algebra
polynomial
modular arithmetic
function
number theory proposed
number theory

Find all polynomials $ f\in\mathbb Z[x]$ such that for each $ a,b,x\in\mathbb N$
\[ a\plus{}b\plus{}c|f(a)\plus{}f(b)\plus{}f(c)\]

[quote="Omid Hatami"]Find all polynomials $ f\in\mathbb Z[x]$ such that for each $ a,b,x\in\mathbb N$
\[ a + b + c|f(a) + f(b) + f(c)
\]
[/quote]
Really nice problem . Here is my solution .
From  $ f(c)\equiv f( - a - b) (\mod a + b + c)$ then we have
\[ a + b + c|f(a) + f(b) + f( - a - b) (\mod a + b + c)
\]
For each pair of $ (a,b)\in \mathbb{N}$ , if $ f(a) + f(b) + f( - a - b)$ is different 0 then when we  choose $ c$ is large enough then the condition $ a + b + c|f(a) + f(b) + f( - a - b)$ is not hold .
Therefore we must have $ f(a) + f(b) + f( - a - b) = 0,\forall a,b\in \mathbb{N}$
Take $ a = b$ then we have :
\[ f( - 2a) + 2f(a) = 0,\forall a\in \mathbb{N}
\]
By compare leading coefficient we have $ \tex{deg} P = 1$
Then easy to check that $ f(x) = ax$ with a is a integer .

here is my solution:
we put $ a \equal{} b \equal{} 0$ and we get $ c |2f(0) \plus{} f(c) \equiv 3a_0 \pmod {c} \implies c |3a_0$.
so the number $ 3a_0$ has infinitely many divisors which means that $ a_0 \equal{} 0 \implies f(0) \equal{} 0$.
now we put $ c \equal{} 0,a \equal{} kb,b \equal{} b \implies (k \plus{} 1)b |f(b) \plus{} f(kb)$.
now let's suppose that $ f(x) \equal{} x.q(x),q(x) \in Z[x]$.
let's assume $ deg(q(x)) > 0$ and $ q(x) \equal{} a_nx^{n \minus{} 1} \plus{} ... \plus{} a_1$.where $ n > 1,a_n \neq 0$.
we must have $ (k \plus{} 1)b | b.q(b) \plus{} kb.q(kb)$.
now we choose $ k$ s.t. $ \gcd(k \plus{} 1,b) \equal{} 1$.
so we must have $ k \plus{} 1 | q(kb) \minus{} q(b) \equal{} b^{n \minus{} 1}.a_{n}(k^{n \minus{} 1} \minus{} 1) \plus{} ... \plus{} b.a_2(k \minus{} 1) \rightarrow k \plus{} 1 | \sum_{i}b^{2i \minus{} 2}.a_{2i}$.
where the summation is done over all positive integer $ i$ s.t. $ 1 \le 2i \le n \minus{} 1$.
then we have just to choose an integer $ b$ for which the summation isn't zero(obviously it exists) and take a sufficient large $ k$ s.t. $ |k \plus{} 1| > |\sum_{i}b^{2i \minus{} 2}.a_{2i}|$.now we have a contradiction and we must have $ deg(q(x)) \equal{} 0$.
we r done!

[quote="earldbest"][quote="TTsphn"]
From  $ f(c)\equiv f( \minus{} a \minus{} b) (\mod a \plus{} b \plus{} c)$ then we have
[/quote]

how did you get this? :([/quote]
as we have proven $ f(0)\equal{}0$
so we get $ 0\plus{}(a\plus{}b)\plus{}c | f(0)\plus{}f(a\plus{}b)\plus{}f(c) \implies f(c) \equiv\minus{} f(a\plus{}b) \pmod {a \plus{} b \plus{} c}$.
but we had $ a\plus{}b\plus{}c | f(a)\plus{}f(b)\plus{}f(c)$
which implies $ a\plus{}b\plus{}c | f(a)\plus{}f(b)\minus{}f(a\plus{}b)$.
the rest can be solved in the same way as TTsphn's solution.
or by getting $ f(a)\plus{}f(b)\equal{}f(a\plus{}b)$ u can consider cauchy's functions because $ f(x)$ is a polynomial and so continious and get directly the result.
but i don't know how we get $ f(c)\minus{}f(\minus{}a\minus{}b) \equiv 0 \pmod {a\plus{}b\plus{}c}$!?

[quote="shoki"][quote="earldbest"][quote="TTsphn"]
From  $ f(c)\equiv f( \minus{} a \minus{} b) (\mod a \plus{} b \plus{} c)$ then we have
[/quote]

how did you get this? :([/quote]

but i don't know how we get $ f(c) \minus{} f( \minus{} a \minus{} b) \equiv 0 \pmod {a \plus{} b \plus{} c}$!?[/quote]

I understand now, $ f(c)\minus{}f(\minus{}a\minus{}b)$ is divisible by $ (a\plus{}b\plus{}c)$ since $ f$ is a polynomial :D

what a stupid ....!! i didn't pay attention to the sign! :mad:  :mad:

$1.$ $a=b=c=n$ $:$ $3n\mid 3f(n) \: \Rightarrow\: n\mid f(n)$ for all positive integer $n$ so $f(n)=ng(n).$

$2.$ $p>f(1)+\cdots +f(n-1)$ and $b=n-a,\, c=p-n$ : $p$ divides $f(a)+f(n-a)+f(p-n).$
    $0<f(a)+f(n-a)<p$ and $f(a)+f(n-a)\equiv -f(p-n)\: \: (mod\: \: p)$ so $f(a)+f(n-a)$ is same for all $1\leq a\leq n-1.$

$3.$ By $2,$ $f(n-1)+f(n+1)=f(n)+f(n) \: \Rightarrow\: f(n)-f(n-1)=f(n+1)-f(n)$ so $f(n)=f(1)+(n-1)c$ for constant $c.$

$4.$ By $1$ and $3,$ $ng(n)=g(1)+(n-1)c.$ $n$ divides $g(1)-c$ for all $n$, so $c=g(1)$ and $g(n)=g(1).$

$5.$ So $g(x)$ is constant and $f(x)=kx$ $(k$ is constant, which is positive integer$).$


Ring Theory
superior algebra
superior algebra unsolved

I don't understand concept " subring of $ \mathbb{C}$ generated by the mth roots of unity" help me.
Thank!

If you have some subset $ A$ of a ring and you are looking for the smallest subring containing $ A$, then you just do:
Add $ 1$ to $ A$ if necessary. Consider all numbers you can reach by a finite number of additions, subtractions and multiplication of elements of $ A$.

This indeed is a ring. Now for $ A$ being the $ m$-th complex roots of unity, you get a cyclotomic extension. In fact, if $ \zeta$ is a primitive such root, then your ring is the smallest object you can reach using $ \mathbb Z$ and $ \zeta$ using addition, subtraction, multiplication. So in other words, it is $ \mathbb Z [\zeta]$, which consists of all finite sums of type $ \sum_{k \equal{} 0}^s a_k \zeta^k$ with $ a_k \in \mathbb Z$; as $ \zeta^m \equal{} 1$, you can substitute $ \zeta^k$ by $ \zeta^{k \minus{} m}$ if $ k\geq m$, giving that these finite sums are those of type $ \sum_{k \equal{} 0}^{m \minus{} 1} a_k \zeta^k$.
You can still kill some summands (e.g. because $ 1 \plus{} \zeta \plus{} \zeta^2 \plus{} ... \plus{} \zeta^{m \minus{} 1} \equal{} 0$) if you want.


Can someone please post the scores of top few teams from NYSML?  Minnesota's ARML team takes this year's NYSML for practice next week, and I'd like to be able to compare scores.

[quote="Sly Si"]Minnesota's ARML team takes this year's NYSML for practice next week, and I'd like to be able to compare scores.[/quote]

By the way, how is Minnesota's ARML team chosen? It seems to draw from quite a diverse set of schools.

NYC A                130
Westchester A  116
NYC S                113
Monroe A           112
NYC N                109
NYC Y                106
NYC C                103

This years test must have been more difficult than last year's because everyone's net improvement was actually a negative number.

Last year's exam had 48 perfect individual scores.  Enough said.

Minnesota's ARML team consists of:
The top 10 individuals in the regular season of the state high school math league
The top 10 individuals at the state tournament
The top 10 scores on AMC12
Since there is always some overlap in these groups, the coaches meet shortly after the state tournament and pick people (based on the sources above as well as AMC10) to fill out the 30 spots, as well as 5 or so younger students (usually grade 10 and below) to serve as alternates and to train with the team for future years.

Our gold (A) team, as well as which alternates get to go to Iowa, is picked on the basis of students' performance during practices.


Triangle ABC inscribled in the circle (O). M moves on the arc BC. I is the midpoint of BC. K is the point such that I is the midpoint of KM. Find the locus of H such that H is the orthocenter of traingle ABK.
Help me!!!

Who can solve this problem? Is it too difficult?

[img]http://img231.imageshack.us/img231/122/picturesp5.png[/img]

[quote="natural"] Find the locus of H such that H is the orthocenter of traingle ABK. [/quote]

Can you explain that more clearly?

[quote="Poincare"][quote="natural"]Find the locus of H such that H is the orthocenter of traingle ABK. [/quote]
Can you explain that more clearly?[/quote]
$ M$ moves on arc $ BC$, so $ H$ can be in different points. The question asks for the locus of $ H$, which is the set of all possible locations of $ H$.

I think this problem is nice. You shoud post full solution.

Maybe I'm getting a bit repetitive, but couldn't H be the [b]barycenter[/b] of ABK? That will make everything definitely more easy, whereas the original problem lead to an absurd locus without a simple formula in cartesian coordinates (well, it seems to be so).

[quote="thanhnam2902"]I tried to done this problem but it's difficult. Used Geosetchpad I see the locus but I can slove it. I will try more.[/quote]
I also used Geometry Sketchpad. This locus is too strange so I can't solve this problem  :( .Who can help me? :?:
[img]http://img91.imageshack.us/img91/2443/pictureqy9.gif[/img]


superior algebra
superior algebra solved

Let A,B,C be in M_n(R) with n>=2 with the property that BA=BC=AC=On. Show that det(In+A^2+B^4+C^6)>=0.

cheers! :D

it is equivalent with
det[(I_n+B^4)(I_n+A^2)(I_n+C^6)]\geq 0... obviously :D :D


induction
algorithm
number theory
Euclidean algorithm

What does the set notation mean in this case?

$(a,b)$ is the greatest common divisor of $a$ and $b$.

[quote="thebrokenlight"]Prove that, for m ≠ n, $(F_m,F_n)=1$, where $F_k=2^{2^{k}}+1$.

Furthermore, using this result, prove that there exist an infinite number of primes.[/quote]
[b]Lemma[/b]
If $m>n$, then $2^{2^n}+1$ divides $2^{2^m}-1$.
[u]Proof of lemma:[/u]
We will use mathmatical induction (by fixing n).
For $m=n+1$, $2^{2^m}-1=2^{2^{n+1}}-1=(2^{2^n})^2-1=(2^{2^n}+1)(2^{2^n}-1)$ is divisible by $2^{2^n}+1$.
Assume the lemma is true for $m=n+k$, i.e. $2^{2^n}+1$ divides $2^{2^{n+k}}-1$.
Now for $m=n+k+1$,
$2^{2^m}-1=2^{2^{n+k+1}}-1=(2^{2^{n+k}})^2-1=(2^{2^{n+k}}+1)(2^{2^{n+k}}-1)$ is divisble by $2^{2^{n+k}}-1$$\Rightarrow$ divisible by $2^{2^n}+1$. Thus the lemma is proved.

Now we go back to the original problem: If $m\neq n$, then $(2^{2^m}+1,2^{2^n}+1)=1$.
[u]Proof[/u]
Without loss of generality, let $m>n$.
By Euclidean Algorithm, $(qx+r,x)=(r,x)$. By our lemma, we can let $x=2^{2^n}+1$ and $qx=2^{2^m}-1$ for some $q$. Then $(2^{2^m}+1,2^{2^n}+1)=(qx+2,x)=(2,x)=1$ as $x$ is an odd number. The conclusion follows.

[b]Corollary[/b]
There exists infinitely many primes.
[u]Proof of corollary[/u]
Since $(F_m,F_n)=1\text{ }\forall m\neq n\in \mathbb{N}$, every $F_m$ is either itself prime or has a prime factor other that the ones of other $F_n$. Therefore, there exists infinitely many primes.

An other way of doing it is to see that $\prod_0^{n-1} F_k = F_n - 2$ (provable by induction) so any common divisor of $F_n$ and $F_p$ ($n \neq p$ must divide 2; since all $F_n$ are odd ....

Argh! I tried to solve that in Intro to Number Theory!

I got pretty close, but not quite there.  Good job!


geometry
analytic geometry
geometric transformation
reflection
trapezoid
geometry solved

Hi, I can't solve this problem maybe you guys can?

The language is in swedish but the question is how big the grey-figure is? Each side of the square is 1mile (10km).

With the notations of the figure, we have:

\[
\begin{array}{l}
 PQ = QR = RP = 1 \\
 \angle PQR = 60^\circ  \\
 \left[ {PQR} \right] = \frac{{\sqrt 3 }}{4} \\
 \end{array}
\]

Therefore:

\[
\left\{ \begin{array}{l}
 A + 2B + C = \left( {\frac{\pi }{6} - \frac{{\sqrt 3 }}{4}} \right) + \frac{\pi }{6} \\
 A + 3B + C = \frac{\pi }{4} \\
  \\
 A + 4B + 4C = 1 \\
 \end{array} \right.{\rm{     }} \Rightarrow {\rm{   }}A = 1 - \sqrt 3  + \frac{\pi }{3}
\]

Best regard
Leon

Denote the square ABCD counterclockwise $a = AB = BC = CD = DA = 1 km$,  and put the coordinate origin at the lower left vertex $A$. Divide the gray area by the square midlines and calculate the area of one of the 4 equal parts. Let $M$ be intersection of the square diagonals and midlines. Let $D'$ be reflection of $D$ in $AB$ $\Longrightarrow$ $DD'$ be diameter of the circle $(A)$ with radius $a = AB = AD$. Let midlines parallel to $AB, BC$ cut the short arc $BD$ of circle $(A)$ at $P, Q$. Let $Q' \in (A)$ be reflection of $Q$ in $AB$. Let $K, L$ be midpoints of $AB, AD$. Area $ \mathcal S$ of the curvilinear $\triangle MPQ$ is equal to quarter the area of the circle $(A)$, take away area of the curvilinear trapezoid $DD'Q'Q$ and add area of the square $AKML$. Area $\mathcal T$ of the curvilinear trapezoid $DD'Q'Q$ is equal to half the area of circle $(A)$, take away third the area of circle $(A)$ and add the area of isosceles triangle $AQQ'$ with the angle $\angle QAQ' = 120^\circ$.

$\mathcal T = \frac{\pi a^2}{2} - \frac{\pi a^2}{3} + \frac{a^2 \sqrt 3}{4} = \frac{\pi a^2}{6} + \frac{a^2 \sqrt 3}{4}$

$\mathcal S =  \frac{\pi a^2}{4} - \mathcal T + \frac{a^2}{4} = \frac{a^2}{4}\left(\frac{\pi}{3} - \sqrt 3 + 1\right)$

and the entire gray area is $4 \mathcal S$.


Let $ a,b,c$ be non negative real numbers. Prove that
$ \frac{a^2(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{b^2(c\plus{}a)}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{c^2(a\plus{}b)}{a^2\plus{}ab\plus{}b^2}\ge 2\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}$

[quote="Sunjee"][quote="dduclam"]Let $ a,b,c$ be non negative real numbers. Prove that
$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}$[/quote]
nice inequality. Your inequality is equivalent to
\[ \sum\limits_{cyc}\frac {ab(a \minus{} b)^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)(b^2 \plus{} bc \plus{} c^2)(a^2 \plus{} ac \plus{} c^2)}\geq 0
\]
if i'm not wrong[/quote]

Very beautyful,Sunjee  :wink:
Equality holds for $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$

We have a stronger inequality:

$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}$

Your last inequality is equivalent to
\[ \sum\limits_{cyc}\frac{ab(a\minus{}b)^2(a^3\plus{}b^3\minus{}c^3\minus{}abc)}{(a^2\plus{}b^2\plus{}c^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ac\plus{}a^2)}\geq 0\]
i'm trying prove that by S.O.S method...

[quote="dduclam"][quote="Sunjee"][quote="dduclam"]Let $ a,b,c$ be non negative real numbers. Prove that
$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}$[/quote]
nice inequality. Your inequality is equivalent to
\[ \sum\limits_{cyc}\frac {ab(a \minus{} b)^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)(b^2 \plus{} bc \plus{} c^2)(a^2 \plus{} ac \plus{} c^2)}\geq 0
\]
if i'm not wrong[/quote]

Very beautyful,Sunjee  :wink:
Equality holds for $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$

We have a stronger inequality:

$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
My solution for the second
by Cauchy Schwarz, we have
\[ \sum \frac{a^2(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\cdot \sum \frac{a^4(b^2\plus{}bc\plus{}c^2)}{b\plus{}c} \ge (\sum a^3)^2\]
It suffices to prove that
\[ (\sum a^3)(\sum a^2) \ge 2\sum \frac{a^4(b^2\plus{}bc\plus{}c^2)}{b\plus{}c}\equal{}2\sum a^4(b\plus{}c)\minus{}2abc\sum \frac{a^3}{b\plus{}c}\]
\[ \Leftrightarrow \sum a^3 (a\minus{}b)(a\minus{}c)\plus{} abc (\sum \frac{2a^3}{b\plus{}c} \minus{}\sum a^2) \ge 0\]
by schur Inequality, we have
\[ \sum a^3 (a\minus{}b)(a\minus{}c) \ge 0\]
and the Am-GM gives us
\[ \sum \frac{2a^3}{b\plus{}c} \ge \sum a^2\]
We have done.
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$ or any cyclic permutations. :)

PS: Nice Inequality, Lam. :)

[quote="can_hang2007"][/quote][quote="dduclam"][quote="Sunjee"][quote="dduclam"]Let $ a,b,c$ be non negative real numbers. Prove that
$ \frac {a^2(b + c)}{b^2 + bc + c^2} + \frac {b^2(c + a)}{c^2 + ca + a^2} + \frac {c^2(a + b)}{a^2 + ab + b^2}\ge 2\frac {a^2 + b^2 + c^2}{a + b + c}$[/quote]
nice inequality. Your inequality is equivalent to
\[ \sum\limits_{cyc}\frac {ab(a - b)^2(a^2 + b^2 + c^2 + ab + bc + ca)}{(a + b + c)(b^2 + bc + c^2)(a^2 + ac + c^2)}\geq 0
\]
if i'm not wrong[/quote]

Very beautyful,Sunjee  :wink:
Equality holds for $ a = b = c$ or $ a = b,c = 0$

We have a stronger inequality:

$ \frac {a^2(b + c)}{b^2 + bc + c^2} + \frac {b^2(c + a)}{c^2 + ca + a^2} + \frac {c^2(a + b)}{a^2 + ab + b^2}\ge 2\frac {a^3 + b^3 + c^3}{a^2 + b^2 + c^2}$[/quote][quote="can_hang2007"]
My solution for the second
by Cauchy Schwarz, we have
\[ \sum \frac {a^2(b + c)}{b^2 + bc + c^2}\cdot \sum \frac {a^4(b^2 + bc + c^2)}{b + c} \ge (\sum a^3)^2
\]
It suffices to prove that
\[ (\sum a^3)(\sum a^2) \ge 2\sum \frac {a^4(b^2 + bc + c^2)}{b + c} = 2\sum a^4(b + c) - 2abc\sum \frac {a^3}{b + c}
\]

\[ \Leftrightarrow \sum a^3 (a - b)(a - c) + abc (\sum \frac {2a^3}{b + c} - \sum a^2) \ge 0
\]
by schur Inequality, we have
\[ \sum a^3 (a - b)(a - c) \ge 0
\]
and the Am-GM gives us
\[ \sum \frac {2a^3}{b + c} \ge \sum a^2
\]
We have done.
Equality holds if and only if $ a = b = c$ or $ a = b,c = 0$ or any cyclic permutations. :)

PS: Nice Inequality, Lam. :)[/quote]

So wonderful,Can  :roll:
Thank you very much!

[quote="Sunjee"]Whats your proof Mr. dduclam? Can you share it us?[/quote]

My proof using SOS method too,but ugly    :(

[quote="can_hang2007"].....It suffices to prove that
\[ (\sum a^3)(\sum a^2) \ge 2\sum \frac {a^4(b^2 \plus{} bc \plus{} c^2)}{b \plus{} c} \equal{} 2\sum a^4(b \plus{} c) \minus{} 2abc\sum \frac {a^3}{b \plus{} c}
\]

\[ \Leftrightarrow \sum a^3 (a \minus{} b)(a \minus{} c) \plus{} abc (\sum \frac {2a^3}{b \plus{} c} \minus{} \sum a^2) \ge 0
\]
[/quote]

Sorry, I don't understand this step.  :blush: Can you please explain it to me.
I think it should be

$ \sum a^5 \plus{} \sum a^3(b^2\plus{}c^2) \plus{}abc\sum a^2 \minus{}2 \sum a^4(b \plus{} c)\plus{} abc (\sum \frac {2a^3}{b \plus{} c} \minus{} \sum a^2) \ge 0$

that is
$ \sum a^3 (a \minus{} b)(a \minus{} c) \plus{}\sum a^3(b^2\plus{}c^2)\minus{}\sum a^4(b \plus{} c)\plus{} abc (\sum \frac {2a^3}{b \plus{} c} \minus{} \sum a^2) \ge 0$

Thank you very much for your kindness.

[quote="manlio"][quote="can_hang2007"].....It suffices to prove that
\[ (\sum a^3)(\sum a^2) \ge 2\sum \frac {a^4(b^2 \plus{} bc \plus{} c^2)}{b \plus{} c} \equal{} 2\sum a^4(b \plus{} c) \minus{} 2abc\sum \frac {a^3}{b \plus{} c}
\]

\[ \Leftrightarrow \sum a^3 (a \minus{} b)(a \minus{} c) \plus{} abc (\sum \frac {2a^3}{b \plus{} c} \minus{} \sum a^2) \ge 0
\]
[/quote]

Sorry, I don't understand this step.  :blush: Can you please explain it to me.
I think it should be

$ \sum a^5 \plus{} \sum a^3(b^2 \plus{} c^2) \plus{} abc\sum a^2 \minus{} 2 \sum a^4(b \plus{} c) \plus{} abc (\sum \frac {2a^3}{b \plus{} c} \minus{} \sum a^2) \ge 0$

that is
$ \sum a^3 (a \minus{} b)(a \minus{} c) \plus{} \sum a^3(b^2 \plus{} c^2) \minus{} \sum a^4(b \plus{} c) \plus{} abc (\sum \frac {2a^3}{b \plus{} c} \minus{} \sum a^2) \ge 0$

Thank you very much for your kindness.[/quote]
Thank you very much, manlio. I had a big mistake in calculation. :(
I'll try it again.

[quote="dduclam"]
We have a stronger inequality:

$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
I'm so sorry about my previous post :(. Please forgive me...

This inequality is not true, for example $ a \equal{} b \equal{} 1,c \equal{} \frac {1}{2}$. But the following is true
\[ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}
\]
And here is my solution for it

PS: My best thanks to manlio for figure out my big mistake.

Thank you very much, can_hang2007 for you beautiful solutions that make me very happy  :)

[quote="dduclam"]I am sorry since my conjecture is wrong  :(  But this problem is also not true,Can:

$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}$

for also example $ a \equal{} b \equal{} 1,c \equal{} \frac {1}{2}$

And i think (but not prove yet) the following inequality is true:
$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\sqrt [4] {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}$

(Note that If RSH=$ \sqrt [3]{...}$ ,problem's also false.)

If it's corects,it will be a hard problem,i think.

Thanks for your posts,Can and manlio.

(finaly,sorry for my very bad English  :( )[/quote]
Dear Lam, I have checked my Inequality again, and I'm sure that It is true. Please check your counter example again, thank you. :)
[quote]$ \frac {a^2(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\ge 2\sqrt [4] {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}$
[/quote]
This inequality is not true since it is not homogenious.


floor function
algorithm
continued fraction

Prove that every infinite continued fraction is irrational.

 (A continued fraction is a fraction of the form $a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \cdots}}$, where $a_0 \in \mathbb{Z}$ and $a_1, ... \in \mathbb{N}$)

[u][i]Lemma.[/i][/u] [i]Every continued fraction is finite iff $x\in\mathbb{Q}$.[/i]

[u][i]Proof.[/u][/i]

Let $x\in\mathbb{R}$, and let $a_0 = \lfloor x\rfloor$. Using an algorithm, we define a finite or infinite series of numbers $\{a_1 , a_2 , . . .\}\in\mathbb{Z_+}$.

$\cdot$ If $x = a_0$ , then $x=\left< a_0\right>$ is a finite continued fraction. Otherwise, $0 < x-a_0 < 1$, and we define $x_1 = \frac{1}{x-a_0}$, noting that $x_1 > 1$ and that $x=\left< a_0, x_1\right>$.

$\cdot$ Let $k\in\mathbb{N}$, and suppose the positive integers $\{a_i\}_{i=1}^{k-1}$ and $x_k\in\mathbb{R} > 1$ have been defined and that $x =\left<a_0, a_1, . . . , a_{k-1}, x_{k}\right>$. Define $a_k = \lfloor x_k\rfloor$.
If $x_k = a_k$ , then $x =\left<a_0, a_1, . . . , a_k\right>$ and the algorithm stops. Otherwise, define
$x_{k+1} = \frac{1}{x_k-a_k}$, $x_{k-1}\in\mathbb{R}> 1$, note that $x_k =\left<a_k, x_{k+1}\right>$, and
$x =\left< a_0, a_1, . . . , x_{k+1}\right>$, and proceed to next step.

from this, we deduce that if the algorithm stops, then $x$ is a finite continued fraction.  This implies that if the algorithm does not terminate, the contined fraction has an infinite sequence $\{a_i\}_{i=0}^n$.

$\therefore$ Every continued fraction is infinite if $x\notin\mathbb{Q}$, and the converse statement follows.

Masoud Zargar

The contrapositive statement is: if a number is rational, then its continued fraction is finite.

Assume we have a positive rational number $z={a\over b}$, where WLOG we can put $1<a<b$ (for $a=1$ the number itself is its continued fraction). Writing $b=k_0a+r_0$ where $0\leqslant r_0<a$, we get $z=\frac{1}{k_0+{r_0\over a}}$. Now writing $a=k_1r_0+r_1$ where $0\leqslant r_1<r_0$, we get $z=\frac{1}{k_0+\frac{1}{k_1+{r_1\over r_0}}}$. Then writing $r_0=k_2r_1+r_2$ where $0\leqslant r_2<r_1$, we get $z=\frac{1}{k_0+\frac{1}{k_1+\frac{1}{k_2+{r_2\over r_1}}}}$, and so on. Since $r_i$ are strictly decreasing non-negative integers, the descent can't be infinite because sooner or later $r_i$ will become zero and the procedure would end. It can be noted that the continued fraction will have at most $r_0+1$ terms.

EDIT: boxedexe types faster :) though I'm not sure if he proved the link between the rationality of a number and finiteness of its continued fraction.

About your question:

$[(p\implies q)\iff(\lnot q\implies\lnot p)]=\text{p implies q if and only if when p is false q is likewise.}$

EDIT regarding your EDIT: A continued fraction is finite if and only if $x$ is rational, since it must end somewhere.  If a continued fraction is infinite, then $x\notin\mathbb{Q}$.  The converse statement would be: If $x\notin\mathbb{Q}$, then the continued fraction is infinite. Hopefully that satisfies you and completes teh explanation of the proof (hopefully) :)

Masoud Zargar

[quote="boxedexe"]About your question:

$[(p\implies q)\iff(\lnot q\implies\lnot p)]=\text{p implies q if and only if when p is false q is likewise.}$

Masoud Zargar[/quote]

Thanks, but I've found the term in the meantime: it's called "the contrapositive statement". And I also added a note at the end of my previous post.

[quote="boxedexe"]EDIT regarding your EDIT: A continued fraction is finite [b]if[/b] and only if $x$ is rational, [b]since it must end somewhere[/b].[/quote]

Sorry for not replying sooner, I missed this :)

The bolded parts are exactly what we want to prove strictly :) intuitively it's clear, but your proof (as far as I can tell) is based on taking the initial statement (or the "if" part of your statement) for granted and then juggling up the parts.


Let $x_{i}$ are integers for $i \in \{1,2, \cdots, n\}$ and $n \geq 2$.
Let $S = \prod_{1 \leq i \leq j \leq n}(a_{i}-a_{j})$ and $T = \prod_{1 \leq i \leq j \leq n}(i-j)$.
Show that $T | S$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=4858

continue discussion there


Let $ x\geq y\geq z$ be real numbers such that $ xy\plus{}yz\plus{}zx\equal{}1$.

Prove that $ xz<\frac{1}{2}$.

Is $ \frac{1}{2}$ the best value?

Assume the contrary, that $ xz\ge\frac 12$. Since $ xz > 0$, either $ x,z > 0$ or $ x,z < 0$. In the first case $ y\ge z > 0\implies x \ge y \ge z > 0$ and in the second case $ y\le x < 0\implies 0 > x \ge y \ge z$. Hence in both cases $ xy,yz,zx > 0$. In the first case, since we have $ xy\ge xz\ge\frac 12$, we get $ 1 \equal{} xy \plus{} yz \plus{} zx\ge\frac 12 \plus{} \frac 12 \plus{} yz > 1$, contradiction. Thus $ xz < \frac 12$.

Again, in the second case $ yz\ge xz\ge\frac 12$ implies $ 1 \equal{} xy \plus{} yz \plus{} zx\ge\frac12 \plus{} \frac 12 \plus{} xy > 1$, contradiction again. Hence in all cases $ xz < \frac 12$.

Yes $ \frac 12$ is the best value. Take $ y \equal{} z\to 0$ and $ x$ large enough so that $ xy \equal{} xz\to\frac 12$. Then we can get $ xz$ arbitrarily close to $ \frac 12$.

This is from meditteranean math contest last year.  I didn't solve the second part.  :oops: The official solution is much longer. I think you need to add some stuff to your proof (supremum needed) though I may be wrong. :maybe:

Ok if you don't like my previous reasoning how about this?

Let $ y \equal{} z \equal{} \epsilon$. Then $ x \equal{} \frac {1 \minus{} \epsilon^2}{2\epsilon}$. Thus $ xz \equal{} \frac {1 \minus{} \epsilon^2}{2}$ and therefore we can take $ \epsilon\to 0$ and $ x$ large enough to find that $ xz\to \frac 12$.

From $ x\geq y\geq z$, we have
\[ (x\minus{}y)(y\minus{}z)\geq 0\Rightarrow xy\plus{}yz\geq y^2\plus{}xz\Rightarrow xz\leq \frac{1\minus{}y^2}{2}\]
This proves $ xz\leq \frac{1}{2}$. If we had equality, then $ xz\leq \frac{1\minus{}y^2}{2}$ would imply $ y\equal{}0$. But then plug into the original $ xy\plus{}yz\plus{}zx\equal{}1$ to get $ \frac{1}{2}\equal{}1$, a contradiction. So the inequality is strict.

To show $ \frac{1}{2}$ is the best constant, just consider $ y\equal{}z$, $ x$ which solves $ xy\plus{}yz\plus{}zx\equal{}1$, and $ y\Rightarrow 0^\plus{}$, as nayel did.


A. If p>1, evaluate the expression (1 + 1/(2^p) + 1/(3^p) + 1/(4^p) + ...)/(1 - 1/(2^p) + 1/(3^p) - 1/(4^p) + ...).

I tried reverse engineering, but that didn't seem to work.

B. Consider the series whose terms are the reciprocals of the positive integers that can be written in base 10 notation without using the digit . Show that this series is convergent and the sum is less than 90.

For A, start with only the odd ones (i.e. 1+1/3^p+1/5^p+...) and keep multiplying by 1/2^p infinitely many times, and then add them all together. If S=1+1/3^p+1/5^p+..., then zeta(p)=1/(1-1/2^p)*S. For the bottom, keep multiplying by 1/2^p, but subtract all the other ones. Then you get (1-1/2^(p-1))/(1-1/2^p), and dividing the two gives 1/(1-1/2^(p-1)).

zeta(p) is just another name for 1 + 1/(2^p) + 1/(3^p) + ...

Well, sort of. That's what it is if Re{p}>1, and then it converges. Go to [url]http://functions.wolfram.com[/url] for information about the zeta function (after which the second part of my username was named) and many others.


My apologies for the poor diagram but find the equivalent resistance between A and B. And don't take anything as obvious, try proving that as well.

that is an arihant sum da  :huh:

the resistance can be made kx equivalently ( i mean the $ \infty$ one)

is the answer $ \frac {x \cdot ( 2 \cdot k \minus{} 1 \plus{} \sqrt {4 \cdot k^2 \plus{} 1} )}{2 \cdot k}$ ?

[quote="pardesi"][hide="Solution"]Let the equivalent resistanec between A and B be $ x \Omega$ now let me define $ k \Omega \equiv 1 \beta$ where $ \beta$ is a suitable unit of my choice which satisfies the properties as given by it's definition
now
consider the next points to A,B as C,D(i think i am clear ...)
now the resistances here are $ R \beta$ and $ R \beta$ which are connected to $ kR \beta$ and $ k R \beta$ and so on
clearly all laws of physics which hold true for resisitance with $ \Omega$ as an unit hold true with $ \beta$ as unit because they are derived independently of an unit
so by the symmetry of the situations  the resisistance between these two points is $ x \beta \equiv kx \Omega$
so we have a cicuit in which $ kx,R$ are in parallel and whole in series with $ R$ so we have
$ \frac {kxR}{kx \plus{} R} \plus{} R \equal{} x$
so solving we get $ R_{eq} \equal{} R(\frac {2k \minus{} 1 \plus{} \sqrt {1 \plus{} 4k^{2}}}{2k})$ [/hide][/quote]

 :thumbup:    :10:

Excellent . Many people say that its very very obvious that its kx but I wanted a good explanation. BTW, have you seen this problem before?

[quote="mathclass"]Another nice question that can be solved like this is that of the infinite Atwood machine. Say you have massless pulleys and strings; every pulley has a mass M on the left string. The top pulley is connected to the ceiling, and every pulley after that has another atwood machine on the string. What is the acceleration of the mass M on the top pulley?

[I can't phrase it very well. Picture this:

http://www.staff.amu.edu.pl/~romangoc/graphics/M3/7-atwood-machine/M3-7-fig1.gif

but with another Atwood machine on the right hand side, which repeats infinitely][/quote]

I believe you are referring to this. http://www.mathlinks.ro/viewtopic.php?p=1065889#1065889. Problem 124

Can you give a nice soln for this? I don't think similar ideas can be used because the frame of reference becomes non-inertial.

I don't think my reasoning is exactly rigorous, but it seems fairly intuitive to me. What I did when I saw this problem was to think of it in terms of equivalent masses. Your first pulley has a mass M on the left and an equivalent mass, let's call it M_e, from the accelerating pulleys on the right string. This is equivalent to a system consisting of two pulleys; the top one has mass M on the left, and another pulley with M and M_e on the right. I don't have access to a drawing program, so I can't really draw it.. but yeah.

Now consider (a) a single mass m on a string accelerating downwards with acceleration a and (b) a pulley with masses M_1 and M_2 accelerating downwards with acceleration a. It's not hard to see that in order to have the same tensions in the strings holding the mass in (a) and the pulley in (b) be equal, m = 4M_1M_2/(M_1+M_2).

Going back to the original problem, we should have M_e = 4M_e * M / (M_e + M). So we get M_e = 3M, and the mass M accelerates at g/2.

[edit]

i just realized you're basically asking for a rigorous solution right? so i'll think about it for now.. aime's on my mind for tomorrow :O


Unknown to the measurer, a one-yard tape measure has been stretched $ \frac{3}{4}$ inches, and then used to measure a rectangular lawn. The area is computed using the measured dimensions which are 20 yards by 30 yards. What is the number of square yards in the positive difference between the actual area and the computed area? Express your answer as a mixed number.

Since every yard measured by the tape measure is stretched $ \frac{3}{4}$ inches, you can just multiply the computed dimensions by $ 1\frac {3}{4}$ to find the real dimensions.

So $ 20\times1\frac {3}{4} \equal{} 35$ and $ 30\times1\frac {3}{4} \equal{} 52\frac {1}{2}$.

So the computed area is $ 20\times30 \equal{} 600$ and the actual area is $ 35\times52\frac {1}{2} \equal{} 1837\frac {1}{2}$

To find the answer you do: $ 1837\frac {1}{2} \minus{} 600 \equal{} 1237\frac {1}{2}$

[quote=Bimikel][quote]Since every yard measured by the tape measure is stretched $ \frac{3}{4}$ inches, you can just multiply the computed dimensions by $ 1\frac {3}{4}$ to find the real dimensions.[/quote]
That was $\frac{3}{4}$ inches stretched, not yards. It would be $\frac{1}{48}$ yards stretched because $\frac{3}{4}$ divided by 36 (conversion from inch to yard) is $\frac{1}{48}$.
Just do the same steps from LLin1's post except have 20 and 30 times $1\frac{1}{48}$ and the results should be $\frac{245}{12}$ and $\frac{245}{8}$.
The product of both fractions is $\frac{60025}{96}$ or $625\frac{25}{96}$ which is the computed area.
$625\frac{25}{96}-20*30=25\frac{25}{96}$.
Can't believe this is about 9 years later...[/quote]

 Me too. I also can't believe this is 9 years later


I was lookig at past AIMEs and came across this problem from 1989 that I cannot solve (#9)
$133^5+110^5+84^5+27^5=n^5$
find $n$.

Answer:
[hide]144[/hide]

Thanks, but I have one question.  How did you get $\frac{3125}{1024}$?

He needed to find a  good upper bound, i.e. one that would let him know the answer.  The step right before that he had $3\cdot 133^5$ so he could sub in anything greater than 3 for 3 and still have an upper bound.  He used $\frac{3125}{1024}$ because those numbers are 5th powers and it's greater than 3.

Thanks, I hope I can see things like this when I take the AIME.


abstract algebra
vector
superior algebra
superior algebra solved

[i]This problem was given in my algebra exam today.[/i]

Let M be a finite Z[i] module, such that 3*x=0 for all x in M. Prove that |M| is a power of 9.

This is an exam problem. How hard can it be?
Belive me my teacher ment this to be the problem that makes a difference, so he wanted only the best of us to solve it. But still he wanted to be sure that the best solve it, so the solution has some ideas, but not that many.
Good luck.

Let me give you my solution.
3*x=0 for all x, so all the elements of the commutative group M, have order 3.

Lemma: M[u]~[/u](Z3) n for some n.
Proof: By induction. Consider we have a set x1,...,xk of linearly independent elements over Z3. Then they generate a group izomorphic to (Z3) k. If we have other elements in M than the ones in <x1,...,xk> then we take one of them x_(k+1) and obtain a new set x1,...,x_(k+1). This construction ends because M is finite.

We identify M to (Z3) n for some n. So M has also a structure of a Z3 vector space.
Let f:M->M, the endomorphism given by f(x)=i*x. Let e1=(1,0...0), e2=(0,1,0...0)... en=(0...0,1) the canonical basis of M over Z3.
Let F=(f_ij)i,j=1..n be in Z3 such that f(ei)= \sum {j=1..n} f_ij*ej. (because M is a vector space having e1..en as a basis over Z3).

f(f(x))=i*f(x)=i*(i*x)=-x, so when going to matrices, we get F 2=-In.

Assume n is odd. Then (det(F)) 2 =-1. The last equality is impossible in Z3, so we get a contradiction.
So n is even, so M[u]~[/u](Z3) 2n' for 2n'=n, so M has 9 n' elements.

This is my solution, but it is not the easiest. One good colleague of mine found an easier proof. Can you?

I now have 4 posts, the only 4 posts in this topic.

This is a simpler solution, probably the official solution. At least 2 of my colleagues found it.

Prove the following:
-K=Z[i]/3Z is a field with 9 elements.
-Prove that M has a Z[i]/3Z vector space structure.
-If x_1,...x_n is a basis of M over K, then M has 9 n elements.


modular arithmetic
induction
number theory unsolved
number theory

Let $ L_n$ denote the $ n$th term of the Lucas squence, defined by $ L_0\equal{}2, L_1\equal{}1, L_{n\plus{}1}\equal{}L_n\plus{}L_{n\minus{}1}$. Let, in addition, $ a_1, a_2, ...$ be a sequence of integers such that

$ \sum_{d|n}a_d\equal{}L_n$

for every positive integer $ n$.

Prove that $ n|a_n$ for all $ n$.

Here is my proof.  I am not sure of its rigor, so please do comment on it if it is wrong.
[hide]It is well known that $ L_n$ ($ n>0$) is the number of subsets, including the empty set, of $ 1,2,...,n$ such that no two elements are consecutive modulo $ n$ (notice that this means that $ n$ and $ 1$ are consecutive).  Consider the infinite sequence of natural numbers $ 1,2,3,...$.  We color each integer red or blue such that no two red integers are consecutive.  Call such a sequence good.  We claim that $ a_n$ is the number of good sequences whose smallest period is $ n$.  We shall prove this by induction.

It is obvious for the base case: $ a_1\equal{}1$ which is indeed the number required.  The only sequence is coloring all integers blue.  Assume the claim is true for $ a_1,a_2,...,a_{n\minus{}1}$.  By the well known fact stated at the beginning, $ L_n$ is the number of good sequences with period $ n$.  But, we are not garunteed that $ n$ is the smallest period.  We must eliminate all sequences with a smaller period.  It is obvious that these sequences must have period dividing $ n$.  Thus, we have $ a_n\equal{}L_n\minus{}\sum_{d|n,d\ne n}a_d$, as desired.

So, we have proved that $ a_n$ is the number of good sequences with smallest period $ n$.  But, shifting such a good sequence over by $ 1\pmod{n}$ still results in a good sequence with smallest period $ n$.  In fact, shifting the sequence over by $ 1,2,...,n\minus{}1\pmod{n}$ still results in a good sequence.  Thus, $ n|a_n$.  [/hide]

Interesting.  I believe your proof would be nicer, though, if phrased in terms of [url=http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula]Mobius inversion[/url], which gives

$ a_n \equal{} \sum_{d | n} \mu(d) L_{n/d}$

from which your result follows by a PIE argument.

Let $\alpha=(1+\sqrt5)/2$ and $\beta=(1-\sqrt5)/2$, so $L_n=\alpha^n+\beta^n$ for $n\ge0$. We have
\[\sum_{n\ge1} a_n\frac{x^n}{1-x^n} = \sum_{n\ge1} L_n x^n = \frac{x\alpha}{1-x\alpha}+\frac{x\beta}{1-x\beta} = \frac{x+2x^2}{1-x-x^2}.\]Dividing both sides by $x$ and integrating, we have
\[\sum_{n\ge1}-\frac{a_n}{n}\ln(1-x^n) = -\ln(1-x-x^2),\]whence letting $b_n=a_n/n$ and exponentiating both sides,
\[1-x-x^2 = \prod_{n\ge1}(1-x^n)^{b_n} = \prod_{n\ge1}\left(1+b_n(-x^n)+\sum_{k\ge2}\binom{b_n}{k}(-x^n)^k\right).\]Now it's clear by induction on $n\ge1$ that $b_n\in\mathbb{Z}$ for all $n$ (just note that $[x^n](1-x-x^2)$ is always an integer).

We can also use [b]t0rajir0u[/b]'s combinatorial arguments from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1579990#p1579990]here[/url].


1) Αποδείξτε ότι η συνάρτηση f, f(E)=[i]X[/i][size=84]E[/size] όπου Ε ένα σύνολο και [i]X[/i][size=84]E[/size] η χαρακτηριστική του συνάρτηση είναι 1-1 και επί

Otan les i xaraktiristiki tu sinartisi???Giati emis den exume kani akoma tetia.......(enois tin taytotiki??...)
na apantiso opos o ma8itis?? :rotfl:  :rotfl:
[b]ONLY for thatin:[/b][hide]ep......giati to anikses???den ipa mono gia ton thatin???e??.....pes mu....den ipa???....
sorry gia x8es ,ala molis ixa kani log out.....
xmmm....kalo mu akugete....
aplos nomizo oti iparxi ena provlima...idka stin proti kinisi...pos 8a pexti prota to S-m an proigumenos den exi pexti to T-l?????
nomizo oti mia kaliteri sira ine i S-l,T-l,S-m,R-m......kai sigura apotelasmata.... :rotfl:  :rotfl:  :rotfl: [/hide]

 :D

και που θέλεις να ξέρω εγώ τι είναι?

[i]Χ[/i][size=84]Ε[/size](x)=τ(x ανήκει E)

Νομίζω ότι εννοεί την

$\\ \chi_{E}(x) = 1$ αν $x \in E$ και $\chi_{E}(x) = 0$ αν $x \notin E$.

σωστά!

[hide="μόνο για το RAKAR"]
sd giati vlepeis ta pm twn allwn?

 πολύ ωραία! τώρα το είδε!![/hide]

Δεν έχεις ορίσει πεδίο ορισμού και πεδίο τιμών.

Αν πάλι εννοείς την συνάρτηση $f : \mathcal{P(X)}\to \{ g, \,\, g: \mathcal{X}\to \{ 0,1 \}\}$

όπου $\mathcal{X}$ ένα σύνολο και $\mathcal{P(X)}$ το σύνολο των υποσυνόλων του

τότε είναι ένα προς ένα και επί .

Αααααντε...ντε....γι αυτο δεν μπορουσα να το λυσω....
Δεν ειχες ορισει πεδιου ορισμου +πεδιοτιμων...... :rotfl:
Τωρα που το διευκρινισε ο Ramanujan,η ασκηση εχει προφανως λυθει καθως οδηγουμαστε στην λυση κατα τετριμμενο τροπο..... :rotfl:  :rotfl:

Mia askisi Analisis protoy examinoy, tin opoia theoro idiatera endiaferousa
Epeidi apo metafrasi skizo :D tin postaro sta agglika.

$1)$

Let ${(a_{p_{n}})},{(a_{q_{n}})}, ... ,{(a_{s_{n}})}$ be subsequences of
${(a_{n})}$ such that the sequences ${(p_{n})},{(q_{n})}, ... ,{(s_{n})}$ are
pairwise disjoint and form the sequence ${(n)}$. Show that, if
$S, S_{p}, S_{q}, ... , S_{s}$ are the sets of all the limit points of the sequences
${(a_{n})},{(a_{p_{n})},{(a_{q_{n}})}, ... ,{(a_{s_{n}})}}$ respectively. then
$S = S_{p}\cup S_{q}\cup \ ... \cup S_{s}$.
Conclude that, if evey subsequence ${a_{p_{n}}},{a_{q_{n}}}, ... ,{a_{s_{n}}}$ converges to $a$, then the sequence ${(a_{n})}$ also converges
to $a$.

$2)$

Is problem $1$ true in the case of infinitely many subsequences?

$\\ 1)$

Από τον ορισμό του οριακού σημείου $S_{p}\cup S_{q}\cup \ ... \cup S_{s}\subseteq S$ .

Έστω $\ell \in S$ .Τότε υπάρχει υπακολουθία $a_{m_{n}}$ της $a_{n}$ τέτοια ώστε $\lim_{n \to \infty }a_{m_{n}}= \ell$ .

Αλλά αφού οι υπακολουθίες ${(p_{n})},{(q_{n})}, ... ,{(s_{n})}$ καλύπτουν την ακολουθία ${(n)}$

η ακολουθία $(m_{n})$ θα συμπίπτει με μιά εκ των  ${(p_{n})},{(q_{n})}, ... ,{(s_{n})}$ για άπειρα $n \in \mathbb{N}$

έστω την  ${(p_{n})}$ .Τότε οι ακολουθία $(a_{m_{n}}) \cap (a_{p_{n}})$ θα συγκλίνει στο  $\ell$ άρα $\ell \in S_{p}$

συνεπώς $S = S_{p}\cup S_{q}\cup \ ... \cup S_{s}$ .

Αν λοιπόν κάθε υπακολουθία συγκλίνει στο $a$ έχουμε $S_{p}=S_{q}= ... = S_{s}=\{a\}$

άρα $S =\{a\}$ ή  $\lim_{n \to \infty }a_{n}= a$ .

Sorry αν υπάρχει λάθος .  :)

Για το $2 )$  νομίζω η απάντηση είναι όχι αλλά δεν βρήκα αντιπαράδειγμα ακόμα .

sef καμμία ιδέα για το $2$  ?  :)

Έχεις την λύση της άσκησης ??

Αν όχι πού την βρήκες  ?

Ο τροπος επιλυσης του 1) που ξερω ειναι διφορετικος, θα ελεγα
πιο αυστηρος. Θα δωσω την αποδειξη για το δυο
παρε σαν $a_{n}$ την ακολουθια

$a_{n}= 0 \ \ ama \ n = 2^{k}, \ k=0,1,2,... \ \ allios \ 1$
Καθε υπακολουθια ${a_{{2}k-1}},{a_{{2}(2k-1)}},{a_{2^{2}(2k-1)}}, \...\ ,{a_{2^{m}(2k-1)}}$ συγκλινει στο 1, ενω η $a_{n}$ αποκλινει.

To βιβλιο δεν το θυμαμαι αλλα υπαρχει στην κεντρικη βιβλιοθηκη του e.m.p. Αμα θες να σου πω ποιο ειναι πες το μου να περασω.

emeis exoume mia askisi pou leei na deiksete oti den uparxoun A, B tetragonikoi pinakes wste AB-BA=I kai o Rakar mia askisi pou leei an A, B tetragonikoi pinakes wste AB-BA=I ndo kati.... :rotfl:
ΕΜΠvsΑΠΘ.... kai exw apodeiksei oti tha nikisoume!!!!....  :D

[hide="mono gia to sef"]btw sef pou eisai? HMMY? kai ti etos?[/hide]

[quote="sef"]

To βιβλιο δεν το θυμαμαι αλλα υπαρχει στην κεντρικη βιβλιοθηκη του e.m.p. Αμα θες να σου πω ποιο ειναι πες το μου να περασω.[/quote]

Θα ήθελα να το μάθω  :)

[quote="thatin"]
emeis exoume mia askisi pou leei na deiksete oti den uparxoun A, B tetragonikoi pinakes wste AB-BA=I kai o Rakar mia askisi pou leei an A, B tetragonikoi pinakes wste AB-BA=I ndo kati.... Rotfl
[/quote]

$Tr(AB)=Tr(BA) \Leftrightarrow Tr(AB-BA)=0 \ \ \forall A,B \in M_{n}(K) \ \ ,(n \in \mathbb{N^{*}}$ και $K$ ένα σώμα$)$.

Επειδή όμως $Tr(I_{n})=n$, προφανώς δεν μπορεί να ισχύει η σχέση $AB-BA=I_{n}$.

......

1-0 λοιπόν!!!
τολμώ να πω αυτογκόλ...!

Pros thatin eimai 1o etos mathimatiko athinas

Pros Ramanujan, to biblio einai se dio tomoys kai legete
problems in  mathematical analysis 1 kai 2
sigg. W.J Kaczor, M.T Nowak ekdoseis AMS.
Parepiptontos 60 euro to ena.

[quote="RAKAR"]Lipon..... :mad:  :mad: o αγωνας δεν εχει τελιωσει ακομα.... w8ing μεχρι να σφιριξει ο διαιτητής.......δεν καταλαβαινω ομως γιατι ειναι 1-0 +μαλιστα αυτογκολ...???
αυτο που εγραψε ο στεδες που προκυπτει απο την ιδιοτητα Tr(AB)=Tr(A)Tr(B) Aπλως αποδεικνυει οτι δεν υπαρχουν τετοιοι πινακες.....
+που το προβλημα με την δικια μου που λεειι παρεπιπτοντως

<<Αν Α,Β, ειναι τετραγωνικοι πινακες με $AB-BA=I$ νδο $A^{n+1}B-BA^{n+1}=(n+1)A^{n}$ >>το οποιο αποδεικνυεται πολυ ευκολα με επαγωγη.....δεν υπαρχει καπιο λα8ος....απλως δεν ισχυει η αρχικη υπο8εση....
π.χ αν $3=5$ τοτε επαγωγικα $3^{n}=5^{n}$... η επαγωγη δεν ειναι λα8ος,.........απλως εχοθμε υπο8εσει κατι <παραλογο>και καταληγουμε σε επισης παραλογο αποτελεσμα.....συνεπως +οι δυο ασκησεις ειναι σωστα δομημενες....αλλα προχειρα κατασκευασμενες........
[hide]αρα....0-0  και παμε στα πεναλντι.....
Η καλυτερα ...ανταλλαξαμε βασιλισες και παμε σε ισοπαλο φιναλε πιονιων......αλλα νομιζω οτι το απ8 εχει την οποζισιον.....[/hide]

 :D  :D[/quote]

hm... και όμως είναι λάθος γιατί μία άσκηση που θεωρεί κάτι λάθος σωστό πρέπει να ορίζει τι ακριβώς είναι σωστό από την προγενέστερη μαθηματική μας ζωή γιατί διαφορετικά καταλήγουμε σε αντίφαση. Πχ: 3=5<=>0=2<=>0=1<=>0=ν για κάθε ν οπότε το $3^{5}$ γίνεται $0^{0}$ που δεν ορίζεται.
Και σε κάθε περίπτωση η δική μας άσκηση δεν είνια πρόχειρη, η δική σας είναι! 101
άρα 1-0 για να μην πω 2-0

re me duleyete???
[b]thatin[/b]:gt otan orisame to synolo ton fisikon ari8mon den kaname kapies paradoxes??kapia aksiomata???vasi ayton ton aksiomaton tu peano domite olo to sinolo ton fisikon ari8mon....e....an ixame ipo8esi katialo 8a proekipte ena entelos diaforetiko sinolo me entelos diaforetikes idiotites...synepos to oti den iparxun pinakes A,B oste na isxii to AB-BA=I ine ipokimeniko ...den iparxun me ton tropo pu orisame tis praksis metaksi pinakon...se ka8e periptosi i askisi fenete na <ksepernai>to empodio ayto gt lei an isxii AB-BA=I (to 8eori dedomeno+den zitai na apodixti i or8otita tu....)ndo........
+o lompatevski ksekinise me kapies paradoxes pu fenotan eksopragmatikes +omos katelikse se mia kenurgia geometria....pu kses....mpori na iparxi katalilos dianosmatikos xoros(me pros8esi +pol/mo)pu na isxii ayto....pantos ontos..ayti pu epeleksan tin askisi ayti gia na mpi sto vivlio...  ligo.....:stretcher:
ara sinexizume na imaste.0-0...kai malista apodikniete oti den 8a iparksi nikitis..... :P  :P
[b]stedes[/b]
[quote="stedes"][hide="προς thatin και ρακαρ"]Εντάξει, μη μαλώσετε κι όλας... . Το σκορ ούτως ή άλλως είναι 0-1 και αυτό φαίνεται από το ποιος έστειλε πρώτος λύση στην άσκηση*... :P .

*τη σωστή άσκηση... :roll: [/hide]

[u]Άσκηση προς προβληματισμό[/u] (...Lol)
Nα βρεθούν (αν υπάρχουν...) τετραγωνικοί πίνακες $A,B$ (με $A$ διάφορο του μηδενικού πίνακα) και φυσικός $n$, τέτοιοι ώστε να ισχύει $A^{n+1}B-BA^{n+1}=(n+1)A^{n}$.

.....[/quote]
ginete na iparxi kapios n fisikos kai pinakes A,B oste na isxii ayto +na min isxii to AB-BA=I??(pu apodix8ike oti den isxii??...ine san na leme oti i ma8imatiki epagogi...exi ...<kena..>...)
plaka me kanis??? :rotfl:  :rotfl:

edit:logiko epakoloy8o.....Tr(A^{n+1}B)=Tr(A)Tr(A)Tr(A).....[N-FORES]Tr(B)=Tr(B)Tr(A^{n+1})...diladi Tr((n+1)A^{n})=0,atopo...to opio  ine anamenomeno...apo kati anali8es...kataligume se kati anali8es....(afu i epagogi...epagi tin ypo8esi...)la8os ipo8esi=la8os apotelesma.....

το θέμα δεν είναι ποιος έλυσε πρώτο την άσκηση γιατί κάποιοι από εδώ την έλυσαν πριν καν σας την πω, αλλά ποιοι έχουν ις καλύτερες-σωστές ασκήσεις. όπως είπα ρακαρ πρέπει η άσκηση να αναφέρει τι ισχύει γιατί αλλιώς θεωρούμε ότι δουλεύουμε στο δεδομένο χώρο. α, και εμείς έχουμε φτάσει μέχρι την πρόσθεση οπότε... w8! τελικά 145=1!+4!+5! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ωραίο?

κάποιοι έλεγαν ότι δεν υπάρχει τέτοιος αριθμός....:^7:  :D
2-0

τελικά 145=1!+4!+5! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ωραίο?

 :rotfl:  :rotfl:  :rotfl: ....kalo.....me programataki...e?
ok...ok...emp>ap8(1).....omos a>e..opote...ap8>emp(2)(isxesi R ine sxesi diataksis...)
apo tis 1+2 pernume...ap8>emp>apu.....atopo.....ara...(blepe lima stedes...)
i ipo8esi ine lan8asmeni...diladi den isxii emp>ap8........tespa...
brikes lisi gia to 100a+10b+c=a^a+b^b+c^c...??

 :D

όχι...... με [b]φίλο[/b]

το προγραμματάκι είπε πως δεν υπάρχει abc=a^a+...

όπως όλοι ξέρουν στην pascal για την αξιολόγηση μιας λέξης λαμβανουμε υπ'οψιν όλα τα γράμματα...

α, δες τα email σου... τι λες για [b]φυτευτό[/b]? (για να ζηλέψουν [b]μερικοί[/b]..... ποιοι άραγε?)

[color=indigo][size=167]και ναι!!!!
ο compiler λέει πως[/size] [size=200] NTUA[b]>[/b][/size]AUTH[size=167]!!!!!!!!!!!!!!!!!!!!!!!!!!![/size][/color]

[quote="RAKAR"]
[quote="stedes"][u]Άσκηση προς προβληματισμό [/u](...Lol)
Nα βρεθούν (αν υπάρχουν...) τετραγωνικοί πίνακες $A,B$ (με $A$ διάφορο του μηδενικού πίνακα) και φυσικός $n$, τέτοιοι ώστε να ισχύει $A^{n+1}B-BA^{n+1}=(n+1)A^{n}$.
[/quote]
ginete na iparxi kapios n fisikos kai pinakes A,B oste na isxii ayto +na min isxii to AB-BA=I??(pu apodix8ike oti den isxii??...ine san na leme oti i ma8imatiki epagogi...exi ...<kena..>...)
plaka me kanis??? Rotfl Rotfl

edit:logiko epakoloy8o.....Tr(A^{n+1}B)=Tr(A)Tr(A)Tr(A).....[N-FORES]Tr(B)=Tr(B)Tr(A^{n+1})...diladi Tr((n+1)A^{n})=0,atopo...to opio ine anamenomeno...apo kati anali8es...kataligume se kati anali8es....(afu i epagogi...epagi tin ypo8esi...)la8os ipo8esi=la8os apotelesma.....[/quote]

Όταν θεωρώντας κάτι αναληθές ως δεδομένο καταλήγουμε σε ένα αποτέλεσμα, δεν σημαίνει απαραίτητα ότι και το αποτέλεσμα δεν ισχύει.
(π.χ. το 3=5 δεν ισχύει,πολλαπλασιάζοντας όμως με 0 και τα δύο μέλη παίρνουμε 0=0 που ισχύει)
Και εν τέλει η άσκηση δεν λέει ούτε ότι υπάρχουν πίνακες, ούτε ότι δεν υπάρχουν. Αυτό που λέει είναι να βρεθούν αν υπάρχουν.
Επίσης όταν λες "Tr((n+1)A^{n})=0,atopo...", πού ακριβώς είναι το άτοπο;;

[quote="stedes"][quote="RAKAR"]
[quote="stedes"][u]Άσκηση προς προβληματισμό [/u](...Lol)
Nα βρεθούν (αν υπάρχουν...) τετραγωνικοί πίνακες $A,B$ (με $A$ διάφορο του μηδενικού πίνακα) και φυσικός $n$, τέτοιοι ώστε να ισχύει $A^{n+1}B-BA^{n+1}=(n+1)A^{n}$.
[/quote]
ginete na iparxi kapios n fisikos kai pinakes A,B oste na isxii ayto +na min isxii to AB-BA=I??(pu apodix8ike oti den isxii??...ine san na leme oti i ma8imatiki epagogi...exi ...<kena..>...)
plaka me kanis??? Rotfl Rotfl

edit:logiko epakoloy8o.....Tr(A^{n+1}B)=Tr(A)Tr(A)Tr(A).....[N-FORES]Tr(B)=Tr(B)Tr(A^{n+1})...diladi Tr((n+1)A^{n})=0,atopo...to opio ine anamenomeno...apo kati anali8es...kataligume se kati anali8es....(afu i epagogi...epagi tin ypo8esi...)la8os ipo8esi=la8os apotelesma.....[/quote]

Όταν θεωρώντας κάτι αναληθές ως δεδομένο καταλήγουμε σε ένα αποτέλεσμα, δεν σημαίνει απαραίτητα ότι και το αποτέλεσμα δεν ισχύει.
(π.χ. το 3=5 δεν ισχύει,πολλαπλασιάζοντας όμως με 0 και τα δύο μέλη παίρνουμε 0=0 που ισχύει)
Και εν τέλει η άσκηση δεν λέει ούτε ότι υπάρχουν πίνακες, ούτε ότι δεν υπάρχουν. Αυτό που λέει είναι να βρεθούν αν υπάρχουν.Και εν τέλει η άσκηση δεν λέει ούτε ότι υπάρχουν πίνακες, ούτε ότι δεν υπάρχουν. Αυτό που λέει είναι να βρεθούν αν υπάρχουν.
Επίσης όταν λες "Tr((n+1)A^{n})=0,atopo...", πού ακριβώς είναι το άτοπο;;[/quote]

[b]
Όταν θεωρώντας κάτι αναληθές ως δεδομένο καταλήγουμε σε ένα αποτέλεσμα, δεν σημαίνει απαραίτητα ότι και το αποτέλεσμα δεν ισχύει.[/b]
+omos simeni.....an odigumaste sto apotelesma me isodinamies......(i epagogi stin usia ine isodinamia....)episis +i apodiksi su me ta ixni einai me isodinamies....eno to 3=5 ,0=0 den ine isodinamia....(otan leo ine isodinamia enoo an kai mono an....)an me iso dinamies ksekinisume apo kati anali8es kai ftasume se kati ali8es.....tote pernontas tin anti8eti fora,,,(mporume na to kanume afu ine isodinamia....)ine san na ksekiname apo kati ali8es kai kataligume se kati anali8es.....opote...i exume katariyi tin ma8imatiki epagogi   i exume kani kapia..mlk.....+ imaste gia..:stretcher:

[b]
Και εν τέλει η άσκηση δεν λέει ούτε ότι υπάρχουν πίνακες, ούτε ότι δεν υπάρχουν. Αυτό που λέει είναι να βρεθούν αν υπάρχουν.[/b]
nai +fono,alaine profanes oti den iparxi ute tetios fisikos ala ute kai pinakes A,B oste na isxii ayto.......

[b]Επίσης όταν λες "Tr((n+1)A^{n})=0,atopo...", πού ακριβώς είναι το άτοπο;;[/b]
epidi Tr(A^{n+1}B)=Tr(A)Tr(A)Tr(A).....[N-FORES]Tr(B)=Tr(B)Tr(A^{n+1}) gia                        ka8e n, 8a ine Tr(A^{n+1}B-BA^{n+1})=0 gia ka8e n pu simeni
Tr((n+1)A^{n})=0 gia ka8e n ,pu simeni Tr (A^{n})=0 gia ka8e n kai ara
o pinakas A prepi na ine o midenikos.....to atopo sririzete sto oti isxii gia ka8e n...prosoxi...to atopo ine gia tin askisi pu ipa ego...(opos iparxi +sto vivlio)..oti diladi kai to simperasma tis epagogis ine anali8es....den anaferete stin <askisi gia provlimatismo>pu evales esi....opos apedikses oti i ipo8esi ine anali8is ...leo afu i ma8imatiki epagogi ine <klisti os pros tis sinepagoges tis>(an mpori na ipo8i kati tetio...)8a ine kai to simperasma anali8es...kai aploseyaxna ton tropo gia napodikso tin anli8ia tou zitumenu....

ntaksi ...ftani...tin analisame arketa ayti tin askisi...
[b]gia deite edo...mia mikri alagi alazi ta panta.....[/b]

esto oi tetragoniki pinakes n epi n A,B.an i pinakes A,AB-BA antimetati8ente ndo $A^{n+1}B-BA^{n+1}=(n+1)(AB-BA)A^{n}$ me $n\geq 2$

 :D

[quote="RAKAR"][b]
Όταν θεωρώντας κάτι αναληθές ως δεδομένο καταλήγουμε σε ένα αποτέλεσμα, δεν σημαίνει απαραίτητα ότι και το αποτέλεσμα δεν ισχύει.[/b]
+omos simeni.....an odigumaste sto apotelesma me isodinamies......(i epagogi stin usia ine isodinamia....)...[/quote]

Και πως προκύπτει δηλαδή ότι η επαγωγή είναι ισοδυναμία; Και αν όντως κάπως προκύπτει τότε τί είναι ισοδύναμο με τί;

[quote="RAKAR"][b]
Και εν τέλει η άσκηση δεν λέει ούτε ότι υπάρχουν πίνακες, ούτε ότι δεν υπάρχουν. Αυτό που λέει είναι να βρεθούν αν υπάρχουν.[/b]
nai +fono,alaine profanes oti den iparxi ute tetios fisikos ala ute kai pinakes A,B oste na isxii ayto.......[/quote]

Επιμένεις ότι είναι προφανές;

[quote="RAKAR"][b]Επίσης όταν λες "Tr((n+1)A^{n})=0,atopo...", πού ακριβώς είναι το άτοπο;;[/b]
epidi Tr(A^{n+1}B)=Tr(A)Tr(A)Tr(A).....[N-FORES]Tr(B)=Tr(B)Tr(A^{n+1}) gia                        ka8e n, 8a ine Tr(A^{n+1}B-BA^{n+1})=0 gia ka8e n pu simeni
Tr((n+1)A^{n})=0 gia ka8e n ,pu simeni Tr (A^{n})=0 gia ka8e n kai ara
o pinakas A prepi na ine o midenikos.....to atopo sririzete sto oti isxii gia ka8e n...[/quote]

Αυτό το  "[b]8a ine Tr(A^{n+1}B-BA^{n+1})=0 gia ka8e n pu simeni Tr((n+1)A^{n})=0 gia ka8e n[/b]"  από που βγαίνει;

Το ότι θα ισχύει για το συγκεκριμένο $n$ για το οποίο θα ισχύει και η δοσμένη ισότητα συμφωνώ, αλλά εσύ λες

"$Tr(A^{n+1}B-BA^{n+1})=0 \ \ \forall n \Leftrightarrow Tr((n+1)A^{n})=0 \ \ \forall n$",

 ενώ ισχύει $A^{n+1}B-BA^{n+1}=(n+1)A^{n}$ για ένα [u]συγκεκριμένο[/u] $n \in \mathbb N$.

Τέλος, η πρόταση "[b]Tr (A^{n})=0 gia ka8e n kai ara o pinakas A prepi na ine o midenikos.....[/b]"
επίσης δεν ισχύει, αφού για τον πίνακα $A=[\{0,0\},\{1,0\}]$ ισχύει
$Tr(A^{n})=0 \ \ \forall n \in \mathbb N^{*}$ χωρίς ο $A$ να είναι ο μηδενικός πίνακας.

Ξαναδές την άσκηση μήπως δεν την κατάλαβες καλά... .

malon den katalaves pu anaferomun.....
[quote]prosoxi..[b].to atopo ine gia tin askisi pu ipa ego...[/b](opos iparxi +sto vivlio)..oti diladi kai to simperasma tis epagogis ine anali8es....[b]den anaferete stin <askisi gia provlimatismo>pu evales esi....[/b][/quote]
Τέλος, η πρόταση "Tr (A^{n})=0 gia ka8e n kai ara o pinakas A prepi na ine o midenikos....."
επίσης δεν ισχύει, αφού για τον πίνακα A=[\{0,0\},\{1,0\}] ισχύει
Tr(A^{n})=0 \ \ \forall n \in \mathbb N^{*} χωρίς ο A να είναι ο μηδενικός πίνακας.

 :blush: ontos...edo pao paso...ara...pi8anon na iparxun kapies lisis....
mperdeytika sto oti i anti8eti tis protasis  not(isxii gia ka8e n)=iparxi tulaxiston ena n oste na min isxii   kai oxi ..den iparxi kanena n oste na isxii.....
 pantos i epagogi (opos tin fantazome diagramatika ego...)tin vlepo san na ine ena monopati..... briskomaste stin arxi...(ipo8esi)kai akolou8ontas to monopati(epagogi) kataligume se kapio apotelesma....ine fanero oti i ipo8esi kai to apotelesma,exun tin idia egkirotita...i isxiun kai ta dio i kanena......

 :D

[quote="RAKAR"]prosoxi...to atopo ine gia tin askisi pu ipa ego...(opos iparxi +sto vivlio)..oti diladi kai to simperasma tis epagogis ine anali8es....den anaferete stin <askisi gia provlimatismo>pu evales esi....
[/quote]
Μα η άσκηση που είπες, το είπαμε εξ'αρχής ότι ήταν λάθος. Ποιός ο λόγος να γράψεις όλα αυτά;  :wink:

Anyway, στην προσπάθειά μου να αποδείξω ότι δεν υπάρχουν τετραγωνικοί πίνακες $A,B$ τέτοιοι ώστε να ισχύει $A^{n+1}B-BA^{n+1}=(n+1)A^{n}$ για κάποιο φυσικό $n$, έβγαλα κάποια συμπεράσματα για τους $2x2$ πίνακες:

Αν $A,B \in M_{2}(\mathbb R)$ τότε ισχύουν τα εξής:

$1)$ $A^{2}B-BA^{2}=Tr(A)\cdot(AB-BA)$

$2)$ αν $A=\{\{a1,a2\},\{a3,a4\}\}$ τότε $A^{3}B-BA^{3}=(a_{1}^{2}+a_{2}a_{3}+a_{1}a_{4}+a_{4}^{2})\cdot(AB-BA)$

Έπεται συνέχεια...  :P

Στέλιος...

που "Έπεται συνέχεια..."  οέο?
μιλώντας για ομιλίες, πάει κανείς από εδώ στο σεμινάριο του
κ. Αγγελόπουλου?

περυσι ελεγε πολυ.εσυ πας ?

nai, exw paei mia fora.
exasa to prwto kai to teleutaio (vasika den katalava to prwto egine telika?)

Έστω g(x) R->R
Nδο ότι η f
f(x)=0 αν x στο R-Q
f(x)=g(x) αν x στο Q
είναι συνεχής στα x που είναι ρίζες της g και παραγωγίσιμη στις ρίζες 2ν βαθμού της g


vector
abstract algebra
linear algebra
linear algebra unsolved

Prove that $I_{4}$ cannot be written as the sum of less than $4$ matrices of rank $1$. ;)

I think we've already discussed this one in the romanian forum. It's just the fact that $rank(a+b)$ is at most the sum of ranks, that's all. Needless to say, this was discovered long before I.Savu.  ;)

[quote="jmerry"]This is a special case of the general theorem that the tensor rank is the same as every other rank for matrices.[/quote]

what is tensor rank?

Yes, harazi i agree with you on this one. Actually I have discovered that lots of problems proposed by I. Savu are discoveered long before . :o

The tensor rank: Interpret the space of $m\times n$ matrices as the tensor product of $\mathbb{R}^m$ and $\mathbb{R}^m$. This is the space of linear combinations of "products" $uv$ of elements in the two spaces respectively, with identifications based on the vector space properties ($c(uv)=(cu)v=u(cv)$ when $c$ is a scalar, and $u(v+w)=uv+uw,(u+v)w=uw+vw$ when $u,v,w$ are vectors of the right types). The tensor rank of an element $a$ of this tensor product is the minimum $r$ for which we can write $a=\sum_{j=1}^r c_ju_jv_j$ with $c_j$ a scalar and $u_j,v_j$ vectors of appropriate type.

That's probably confusing, so here's a slightly different version. Use the tensor product of the dual spaces $(\mathbb{R}^m)^*$ and $(\mathbb{R}^n)^*$; this is the space of bilinear maps from $\mathbb{R}^n\times\mathbb{R}^n$ to $\mathbb{R}$. The tensor rank of an element $a$ is the minimum $r$ for which we can write $a(x,y)=\sum_{j=1}^r c_j u_j(x)v_j(y)$ with $c_j$ scalars and $u_j,v_j$ linear maps.
That's obviously isomorphic to the space of matrices, and we can see that those products of linear maps are precisely the rank one matrices.


A computer is programmed to randomly generate a string of six symbols
using only the letters A,B,C. What is the probability that the string will not
contain three consecutive A’s?

it would probably be easier to count the number of strings that DOES have three consecutive A's, and then count the total number of possible strings and subtract.

i would do the first part by taking three consecute A's and counting how many ways i can place the remaining three symbols 'around' the three A's.

That runs the risk of over-counting -- I'd split it up into the cases where there's a block of six As, a block of five (but not six) As, a block of four (but not five) As and a block of three (but not four) As.

This problem is pretty easy.  I'm pretty sure I did it right.  Here it is:

There are 4 ways that three A's can be in a row in a list of 6 letters (for now ignoring the other three letters):

A A A _ _ _  (option 1)
_ A A A _ _  (option 2)
_ _ A A A _  (option 3)
_ _ _ A A A  (option 4)

Now, looking at the first example (option 1), each of the three blanks can be filled in with either A, B, or C.  Since there are three blanks, and three possibilities for each blank, there are 27 (3x3x3) possible chains for this possibility.

For option 2, the last two blanks still have three letter possibilities, but the first blank can only be B or C, since if it was A, it would be repeating some of the chains made by option 1.  So, 2x3x3 is 18, which is the number of possible chains for option two.

Option 3 is similar to option 2.  The first and third blanks can be filled in with any of the three letters, but the second blank cannot be A, since that would create some of the chains made in option 2.  Also, for option 4, the first and second blanks can be any letter, while the third blank causes repeats when it is A.  So, option 3 and option 4 both have 18 possible chains.

Now, how many total possible chains are there?  Each of the six positions can be one of the three letters, so the total possible chains is 3^6, or 729.  So, the probability that a chain WILL have a chain of three or more is 81 (27+18+18+18)/729, or 1/9.  That makes the probability that there will NOT be a chain of three or more A's to be 8/9!!!!!

Let me know if I made a mistake


function
quadratics
algebra
quadratic formula

1) A man stacks balls so that the base is a hexagon with side 10.

he keeps on stacking the balls untill there is only 1 ball left at the top.

How many balls are there?

2) A person is burning things with a magnify glass.

Let the radius of the glass be x units and y be the temperature it takes to burn something

Assuming that the function $y=x^2+7x+3$ is true, how large does the raius have to be to:

(a) burn an ant;100 degrees
(b) burn a blade of grass; 200 degrees
(c) burn a piece of paper;451
(d) start a forest fire; 850

approximate your answers to the nearest 100th

Hm the 2nd one's quite trivial, however, I drew out the 1st one, with a hexagon side 3 first, it seems that to stack on top of the one with 3, it wouldn't actually turn into a hexagon side 2, heh.

OK, I'm quite privileged to have a lot of little metal spheres(they're from my lil bro's construction set I think, hehe) After trying to stack it, there's quite a many ways to stack the balls.
Are we to assume that each layer is a hexagon, at least till where it's sides of 2, then a triangle made of 3 balls, then the single ball?

If it is that way, Then we can solve it like this

The first layer would have:
10+11+12+...+18+19+18+...+11+10
The second would have:
9+10+11+...+16+17+16+...+10+9

Continueing in this way, till we get to
3+4+5+4+3
2+3+2

We can clearly see the sum is 2(2+18)+3(3+17)+4(4+16)+...+9(9+11)+10*10+19=(20)(2+3+4+...+9)+10*10+19=20*44+100+19=880+100+19=999.
Adding 3 for the triangle, and 1 for the single ball at the top, we'd get 1003.

P.S. Seeing how the sum is nicely 999, I just know I did it way too long... heh

[hide="number 2"]

a) $x^2+7x+3=100$
$x^2+7x-97=0$.
$x=6.95$

Every time only the c term changes. The quadratic formula only has one c in it. Let's play around with c values only.
to get,

b) $x=10.97$

c) $x=17.95$

d) $x=25.81$ [/hide]


i'm new here, please help me with this.

x, y, z are pozitive reals. prove

$x+y+x^{2}y^{2}\geq x^{2}y+xy^{2}+1$

[quote="N.T.TUAN"]$\Longleftrightarrow (x^{2}y^{2}-1)+x+y-xy(x+y)\geq 0\Longleftrightarrow (xy-1)(xy+1)+(x+y)(1-xy)\geq 0 \Longleftrightarrow (1-xy)(x+y-1-xy)\geq 0 \Longleftrightarrow (1-xy)(x-1)(1-y)\geq 0$.[/quote]

which is wrong for $x,y<1$ :wink:


Hii,

I have posted in "games" a post titled  "Concatenation : prime HVD" and I still do not understand why it was locked.
What was wrong?
Thank you for any answer.

Since completely stupid games were forbidden, games of your type (pseudo math games without any reasonable content) started occurring. At first, I tolerated them for the sake of alternatives, and believing that they would not get too much. But now there are already more than enough of them. Most of them lack creativity, too (no, it's not interesting to find those $ \{a,b,c,d,e,f,g,h,i\} \equal{} \{1,2,3,4,5,6,7,8,9\}$ such that $ a^{b\minus{}c} \cdot d \plus{}e \cdot f \minus{} g^{h\plus{}i}$ gets as big as possible). If it's a true math problem, put them to the math forums. If it's a game that needs some kind of creativity, put it into Games. But if it's lacking both, put it nowhere.
In your case, that's just a brute force problem. What's the creativity in gaming¿

Also note that games should (if possible) be interesting to outsiders, too. This isn't a necessary requirement, but be sure that the probably to be accepted is higher then (e.g. The Mole is a good one, math games are bad ones).

Thank you for your answer.
So you can remove any post that you as moderator feel it lacks creatitvity.
I agree with your explanations.
Have a good day.


Determine todos los números naturales $n$ tales que $n^{n}+3n^{3}+81$ es un cuadrado perfecto.

[color=blue]¿Consideras al $0$ entre los naturales?[/color]

No, en general no considero al $0$ natural.

[color=blue]Yo tampoco, pero nunca está de más preguntar. Uno nunca sabe…[/color]

Claro, todo depende de la definicion de naturales que más te guste.

Dividiremos en dos partes, cuando $n$ es par, o impar.
I. CASO.-
$n$, impar, analisando modulo 8.
$n^{n}+3n^{3}+81=n.n^{n-1}+3n.n^{2}+1=n+3n+1=5 mod 8$, pero los restos cuadraticos de 8 son, 0,1,4,1,0,1,4,1, pero ninguno es 5, por lo tanto para $n$ impar no hay solucion.
II. CASO.-
$n$, par.
$n=2k$, --->acotaremos entre dos potencias.
$n^{n}+3n^{3}+81>n^{n}$, ahora si n>4=>
$[(2k)^{k}+1]^{2}=(2k)^{2}k+2(2k)^{k}+1>(2k)^{2}k+128k^{3}+1>(2k)^{2}k+24k^{3}+81$, entonces solo falta analisar para $n$ 2,4, cumpliendo asi solo para $n=4$-->$n^{n}+3n^{3}+81=4^{4}+3(4^{3})+81=23^{2}$.
---
.'. la unica solucion es $n=4$.


function
real analysis
real analysis unsolved

Let $ f$ be continuous on $ [a,b]$ such that $ f$ takes each of its values finitely many times. Prove that for some $ c$ cardinality of $ \{x\in [a,b]: \ f(x)\equal{}c\}$ is the odd number.

let me say an idea: i think we can prove that  every time f change its monotonicity some values always have odd cardinality depending on max and minimum of f.

If you can formalize the notion of "odd" and "even" for infinity, your idea would work.
In this form, it works only if the values $ f(a), \ f(b)$ at the endpoints are different: then, choose the line $ y \equal{} c\in [f(a),f(b)]$ and it will give the odd number of intersections(the number of strict local extremes is at most countable, so we can choose the line not containing any).
If $ f(a) \equal{} f(b)$, this does not work any more.


Is 0 meters equal to 0 seconds?

[This question may seem ridiculous, but I'm not sure what my answer is.  I genuinely want to hear what other folks think.]

Haha.  Or in other words, "what is the algebraic structure of 'numbers with units?'"  I think the answer is probably (but not definitely) no: unless you want to introduce quantities like "1 second + 3 meters/second," (which you could do, and turn it all into a big algebra over the reals, in which case the answer would be "yes"), the objects "1 second" and "1 meter" don't belong to a space with an additive structure.  The fact that $ 0 \times x \equal{} 0$ for all $ x$ in "nice" structures follows from axioms of distributivity and so on, i.e. they're a consequence of the fact that we can add and multiply arbitrary elements and the fact that these operations interact nicely.  Since in the structure of "numbers with units" we have arbitrary multiplications but not arbitrary additions, it seems unlikely that a reasonable collection of axioms describing this space would have a single additive identity.  Now, one thing you could try (if you don't have plans this weekend ;) ) is to write up a bunch of axiomatizations of the collection of "numbers with units" and decide which one is best.  If the best one has a single additive identity, the answer is "yes;" if it has various different additive identities, one for each collection of units, then the answer is "no."

Good points.

A related question is: what is the value of $ 1 \textrm{ meter} \plus{} \minus{}(1 \textrm{ meter})$?  The usual axiom $ x \plus{} \minus{}x \equal{} 0$ would suggest an answer of 0.  But maybe the answer should be 0 meters (treated as different from 0).

Let me raise another question.  Let $ A$ be the set of prime numbers that are multiples of 4.  Let $ B$ be the set of ordered pairs $ (x, y)$ of real numbers such that $ x < y < x \minus{} 1$.  According to the equality axiom of set theory, $ A \equal{} B$.  (They contain the same elements, because they are both empty.)

That seems a little strange to me.  I almost feel like it is a type error to compare $ A$ and $ B$.  (One is a set of integers, the other is a set of ordered pairs.)  Maybe there is a kind of typed set theory that wouldn't allow such comparisons.  In that case, maybe that kind of typed set theory would be relevant to the "numbers with units" question.

I don't think 0 meters is equal to 0 seconds because 0 meters is a relative measure of distance. if you measured from -1 meters, then 0 meters is 1 meter instead.
That's why the meters has to remain.

It's like how 0 volts is not equal to 0 seconds because voltage is a relative thing.

This also dies if we consider something like 0 decibels, since bels are on a logarithmic scale.

[quote="Ravi B"]A related question is: what is the value of $ 1 \textrm{ meter} \plus{} \minus{} (1 \textrm{ meter})$?  The usual axiom $ x \plus{} \minus{} x \equal{} 0$ would suggest an answer of 0.  But maybe the answer should be 0 meters (treated as different from 0).[/quote]
If we had a unique additive identity $ 0$, then I certainly believe that we would have the equality $ 1 \textrm{ meter} \plus{} \minus{} (1 \textrm{ meter}) \equal{} 0 \textrm{ meters} \equal{} 0 \equal{} 0 \textrm{ seconds}$.  And if you're willing to tolerate the quantity $ 1\textrm{ meter} \plus{} 3\textrm{ seconds}$ as some abstract member of this algebra lacking physical meaning, then I think that will probably work out fine.  Actually, I think I'm beginning to like that as a solution to this problem, although I suspect it wouldn't satisfy my engineer or physicist friends.  If we have different additive identities for each unit then certainly $ 0 \textrm{ meters} \neq 0$, where here $ 0$ is not special at all (it's unitless, i.e. its units are the multiplicative identity in the group of units, and so it isn't any different than any other quantity with units -- the only "special" number in this regime would be 1 (with no units) which is the multiplicative identity because 1 is the multiplicative identity for numbers and "no units" is the multiplicative identity for units).

[quote]Let me raise another question.  Let $ A$ be the set of prime numbers that are multiples of 4.  Let $ B$ be the set of ordered pairs $ (x, y)$ of real numbers such that $ x < y < x \minus{} 1$.  According to the equality axiom of set theory, $ A \equal{} B$.  (They contain the same elements, because they are both empty.)

That seems a little strange to me.  I almost feel like it is a type error to compare $ A$ and $ B$.  (One is a set of integers, the other is a set of ordered pairs.)  Maybe there is a kind of typed set theory that wouldn't allow such comparisons.  In that case, maybe that kind of typed set theory would be relevant to the "numbers with units" question.[/quote]
I was going to say that this doesn't bother me at all: I don't have any problem seeing "0" as different from "nothing."  A set with 0 elements contains nothing -- it doesn't matter whether we call the things it doesn't contain elements, integers, ordered pairs or seconds: it doesn't contain any of them.  But now you've got me thinking about it and maybe I'm going to have to go wrap my head around the question of what "0 seconds" actually [i]is[/i].  (My problem here: I want to say that 0 seconds really is something.  But then surely 0 seconds and 0 meters are different, because they're two things of entirely different sorts.  But then this suggests that the solution I was becoming happy with above is wrong.  Bah.)

ProtestanT, your last line begs the question.  Also, I don't buy your relative measure talk: the point we label 1 meter on a graph is relative to the point we chose to label 0 meters, but the actual distance between these points (1 meter) is not relative.  As long as it and I sit in the same reference frame, it doesn't matter where the meter is: it'll still look just like 1 meter to me.

Hamster, it's not at all clear to me what you mean.  Could you elaborate?

decibels are the units of sound, and it's a logarithmic scale. As a result, a sound measuring at 0 dB can still be heard - it could not be regarded as a $ 0$.

Similar things happen with $ ^{\circ}F$ and $ ^{\circ}C$, as well as any logarithmic scale.


Podijelio bih s vama zanimljivu vijest da sutra (30. julija) počinje internetska liga Puzzleup ( http://www.puzzleup.com/ ), što je u biti rješavanje kombinatornih matematičkih zadataka.

Zadaci nisu baš poput onih na natjecanjima, ali su jako interesantni a neki su i prilično teški. Idemo za Balkan osigurati bar 5 mjesta u top10!  :D

Ajde, ja sam od preksutra u Istanbulu, ali dobro... u idealnom slucaju cu imati laptop i pogledat cu.

Prosli put sam se bas na vrijeme prijavio (pred kraj sezone) :)

Sad cu valjda biti azurniji...  :lol:

prvi zadatak mi izgleda jako ružan i težak...

Ruzan jeste, ali odvoji malo vremena za njega, zabavi se. Radio sam ga od objavljivanja do maloprije-ali sam ga uradio  :lol: , dakle moze se. Jest da sam raspisivao milion nekih slucajeva, ali isplatilo se  :wink:

By the way, preporucujem ovaj zadatak kao zgodnu vjezbu programiranja (naime, jutros mi je bilo dosadno pa sam napisao programcic u c++ koji rjesava zadatak-svega 30 linija koda, a mozda moze i malkice krace).

Ovaj drugi sam, nadam se tocno, rijesio programiranjem.

Buduci da nikad necemo saznati tocnost nasih rezultata kao ni rjesenja, bilo bi dobro da po zavrsetku lige usporedimo svoje rezultate ili rjesenja.

Nema frke, ova dva programa su sigurni na mom hard disku. ;)

A ako se onaj prvi i mogao rijesiti "pjesice", ovaj ne moze nikako... Dakle, rekurzija+cpp  :lol:

P. S. kolekcija rjesenja puzzleup lige ove godine ce prije biti zbirka rijesenih zadataka iz programiranja, nego zbrka (sic!) zadataka iz matematike  :)

Malo sam razocaran novim zadatkom. Rjesenje bukvalno stane u pola reda, a tezina je otprilike kao kantonalno takimcenje za prve razrede... Ako ne i ispod. Ali barem ne moram programirati  :lol:

Moram priznati da su zadaci ove godine puno lakši nego lani, jer sam dosad (nadam se točno) riješio sve osim jednog.

[b]edit:[/b] Ipak mislim da sam dosad riješio svih 7. Gotovo svaki zadatak je ili vrlo lagan, ili ga treba pametno isprogramirati.

Prije edita bih i pitao koji je to ostao nerijesen. A ovako se mogu samo sloziti sa konstatacijom.  :)

jupi! na trenutnoj score tabli imam samo 2 boda manje od maksimuma. ali kad bi se gledali svi zadaci zaostajao bih i više jer sam izgubio desetak bonus poena.

Zanimljivo... ja zaostajem 2 boda za maksimumom, a upravo dva boda svega sam i izgubio (po svojoj racunici), i to na ovom posljednjem zadatku.

i to nije pokazatelj da smo dobri nego da su zadaci loši. još ni za jedan zadatak (osim možda prvoga) nisam trebao dugo razmišljati.

Ja jos uvijek razmisljam o ovom posljednjem zadatku, i to ne o tome kako ga rijesiti (to sam vec zavrsio) nego zasto su bas kraljice na ploci :)

Valjda sam navikao da zadaci imaju smisla, a ovaj ga bas i nema u prevelikim kolicinama  :)

prije svakog kola nadam se da će konačno doći pravi zadatak, i onda se razočaram.

i što se tiče bonus poena: ako oni budu odlučivali o prvima ( što će se dogoditi ako se nastavi ovakav trend ) onda će to biti stvarno loša sezona.

Idem večeras u Makedoniju na Balkanijadu iz informatike, jej! :D
Tako da zadatak koji je upravo objavljen neću imati vremena ni volje rješavati, a k tome je i užasno ružan. Dakle gubim još nekoliko bonus poena. :(

Ma nije bas nesto ruzan, samo je ubitacno nezanimljiv. K tome, vise je enigmatske prirode nego matematicko/informaticke.  :sleeping:

(pod ovim mislim da bi se prije mogao pojaviti u nedjeljnim novinama nego na kakvom takmicenju :D )

Uzivaj u FYRoM!!  :lol:

I završila je još jedna sezona PuzzleUpa... Kako je No. 12 bio posljednji zadatak na koji sam uopće poslao rješenje, nije iznenađenje da me nema među prvih 50 (završio sam na 64. mjestu, sa okruglo 1500 poena, koliko si ti imao na kraju, djuro?).

Bilo je zabavno, samo da nije fakultetskih obaveza...  :)

Meni nije bilo zabavno. Jer kao što smo već komentirali, zadaci su razočaranje. Ja sam prekinuo nakon 17. zadatka, s time da nisam rješavao 10. ni 16.  Nalazim se na 61. mjestu sa 1611 bodova i zaključujem da sam prema tome dosta toga fulao. Nema veze.

Kad kažem da je bilo zabavno, poredim svoj rad na Puzzleupu sa senzorima razine tečnosti kojima se trenutno bavim... pa kažem da je bilo zabavno. A da su zadaci bili lošiji no prošlih godina, tačno. No dobro, slijedeći izazov mi je Project Euler, čim se dokopam slobodnog vremena.

22.07. kreće nova sezona. Ne pada mi na pamet da se takmičim ove godine, ali eto, koga zanima...


induction
number theory proposed
number theory

how can I prove that $ r_0\equal{}1\equal{}5^0$ is the  only integer power of 5 in the sequence

$ r_0\equal{}1$

$ r_1\equal{}7$

$ r_{n\plus{}2}\equal{}8r_{n\plus{}1}\minus{}r_n$

??

Yes, I forget 3.
$ r_n\equal{}Rat (1\plus{}\sqrt{\frac{3}{5}})(4\plus{}\sqrt{15})^n$, were $ Rat (a\plus{}b\sqrt{15})\equal{}a,a,b\in Q$.
$ 5|r_n\to n\equal{}n_0\mod 5, n_0\equal{}2$. Let $ m_k\equal{}n_0\plus{}5*n_1\plus{}5^2*n_2\plus{}...\plus{}5^{k\minus{}1}n_{k\minus{}1}$ is minimal n, suth that $ 5^k|r_n.$
We find $ r_{m_k}\equal{}Rat(a_k*5^k\plus{}b_k\sqrt{\frac{3}{5}})$, were $ a_k*5^k\plus{}b_k\sqrt{\frac{3}{5}}\equal{}(1\plus{}\sqrt{\frac 35})(4\plus{}\sqrt{15})^{m_k}$. By induction $ (5,a_k)\equal{}(5,b_k)\equal{}1, f_k,b_k\in Z$.
Then we calculate $ n_k$, suth that $ 5^{k\plus{}1}|a_{k\plus{}1}$, because $ (5,b_k)\equal{}1$ we get $ n_k\not \equal{}0$, therefore $ m_{k\plus{}1}>5^k$.


geometry
3D geometry
sphere
calculus
calculus computations

Suppose that a droplet of mist is a perfect sphere and that, through condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that under these circumstances the droplet's radius increases at a constant rate.

Suppose that a droplet of mist is a perfect sphere (so $ V(r)\equal{} \frac{4}{3}\pi r^3$) and that, through condensation, the droplet picks up moisture at a rate proportional to its surface area (i.e., $ \frac{dV}{dt}\equal{}kS(r)$ for some $ k\in\mathbb{R}$). Show that under these circumstances the droplet's radius increases at a constant rate.

We have that

\[ \frac{dV}{dt}\equal{}\frac{d}{dt}\big( \frac{4}{3}\pi r^3\big) \equal{} 4\pi r^2\frac{dr}{dt}\equal{}S(r)\frac{dr}{dt}\]

but $ \frac{dV}{dt}\equal{}kS(r)$, and hence

\[ kS(r)\equal{}S(r)\frac{dr}{dt}\Rightarrow \frac{dr}{dt}\equal{}k\]

as required.


A machine is a device that do some task.

A magnet can attract metal object.
device=magnet
task=attract metal object

Therefore a magnet is a machine, a very simple machine. Yes or no?

Now that's an ironic question :).

A machine's sole purpose is to do work, while a magnet does NO work at all. In my completely non-professional opinion they aren't the same thing :o.

That's not what a machine is. I believe a machine is a piece of equipment that makes a job easier, usually by decreasing force per time and increasing distance. For a machine to be useful, you must do work on it.
A magnet does not fit this description. It creates a magnetic field and objects in the field "react" to it. A magnet converts an object's magnetic potential energy to kinetic energy, analagous to how the Earth converts an object's gravitational energy to kinetic energy.

From Dictionary.com

ma·chine    ( P )  Pronunciation Key  (m-shn)
n.

1. a) A device consisting of fixed and moving parts that modifies mechanical energy and transmits it in a more useful form.
    b) A simple device, such as a lever, a pulley, or an inclined plane, that alters the magnitude or direction, or both, of an applied force; a simple machine.

2.  A system or device for doing work, as an automobile or a jackhammer, together with its power source and auxiliary equipment.

There are other definitions, some relate to biology etc. The word "machine" is definitely too vague for this type of question. No matter how you look at it though, a magnet is not a machine, and I guess that's all that really matters.

From your definition , yes.
But looking at a general aspect of it , a magnet is not considered as a machine since you cant say it was made by man to do some specific work {which is the purpose of all machines.}

Look a machine is just a name and because it doesn't have a well agreed definition why does it matter????????????????? :?:

magnet = not machine... because it doesnt exert any force (i could be wrong, i dont really do physics), its pull relies on the earth's magnetic field. so the magnet having any pull isnt the magnet's "fault"

that is absolutely wrong, unfortunately. Magnets do exert a force. In the static case, it is called the Biot-Savart law. Magnets on the otherhand Do No Work. Also, there are many kinds of magnets, but I don't believe I know of any that depend on the earth's magnetic field. Magnets occur in nature from the proper alignment of dipoles within the substance.

[quote] Magnets do no work. [/quote] True as a classical approximation in arguments about 'perpetual motion' machines, but actually false.

Of course Magnets exert an electromagnetic force upon nearby charges in its field, such as electrons.  This diverts their trajectories, causes spin flipping, and changes the size and spacial configuration of magnetic domains.

Magnets do work in 'magnetizing' other objects, creating local order and lowering entropy, altering magnetic domains, and countering opposing forces on the macro-scale.

Work is performed by an electromagnet for instance, in holding a car in the air above gravity, since there is never a static balance of forces, but a constantly changing and adjusting one.  Also the strength of the magnet is usually deliberately set to be far stronger than the gravitational force, so excess work is done at the surface as metal is crushed and electromagnetic energy is converted to heat.

I presume that magnets are sort of machines,if we have to go by the definition of rthe dictionary.Since magnets can do work and are always in a capactiy of doing work so why not a magnet a machine.Magnets are capable of producing electricity so shoud it not be considered a machine.


[quote="daermon"]Does anybody know of any resources that I could give/use to tutor an advanced third grade student?  thanks :D[/quote]
AoPS vol. 1
If you think that's too hard make sure they completely comprehend and are comfortable with negatives and go with the intro to number theory.

I think third grade is a bit early for AoPS volume 1  :P
Number theory might work :D

My teacher was really stupid in third grade...
So this one kid, Vincent, and I just did our own advanced math because we were just so smart!

[quote="Go Around the Tree"]I wish I could have had a tutor in 3rd grade. I was a noob then.[/quote]

our third grade class was so advanced we did 6th grade basics midway through the year.

anyway, if I were you, daermon, I would take a look at the curriculem that third grader has and enhance it. once he is 100% certain about the material, I would then go to fourth grade material...=)

hey daermon,

if you have KUMON in your state, perhaps that'll help.... :P

 to 7h3.D3m0n.117 -

yah, i know this school where 7th graders can study 9th grade/ 10th grade math......


inequalities
function
algebra
polynomial
inequalities proposed

Let $ a,b,c$ be real numbers such that: $ a^2\plus{}b^2\plus{}c^2\equal{}3$ .Prove that:
$ a\plus{}b\plus{}c\minus{}abc \le \sqrt{\frac{305\plus{}13\sqrt{13}}{54}}$

[quote="quykhtn-qa1"]Let $ a,b,c$ be real numbers such that: $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ .Prove that:
$ a \plus{} b \plus{} c \minus{} abc \le \sqrt {\frac {305 \plus{} 13\sqrt {13}}{54}}$[/quote]
Let $ a \plus{} b \plus{} c \equal{} p,$ $ ab \plus{} ac \plus{} bc \equal{} q,$  $ abc \equal{} r$ and
$ f(r) \equal{} \frac {p(p^2 \minus{} 2q)}{3} \minus{} r\left(\sqrt {\frac {p^2 \minus{} 2q}{3}}\right)^3.$
Hence, $ f$ is a linear function of $ r.$ Thus, $ f$ gets the maximum value on bounders of $ r.$
But $ r$ is a free term of  degree three polynomial with three real roots.
Id est, we obtain the bounders of $ r,$ when two of $ a,$ $ b$ and $ c$ are equal.
Hence, it remains to prove that $ a \plus{} 2b \minus{} ab^2\leq\sqrt {\frac {305 \plus{} 13\sqrt {13}}{54}},$ where $ a^2 \plus{} 2b^2 \equal{} 3,$ which is indeed true.
The equality holds when $ a \equal{} \frac {3 \minus{} \sqrt {13}}{\sqrt {10 \minus{} 2\sqrt {13}}}$ and $ b \equal{} c \equal{} \frac {2}{\sqrt {10 \minus{} 2\sqrt {13}}}.$


Find the minimum value of $ x\plus{}y$ if $ x$ and $ y$ are positive integers that satisfy the equation $ (x\minus{}1)^3 \plus{}153(y\minus{}3)^3\equal{}63$

[hide]
$ (x\minus{}1)^3\equal{}a^3 , (y\minus{}3)^3\equal{}b^3$
6 and -1 work for a and b (I guessed that on my first try w00t)
They correspond to an x and y of 7 and 2, total is 9.

Well since I guessed I think I should include a way to do it regularly.
Note that x and y are positive.  That means, for the most part, both the cubed values will be positive and thus will not work.  So we have to find a pair such that either but not both $ x\minus{}1$ and $ y\minus{}3$ is negative.
[/hide]


A circular track is 250m long. John and Paul run in opposite directions around it to practice for a competition. If John and Paul's speed are in the ratio of 3:2 and they meet again after 4 minutes, how fast is John in kph?

[hide="Solution"]

For every $ 3$m John runs, Paul runs $ 2$m. Therefore John runs $ \frac{3}{5}$ of the $ 250$ metres, that is, $ 150$ m $ \equal{}0.150$ km. He does this in $ 4$ minutes which is $ \frac{1}{15}$ of an hour. So he runs $ \frac{0.150}{\frac{1}{15}} \equal{} 2.25$ km per hour. Seems very slow.

[/hide]

Exactly how I solved it!

:starwars:
:bruce:
:wallbash_red:


induction
number theory
greatest common divisor
number theory proposed

What does that mean? Do you mean to say that 47^... divides 48^...?

For which positive integers n is 48n - 1 divisible by 47[n/2]?

Yes ,that 's right.
I'm very sorry ....
Find all positive integer n satisfy:
     (47^[n/2]) divides (48^n)-1.

Lemma: p>2 prime. Then pm| (hit Shift+\ for the "|" symbol. c'mon it's not that hard) (p+1)n-1 implies pm-1|n, for all m>0.

Proof. Induction. I'll prove that for all m, pm|(p+1)p^(m-1)-1, and pm+1 does not divide (p+1)p^(m-1)-1.

For m=1 it is obvious that p|p+1-1 and p^2 dows not divide p.

Suppose we proved it for m. then (p+1)p^m-1=[(p+1)p^(m-1)-1][(p+1)(p-1)*p^(m-1)+...+(p+1)p^(m-1)+1].
It is easy to finish from here because p|[(p+1)(p-1)*p^(m-1)+...+(p+1)p^(m-1)+1] and p^2 does not.

Now, suppose p^m|(p+1)^n-1.
From what was proved earlier we get p^m|(p+1)p^(m-1)-1.
(p+1)^k-1 is a Mersenne sequnce, so p^m|(p+1)gcd(n,p^(m-1))-1.
Again from what we proved earlier, since gcd(n,p^(m-1))=p^a for some a \leq m-1, we get a=m-1, so p^(m-1)|n.
-----------------------------------------------------------------------------------
Take p=47. For m=[n/2] and n=n, in the lemma.
If m=0, then n is 1 and the problem checks.
If m>0, from the lemma we get 47[n/2]-1|n. From this we get n=2, because if n>2, then 47^([n/2]-1)>n and the divisibility is impossible.

It's nice solu.
I think we can solve the more generized problems
   Find all m,n>0 are integer such that  47^n| 48^m -1
Or more replace 47 by a prime number p(odd but the case p=2 is not dificult to solve)and 48 by an integer  a.In this mail,i solve the first problem and the second may be the same

At first we have the following lemma: we denote

   Lemma:  47^(n+1) divides by 48^(47^n) - 1 but 47^(n+2) for n =0,1,2,...

This can be done by induction!(As i knew the lemma is called Dickson Lemma,the one wrote the fausmost book :"History of Number theory")

Now we fix n and we assume that d is the smallest positive integer number such that 47^n| 48^d -1.Since the fact that
              gcd(a^m-1,a^n-1 )=a^gcd(m,n)-1
or DIvisible Algorithe and the defination of d we have
                   d= gcd( d, 47^(n-1) )
so d=47^t where t<n .If t<n-1 then 47^(t+2) | 47^(n) | 48^d-1 = 48^(47^t) -1 which contract with the lemma!!
So t=n-1.And by the defination of d , d=ord(47)48 so 47^n | 48^m -1 iff
m is divisible by 47^(n-1).
This solve the problem
In the orginal problem of cause no such n exist!

Ps: Hi Thanh Binh,you are Vietnamese,ain't you?


geometry
inequalities
triangle inequality

Well, as many of you already know, the PSAT was today, and I'm pretty sure we can discuss it (since people have already started discussing it on many other forums). But if not, a mod/admin can lock this.

So... how did you do? Any questions you were unsure of? If so, which?
As for me, I think (or more like hope  :P ) I got an 80 in Math, missed maybe 5-10 questions in Critical Reading  :( , and missed maybe 2-5 questions in Writing.
I really thought that the Critical Reading was pretty hard.  :wink: At least, I know I messed up on it. How about you?

i just mess up anyways >.<
the math is always easy, i just hope i didn't read the question too fast. but i checked everything like 3 times and then started playing tetris on my calculator cuz it was really boring
for the other sections i finished 2-.5 minutes before the time is up

i think i did pretty bad, probably not gonna get the nation merit thing. The questions didn't click like when i took the practice test.

I think I did reasonably well.  I think I got an 80 on the math section, although I'm still hoping [color=red][censored just in case][/color] on that that train question... :blush: .

As far as writing, I want an 80 there too, but I guessed on 3 questions after eliminating 3 choices on each, so I don't know what my score will be there.

Verbal was much worse, I guesses on about 8 or 9 questions after eliminating 2 answers.  I also probably missed a few questions, so I'm just hoping for a 75, which is unrealistically high.

Target Score: 235
Actual Score: probably around 225
Last Year's Score: 220 (76 on math due to ONE careless error on question designated easy)

Please cease discussing the problems. I am sorry to say our high school will administrate this test next Monday. But I am a senior anyway.

Okay we won't talk about specifics
until monday (I checked and that is the only other date possible)
This is my first time taking them, and I'm a sophomore... so I felt kind of stupid
I'll say more after the second test is done with..

I'm hoping for a 80/68/77

Critical reading was pretty lame.

[quote="Tiks"]Who can explain me what is PSAT(I know what is SAT :blush: ) and is it something necessary for college application?  :maybe:[/quote]
Hope [url=http://www.collegeboard.com/student/testing/psat/about.html]this[/url] helps. I don't think you have to worry about it.

Our proctor kept stopping us with about 3 min left for CR...and kept making up for it in the math.  So annoying...that actually affected my score, because I didn't get time to check my answers for cr!

Well anyways, the vocabulary kills me.  I also kinda hate how in the CR, you have to find a balance between taking this literally and figuratively.  I never seem to be able to find that balance.  Anyways, I'm predicting

Math: 80
Writing: 80
CR: 65

Wow, there seems to be a inverse relationship between math scores and reading scores lol.    :wink:   I think I only scored about 700 on reading.

Looking at my post above (the first one), what would my actual scores look like?
Also, how do you think they'll curve the Critical Reading section? I'm thinking (maybe more like hoping  :oops: ) that they'll curve it pretty leniently, since it seems to me like it was slightly harder than usual.

I don't know about you guys, but we took it Wed. , and I thought the whole test itself (not just the math) was very easy.  I always love when they have technical/mechanical-type passages, as I have a better time analyzing those rather than literary-type passages.

I don't know if its allowed or not to talk about questions, but they give different forms when they give it on different days. The train one on wednesday was funny, I believe that was form W. From people who took it on different days, what forms were given? (hopefully not W lol)

Go Beyond,

Tests given on different dates are different versions.  If Wednesday's test was given on a Monday, the scores from the school will probably be disqualified.  As soon as Wednesday was over, College Confidential opened up their discussion boards, and a bunch of questions from form W were discussed.

I thought the CR section was a bit harder than I remember when I took it last year, and last year I had to struggle to stay awake during the test - I'm usually still sleeping at 8 o'clock on a Saturday morning  :lol: The rest of the test was... meh. Never liked the Writing section.

CR 80
M 80
W 77

237

does psat actually matter for anything besides national merit?

Not really. Although you get a decent amount of recognition from colleges.

80 math
80 CR :!:
65 writing  :oops:

225, i am happy.

I hate critical reading.  :(

Just watch me make a 60 on PSAT in CR and a 600 in CR on the SAT I.

.. I have a 550 in 8th grade SAT....

Anyone notice the decent curve on CR? I got 1 wrong (#21.. second guessed myself :/) but still got an 80.

[quote="Phelpedo"]A surprisingly high number of my friends got 80's on the Critical Reading  :o[/quote]

How do you guys do so well in Critical Reading?

Do you have any advice?

I am only scoring in the 650s-700s.

[quote="lfm"][quote="mathgeniuse^ln(x)"]I will start doing that.[/quote]

Try reading the editorial page of the New York Times.  It is written at a very high level. It is a great way to boost your reading skills. Don't be discouraged if you cannot understand everything at first.[/quote]

Does Wall Street work as well?

Are there any good books that cover SAT/PSAT type vocab. It is pretty early for me to start studying for the SATs but I just don't want to get a really bad score considering my vocabulary isn't too good.

I agree that the best way to improve vocab, CR, and W is to read more and often. And reading the "right" stuff probably is just as important. Reading some of the online publications or Da Vince Code, for example, may not be as helpful as reading Thomas Friedman's op-ed.

Improving vocab is also very helpful. There are tons of stuff on the market for vocab. I used this little one: "Verbal Power: rootword flashcards" (actually a little book) by Nanette Andrews, ISBN 0963071505.

I like it; others might not. So, take a look at it first at the store and see if the format suits you.

We'll need good vocab, CR and W skills for GRE too :P

Just got my score back.

Reading: 75
Math: 80
Writing: 70
Total: 225

Lower than the 231 I got last year, but 225 should be enough to make National Merit semifinalist anyways.

Math:800
CR:790
Writting:800

and... i don't believe i missed omitted on CR but somehow I omitted one and missed one.. I asked for my test back to see..

one missed won't give a 800 still, would it? Just hoping.. just in case

Interesting.
I misbubbled on the writing.
So a 233.
Not a 236. And certainly not 240, which my friend got with ease. I kind of despise him for that.  :roll:


When I typed the formula:
[code]
\begin{eqnarray}volume & = & 1 + (\frac {6}{27} + \frac {30}{729} + \cdots)\nonumber \\
& = & 1 + \frac {\frac {6}{27}}{1 - \frac {5}{27}}\nonumber \\
& = & \frac {14}{11}\nonumber\end{eqnarray}[/code]

In this forum, It display  correct.  It  display this image:\begin{eqnarray}volume & = & 1 + (\frac {6}{27} + \frac {30}{729} + \cdots)\nonumber \\
& = & 1 + \frac {\frac {6}{27}}{1 - \frac {5}{27}}\nonumber \\
& = & \frac {14}{11}\nonumber\end{eqnarray}

But,when  I typed  the same formula  In  my  computer,It display  other image:

Please tell me why?

Why different results?

I use  MiKTeX 2.7+ImageMagick 6.3.8 Q16.

thank  you  very much!

[quote="stevem"][quote="yanglei1979"]I use  MiKTeX 2.7+ImageMagick 6.3.8 Q16.[/quote]ImageMagick is used in LatexRender [b]not[/b] in a document so precisely what are you doing? There should not be a problem in a normal document.

If you are using LatexRender you will get the addtion of "to". That's because \begin{eqnarray} should not be in maths mode and latex gets confused. The workaround is to add \par before \begin{eqnarray} and all will be well. This forum used to show the same error but it won't do now as it fixes things behind the scenes.

Some other comments:
1. You should not be using eqnarray as it has been replaced by align - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url] 5 (but still add \par in Latexender);
2. volume is not v times o times l ... so you need to use \text{volume}.[/quote]

Thank  you very  much!

You are a great professor!

Can you tell me how  can  I   repair  my computer's  programe?

The same as this forum.

[quote="stevem"][quote="yanglei1979"]Can you tell me how  can  I   repair  my computer's  programe?

The same as this forum.[/quote]What program? You'll have to explain what you mean by 'repair' - what is wrong with it?[/quote]

Thank you very much!

I use PHP language Development website.

When I  add  \par  before  \begin,  This issue is resolved.

I very  very love you! :rotfl:

[quote="yanglei1979"]When I  add  \par  before  \begin,  This issue is resolved.[/quote]
The [b]\eqnarray[/b] command is deprecated for several reasons; instead, use [b]\align[/b]. With [b]\align[/b], the additional [b]\par[/b] is not necessary.