

normally top 4 is obvious, or we just take a few more tests to decide.

alternate is the 6th grader if one makes it, or whichever 7th grader we want in the hotel at state with us.  :P[/quote]

lucky!, you get a hotel!

-jorian[/quote]
Everyone in his state does because his state is so big that some people have to drive like 5 hours. You wanna wake up at 2 to drive 5 hours for a competition?[/quote]

yah, it takes us like 5 hours,  so we actually spend TWO nights in the hotel.

[quote="Walk Around The River"][/quote][quote="bpms"][quote="jhredsox"][quote="Walk Around The River"]we use...

the school sprint round
a random chapter sprint round

and avg those.  highest 9 are on team, but the team must have at least 3 7th/6th graders.

normally top 4 is obvious, or we just take a few more tests to decide.

alternate is the 6th grader if one makes it, or whichever 7th grader we want in the hotel at state with us.  :P[/quote]

lucky!, you get a hotel!

-jorian[/quote]
Everyone in his state does because his state is so big that some people have to drive like 5 hours. You wanna wake up at 2 to drive 5 hours for a competition?[/quote][quote="Walk Around The River"]

yah, it takes us like 5 hours,  so we actually spend TWO nights in the hotel.[/quote]
Okay, that seems pointless. The competition ends at like 4:00. Anyways, are the two nights paid for by fleng, your school, or you guys? I know fleng pays for one night for the people from our county.

[quote="bpms"][/quote][quote="Walk Around The River"][/quote][quote="bpms"][quote="jhredsox"][quote="Walk Around The River"]we use...

the school sprint round
a random chapter sprint round

and avg those.  highest 9 are on team, but the team must have at least 3 7th/6th graders.

normally top 4 is obvious, or we just take a few more tests to decide.

alternate is the 6th grader if one makes it, or whichever 7th grader we want in the hotel at state with us.  :P[/quote]

lucky!, you get a hotel!

-jorian[/quote]
Everyone in his state does because his state is so big that some people have to drive like 5 hours. You wanna wake up at 2 to drive 5 hours for a competition?[/quote][quote="Walk Around The River"]

yah, it takes us like 5 hours,  so we actually spend TWO nights in the hotel.[/quote][quote="bpms"]
Okay, that seems pointless. The competition ends at like 4:00. Anyways, are the two nights paid for by fleng, your school, or you guys? I know fleng pays for one night for the people from our county.[/quote]

Fleng is Florida Engeneering.  they sponsor our mathcounts.

and bpms, we dont pay for any of it. so i guess fleng does.  all the big bend teams stay two nights.

[quote="mathgeniuse^ln(x)"][quote="Ubemaya"]Well CA is split into two competitions and people drive up to like 7 hours. We only have to drive 3![/quote]

Why were is yours held?

Seems to me for you it is UC-Davis, but I don't know.[/quote]

Yeah for people living in the Bay Area, States are at UC Davis.

As for the original question. First we have a preliminary test to allow 20 people to attend meetings. ( 200 people try out) After that the coach uses the school round (only sprint and target) to determine the top 8.

We're lucky if we get 8...this is my schools second year of MC-I pushed for it in 7th grade-last year, we had 4, and did awfully. MC attracted more people this year(about 15) b/c I went to nats last year, and others were "inspired". Our team is determined by performance on about 5 warmups and workouts from the handbook, the entire individual school comp., and scores on 5 team rounds.(2 chap., 2 state, and 1 nat.)

My school just uses any random mathcounts problem they can find. A lot of years some questions are the same.

[quote="perfect628"]You're judged on a single problem?[/quote]
I think he meant sets of problems...

[quote="laughinghead505"]I don't know how we're doing it this year, but last year we used sprint and countdown. The year before that I think we just did countdown.

My school competition is on Wednesday! I'm so nervous...[/quote]
I just did my school competition today (probably failed... :P ) and we used some team rounds, sprint rounds and target rounds.


The natural numbers $2^1$ through $2^{25}$ are placed on a 5 x 5 magic square such that the product of all rows are equal. What is this product?

[hide]
Why do you do
$5 * 13$ ?
Is it because the median is 13?  Why does that work?

I was thinking this problem could be proken down into a Magic Square Sum since 2 is the base of every number, all of the exponents must add up to be the same in each row.  So instead of a product it can be simplified to a Magic Square Sum placing the numbers 1 through 25 in 5 rows that add up to the same number.  I wasn't really sure where to go from there though.

Thanks
[/hide]

Good answer. Think of it like this:
call the products of the rows $a$

Therefore the first row multiplies to $a$
                the second row multiplies to $a$
So on
                the fifth row multiplies to $a$

Therefore, if we multiply all these rows together, we get $a^5$
But also, we will have multiplied all the elements of the matrix once. (think about it)
This is equal to $2^{\frac{25.26}{2}}=2^{13.25}=a^5$

So we get $a=2^{13.5}=2^{65}$

[hide]we know that the product of the numbers from 2^1-2^25 are in the magic square so the product of all the numbers is 2^(26*25)/2=2^325. Since there are 5 rows in a 5x5 square and we know each row is equal to the others, the product of an individual row is 2^325/5=2^65. [/hide]

[hide]

there are five rows. all the rows multiply to be the product of ALL THE NUMBERs, $2^{13*25}$.

so it is $2^{13*25/5} = 2^{65}$
[/hide]


See here: http://www.tpub.com/math1/10a.htm

or here: http://www.themathpage.com/alg/algebraic-expressions.htm

It is an expression consisting of one or more numbers and variables along with one or more arithmetic operations.

It is an expression containing variable(s) that does not have an equals sign to it, or any other signs in that family (like inequalities).   :wink:

An algebraic expression is a value.  Any value.

If you make a statement comparing two algebraic expressions, it becomes an equation or inequality.
For example,

2x+3 is an algebraic expression
4 is an algebraic expression
2x+3=4 is an equation
2x+3>4 is an inequality


A polynomial in $ n$ variables has the property that if each variable is given one of the values $ 1$ and $ \minus{}1$, then the result is positive whenever the number of variables set to $ \minus{}1$ is even and
negative when it is odd. Prove that the degree of the polynomial is at least $ n$.

Do not double post: http://www.mathlinks.ro/viewtopic.php?p=1386752#1386752


An airline reported a total of 2500 flight departures and, of those, 1760 departed on time. What percent of the
airline's flight departures were on time? Express your answer to the nearest tenth.

[hide="Solution"]
Divide to get answer:
$ \frac{1760}{2500}\equal{}.704\approx70 \text{ percent}$
P.S. It looks like you are an AoPS bot, right?
[/hide]

[quote="minicon"][hide="Solution"]
Divide to get answer:
$ \frac {1760}{2500} \equal{} .704\approx70 \text{ percent}$
P.S. It looks like you are an AoPS bot, right?
[/hide][/quote]

Read carefully  :wink:

[quote="GameBot"]Express your answer to the nearest tenth[/quote]

Oops, then it would be [hide="this."]$ 70.4 \text{ Percent}$[/hide]


probability
LaTeX
logarithms
calculus
irrational number

An urn contains $100$ balls numbered $1$ through $100$. Repeat this experiment $n$ times: pick a ball at random from the urn and then return it to the urn.

Compute the smallest $n$ for which the probability is greater than $\frac{1}{2}$ that the same numbered ball is picked at least two times.

[hide="Solution, possibly wrong; I am not great at the involved probability problems."]The number of (order matters) ways to select the same ball twice in a sequence of $n$ removed balls is $100^{n-2}(100)(1)$ times $\binom{n}{2}$ places in the sequence you can take out the ball twice. This is out of $100^{n}$ ways to select (order matters) the sequence of 100 balls, so the probability that the sequence satisfies the condition for $n$ repetitions is
\begin{eqnarray*}\frac{100^{n-1}\binom{n}{2}}{100^{n}}&=&\frac{\binom{n}{2}}{100}\\ &=&\frac{n(n-1)}{200}\end{eqnarray*}
$\begin{eqnarray*}\frac{n(n-1)}{200}&>&\frac{1}{2}\\ n(n-1)&>&100\\ \end{eqnarray*}\boxed{n>10}$[/hide]

And I am officially ready to rip apart the next dollar sign I see because this LaTeX just isn't cooperating.

I'm not sure, but I think
\[100^{n-2}*100*\binom{n}{2}\]
doesn't correctly count the number of cases. ex: it will overcount the cases where there are 3 balls of the same color. I think you have to use PIE here.

well if you do the experiment twice there's a $(\frac{100}{100})(\frac{99}{100})$chance that no numbers are repeated.

thrice is $(\frac{100}{100})(\frac{99}{100})(\frac{98}{100})$

four times is $(\frac{100}{100})(\frac{99}{100})(\frac{98}{100})(\frac{97}{100})$

man i see a pattern

and it goes down to 11 times, that's when that expression is greater than $\frac12$

so i'd say it's [b]11[/b]

The problem is very famous often asked as :    about  "probability of  (at least ) two people in a group of $n$  having the same birthday"

(If "100  balls", is changed to "365 days" the problem, often asked what the size of the group should ($n$) be that the probability is > (.5)

A Chalange question: Can some show that the rough estimate for the answer for 'x' balls is about  (actually  less than ) $\sqrt{1.4 \cdot x }$ (for large x's) (In the case of 100 balls the answer is   sqrt(140)  or 11 (taking the smaller integer part..

well, just looking at it, i'd say it has to do with the fact that $1.4$ is really close to $\sqrt2$

Well if you have to guess this irrational number ....here's  a hint,  the constant is  not sqrt(2) but    closer to $1.38629436$

Edited later:

Better approximation for n for $x$ balls:
$\approx (1/2)+\sqrt{1/4+2 \ln (2) \cdot x}$

Ugh... blasting through some pretty BSed calculus gives me $\ln 4$  :maybe: so that seems to be consistent with what you have. I'm curious to see the actual proof of that.

[quote="The Zuton Force"]ives me $\ln 4$  :maybe: so that seems to be consistent with what you have. I'm curious to see the actual proof of that.[/quote]
Yep that's right..
Okay, here is a proof, hope it is simple enough. I am assuming one is familiar with:

[hide="hnt"]
When $n$ is large,
 $(1+1/n)^{n}\approx e$

[hide="Rest is easy"]So Treething's part above becomes:
$(1-1/100)(1-2/100)(1-3/100) \dots (1-(n-1)/100)$
$\approx e^{-1/100}\cdot e^{-2/100}\dots = e^{-n(n-1)/200}\approx e^{-n^{2}/200}$

(Again we assume n is large enough that $n(n-1) \approx n^{2}$ will work)

So we have to solve

$e^{n^{2}/200}= (1/2) = e^{-ln(2)}$
QED[/hide] [/hide]

I'm confused, don't you want the smallest $n$ so that
\[1-(\frac{100}{100})(\frac{99}{100})(\frac{98}{100})\cdots \]
is greater than $\frac{1}{2}$? Because I keep getting 13 for some reason  :maybe:

Yes, the answer is $13$,.

 For $12$ probability is about 0.497, for 13 it is  about .557

Actually the formula I gave,(for approximation)   (taking n(n-1) instead of n^2) one gets  the probability values for 12 as about  (0.48)( and for 13 as (0.54) close enough to  actual answers ..

"Better' answer than just $\sqrt{139}$ is given by  (solving $n^{2}-n =139$) or

$n = (1/2)+\sqrt{(1/4)+139}$

As mentioned before : This, is  not exact,  just an approximation, but easier to calculate than 12 or 13 multiplications and close enough as long as the number of balls are high)


Haha I think I know the answer.  But i don't think I wanted to know the answer :(

[quote="lightrhee"][quote="Elyot"]I'm just learning LaTeX, so I thought this would be a good chance for me to practice it...

If you get this, please don't post the solution/answer... I want to see how many of you actually figure this out... the clue is "Take me to the limit!"

$\displaystyle\lim_{u\to\infty}\int e^x = f(u^n)$

Have fun with this one boys and girls![/quote]

What is the thing that we need to get? Are we looking for what [i]f[/i] is?[/quote]

You're supposed to figure out what the question it.   :harhar:

[quote="Sunny"][quote="Elyot"]It's kinda a riddle.  Just figure it out.  I'm not gonna say anymore.

We should place bets on how long until David figures it out...[/quote]

2 days.  He'll go crazy meantime.[/quote]

I'd vote for at least a week.  I actually saw this on a T-shirt worn by a former math teacher of mine... the shirt was given to him by his SON-IN-LAW!!!

I take it you got this one pretty quickly though!  Go back and read my original post now and you'll realize all the subtle hints I gave...

[quote="Elyot"][quote="Sunny"][quote="Elyot"]

I'd vote for at least a week.  I actually saw this on a T-shirt worn by a former math teacher of mine... the shirt was given to him by his SON-IN-LAW!!!

I take it you got this one pretty quickly though!  Go back and read my original post now and you'll realize all the subtle hints I gave...[/quote]

Oh wait...knowing David...i'll move my bet to 2 weeks.  Yeah...maybe its not a good thing I saw it so quickly...corrupted by Elyot Grant.  lol.[/quote]

;)  And the fact that the script is called LaTeX!! LMAO![/quote]

Wow elyot...just Wow.  :)

[quote="Elyot"][quote="Sunny"][quote="Elyot"]

;)  And the fact that the script is called LaTeX!! LMAO![/quote]

AH HAHAHAHA.  Reader Discretion is advised in this thread.  Ah hahaha...[/quote]

Shhh!!!  You're gonna give it away![/quote]

Nah, she'll probably just make Dave more frustrated.

Sorry dave, but I'm thinking you won't get it until next year's seminar time...when you turn 18.

[quote="TheDreamer"]

Sorry dave, but I'm thinking you won't get it until next year's seminar time...when you turn 18.[/quote]

Heh...i doubt he'll get it even then...so innocent....

[quote="pengshi"]wow, this thread has grown so quickly! for those who are trying to grunge this integral with maple, here are 2 vital hints.

hint 1: this puzzle contains an undisputed and universal fact of life.

hint 2: integral=Sum[/quote]

Wow, thanks!  I finally got it after getting those hints.  No wonder why my original answer didn't make sense.

[quote="pengshi"]this puzzle contains an undisputed and universal fact of life.
[/quote]

Do you truly know this for a FACT?  Got something to tell us???

;)

just curious is the solution for this puzzle kind dirty?

[quote="Sunny"][quote="opple"]is it saying that[hide] limitless sex is fun[/hide]?[/quote]

Heh, you got it.

Er...not many posts while I was away........[/quote]

No kidding, considering 90% of the posts' makers were gone...

Anyways welcome back guys, for the short time, anyway ^_^

[quote="Sunny"][quote="Elyot"]Bwahahahah!  Yes, that's the answer.  Good to see you're back posting on AOPS so soon, Sarah.

I think there was a famous mathematician that said you can find the most fundamental of all truths through mathematics.  :P[/quote]

Heh, I'm waiting for you to post times you're at stampede! And if you have grandstand tickets, or just ride tickets.[/quote]

Felix says he'll get us all that info tomorrow (hopefully).
I'll let you know ASAP.

[quote="Sunny"][quote="Elyot"]
Felix says he'll get us all that info tomorrow (hopefully).
I'll let you know ASAP.[/quote]

Awesome.  Just hope its not raining when you go.[/quote]

You'd come anyway, right?  I mean... this will be the last time you get to see us for like... a year!

[quote="Elyot"]

You'd come anyway, right?  I mean... this will be the last time you get to see us for like... a year![/quote]

If its raining, are you going?  Of course, if your in Calgary again I'm there!

[quote="Sunny"]
If its raining, are you going?  Of course, if your in Calgary again I'm there![/quote]

Probably... I don't know... there's no real concrete plans... Felix is really handwavy with the plans... but they do have the outings definitively planned.... I think.  Whatever.  If we don't show up, then you'll get left out in the rain perhaps?  I'm sure somebody has a cell phone or something maybe.  Idunno.  I'll talk to Felix tomorrow.


Prove that if $a,b,c>0$ we have: $ \sum \frac{ab}{a+b} \leq \frac{a+b+c}{2}$.

Here is what I found:
we have that:
$ \sum \frac{2ab}{a+b} \leq \sum \frac{a+b}{2}=a+b+c$.

do you have other sols? :D :D
cheers!

well I guess that's the most obvious solution, obvious enough not to look for others

Honestly, I think you should not post the sols right away with the problems :D


floor function
limit
number theory unsolved
number theory

Let $ \theta$ be real and $ 0 < \theta < 1$. Define

$ g_n = \left \{\begin{array}{cc} 0 & \; \text{if} \; \lfloor n\theta \rfloor = \lfloor (n-1)\theta \rfloor  & 1 &  \; \text{otherwise}  \end{array} \right \;$

Prove that

$ \displaystyle\lim_{n\to\infty}\frac{g_1+g_2+...+g_n}{n} = \theta$

[hide]
g_1 + g_2 + ... + g_n = the number of positive terms in the sequence g_i where i is a positive integer that is less than or equal to n. This number if the floor of theta times n, and the limit of this/n as n approaches infinity is theta. [/hide]

Could you explain how you got to here:

"This number if the floor of theta times n, and the limit of this/n as n approaches infinity is theta."