What is the sum of the ordinates and abscissas of all intersection points between $ f(x)\equal{}x^{3}\plus{}3x^{2}\plus{}2x\minus{}1$ and its inverse function?

[hide]Inverse=> y=x
So $ x\equal{}x^{3}\plus{}3x^{2}\plus{}2x\minus{}1$
$ x^{3}\plus{}3x^{2}\plus{}x\minus{}1\equal{}0$
So sum of x's is $ \frac{\minus{}3}{1}\equal{}\minus{}3$
and since its about y=x, the sum of ordinates is the same
so -3(2)=-6.[/hide]

Abscissa is a fancy term for the x-coordinate. Ordinate is the same for y.


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Prove that every finite group is a Galois group of some extgension of fields.

Embed the finite group in $ S_n$. Show that $ \mathbb{Q}(x_1,\cdots,x_n)$ is Galois over a subfield ($ \mathbb{Q}$ adjoin elementary symmetric polynomials.) So every subgroup can be realized as a Galois group through Galois correspondence.


Solve for $x$:

\[ \arctan (3x) +\arctan (5x) =\arctan (7x) +\arctan (2x) \]

[hide]Take the tangent of both sides. \begin{eqnarray*} \tan (\arctan (3x) + \arctan (5x)) &=& \tan (\arctan (7x)+\arctan (2x))\\ \frac{8x}{1-15x^2} &=& \frac{9x}{1-14x^2} \end{eqnarray*} And from here the solution is simple.[/hide]


linear algebra
matrix
vector
analytic geometry

$ f: R^{2n}\longrightarrow R^n$
and determined that:
$ f \begin{bmatrix}x_1\\x_2\\x_3\\.\\.\\x_{2n}\end{bmatrix}\equal{} \begin{bmatrix}x_1\plus{}x_{n\plus{}1}\\x_2\plus{}x_{n\plus{}2}\\x_3\plus{}x_{n\plus{}3}\\.\\.\\x_n\plus{}x_{2n}\end{bmatrix}$

find basis of ker $ f$

the $ n$ vectors in $ \mathbb{R}^{2n}$, with $ i$-th coordinate 1, $ i \plus{} n$-th coordinate -1, and the rest 0, $ i\equal{}1, ..., n$

Note that the matrix representation of this operator is the $ n\times 2n$ matrix with block structure $ \begin{bmatrix}I&I\end{bmatrix}.$ Note also that this is already in reduced row echelon form, from which it is straightforward to compute a basis for the null space - which pleurestique has done. The rank of this matrix is $ n$ and the dimension of the null space is $ n.$


Let $ a$, $ b$ and $ m$ be positive integers. Prove that there exists a positive integer $ n$ such that $ (a^n \minus{} 1)b$ is divisible by $ m$ if and only if $ \gcd (ab, m) \equal{} \gcd (b, m)$.

Let $ gcd(b,m)\equal{}k$ , where $ k$ is a positive integer,
the problem becomes:
There exists a $ n$ such that $ \frac{m}{k}|(a^n\minus{}1)\frac{b}{k}$ $ \Leftrightarrow$ there exists a $ n$ such that $ a^n\equal{}1(mod \frac{m}{k})$ ,
[b]Sufficiency[/b]
when $ gcd(ab,m)\equal{}gcd(b,m)$, $ gcd(a,\frac{m}{k})\equal{}1$
taking $ n\equal{}\phi(\frac{m}{k})$ and the sufficiency part is done.

[b]Necessity[/b]
when $ gcd(ab,m) \neq gcd(b,m)$, $ gcd(a,\frac{m}{k}) \neq 1$, so $ a^n\equal{}1(mod \frac{m}{k})$ is impossible for any $ n$.


calculus
integration
algebra
polynomial
real analysis
real analysis unsolved

Let $ P$ be a polynomial of degree $ n$ such that
$ \int_{0}^{1}x^{k}P(x)dx\equal{}0$
for $ k\equal{}1,2,\ldots ,n$.

Prove (or Disprove) that for a positive integer $ m$,
$ \int_{0}^{1}\left [ P(x) \right ]^{m}dx\equal{}(n\plus{}1)^{m}\left ( \int_{0}^{1}P(x)dx \right )^{m}$.

Put $ m \equal{} 1$, then the proposition is false unless any solution to the given condition also satisfies $ \int_{0}^{1} P(x) \, dx \equal{} 0$.

But it is clear that $ P(x) \equal{} x \minus{} \frac{2}{3}$ satisfy the given condition but $ \int_{0}^{1} P(x) \, dx \equal{} \minus{} \frac{1}{6}$, so the proposition is false.

But you can prove that it is true for $ m \equal{} 2$. Using [url=http://en.wikipedia.org/wiki/Legendre_polynomial#Shifted_Legendre_polynomials]the shifed Legendre polynomials[/url], we see that if $ P(x)$ satisfies the given condition, then

$ P(x) \equal{} C \sum_{k\equal{}0}^{n} (\minus{}1)^k (2k\plus{}1) \tilde{P}_k (x)$

where $ C \equal{} \int_{0}^{1} P(x) \, dx$. Note that in particular we have $ P(0) \equal{} C (n\plus{}1)^2$. So

$ \int_{0}^{1} P(x)^2 \, dx \equal{} P(0) \int_{0}^{1} P(x) \, dx \equal{} (n\plus{}1)^2 C^2$

as desired.

Solution for $ m\equal{}2$.

We let $ P(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots \plus{}a_{0}$ and assume that
$ \int_{0}^{1}\left [ P(x) \right ]^{2}dx\equal{}a_{0}\int_{0}^{1}P(x)dx$.

Also, it is easy to show that
$ \int_{0}^{1}x^{k}P(x)dx\equal{}\frac{a_{n}}{n\plus{}k\plus{}1}\plus{}\frac{a_{n\minus{}1}}{n\plus{}k}\plus{}\cdots \plus{}\frac{a_{0}}{k\plus{}1}$
for $ k\equal{}0,1,\ldots,n$.

The problem now asserts that
$ \frac{a_{n}}{n\plus{}k\plus{}1}\plus{}\frac{a_{n\minus{}1}}{n\plus{}k}\plus{}\cdots \plus{}\frac{a_{0}}{k\plus{}1}\equal{}0$
for $ k\equal{}0,1,\ldots,n$.

Further, there exist a polynomial $ F(k)$ of degree at most $ n$ such that
$ \frac{a_{n}}{n\plus{}k\plus{}1}\plus{}\frac{a_{n\minus{}1}}{n\plus{}k}\plus{}\cdots \plus{}\frac{a_{0}}{k\plus{}1}\equal{}\frac{F(k)}{(k\plus{}1)\cdots (n\plus{}k\plus{}1)}$. =====> $ (1)$

Since $ F(k)\equal{}0$ for $ k\equal{}0,1,\ldots,n$, there is a constant $ C$ such that
$ F(k)\equal{}C(k\minus{}1)(k\minus{}2)\cdots (k\minus{}n)$.

Putting $ k\equal{}0$ in $ (1)$, we have
$ \int_{0}^{1}P(x)dx\equal{}\frac{a_{n}}{n\plus{}1}\plus{}\frac{a_{n\minus{}1}}{n}\plus{}\cdots \plus{}\frac{a_{0}}{1}\equal{}(\minus{}1)^{n}\frac{C}{n\plus{}1}$.

Now, we multiply $ (1)$ by $ (k\plus{}1)$, it is easy to show that $ a_{0}\equal{}(\minus{}1)^{n}(n\plus{}1)C$ with $ k\equal{}\minus{}1$.

So, we now have
$ a_{0}\equal{}(n\plus{}1)^{2}\int_{0}^{1}P(x)dx$
which satisfies the desired equality.                                                                              $ \square$


superior algebra
superior algebra solved

Is this problem correct ? I believe not:

Let $G$ be a finite group operating on a finite set $S$.
For each $s \in S$ show that
\[ \sum_{t \in Gs} \frac{1}{\left. \# ( Gt \right)} = 1  \]

what is $G_t$. If it is the orbit of $t$, or the stabilizator of of $t$, it is true.

Yes $Gt$ is the orbit of $t$.
Can you show your proof ? :)

For any $s\in S$, $G$ acts transitively on $Gs$. This means that for every $t\in Gs$ we have $Gt=Gs$.
The problem is just smth like $\underbrace{\frac 1n+\ldots+\frac 1n}_{n\ times}=1$.

I don't see why $G$ should be finite. $Gt$ is anyway finite because it is a subset of $S$.
Or if $G$ is finite, then $S$ need not be finite.


Problem 1
Five masses are weighed in pairs with all the possible combinatios. If the 10 weighings were 72, 73, 76, 77, 79, 80, 81, 83, 84 and 87 find the weight of each mass.

Problem 2
In algebra the following "cancel out" is not allowed

$\frac{8\alpha}{\alpha+\beta}= \frac{8}{\beta}$

If $\alpha$ and $\beta$ are positive integers, $\alpha,\beta\neq0$ show that the previous equation is valid if and only if $\alpha=\beta=2$

Problem 3
Given an isosceles triangle $AB\Gamma$ $(AB=A\Gamma)$ and $M$ arbitrary point on the side $AB$. In the projection of $B\Gamma$ (towards $\Gamma$) we take $\Gamma N=BM$ and $\Gamma\Sigma=AB$. Show that the midpoints of $B\Gamma$, $NM$ and $A\Sigma$ are collinear.

Problem 4
A Mathematician told one of his students to think a 3-digit number $\alpha\beta\gamma$ where $\alpha, \beta, \gamma$ are natural numbers $\alpha,\beta,\gamma \in \{1,2,3,...9\}$. Then he asked his student to write $\alpha\gamma\beta$, $\beta\alpha\gamma$, $\beta\gamma\alpha$, $\gamma\alpha\beta$ and $\gamma\beta\alpha$ on his textbook, then to add then and finally to announce (to the other students in the class and to him) the sum $\Sigma$. When the student announced the sum $\Sigma$, the Mathematician immediatetly announced the initial number $\alpha\beta\gamma$. If the sum $\Sigma=3194$, find the number $\alpha\beta\gamma$.

[hide="1"]$\frac{8\alpha}{\alpha+\beta}= \frac{8}{\beta}\Leftrightarrow \alpha\beta = \alpha+\beta$.

If $(\alpha, \beta) = d$, then $\alpha = dx$ and $\beta = dy$ with $(x, y) = 1$.

So $dxy = x+y$. Hence $x \mid x+y$. From which $x = y = 1$.

Finally $d = 2 = \alpha = \beta$.[/hide]

[hide="2"]The problem condition is equivalent to $122\alpha+212\beta+221\gamma = 3194$.
From here we see that $\gamma = 2k$ with $k \in \left\{1, 2, 3, 4\right\}$.
Then the four cases are:
I. $k = 1$. Hence $61\alpha+2.53\beta = 1597-221 = 1376 \equiv 51 \pmod{53}\Rightarrow 8\alpha \equiv 51 \pmod{53}$. Impossible.
II. $k = 2$. Hence $8\alpha \equiv 42 \pmod{53}$. Impossible.
III. $k = 3$. Hence $8\alpha \equiv 33 \pmod{53}$. Impossible.
IV. $k = 4$. Hence $8\alpha \equiv 24 \pmod{53}$, which gives $\alpha = 3$, $\gamma = 8$ and $\beta= 5$.
So the number is $358$.[/hide]

I think there's something wrong in the text of problem 3.
Also you should post the problems in National Olympiads, and every problem in a different topics in Pre-Olympiad or Intermediate for example.  :)

[quote="Andreas"]
I think there's something wrong in the text of problem 3.
Also you should post the problems in National Olympiads, and every problem in a different topics in Pre-Olympiad or Intermediate for example.  :)[/quote]

I 've correct it.  It was a typo in latex.  I am sorry for posting all the four problems in the same post. I will post the next topics in Pre-Olympiad. Shall I post the previous problems (2005 TST Junior) again in Intermediate?

For #2, couldn't you just say [hide] \[\frac{8\alpha}{\alpha+\beta}=\frac{8}{\beta}\iff \alpha\beta=\alpha+\beta \iff (\alpha-1)(\beta-1)=1 \iff \alpha=\beta=2\] [/hide]?

For #1, do we know which pairs of masses sum to which weights?  If so, we can take the sum of our 10 weighings and divide by 4 to get the sum of the 5 masses.  Then we can subtract out the sum of four of the masses to get that fifth mass.
Otherwise...

[hide="1"]Say the masses are $x_{1}< x_{2}< \ldots < x_{5}$. We know that the two smallest sums are $x_{1}+x_{2}$ and $x_{1}+x_{3}$. We also know the largest two are $x_{5}+x_{4}$ and $x_{3}+x_{5}$. But there are only 5 variables, so these 4 equations combined with the sum of all the masses give us enough information to solve.
[/hide]


Find the sum to n terms of the following progression

1/(1x3x5)+1/(3x5x7)+1/(5x7x9)+...

[quote="hotheadhacker_mathslover"]Find the sum to $ n$ terms of the following progression

$ \frac{1}{1\times3\times5}\plus{}\frac{1}{3\times5\times7}\plus{}\frac{1}{5\times7\times9}\plus{}\cdots$[/quote]

You must evaluate the finite sum
$ \sum_{k \equal{} 1}^{n}\frac {1}{(2k \minus{} 1)(2k \plus{} 1)(2k \plus{} 3)}.$

Use the decomposition into partial fractions
$ \frac {1}{(2k \minus{} 1)(2k \plus{} 1)(2k \plus{} 3)} \equal{} \frac {\frac {1}{8}}{2k \minus{} 1} \minus{} \frac {\frac {1}{4}}{2k \plus{} 1} \plus{} \frac {\frac {1}{8}}{2k \plus{} 3}$
to find (by telescoping) that
$ \sum_{k \equal{} 1}^{n}\frac {1}{(2k \minus{} 1)(2k \plus{} 1)(2k \plus{} 3)} \equal{} \frac {1}{12} \minus{} \frac {\frac {1}{4}}{(2n \plus{} 1)(2n \plus{} 3)}.$

M@ri@n --can u explain ur solution further

[quote="binomial_4eva"]something is wrong with this answer[/quote]

Can you explain what is wrong? My answer is correct because
$ \equal{}\cdots \equal{}\frac{1}{8}\left (\frac{1}{2k\plus{}1}\minus{}\frac{1}{2k\plus{}3}\right )_{k\equal{}0}\plus{}\frac {1}{8}\sum_{k \equal{} 1}^{n \minus{} 1}\left (\frac {1}{2k \plus{} 1} \minus{} \frac {1}{2k \plus{} 3}\right ) \minus{}\\
\minus{} \frac {1}{8}\sum_{k \equal{} 1}^{n\minus{}1}\left (\frac {1}{2k \plus{} 1} \minus{} \frac {1}{2k \plus{} 3}\right ) \minus{}\frac{1}{8}\left (\frac{1}{2k\plus{}1}\minus{}\frac{1}{2k\plus{}3}\right )_{k\equal{}n}\equal{}\\
\equal{}\frac {1}{12} \minus{} \frac {\frac {1}{4}}{(2n \plus{} 1)(2n \plus{} 3)}.$

Check your answer by plugging in small values of n.

Also, if you believe something is wrong, be more specific than "That's wrong." The solution's writer can't be expected to defend his/her solution without knowing what your complaints are.

The fact that
$ \sum_{k=1}^{n}\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{1}{12}-\frac{\frac{1}{4}}{(2n+1)(2n+3)},\qquad\forall n\in\mathbb{N},$
can be also proved easy by induction. You can verify it.

hey guys any other methods to solve this question without having to use summation?Im nt quite familiar with it yet.

Actually the expression looks more easier to evaluate in the form of summation. Also, he just expressed the denominators into factors in the general case for every input of $ n \in \mathbb{N}$. Notice that as we increase $ n$ infinitely, it converges to the value $ \dfrac{1}{12}$


The Sprint Round competition consists of 30 problems with a time limit of 40 minutes. You completed 20 Sprint Round problems in 25 minutes. On average, how many times longer will you be able to spend on each remaining problem than you did on each of the first 20? Express your answer as a common fraction.

The first questions take $ \frac{25}{20}$ or $ \frac{5}{4}$ minutes per problem. The last ten take $ \frac{15}{10}$ or $ \frac{3}{2}$ minutes per problem. This basically gives you this equation:

$ \frac{5}{4} \cdot x \equal{} \frac{3}{2}$

Solving for $ x$ gives you $ x \equal{} \boxed{\frac{6}{5}}$


From note to self:

APost topic "Logical Jounry of Zoombinis." is it useful? or is learning mathematics directly more useful? (since Zoombinis is really only designed to be entertaining)

Well, it's a use of spare time with good PR, which always gives it merit.

It's okay I've owened it for a while. When you get to hardest diffuculty it can be really challenging.


function
trigonometry
complex analysis
complex analysis unsolved

Hello everyone!

It's becoming an habit to reach for all your help :D!

I have that:
\[ B_{n}\equal{}\minus{}\frac{1}{(\beta\minus{}n)!n}\left[\frac{d^{\beta\minus{}n}}{dx^{\beta\minus{}n}}\left(f'(x)\theta(x)(x\minus{}a)^{\beta\plus{}1\minus{}n}\right)\right]_{x\equal{}a}\]
where $ \beta$ is the multiplicity of the pole $ a$ of the function $ f(x)$.

There is also the information that $ \Theta(x) \equal{}\theta(x)(x\minus{}a) \equal{} t$, where $ t$ is a number given.

In a particular example,
\[ f(x)\equal{}\frac{1}{x\minus{}a}\]
And it is considered the Kepler's equation
\[ x\equal{}a\plus{}e\sin x.\]
This can be manipulates to the form:
\[ \frac{1}{\sin x}(x\minus{}a)\equal{}e\]
so we can make $ \Theta(x) \equal{}\frac{1}{\sin x}(x\minus{}a)$ and $ \theta(x) \equal{}\frac{1}{\sin x}$.

It is stated that
\[ B_{1}\equal{}\frac{1}{\sin a},\quad B_{2}\equal{}B_{3}\equal{}\ldots\equal{}0\]
How can I get this last result? I've tried inumerous diferent ways and can't reach it :(.

Thank you once again! :)

Just to let you know, what I get is:

\[ \frac{1}{\sin x (x\minus{}a)}\]  :maybe:


inequalities
algebra
system of equations
calculus
calculus computations

How would you solve this problem by using legrange multipliers.
Find the minimum value of $xy+yz+xz$ given $x^{2}+y^{2}+z^{2}=1$
i know that the answer is $\frac{-1}{2}$ by the trivial inequality $(x+y+z)\geq{0}$

[hide="what i have gotten"]
$f(x,y,z)=xy+yz+xz$ and $g(x,y,z): =x^{2}+y^{2}+z^{2}=1$

$\triangledown f(x,y,z)= \left<y+z, x+z, x+y\right>$
$\triangledown g(x,y,z)= \left<2x, 2y, 2z\right>$

solving $\triangledown f(x,y,z)=\lambda \triangledown g(x,y,z)$

what i am having trouble with is solving this.

$2x\lambda=y+z$
$2y\lambda=x+z$
$2z\lambda=x+y$
[/hide]

It's just a system of equations, and you should solve it in any way that's possible.  The fact that you already know what the answer is should make it easy.  One fairly straight-forward approach is just to add the three equations.


Euler
geometry
3D geometry
pyramid
sphere
national olympiad

Day 1
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=246409]Problem 1.[/url] Let $a_{1},..., a_{n},...$ be positive real numbers such that $\frac{a_{n}+a_{n+2}}{2}< a_{n+1}$ for every $n \geq 1$. Show that $a_{n}< a_{n+1}$ for every $n \geq 1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=246410]Problem 2.[/url] An acute triangle $ABC$ is given. Find the locus of points $M$ in the interior of $ABC$ such that $AB-FG = \frac{MF \cdot AG+MG \cdot BF}{CM}$ where $F$ and $G$ are the feet of the perpendiculars from $M$ to $BC$ and $AC$ respectively.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=246413]Problem 3.[/url] Find all positive intergers $n$ such that $2^{\phi(n)}-1$ divides $n^{n}$, where $\phi$ is the Euler $\phi$-function.

Day 2
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=247004]Problem 4.[/url] Let $n \geq 2$ be a natural number. A pyramid $P$ has base $A_{1}A_{2}...A_{2n}$ and apex $O$. The polygon $A_{1}A_{2}...A_{2n}$  is regular and the point $C$ is its centre. The line $OC$ is perpendicular to the plane of the base of $P$. A sphere passes through $O$ and meets each of the line segments $OA_{i}$ internally. For each $i = 1, 2,..., 2n$, let $X_{i}$ be the point (other than $O$) where the sphere meets $OA_{i}$. Prove that $OX_{1}+OX_{3}+...+OX_{2n-1}= OX_{2}+OX_{4}+...+OX_{2n}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=247003]Problem 5.[/url] Find the number of subsets $B$ of $\{1,2,3,...,2005\}$ such that the sum of the elements of $B$ is congruent to $2006 \mod 2048$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=247002]Problem 6.[/url] Let $n \geq 3$ be an integer. Consider positive real numbers $a_{1},..., a_{n}$ such that their product is 1. Show that: $\sum_{i=1}^{n}\frac{a_{i}+3}{(a_{i}+1)^{2}}\geq 3$

PS: Can this be a sticky and/or be listed in the front?

Disastrous, I solved 3 on the first day and none on the next.

[quote="kueh"]PS: Can this be a sticky and/or be listed in the front?[/quote]Yes it can :D When will be the results available? :)

The team for Mexico is

Saul Glasman (myself)
Nathan Kettle
Andre Kueh (kueh)
Matthew Lee
Martin Orr
Jack Shotton

Yet two more MathLinkers present at the IMO :D Congrats guys! :)

It seems we must provide all the ML-shirts at the beginning of the IMO opening ceremony so that the MLer have enough time to get to know each other.  :)

Well, I have now joined, to make it three (I am Jack)

Congratulatinos to all UK students on the team

congratulations to you! thanks valentin (=

All except for the solid geometry (which was plucked from the air) was from the Balkan Olympiad Shortlists.

You all from England,  me too :D  :D  :D  :D   for a moment I thought.... nevermind. :roll:


quadratics
calculus
integration
algebra
number theory proposed
number theory
Ireland

Let $ x,y$ be positive integers with $ y>3$ and $ x^2\plus{}y^4\equal{}2((x\minus{}6)^2\plus{}(y\plus{}1)^2).$ Prove that: $ x^2\plus{}y^4\equal{}1994.$

A ugly exercise..
It is a quadratic equation in $ x$ with discriminant $ \Delta: \equal{}y^4\minus{}2(y\plus{}1)^2\plus{}72$, but for all $ y>5$ we have $ (y^2\minus{}2)^2<\Delta<y^4$; now $ \Delta\equal{}(y^2\minus{}1)^2$ has no integral solution. So $ y \in \{4,5\}$, and it is only computational to verify that the unique solution is $ (x,y)\equal{}(37,5)$.


quadratics
algebra
geometry unsolved
geometry

If in a quadrilateral a circle of radius $r$ can be inscribed and the quadrilateral itself can be inscribed in a circle of radius $R$ then Find the distance between the centres of the two circles(only in terms of $r$ and $R$).

Denoting d = IO, then

$\frac{1}{r^2} = \frac{1}{(R - d)^2} + \frac{1}{(R + d)^2}$

(see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34217]Concurrency problem related to Poncelet's Theorem[/url]). The root of this quadratic equation for $d^2$ such that $d^2 < R^2$ is

$d^2 = r^2 + R^2 - \sqrt{r^4 + 4r^2R^2}$


Pretty personal question...I would say 16+...depending on how good of a friend you mean

I voted 16 or more. I never hang out with my mathematical friends.... and in my One Word poll I voted social.... I just dont like being around my mathematical friends all that much because all they talk about is Rubiks cube times and what place they got at what tournament (and I beat them still every time XD). They even bring their DS's to math tournaments! I do have some friends that I hang out with that are on Math Team but they really suck! I only have one friend that I hang out with that is really good at math team who just about saved me from killing myself in math seminar this year... None of my P.E. friends are even in advanced math (while most of my mathematical friends are in double-advanced)!!! At parties I never see my mathematical friends (unless its one of their parties XD!!!!!) ...  and only two of my girlfriends ever have been on any math team... and only once have I seen a math friend at the movies...  so i guess I'm really anti-math-social.

i hav only one real friend. there's these 2 other ppls that i hang out with, but not that often. i used to think i had another friend, but now it turns out that she hates me.

One across the street. That's it.

The rest... uuhhh.. Some are annoying [size=0]BlueMouse[/size], and some are nice.

Um...not really friends, but about 3-4. Most of my friends are mathy i guess. We really dont talk about math though.

16+. The whole fifth grade are my friends. Not a lot of them are really super math-y though.

16+ for me. Why are the poll options' numbers so low?  :P

I guess some people underestimate math people to have other friends who are math-y.

16+ friends?! I didn't think that anyone could have 20 friends total. Either I'm extremely antisocial, or everyone's definition of "friend" is too loose. I'd like to think it's the latter. It seems we here are sharply divided among two groups - those with a tight definition of "friend" and those with a loose definition.

"Friend" is pretty hard to define, I guess.

Are you jealous?  :P Just-kidding. I guess, if you can get along with anyone and share a few laughs then they are considered your friend.

i have like 1.00000000000000123123412344125696498345948523432424 total friends and 1 is mathematical and the other.......well......erm yeah. im not a very social kid.

16+ I think everyone that signed my yearbook is a pretty good friend (random people don't sign mine). I'm not sure what the total number is, but probably 16+.

Oh, really? I sign anyone's yearbook if they ask. (I usually write "You will most definitely forget me" in Chinese  :lol: ) I've signed yearbooks that I don't know the name of the owner.

[quote="Carbon57"]Okay...well I like sign a lot of people's yearbooks, and my yearbook's signed by a lot of other people...most of which I don't consider friends...I guess friends would be considered people you generally hang out with and talk to a lot [b]outside[/b] of class.[/quote]

I like to keep mine within my friends group... random people don't sign mine.

I live everyone in my grade, or at least I've known and come in contact with (which is everyone) gets to sign. They made up an important year  :wink:

if ur talking about people I kinda know, about 50, this includes my math friends.
Like dynamo, I rarely play with my math friends, maybe a party every couple of months or so. If you are talking about friends I regularly play with (at least once every 2 or so weeks), then about 20 or so, only 1 or 2 of them are from math, and the rest are relatively smart, but not in double advanced math like I am. One good thing is it turns out that most of the ct mc team is smart and athletic/fun to play around. So although we live far away, we've made plans to meet once a month.
The majority of the people in my math class don't really seem weird when you first meet them, but most are antisocial, and either have a trouble having a good conversation, or completely embarrass themselves when they do.
My girlfriend is in advanced math, but is not a genius.

I'm going with the middle ground here and I'm gonna say around 8, using a rather tight definition of friend, meaning people I hang out with (or talk to, if I don't go to the same school) at least three or four times a week. Of course, if you loosen it to "people I'm on good terms with", then it increases to 30+.

And I don't really have many "math friends". Maybe one or two, but I don't really think of them as "mathy", they just happen to be good at math.

[quote="moogra"]16+ I think everyone that signed my yearbook is a pretty good friend (random people don't sign mine).[/quote]

You [i]don't[/i] want to know what happened to my 8th grade yearbook.

I usually define a "friend" as someone I talk to or hang out with at some time. Whether this be in real life, IM, or Xbox LIVE, it doesn't really matter.

what happened to your 8th grade yearbook?

I voted for 16+ .  I have only two friends who are good at math.  That's because  here are no many kids  in Georgia  who like math  and it's too difficult to find people who are interested in math :(

i put 16+ all my friend think i'm too mathy in fact i only have about 1 math friend... :D

about 90% of my friends are non-mathy

but I got that 10% :D

Most of my friends are not really into math. Classmates, mock trial teammates, friends from crew, kids down the street...

I find that even my friendships with people who are into/are good at math really don't revolve around math at all; I would be hesitant to call even these friends my "math friends".


Let $ a,b,c\geq\ 0$, a+b+c=k(k>0)
Find max-min of $ a^2\plus{}b^3\plus{}c^5$

Assuming you do mean "max,min", it was trivial to find using Lagrangian multipliers that the minimum occurs where $ b\equal{}\sqrt{\frac{2}{3}a}$ and $ c\equal{}\sqrt[4]{\frac{2}{5}a}$. I'm not sure where to go from there. The best I can think of is to use simple variable substitution $ u\equal{}\sqrt[4]{a}$ to give us a quartic equation which can be solved by the formula or other methods.

It's also obvious that, beyond a certain point, the maximum is where $ c\equal{}k$, though I'm not sure where. (And I'm really supposed to be working on college apps instead of doing math, so I shouldn't try to find out.)

[quote="TRAN THAI HUNG"]Let $ a,b,c\geq\ 0$, a+b+c=k(k>0)
Find max-min of $ a^2 + b^3 + c^5$[/quote]
The MIN is not difficult, just use A-G ( but need to be patient)  :rotfl:
This is the way to find out MAX
If k>=1:
$ a^2 \le ka \le k^3 a \\
b^3 \le kb^2 \le k^2 b \le k^3 b \\
c^5 \le kc^4 \le k^2 c^3 \le k^3 c^2$
Then
$ {\rm{Then A = }}a^2 + b^3 + c^5 \le k^3 (a + b + c^2 ) = k^3 (k - c + c^2 ) = k^3 [ {c^2 - c - k(k - 1)} ] + k^5$
$  \Rightarrow A = k^3 (c - k)(c + k - 1) + k^5 \\
k > 0,c - k \le 0{\rm{ , }}c + k - 1 \ge 0{\rm{ }}$
$ \Rightarrow A \le k^5$
Equality when a=b=0, c=k
If 0<k<1
$ x^2 \le x,y^3 \le y,z^5 \le z \Rightarrow A \le x + y + z = k$
Equality when (a,b,c)=(0,0,k),(0,k,0),(k,0,0) :D

I found that can make it stronger :
With the above condition, find out max
$ ma^x \plus{} nb^y \plus{} pc^z$ ( $ x \le y \le z$)
I guess that it has . I am trying to solve it. Did I forget anything ???
 :|
But afterall, i 'm sure that when m=n=p=1, max is always $ k^z$ if k>=1 and k if 0<k<1
 :rotfl:

[quote="Darmani"][quote="TRAN THAI HUNG"][quote="Darmani"]I think you mean "max,min." Since $ k$ is unbounded above, there obviously is no max-min.[/quote]
k is a constant.[/quote]

The minimum is a function of k. If k is held constant, then finding the maximum of the minimum function is ridiculously trivial, which makes finding the max-min a bit of a nonsensical statement.[/quote]
I can't understand what you said. With each k, there is a different problem. Check it with k=7 then, the problem will be:
a+b+c=5, a,b,c >= 0 . Find out max of B= $ a^2\plus{}b^3\plus{}c^5$.
Of course, the max will be 5^5. You can find the solution in my post with k>1  :roll:  .
What do you want????

[quote="Darmani"]The way the question is worded, it sounds like he's not asking for either the maximum or the minimum, but the maximum minimum.

The minimum of $ a^2 \plus{} b^3 \plus{} c^5$ is a continuous function of $ k$. Any continuous function can be maximized. I don't see what's so hard to understand about the phrase "maximum minimum."

The concept of maximizing the minimum is extremely important in game theory. See http://en.wikipedia.org/wiki/Minimax[/quote]
Hah Hah Hah
He is my close friend.
That means he ask to find out the Minimum and the Maximum of $ a^2 \plus{} b^3 \plus{} c^5$ . Not the maximum of minimum.
Why did I know that?? Just because I had asked him to find out max of $ a^m \plus{} b^n \plus{} c^p$  :D
Why didn't you come to my topic " Find out max" to try your best.  :oops:


I'm currently trying to understand the mechanics which I was supposed to learn back in 9th grade, so ve indulgent, the concept of energy is very fuzzy in my head.

So, here it goes: An object is thrown under angle $\alpha$ with the horizontal, giving it speed of $v_0$. (on $Ox$ the speed in considered constant, while on $Oy$ variable because of $g$)

First of all, is it true that in this case we have $KE \ + \ PE \ = \ \textrm{constant}$?

I was trying to obtain the maximum height from that relation. this can also be done on other ways, which yield $h_{max} = \frac{(v_0 \sin \alpha)^2}{2g}$.

Supposing the above holds, we have $KE$ (as the object is being thrown) = $PE$ ($mgh$, where $h$ is the maximum height), so \[ KE \ = \ mgh = mg \frac{v_0^2 \sin^2 \alpha}{2g} = m \frac{v_0^2 \sin^2 \alpha}{2} .  \]
This seems different from $KE \ = \ \frac{mv^2}{2}$.

Is this good or bad?

I'm really hoping I'm not speaking non-sense, although there is a big probability that the opposite happens.

I think you are neglecting that even when the object attains its highest point, it still has kinetic energy.

Thanks :blush:  :rotfl:

And another question:

\[ E_k = \int \overrightarrow{F} \cdot d \overrightarrow{s} = \int \overrightarrow{v} \cdot d \overrightarrow{p} = \frac12 mv^2  \]

How do you actually obtain the last value from the ones before ?

Well, saying $E_k=\int\overrightarrow{v}\cdot\overrightarrow{\text{d}p}$ is not exactly correct - I'll just hope that you understand what it really means. Anyway, you only need to realize that $m\overrightarrow{v}=\overrightarrow{p}$:

$\int\overrightarrow{v}\cdot\overrightarrow{\text{d}p}=\frac{1}{m}\int\overrightarrow{p}\cdot\overrightarrow{\text{d}p}=\frac{1}{2m}p^2=\frac{1}{2}mv^2$


Hello. I was reviewing old HW assignments, and I seemed to have forgotten how to solve these three problems. They are on the link below, but for convenience I will also copy them here.

http://people.uleth.ca/~k.vos/courses/Physics2000/ass13.htm

••11. A rectangular coil of N turns and of length a and width b is rotated at frequency f in a uniform magnetic field of magnitude B, as indicated in Fig. 30-42. The coil is connected to co-rotating cylinders, against which metal brushes slide to make contact. The emf induced in the coil is given (as a function of time t) by

ε = 2pfNabB sin (2pft) = ε0 sin (2pft).

This is the principle of the commercial alternating-current generator. What value of Nab gives an emf with ε0 = 171 V when the loop is rotated at 48.8 rev/s in a uniform magnetic field of 0.712 T?

••15   In Fig. 30-44, a stiff wire bent into a semicircle of radius a = 4.8 cm is rotated at constant angular speed f =57 rev/s in a uniform 19 mT magnetic field. What are the (a) frequency and (b) the amplitude of the emf induced in volts in the loop?

••32. In Fig. 30-54, two straight conducting rails form a right angle. A conducting bar in contact with the rails starts at the vertex at time t = 0 and moves with a constant velocity of 7.11 m/s along them. A magnetic field with B = 0.456 T is directed out of the page. Calculate the magnitude of (a) the flux through the triangle formed by the rails and bar at t = 4.37 s and (b) the emf around the triangle at that time. If the emf is ε = atn, where a and n are constants, what are the values of (c) a and (d) n?

[quote] A rectangular coil of N turns and of length a and width b is rotated at frequency f in a uniform magnetic field of magnitude B, as indicated in Fig. 30-42. The coil is connected to co-rotating cylinders, against which metal brushes slide to make contact. The emf induced in the coil is given (as a function of time t) by

ε = 2$\pi$fNabB sin (2pft) = ε0 sin (2pft).

This is the principle of the commercial alternating-current generator. What value of Nab gives an emf with ε0 = 171 V when the loop is rotated at 48.8 rev/s in a uniform magnetic field of 0.712 T? [/quote]

[color=blue]I don't think this question is difficult.
Obviously, $\varepsilon_0=2\pi fNabB$
You've got $f$ $B$ and $\varepsilon_0$, so you can easily find out $Nab$[/color]

[quote]15 In Fig. 30-44, a stiff wire bent into a semicircle of radius a = 4.8 cm is rotated at constant angular speed f =57 rev/s in a uniform 19 mT magnetic field. What are the (a) frequency and (b) the amplitude of the emf induced in volts in the loop? [/quote]
[color=blue]$d\phi=BdS\cos(2\pi ft)=B\frac{\pi R^2}{4}\cos(2\pi ft)$
$\large\epsilon_0=-\frac{d\phi}{dt}=\frac{\pi^2R^2}{2}Bf\cos(2\pi ft)$[/color]

[quote]32. In Fig. 30-54, two straight conducting rails form a right angle. A conducting bar in contact with the rails starts at the vertex at time t = 0 and moves with a constant velocity of 7.11 m/s along them. A magnetic field with B = 0.456 T is directed out of the page. Calculate the magnitude of (a) the flux through the triangle formed by the rails and bar at t = 4.37 s and (b) the emf around the triangle at that time. If the emf is ε = atn, where a and n are constants, what are the values of (c) a and (d) n? [/quote]

[color=blue]Do the same as the solution above[/color]

Thank you. For some reason it seemed more complicated.


$ a,b,c > 0,a \plus{} b \plus{} c \equal{} 2$ . Prove that :
$ \sum \frac {1}{a^2\plus{}b^2} \ge \frac {5}{2}$ .

[quote="tin_124"]$ a,b,c > 0,a \plus{} b \plus{} c \equal{} 2$ . Prove that :
$ \sum \frac {1}{(a \plus{} b)^2} \ge \frac {5}{2}$ .[/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{2}{3}.$ :wink:

[quote="tin_124"]$ a,b,c > 0,a \plus{} b \plus{} c \equal{} 2$ . Prove that :
$ \sum \frac {1}{a^2 \plus{} b^2} \ge \frac {5}{2}$ .[/quote]
WLOG assume $ a\equal{}max(a,b,c)$
we have $ a^2\plus{}b^2\le (a\plus{}\frac{c}{2})^2\plus{}(b\plus{}\frac{c}{2})^2$
$ b^2\plus{}c^2\le (b\plus{}\frac{c}{2})^2$
$ c^2\plus{}a^2\le (a\plus{}\frac{c}{2})^2$
$ \Rightarrow \sum \frac {1}{a^2 \plus{} b^2} \ge \frac{1}{(a\plus{}\frac{c}{2})^2\plus{}(b\plus{}\frac{c}{2})^2}\plus{}\frac{1}{(b\plus{}\frac{c}{2})^2}\plus{}\frac{1}{(a\plus{}\frac{c}{2})^2}$
setting $ x\equal{}a\plus{}\frac{c}{2},y\equal{}b\plus{}\frac{c}{2} \Rightarrow x\plus{}y\equal{}2$
Thus, we need prove that
$ \frac{1}{x^2\plus{}y^2}\plus{}\frac{1}{x^2}\plus{}\frac{1}{y^2}\ge \frac {5}{2}$
$ \Leftrightarrow (\minus{}5x^4\plus{}20x^3\minus{}20x^2\plus{}16)(x\minus{}1)^2\ge 0$

[quote="bunhiacovski"]
Thus, we need prove that
$ \frac {1}{x^2 \plus{} y^2} \plus{} \frac {1}{x^2} \plus{} \frac {1}{y^2}\ge \frac {5}{2}$
$ \Leftrightarrow ( \minus{} 5x^4 \plus{} 20x^3 \minus{} 20x^2 \plus{} 16)(x \minus{} 1)^2\ge 0$[/quote]

It isn't the best way. Before, please using Am-Gm to solve my old problem:
$ \frac{1}{a^4\plus{}b^4} \plus{} \frac{1}{b^4\plus{}c^4} \plus{} \frac{1}{c^4\plus{}a^4} \geq\ \frac{40}{(a\plus{}b\plus{}c)^4}$
With $ a, b, c \geq\ 0.$
 :)

[quote="bunhiacovski"][quote="tin_124"]$ a,b,c > 0,a \plus{} b \plus{} c \equal{} 2$ . Prove that :
$ \sum \frac {1}{a^2 \plus{} b^2} \ge \frac {5}{2}$ .[/quote]
WLOG assume $ a \equal{} max(a,b,c)$
we have ...
$ b^2 \plus{} c^2\le (b \plus{} \frac {c}{2})^2$

[/quote]
I think $ a\geq b\geq c$ or $ c\equal{}\min\{a,b,c\}$ are better.  :wink:


We have 111 coins. They are to be placed in unit cells of the table $ n\times n$ so that the number of coins in any two neighboring (i.e. sharing a common side) cells differs exactly by 1 (a cell may contain several coins and may be empty, too). For which maximal $ n$ is this possible?

Lemma $ 1$) $ n$ cannot be even.
Assume that $ n$ is even, color the table like a chessboard in black and white, notice that all the white (black) cells have the number of coins with the same parity and there are an even number of them, then the total number of coins is even. Contradiction.

Lemma $ 2$) In any row size $ 1 \times 2k$ there must be at least $ k$ coins.
For each pair of neighboring cells there must be at least one coin.

if $ n\ge 15$ then there are $ 15$ rows of $ 1 \times 14$ with at least $ 15 \times 7 \equal{} 105$ coins and an aditional column of $ 14 \times 1$  with at least $ 7$ coins.
for a minimum of $ 112$ coins. Contradiction.

Therefore $ n \le 13$

Example with $ n \equal{} 13$
Color the table like a chessboard in black and white with black on the vertices,
for a coloring with $ 85$ black cells with one coin on each one.
And then place in $ 13$ white border cells $ 2$ on each one.


1)Show that:

sum (cos^2(A-B))>=24cosA*cosB*cosC.

(in a triangle)

bye bye

Solution:

we know that (cosA+cosB+cosC)^2>=8cosAcosBcosC so we replace cos A with -cos(B+C) and we get that cos^2(A-B)>=8cosAcosBcosC and the other two simmilar inequalities!

cheers! :D :D


all i notice is that the lines go in palindromic (is that a word?) order. I say post hint.

A flat out guess: is it a song?

It is not a song.

Hint: [color=white]Convert to binary.[/color]

You convert to binary, and let 1 be white and 0 be red. Color.

OH! I am stupid...... and you even said it earlier!

Binary? I somehow don't get it. Isn't that 1010101110 etc?


I have a question with an answer that is probably pretty simple:

The magnetic force acting on a charged particle can never do work because at every instant the force is perpendicular to the velocity.  The torque exerted by a magnetic field can do work on a current loop when the loop rotates.  How can these seemingly contradictory statements be reconciled?

Thanks for any help/info!

Answer the following questions?

1) What is the definition of work made by a force?

2) What is a torque?

1) Work $ W$ done by a magnetic force $ \vec{F}_B$ in the change of position $ s$ from $ s_i$ to $ s_f$ is given by:

$ W \equal{} \int_{s_i}^{s_f} \vec{F}_B \cdot d\vec{s}$

The magnetic force $ \vec{F}_B$ due to a particle of charge $ q$ moving at velocity $ \vec{v}$ in a magnetic field given by $ \vec{B}$ is described by:

$ \vec{F}_B \equal{} q(\vec{v} \times \vec{B})$

So, work done by a magnetic force is $ W \equal{} 0$.

2) Torque $ \vec{\tau}$ in this case is given by:

$ \vec{\tau} \equal{} \vec{\mu} \times \vec{B}$

where $ \vec{\mu}$ is the magnetic dipole moment of the loop.

I think the magnetic potential energy $ U$ associated with this torque is then given by:

$ U \equal{} \minus{}\vec{\mu}\cdot\vec{B}$

So, work done by the magnetic torque is $ W \equal{} U_f \minus{} U_i$.

However, I'm not quite sure where to go from here...

[quote="Carcul"][quote="Zakary"]I think the magnetic potential energy U[/quote]

That concept does not exist, since we can only define a potential energy function for a conservative field, and the magnetic field is not conservative.[/quote]
that's right Carcul but what he means is the potential energy of the body by the virtue of it's angle made with the magnetic field and that's due to torque not due to the field


function
trigonometry
group theory
abstract algebra
algebra proposed
algebra

[quote="DCTPKTCSPHN"]Find all functions $ f: \mathbb Q\rightarrow \mathbb Q$ such that for any $ x\in\mathbb Q$ and $ y\in\mathbb Q$, we have $ f\left(x \plus{} y \plus{} f\left(x\right)\right) \equal{} x \plus{} f\left(x\right) \plus{} f\left(y\right)$.[/quote]
I found this problem rather complex. There are infinitely many solutions. Some of them :
$ f(x) \equal{} x \plus{} p$, $ p\in\mathbb{Q}$
$ f(x) \equal{} \minus{} x$
$ f(x) \equal{} 2[x] \minus{} x \plus{} n$, $ n\in\mathbb{Z}$, where $ [x]$ is the integer part of $ x$
$ f(x) \equal{} [5\sin(\pi(x \minus{} [x]))] \plus{} 2[x] \minus{} x$

... and a lot of other functions.

Here is a general solution :

1) The solution :

Let $ \mathbb{A}$ any additive subgroup of $ \mathbb{Q}$
Let $ \sim$ the equivalence relation defined by $ x\sim y \iff x \minus{} y\in\mathbb{A}$ and $ r(x)$ any choice function which associates at each rational number a representent of its equivalence class (unique per class).
Let $ m(x)$ any function from $ \mathbb{Q}$ $ \longrightarrow$ $ \mathbb{A}$

Then $ f(x) \equal{} m(r(x)) \plus{} x \minus{} 2r(x)$

2) Demonstration that any function in the form given in (1) is a solution :
$ \forall x\in\mathbb{Q}$ : $ x \plus{} f(x) \equal{} m(r(x)) \plus{} 2(x \minus{} r(x))\in\mathbb{A}$
So $ y \plus{} x \plus{} f(x)\sim y$ and $ r(y \plus{} x \plus{} f(x)) \equal{} r(y)$
So $ f(x \plus{} y \plus{} f(x))$ $ \equal{} m(r(x \plus{} y \plus{} f(x)) \plus{} x \plus{} y \plus{} f(x) \minus{} 2r(x \plus{} y \plus{} f(x))$ $ \equal{} m(r(y)) \plus{} x \plus{} y \plus{} f(x) \minus{} 2r(y)$ $ \equal{} x \plus{} f(x) \plus{} f(y)$
QED

3) Demonstration that any solution may be written in the form given in (1) (so we have "general" solution) :
Let $ f(x)$ such that $ f(x \plus{} y \plus{} f(x)) \equal{} x \plus{} f(x) \plus{} f(y)$
Let $ \mathbb{A} \equal{} \{x\in\mathbb{Q} : f(x \plus{} y) \equal{} f(y) \plus{} x$ $ \forall y\in\mathbb{Q}\}$
3.1) $ \mathbb{A}$ is an additive subgroup of $ \mathbb{Q}$
$ 0\in\mathbb{A}$
If $ x\in\mathbb{A}$, we have $ f(x \plus{} y) \equal{} f(y) \plus{} x$ $ \forall y\in\mathbb{Q}$. Let then $ y \equal{} y_1 \minus{} x$, we have $ f(x \plus{} y_1 \minus{} x) \equal{} f(y_1 \minus{} x) \plus{} x$ and so $ f(y_1 \minus{} x) \equal{} f(y_1) \minus{} x$ $ \forall y_1\in\mathbb{Q}$ and so $ \minus{} x\in\mathbb{A}$
Let then $ x\in\mathbb{A}$ and $ x_1\in\mathbb{A}$ : $ f(x \plus{} x_1 \plus{} y) \equal{} f(x \plus{} (x_1 \plus{} y)) \equal{} f(x_1 \plus{} y) \plus{} x$ since $ x\in\mathbb{A}$. But $ f(x_1 \plus{} y) \equal{} f(y) \plus{} x_1$ since $ x_1\in\mathbb{A}$, and so $ f(x \plus{} x_1 \plus{} y) \equal{} f(y) \plus{} x \plus{} x_1$ and so $ x \plus{} x_1\in\mathbb{A}$
Q.E.D
3.2) $ \forall x\in\mathbb{Q}$ : $ f(x) \plus{} x\in\mathbb{A}$
This is an immediate result of $ f(y \plus{} (x \plus{} f(x))) \equal{} f(y) \plus{} (x \plus{} f(x))$
3.3) Final result :
Since $ \mathbb{A}$ is a group, $ \sim$ : $ x\sim y \iff x \minus{} y\in\mathbb{A}$ is an equivalence relation.
Let $ r(x)$ any choice function which associates at each rational number a representent of its equivalence class (unique per class).
Let $ m(x) \equal{} f(x) \plus{} x$. Since $ f(x) \plus{} x\in\mathbb{A}$, $ m(x)$ is a function from $ \mathbb{Q}\longrightarrow\mathbb{A}$
Then :
$ x \equal{} r(x) \plus{} (x \minus{} r(x))$.
Since $ x \minus{} r(x)\in\mathbb{A}$ ($ x$ and $ r(x)$ are in the same equivalence class), we have :
$ f(x) \equal{} f(r(x) \plus{} (x \minus{} r(x)))$ $ \equal{} f(r(x)) \plus{} x \minus{} r(x)$ $ \equal{} m(r(x)) \plus{} x \minus{} 2r(x)$
Q.E.D

4) Application : some examples

4.1) $ \mathbb{A} \equal{} \mathbb{Q}$, $ r(x) \equal{} 0$ and $ m(x) \equal{} p$ give the solution $ f(x) \equal{} x \plus{} p$

4.2) $ \mathbb{A} \equal{} \{0\}$, $ r(x) \equal{} x$ and $ m(x) \equal{} 0$ give the solution $ f(x) \equal{} \minus{} x$

4.3) $ \mathbb{A} \equal{} \mathbb{Z}$, $ r(x) \equal{} x \minus{} [x]$ and $ m(x) \equal{} n$ give the solution $ f(x) \equal{} 2[x] \minus{} x \plus{} n$

4.4) $ \mathbb{A} \equal{} \mathbb{Z}$, $ r(x) \equal{} x \minus{} [x]$ and $ m(x) \equal{} [5\sin(\pi x)]$ give the solution $ f(x) \equal{} [5\sin(\pi(x \minus{} [x]))] \plus{} 2[x] \minus{} x$

A lot of other examples (for example consider $ \mathbb{A}$ as the set of rational numbers whose least denominator is a power of $ 2$) may be given.

[quote="Maharjun"]Hi, im a little new to group theory. what exactly does additive subgroup imply? and
why does this statement hold true?

[quote="pco"]
If $ x\in\mathbb{A}$, we have $ f(x \plus{} y) \equal{} f(y) \plus{} x$ $ \forall y\in\mathbb{Q}$
[/quote][/quote]
1) An additive subgroup $ U$ of $ \mathbb Q$ is a subset of $ \mathbb Q$ such that $ (U,\plus{})$ is a group, that's to say :
$ 0\in U$
$ x\in U$ $ \implies$ $ \minus{}x\in U$
$ x,y\in U$ $ \implies$ $ x\plus{}y\in U$

2) Why "If $ x\in\mathbb{A}$, we have $ f(x \plus{} y) \equal{} f(y) \plus{} x$ $ \forall y\in\mathbb{Q}$" ?
This is true just due to definition of $ A$, given some lines above :

Let $ \mathbb{A} \equal{} \{x\in\mathbb{Q} : f(x \plus{} y) \equal{} f(y) \plus{} x$ $ \forall y\in\mathbb{Q}\}$

@pco

you have taken $ m(x) \equal{} f(x) \plus{} x \ \forall x \in \Bbb Q$
$ f(x) \equal{} m(x) \minus{} x$
 i'll take your example.

$ \mathbb{A} \equal{} \mathbb{Q}$, $ r(x) \equal{} 0$ and $ m(x) \equal{} p$ give the solution $ f(x) \equal{} x \plus{} p$
but we have $ f(x)\equal{}m(x)\minus{}x \equal{} p\minus{}x$

Why the contradiction???

thanks for your help

[quote="Maharjun"]@pco

you have taken $ m(x) \equal{} f(x) \plus{} x \ \forall x \in \Bbb Q$
$ f(x) \equal{} m(x) \minus{} x$
 i'll take your example.

$ \mathbb{A} \equal{} \mathbb{Q}$, $ r(x) \equal{} 0$ and $ m(x) \equal{} p$ give the solution $ f(x) \equal{} x \plus{} p$
but we have $ f(x) \equal{} m(x) \minus{} x \equal{} p \minus{} x$

Why the contradiction???

thanks for your help[/quote]
I'm sorry but there is no contradiction :

In paragraph $ 3)$, I show that for any solution of the equation, we can find a subgroup $ A$, a choice function $ r(x)$ and a function $ m(x)$ such that $ f(x) \equal{} m(r(x)) \plus{} x \minus{} 2r(x)$ so that the solution I gave was indeed a general solution.

In paragraph $ 4)$, I gave direct interesting (IMHO) examples.
In subparagraph $ 4.1)$, I choosed $ A \equal{} \mathbb Q$, so I had to choose $ r(x) \equal{}$constant (only possible choice) and so I choosed $ r(x) \equal{} 0$ and I could have choosen any function $ m(x)$ from $ \mathbb Q\to A \equal{} \mathbb Q$ and I choosed $ m(x) \equal{} p$ and so I got $ f(x) \equal{} x \plus{} p$ where $ p\in\mathbb Q$

If we try to use paragraph $ 3)$ with function $ f(x) \equal{} x \plus{} p$, we get :
$ A \equal{} \mathbb Q$
$ r(x) \equal{} c$ for any $ c\in\mathbb Q$
$ m(x) \equal{} f(x) \plus{} x \equal{} 2x \plus{} p$

And we indeed get $ f(x) \equal{} m(r(x)) \plus{} x \minus{} 2r(x)$

What maybe disturb you is that a solution may be obtained with different triple $ (A,r(x),m(x))$.
So, for example, $ f(x) \equal{} x \plus{} p$ may be obtained thru $ (\mathbb Q,0,p)$ as well  as thru $ (\mathbb Q, c, 2x \plus{} p)$ as well as thru many other triples.

I think this is because your definition of $ m(x) \equal{} f(x) \plus{} x$ does not hold for all $ x \in \Bbb Q$ and only the following can hold

\[ f\left( {r\left( x \right)} \right) \equal{} m\left( {r\left( x \right)} \right) \plus{} r\left( x \right)\]

i have tried it in all of your examples and the result is coming out to be true. how does one prove this? the actual statement of the required proof is -

if $ r\left( x \right)$ is a choice function such that $ x \sim  r(x) \implies \pm(x\minus{}r(x)) \in \Bbb A$ and $ m(x)$ is any function from $ \Bbb Q \minus{}> \Bbb A$ ($ \Bbb A$ is defined as in your proof) then -

\[ f\left({r\left( x\right)}\right) \equal{} m\left({r\left( x\right)}\right)\plus{}r\left( x\right)\]

i would like some help in proving this though. havent got a clue as to how.

thanks in advance

[quote="Maharjun"]I think this is because your definition of $ m(x) \equal{} f(x) \plus{} x$ does not hold for all $ x \in \Bbb Q$ and only the following can hold
\[ f\left( {r\left( x \right)} \right) \equal{} m\left( {r\left( x \right)} \right) \plus{} r\left( x \right)\]
[/quote]

You are completely wrong.
The definition $ m(x)\equal{}f(x)\plus{}x$ is true $ \forall x\in\mathbb Q$

I had to show in paragraph $ 3)$ that it exists a tripple $ (A,r,m)$ such that $ f(x)\equal{}m(r(x))\plus{}x\minus{}2r(x)$
And I show that such a tripple exists.

I dont have to find all tripples.

And in paragraphe $ 4)$ I sometimes show some other tripples.

That's all.

the question says find all functions. thus generality of solution becomes very important. but you see, generality of the solution boils down to the fact that $ m(x)$can be chosen arbitrarily. however, we see that that is not the case. the $ m(x)$ that you have chosen in the first and third examples does not live up to the fact that

\[ m(x)\equal{}f(x)\plus{}x \ \forall x \in \Bbb Q\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}(i)\]

and as you showed me in your later post, it most definitely shows that your solution is correct. however, you see, you had to land upon the case of $ m(x) \equal{} 2x\plus{}p$ for $ m(x)$ to satisfy $ (i)$. hence we cannot say that the selection of $ m(x)$ was completely arbitrary. hence as of now, i am inclined to belive there is a compromise on the generality of your solution. experimentation with your result undoubtedly shows that it is most possibly true for arbitrary selections of $ m(x)$ but certain selections of $ m(x)$ do not satisfy $ (i)$ and hence your proof is not sufficient to say anything regarding such $ m(x)$. the reason i think the(slightly incomplete) logic worked out was because the following result is very probably true.

\[ f\left({r\left( x\right)}\right) \equal{} m\left({r\left( x\right)}\right) \plus{} r\left( x\right)\]
(see previous post for complete statement)

i would really like some help in proving this though. havent got a clue as to how.
please help. as i am really amazed by your solution and want to see it complete

thanks in advance

[quote="Maharjun"]the question says find all functions. thus generality of solution becomes very important.[/quote]
Pleaaaaaase, read my post !

I found a formula and I showed that :

1) all functions respecting this formula are a solution (see paragraph 2)

2) all solutions may be written in such a formula (see paragraph 3)

So I found a general solution involving all the solutions.

In paragraph 4, I used paragraph 2 to give some examples.

You compared the tripples given in paragraph 4 with those given in paragraph 3, which is of no interest.

If you prefer, you can find some other tripples for each function in paragraph 4 which match paragraph 3.

For example, $ f(x)\equal{}x\plus{}p$ may be written with the tripple $ (A,r,m)\equal{}(\mathbb Q,0,p)$ as I stated in subparagraph 4.1
But it also may be written with $ (A,r,m)\equal{}(\mathbb Q,c,2x\plus{}p)$ if you prefer, as in paragraph 3.

Now, I'm tired and if you disagree, please :
either find a function in the form I suggested and which is not a solution.
either find a solution which can not be put in the form I suggested.

[quote="Maharjun"]Hey sorry about wasting ur time(and math-links server space) pco. i fully understand your proof now. [b]awesome proof[/b]. just one last question though.

if i have understood your proof correctly, you have defined $ r(x)$ such that it returns [b]only one[/b] unique value for every equivalence class. such that, taking $ r(x) \equal{} x \minus{} 2[x]$ would be wrong right? hope u bear with me this one last time you see im still a little raw with the concepts you have used[/quote]

Yes. Choose one element in each equivalence class and then $ r(x)$ is the application which associates to each rational the unique representant (previously choosed) or its class.

And I cant answer your second question about $ r(x) \equal{} x \minus{} 2[x]$ because it depends on the set $ A$ you choosed.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

[color=darkred]Let $ABC$ be an acute triangle and let $D$ be an interior point of the side $[BC]\ .$

Denote the orthocentre $H\ ,$ the incircle $C(I,r)$ and the circumcircle $C(O,R)$

for the given triangle. Prove that :

$1.\blacktriangleright OI\perp AD$ $\Longleftrightarrow$ $\frac{DB}{AB}+\frac{DC}{AC}=1\ .$

[u]A particular case.[/u] If the point $M$ is the middlepoint of the side $[BC]\ ,$
then $OI\perp AM$ $\Longleftrightarrow$ $\frac 1b+\frac 1c=\frac 2a\ .$

$2.\blacktriangleright OH\perp AD\Longleftrightarrow c^2\cdot DB+b^2\cdot DC=a^3\Longleftrightarrow AD_1^2+DB\cdot DC=a^2\ ,$

where the point $D_1$ is the reflection of the point $D$ w.r.t. the middlepoint of the $[BC]\ .$

[b]Remark.[/b] Find a nice characterization for $IH\perp AD\ !$[/color]

[hide="Here is what I found."]
$3.\blacktriangleright IH\perp AD$ $\Longleftrightarrow$ $(a-c)\cdot \frac{DB}{p-c}+(a-b)\cdot \frac{DC}{p-b}=0\ .$[/hide]

(1).Assume c > b.
Draw a line parallel to BC through O and a line perpendicular to BC through I, these lines intersecting at U.
Triangle IUO is right angled at U.
Denote angle OIU by x.
Follows angle ADB is also x (since OI perp AD and IU perp to BC)
Tan x = OU / IU = (a/2 - (s - c) ) / (r - R Cos A) = (c - b) / 2 (r - R Cos A)
(r - R Cos A) = 4R Sin (A/2) Sin (B/2) Sin (C/2) - R Cos A
                    = R (Cos A + Cos B + Cos C - 1) - R Cos A
                    = R ( Cos B + Cos C - 1)
(DB / AB) + (DC / AC) = (Sin BAD / Sin ADB)+ (Sin DAC / Sin ADC)
                                  = [(Sin (B + x) + Sin (x - C))/ Sin x]
                                  = [Sin x ( Cos B + Cos C) - Cos x ( Sin C - Sin B)] / Sin x
                                  = (Cos B + Cos C) - Cot x ( Sin C - Sin B)
                                 = [(r – R Cos A) / R ] + 1 + [2 (r – R Cos A) /(c – b)] (c – b) /2R
                                 = 1
If D coincides with the midpoint M of BC, we have BD = DC = a / 2
So a / 2c + a / 2b =1 , 1/b + 1/c = 2/a
[size=18][/size][/size][/size]


hello, your first equation is a so-called Lienard equation, google will help you.
Sonnhard.

hello, your second equation has the solution
$ y(x)\equal{}C_1(x\cos(x)\minus{}\sin(x))\plus{}C_2(\cos(x)\plus{}x\sin(x))$.
Sonnhard.


function
integration
trigonometry
real analysis
real analysis unsolved

given that $ f,g$ are 2 [i]continuous [/i]functions of $ u$ defined over
the interval $ [0,1]$, and that

$ \left(\frac{d}{du}(f)\right)\cdot g\;\;\equal{}\;\;\frac{1}{1\plus{}u^{2}}\;\;\;\quad\;\;\;\ldots\;\;\;\ldots\;\;\;\ldots(1)$

and

$ f\cdot\left(\frac{d}{du}(g)\right)\;\;\equal{}\;\;\frac{f^{2}\cdot g^{2}}{1\plus{}u^{2}}\;\;\;\quad\;\;\;\ldots\;\;\;\ldots\;\;\;\ldots(2)$

can you [b][u]explicitly[/u][/b] solve for $ f,g$ ? :| :D

[quote="misan"]given that $ f,g$ are 2 [i]continuous [/i]functions of $ u$ defined over
the interval $ [0,1]$, and that

$ \left(\frac{d}{du}(f)\right)\cdot g\;\; \equal{}\;\;\frac{1}{1\plus{}u^{2}}\;\;\;\quad\;\;\;\ldots\;\;\;\ldots\;\;\;\ldots(1)$

and

$ f\cdot\left(\frac{d}{du}(g)\right)\;\; \equal{}\;\;\frac{f^{2}\cdot g^{2}}{1\plus{}u^{2}}\;\;\;\quad\;\;\;\ldots\;\;\;\ldots\;\;\;\ldots(2)$

can you [b][u]explicitly[/u][/b] solve for $ f,g$ ? :| :D[/quote]

Add (1) and (2) to get

\[ \frac{d}{du}\left(f\cdot g\right) \equal{}\frac{1\plus{}f^{2}\cdot g^{2}}{1\plus{}u^{2}}\]

which is

\[ \int\frac{d\left(f\cdot g\right)}{1\plus{}f^{2}\cdot g^{2}}\equal{}\int\frac{du}{1\plus{}u^{2}}\]

\[ \tan^{\minus{}1}\left(f\cdot g\right) \equal{}\tan^{\minus{}1}u\plus{}C\]

somebody else pick this up from here...


geometry
geometric transformation
combinatorics unsolved
combinatorics

n > 2. Show that there is a set of  2^n-1 points in the plane, no three collinear such that no 2n form a convex 2n-gon.

[quote="mathteam"]Asu este problema se ve bonito de donde es este problema???[/quote]

Replying to an Iranian in Spanish, looks like a good idea...

  darij

Yep. Brilliant.

A translation would be "this problem looks beautiful. Where is it from?".

I thought I would not need to endure Spanish here too, but this language chases me.

If you mean $ 2^{n\minus{}1}$, then ISL 2004 C7
http://web.archive.org/web/20041029111433/www.kalva.demon.co.uk/short/soln/sh94c7.html


geometry
perimeter
power of a point
Pythagorean Theorem

Hey there guys, could you solve these two problems for me? They're from the AoPS Wiki page about the Power of a Point Theorem, and I can't do them for some odd reason, but this is the first time I learn about this theorem so I was hoping that some of you more experienced people could solve them.
Thanks in advance.

Intermediate
Problem 1
Two tangents from an external point P are drawn to a circle and intersect it at A and B. A third tangent meets the circle at T, and the tangents PA and PB at points Q and R, respectively. Find the perimeter of triangle PQR.

Problem 2
Square ABCD of side length 10 has a circle inscribed in it. Let M be the midpoint of side AB. Find the length of that portion of the segment MC that lies outside of the circle.

For 1, by Power of a point, the two tangents from any point to the circle are equal (also called two-tangent theorem). So PA=PB, QA=QT, and RT=RB. The perimeter is just twice the sum of one of the lengths each of the pairs.

For 2, let N be the midpoint of CD, then CN=5, and let MC intersect the circle at K. By power of a point, $ CN^2\equal{}25\equal{}CK \cdot CM$. But by the Pythagorean theorem, $ CM\equal{}5\sqrt{5}$ So $ CK\equal{}\sqrt{5}$, so 1/5 of CM is outside the circle.

Thanks a lot the solutions were very clear to me thanks. To tell the truth I didn't try the second one a lot but I struggled with the first one. So basically, perimeter of triangle PQR equals 2(PA+AQ+BR), and if so then why? I'm sorry for all the questions but I suck at geometry and I'm trying to get better.


A plane $ \pi$ contains the line $ \frac{x\minus{}1}{2}\equal{}\frac{y\minus{}1}{3}\equal{}\frac{z\minus{}5}{6}$ and the point $ (1, \minus{}2, 3)$.

a) Find the equation of $ \pi$

b) find the distance of plane $ \pi$ from origin.

If $ P_{1}\equal{}(x_{1},y_{2},z_{1})$, $ P_{2}\equal{}(x_{2},y_{2},z_{2})$ and $ P_{3}\equal{}(x_{3},y_{3},z_{3})$ are three points in the plane, then $ \overline{P_{1}P_{2}}\times\overline{P_{2}P_{3}}$ gives the normal line to the plane.  You can go from there.


trigonometry
geometry
trig identities
Law of Sines
Law of Cosines

If $sinA: sinB: sinC=4: 5: 6$ and $cosA: cosB: cosC=x: y: 2$,

what ordered pair $(x,y)$ satisfies this extended proportion?

(Source: 2001 Georgia Varsity Math Tournament)

[hide="hint"] This probably isn't the best way to do this, but you could solve this geometrically. Construct a triangle using the Law of Sines then solve with the Law of Cosines.[/hide]
Actually, this is a fast way.


linear algebra
matrix
vector
algorithm
function

Let $ \ A$ and $ \ B$ be square matrix. Prove that $ \ AB\equal{}I\Longleftrightarrow BA\equal{}I$

[hide="A side query"]
I have a doubt that in most of the text books for inverse of a square matrix $ \ A$ , it is written as "If $ \ AB\equal{}BA\equal{}I$, then and only then $ \ B$ is called inverse of $ \ A$".

Can we say simply that "for a square matrix $ \ A$ if there exist a square matrix $ \ B$ such that $ \ AB\equal{}I$ , then $ \ B$ is called inverse of $ \ A$".[/hide]

Yes we can, but this isn't true for non-square matrices and it's also not true in a general ring.

Yes, that's right. Here's a sluggish argument. Let A,B be square matrices of order n such that AB = I  (1).

AB = I => (BA)B = B => (BA-I)B = O (2)

Let $ l_i$ be the line vectors of BA-I. Then (2) is equivalent to the n linear systems of equations

$ B^Tl_i^T \equal{} 0$.

From (1) we know that $ B \neq O$. If $ l_i^T \neq 0$ then the columns of $ B^T$ would be linearly dependent and we would have $ \det(B) \equal{} \det(B^T) \equal{} 0$, which is a contradiction, since from (1) we have

$ \det(A) \det(B) \equal{} 1$.

Therefore, $ l_i^T \equal{} 0$, which means BA-I = O and so BA = I.

I've given this argument before. It's very much specific to square matrices, and is a low-level argument that uses the most important computational algorithm for matrices, namely Gaussian elimination.

Start with an $ n\times n$ matrix $ A.$ Row reduce it by the Gaussian elimination argument. There are exactly two possibilities (and the proof of this is basically an act of chasing the algorithm):

(1) $ A$ is row equivalent to a matrix with a row of zeros across the bottom. Various consequences follow from that: the null space of $ A$ is nontrivial, the function $ x\mapsto Ax$ is not one-to-one, the function $ x\mapsto Ax$ does not map $ \mathbb{F}^n$ onto $ \mathbb{F}^n,$ and so on. Any one of these consequences shows that $ A$ is not invertible. In particular, we cannot have $ BA\equal{}I,$ because we can't have $ Ix\equal{}0$ for any $ x\ne 0.$ Hence, if $ BA\equal{}I,$ we must be in case (2).

or

(2) The Gaussian elimination algorithm row-reduces $ A$ to the identity matrix $ I.$

Now, what do we get from (2)? We have a sequence of elementary matrices (each such matrix generating an elementary row step) such that $ E_kE_{k\minus{}1}\cdots E_2E_1A\equal{}I.$ Elementary row steps are invertible, so elementary matrices are invertible - in fact, we can quite simply characterize the inverses of each of the three types of elementary matrix. Left-multiply this by the appropriate elementary inverses in the appropriate order to get $ A\equal{}E_1^{\minus{}1}E_2^{\minus{}1}\cdots E_{k\minus{}1}^{\minus{}1}E_k^{\minus{}1}.$

But then we have $ BA\equal{}I$ where $ B\equal{}E_kE_{k\minus{}1}\cdots E_2E_1$ and using what we've just proved, we have $ AB\equal{}E_1^{\minus{}1}E_2^{\minus{}1}\cdots E_{k\minus{}1}^{\minus{}1}E_k^{\minus{}1}E_kE_{k\minus{}1}\cdots E_2E_1\equal{}I.$


Hi I have been prove to myself the definition of Dynkin System.

Given a mesurable space \[(X, \mathbb{E})\] where \[\mu\] and \[v\] are messures on  \[\mathbb{E}\] and which implies that

\[\mu(X) = v(X)<\infty\]

Then I will like to show that the Dynkin system

\[\mathbb{D}=\{E \in \mathbb{E}| \mu(E) = v(E)\}\]

Is according to the definition a Dynkin system.
Which is also known in my textbook as an omega-class.

Solution:

\[\mathbb{D}\] is dynkin system:

If there exist a mesurable space, and corresponding messures, if and only if the following axioms are meet:

1)

\[X \in \mathbb{D}\] Since X is the set of messurable subsets in the space, and that \[E \in \mathbb{E}\] then condition (1) is meet.

2)

\[\complement E \in \mathbb{D}\] since according to condition (1) this is a consequence of \[E \in \mathbb{D}\] and therefore since \[E \in \mathbb{E}\]

3)
\[\cup_{(n \in \mathbb{N})}E_{N}\in \mathbb{D}\], for \[n \geq 1\] is a sequence of pairwise diskjoined sets of \[\mathbb{D}\]

This mean that \[E_{1}\cap E_{2}\cap \ldots = \emptyset\] which only valid if and only if \[\mathbb{D}\] is an Dynkin system!

Have I understood the Dynkin system definition correctly?

Best Regards
Fred

Hmm...Am I wrong or posting this in the number theorey proposed and own problems section is not appropriate? :wink:

So let's move it!

Pierre.


We have two sequences:
$x_1 \geq x_2 \geq\cdots \geq x_n$  $y_1 \geq y_2 \geq \cdots \geq x_n$
Then the inequality holds:

$x_1y_1+x_2y_2+\cdots+x_ny_n\geq x_1z_1+x_2z_2+\cdots+x_nz_n$
Where $z_i$ is a permutation of $y_i$. My question is what is the name of this inequality?

Rearrangement inequality.
(you also have $x_1y_1+x_2y_2+\cdots+x_ny_n\geq x_1z_1+x_2z_2+\cdots+x_nz_n \geq x_1y_n+x_2y_{n-2}+\cdots+x_ny_1$)

[quote="dondigo"]Thanks, one more question.
Is is true, that:
$x_1y_1+x_2y_2+\cdots+x_ny_n=x_1z_1+x_2z_2+\cdots+x_nz_n$ if and only if:
$y_1=z_1$ $y_2=z_2$ $\cdots$ $y_n=z_n$[/quote]

That and if you have $y_i$ all equal, equality holds as well (for all permutations).


group theory
abstract algebra
superior algebra
superior algebra unsolved

not hard nut nice result: if $G$ is nilpotent group and it order is divisible by prime $p$ so does its center.

Also see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=nilpotent&t=82991]this[/url].

A finite group is nilpotent precisely when it's a direct product of $p$-groups, for various primes $p$. In particular, its Sylow subgroups are normal and commute with one another. It follows from this that its center is the direct product of the centers of its Sylow subgroups, and the center of a finite $p$-group is non-trivial and hence of order divisible by $p$.


induction
function
limit
real analysis
real analysis solved

Let (a_n)( n natural) be a bounded sequence of real numbers which verifies the relation: 11a_(11n)+17a_(17n)=1984a_n for any n natural. Show that a_n=0 for any n natural.

Let M>0 such that |a_n|<M for every n \in N.
from the recurency we get that 1984|a_n|<=28M=>
|a_n|<(28/1984)M for every n natural
by induction we get that |a_n|<M(28/1984)k for every k natural... now if k->oo results |a_n|<=0 for every n natural=>a_n=0 for every natural n. :)

Nice proof Kuba,

Here another version of this problem

Let p,q,r positive integers such that (p+q)/r < 1
f,g:N-->N functions such that for every n \in N
p.a_(f(n)) + q.a_(g(n)) = r.a_n where a_n is a bounded sequence of real numbers.
Then a_n = 0 is 0 for every n \in N
---------------------------------------------------------------------------
Proof: Just follow the idea of Kuba
M >0 s.t |a_n| \leq M from the relation we have
|a_n| \leq (p+q)/r.M
by induction |a_n| \leq ((p+q)/r)^k.M for any k \in N
k-->+oo we get |a_n|=0 for every n \in N

Kuba, I don't understand why:
[quote]now if k->oo results $|a_n|leq0$[/quote]
I think we can just say that $\lim_{n\rightarrow+\infty} a_n=0$ but we can't say that $a_n=0$ for a fixed $n$
There is something I have missed to understand you, can you explain to me ?

[quote="Kuba"]$|a_n|<M(28/1984)^k$ for every k natural[/quote]

If k tends to +oo you get for any n integer $|a_n|\leq 0$ so the sequence $|a_n|$ is

identically zero

Oups !!!!
Don't why but I have take the $k$ for the $n$  :blush:


Hint:  log27(base x) can also be written as 3/logx(base 3).

[hide]

After doing that, move everything to the left side, and get a common demonimator.

Then, substitute y for logx(base 3), and factor.  You should get:

(2y+3)(y-4)/y <=0

From that we can see:

{y | (negative infinity, -3/2] union (0, 4]}

or

{logx(base 3) | (negative infinity, -3/2] union (0, 4]}

Which means the solution is:

{x | (0, 3^(-3/2)] union (1, 81]}

(remember that the base only be positive, and it cannot be 1)

Q.E.D.

[/hide]

[quote="Altheman"][hide="solution"]let $y=\log_3 x$, then we get

$2y-\frac{12}{y}\le 5$
$y\ne 0$   $\star$

$y(2y^2-5y-12) \le 0$
$y(2y+3 )(y-4) \le 0$

by interval testing, $y \in \left(-\infty,-\frac{3}{2}\right]\cup(0,4]$

then $3^y=x$, so we have $x \in (0,3^{-\frac{3}{2}}]\cup(1,81]$[/hide][/quote]

Your answer is correct. :)


Three friends want to go from Derby to Kent, 32 miles. They must leave and arrive at the same time. Andy has a car, speed 8 mph; Bill has a bike, speed 6 mph; and Charlie leaves on foot, 4 mph. If one friend drops off the car or bike, a trailing friend may pick it up and begin to ride it, but modes of transportation can only be exchanged, dropped off, or picked up on the hour. What is the minimum time in hours that it will take the friends to arrive at Kent?

[quote="dzugash"]Andy has a car, speed 8 mph[/quote]
LOL, what a rattletrap.


1)show that         sin3x.sin^3x  + cos3x.cos^3x = cos^3(2x)
x is the angle we are taking sin and cos of.

2)show that cos^2x  +  cos^2(2pi/3+x)  +  cos^2(2pi/3-x)  =  3/2


Proved that the amount

$3^{2n}-8 n+63$

 apportioned to accept(	Divided on) $64$

You want to prove that $a_{n}: =9^{n}-8n-1$ is divisible by $64$. This is evident for $a_{1}$. Also, $a_{n+1}-9a_{n}=64n$.


Find all functions $f: \mathbb R^{+} \to \mathbb R^{+}$ that satisfy \[ f(xf(y))\plus{}f(yf(x))\equal{}2xy\] for all $x,y >0$.

Let $ f(1)\equal{}a$.
Substituting $ (x,y)\equal{}(1,1)$, then $ f(f(1))\equal{}1$ so $ f(a)\equal{}1$.
Substituting $ (x,y)\equal{}(a,a)$, then $ f(a)\equal{}a^2$ so $ a^2\equal{}1$ so $ a\equal{}1$.
Thus $ f(1)\equal{}1$.
Substituting $ (x,y)\equal{}(x,1)$, then $ f(f(x))\plus{}f(x)\equal{}2x$ for all $ x \in \mathbb{R}_\plus{}$.
Now, by recursion we have that $ f(x)\equal{}x$ for all $ x \in \mathbb{R}_\plus{}$.
Done.

I think, Raja Oktovin, that your phrase "by recursion, we have ..." is a little bit too short. Here is a more precise version of your demo (IMHO) :

Let $ P(x,y)$ be the assertion $ f(xf(y)) \plus{} f(yf(x)) \equal{} 2xy$

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

Using then the sequence $ a_0 \equal{} x$, $ a_1 \equal{} f(x)$ and $ a_{n \plus{} 2} \equal{} \minus{} a_{n \plus{} 1} \plus{} 2a_n$, we get $ f^{[n]}(x) \equal{} a_n \equal{} \frac 13(2x \plus{} f(x) \plus{} (x \minus{} f(x))( \minus{} 2)^n)$

And now, we can say that if $ f(x)\ne x$ for some $ x$, then $ RHS$ will be negative for some $ n$ great enough and so $ f^{[n]}(x) < 0$ for some $ n,x$, in contradiction with the fact that $ f(x)$ is from $ \mathbb R^ \plus{} \to\mathbb R^ \plus{}$

Hence your [good] result $ \boxed{f(x) \equal{} x}$ $ \forall x>0$


function
probability
number theory
totient function
geometric series

I haven't solved this problem yet since I just came up with it, so I don't know how hard it is. Find inf(phi(n)/n), where n is an integer >= 2. (Phi is Euler's totient function, or the number of positive integers less than n coprime to n.)

Alternatively, we could make this problem WAY more interesting by defining phi_2(n) to be the number of integers less than n^2 that don't share factors that are perfect squares (other than 1) with n^2. Find inf(phi_2(n)/n^2).

If you want to go further, define phi_k(n) analogously, and then find inf(phi_k(n)/n^k).

Now I think it's actually an interesting problem.

I dunno, I thought the first problem was rather interesting. 3072/17017 (which is roughly 0.18) was the smallest small value mathematica came up with when I did a quick progam to test up to 1,000,000.

Right. 3072/17017 is the best when n<1000000. That's achieved at n=510510. I need to learn how to program with Mathematica now that I have it running again.

This program was pretty easy -- if it deserves to be called a program.

Min[Table[EulerPhi[n]/n, {n, m, M}]] will find the value for every n between m and M and then report the minimum value to you.  I don't offhand know how to make it tell you what number gave that value.

Further calculation shows that 55296/323323 :approx: .171024 is the best up to ten million, achieved somewhere greater than five million.  My inclination would be that it eventually approaches 0, very, very slowly.  Of course, that's not based on anything at all, really.

In fact, a slight improvement on that intuition:  I bet the minimal cases look like 2*3*5*7*11*13*... up to some p.  This is supported by testing many small cases.  In fact, that value I quoted above is achieved at n = 2*3*5*7*11*13*17*19.  Very interesting.  I'm so tired . . .  I'm sure this makes sense, but I can't sort it out now.

Yes, for the original problem, the infimum is 0.  This is a known result.  The minimal cases, as JBL observed, are when n is the product of the first few primes.  We can analyze those cases using the characterization of phi(n)/n as the product of 1 - 1/p over all primes p dividing n, and the prime number theorem.   For example, by the prime number theorem, the product of the first k primes is about (k ln k / e)k, and the sum of the reciprocals of the first k primes is about ln ln k.  Using those formulas, when n is the product of the first few primes, we get that phi(n)/n is on the order of 1 / lnln n, which approaches 0 very slowly, as JBL conjectured.

Going the other way around, it is known that phi(n) is always at least c n / ln ln n, where c is a positive constant.

Joel and Ravi B are both correct. What about the second problem? It's no harder than the first, but the answer is somewhat amusing. I was disappointed when I figured out that inf(phi(n)/n)=0, but when I found the solution, I thought about how one could construct other functions similar to phi(n) to get different answers.

I believe that phi2(n)/n2 is the product of 1 - 1/p2 over all primes p dividing n.  So the infimum should be the infinite product of 1 - 1/p2 over all primes p.  That product converges to a positive constant.

Yes, but you have to find the positive constant.

Oh right, at first I didn't think the product had a closed-form solution, but now thanks to your suggestion I see.  We can express 1/(1-1/p2) as the geometric series 1 + 1/p2 + 1/p4 + ....  When we multiply over all primes p, we get the infinite sum 1/12 + 1/22 + ..., which is well known (zeta function!) to be :pi:2/6.  Since the product of 1-1/p2 is the reciprocal of the product of 1/(1-1/p2), the final answer is 6/:pi:2.

Cute problem!

That last result you posted can be used to solve this:

If we choose 2 random positive integers, what's the probability that they're coprime? Sounds nice doesn't it (you probably know this problem anyway  :D )

You solve that problem like that? The solution I know shows that it's sum_n mu(n)/n^2, which can easily be shown to be 1/zeta(2).

A related problem (at least by answer) is: What is the density of the set of positive integers that are square-free?

I like that. It's a much simpler problem that has exactly the same solution.


Could some one please upload the problems as soon as you can;

   Saludos a todos, espero se la pasen chingonsisimo, pero suban los problemas.

           Andres Campero-MEx4 en la imo, no quede para la ibero,  :maybe:

[color=green]Cuando termine la prueba, y se conecten los argentinos, los mando

saludos!
CVaneG[/color]

1) Se escriben los numeros del 1 al 2008² en un tablero de 2008x2008. Para cada fila y columna se calcula la diferencia entre el minimo y el maximo. Sea S la suma de los 4016 resultados. Hallar el maximo posible de S.

2) En un triangulo ABC, sea r la bisectriz exterior de $ \angle{ABC}$. Sean P y Q los pies de las alturas de A y C a r. CP intersecta a BA en M y AQ intersecta a BC en N. Demostrar que MN, r y AC concurren en un punto.

3) P(x)= x³ +mx +n
donde m y n son enteros, tal que
si P(x) - P(y) es divisible por 107, entonces x-y es divisible por 107. Demostrar que 107 divide a m.

1) Numbers from 1 to 2008² are written in a 2008x2008 board. For each row and column, the difference between the gratest and the smalest number is calculated. Let S be the sum of this 4016 results. Find the greatest possible value for S.
2) Let ABC be a triangule such that r is the exterior bisector of $ \angle{ABC}$. Let AP and CQ be the distance from A and C to r. CP intersects AB in M and AQ instersects BC in N. Prove that MN, r and AC have an only point in common.
3) P(x)= x³ +mx +n
where m and n are integers, such that
if P(x) - P(y) is divisible by 107, then x-y is divisible by 107. Prove that 107 divides m.

Sorry, I'm still learning English...

gracias loco los estaba buscando..   :wink:   habrá q resolver y esperar los resultados, haber como andamos :)

Does anyone know the results???..... i.e. medals, cutoffs etc...

[quote="pelao_malo"]the same question
i want to know the results of my country
please can someone post the page of the results when they're out?
bye  :D[/quote]

Chi 1 : 7 7 7 7 7 2(silver)
Chi 2 : 7 4 3 7 0 0(bronze)
Chi 3 : 7 1 0 0 0 0(HM)
Chi 4 : 7 0 0 5 0 0(HM)


function
real analysis
real analysis unsolved

$||x||=min(x-E(x); w(x)-x)$ is the distance to $x$ to the near integer

$E(x)=max(n \in Z, n\leq x)$
$w(x)=min(n \in Z, n\geq x)$

$f(x)=\sum_{k\in Z}2^{k}||\frac{x}{2^k}||^2$

Find $f$

Moubi, do you have a solution? I know solution for this problem, but I don't want to spoil this beautiful problem. So, if it is unsolved, tell us and I will post solution. If not, I will let the others solve it.

Come on, Moubi,  of course they will solve it. And please don't call me professor harazi, since I'm far too dumb to deserve such title.  Anyway, I find it very very beautiful problem.

[quote="Moubinool"]If none solve this one, post your solution in a week professor harazi[/quote]

Yes, please [i]Professor[/i] Harazi!  :D

By the way, it seems the problem originates from AMM, created by R.L.Graham and D.E.Knuth.


calculus
integration
trigonometry
calculus computations

Help solve$ \int         \frac{x}{1\plus{}\cos^2x}dx$ :)

I can calulate when we intagrade in some fixed interval by using ($ \frac {x}{1 \minus{} C^{2}} \equal{} x\sum ( \minus{} 1)^{i}C^{2i}$ where $ C$ and $ S$ is $ \cos$ and $ \sin$. but I don't think that there exist some closed form without fixed interval. I think there is not closed form  :(

[b]Maybe[/b]

$ I \equal{} \int_0^{\pi} \frac {x}{1 \plus{} \cos^2x}\ \mathrm {dx} \equal{} \int_0^{\pi}\frac {\pi \minus{} x}{1 \plus{} \cos^2x}\ \mathrm {dx}\implies$ $ I \equal{} \frac {\pi}{2}\cdot\int_0^{\pi}\frac {1}{1 \plus{} \cos^2x}\ \mathrm {dx}$ ...

sorry I was wrong. this integral has closed form and is equalt ot $ \frac{Arctan (\frac{tan x}{\sqrt{2}})}{\sqrt{2}}$

How did you get it? Are you sure it is the answer?

I think Extremal is referring to $ P \equal{} \int \frac{1}{1 \plus{} \cos^2x}\,dx$, which is not what the original poster asked. To compute P, make the substitution tan x = t to get

$ P \equal{} \int \frac{1}{1 \plus{} \frac{1}{1\plus{}t^2}} \cdot \frac{1}{1\plus{}t^2}\,dt \equal{} \int \frac{1}{2\plus{}t^2}\,dt \equal{} \frac{1}{2} \int \frac{1}{1 \plus{} (t/\sqrt{2})^2}\,dt$

$ \equal{} \frac{1}{\sqrt{2}} \arctan(t/\sqrt{2}) \plus{} C \equal{} \frac{1}{\sqrt{2}} \arctan\left(\frac{\tan x}{\sqrt{2}}\right) \plus{} C$.

If it may be solved - the only thing that come to my mind is many times integration by parts. It also may be useful to write $ cos^{2}x\equal{}\frac{1\plus{}cos2x}{2}$ and to calculate simultaneously two integrals.

$ \int \frac{x^{2}}{1\plus{}cos^{2}x} \, dx \equal{} \frac{x^{3}}{1\plus{}cos^{2}x} \minus{} \int x\left(\frac{2x(1\plus{}cosx)\minus{}x^{2}(1\plus{}cosx)^{'}}{(1\plus{}cos^{2}x)^{2}}\right) \,dx \equal{}$
$ \frac{x^{3}}{1\plus{}cos^{2}x} \minus{} 2\int \frac{x^{2}}{1\plus{}cos^{2}x}\,dx \plus{} \int \frac{x^{3}}{(1\plus{}cos^{2}x)^{2}}\,d(1\plus{}cos^{2}x) \equal{}$
$ \frac{x^{3}}{1\plus{}cos^{2}x} \minus{} 2 \int \frac{x^{2}}{1\plus{}cos^{2}x}\,dx \plus{}  \int \,d\left(\frac{x^{4}(1\plus{}cos^{2}x)^{\minus{}1}}{4(\minus{}1)}\right) \equal{}$
$ \frac{x^{3}}{1\plus{}cos^{2}x} \minus{} 2\int \frac{x^{2}}{1\plus{}cos^{2}x}\,dx \plus{} \frac{x^{4}}{(\minus{}4)(1\plus{}cos^{2}x)}$

If follows that:
$ 3\int \frac{x^{2}}{1\plus{}cos^{2}x}\,dx \equal{} \frac{x^{3}}{1\plus{}cos^{2}x} \plus{} \frac{x^{4}}{(\minus{}4)(1\plus{}cos^{2}x)}$
We may conclude - the final answer is:
$ \frac{4x^{3}\minus{}x^{4}}{12(1\plus{}cos^2{x})} \plus{} C$

I want to ask ... is my solution correct? How to prove it? If it is not correct where is the mistake?

[quote]I want to ask ... is my solution correct ? How to prove it ?[/quote]

[b]Very simple : [/b]$ \left[\frac {4x^{3} \minus{} x^{4}}{12(1 \plus{} cos^2{x})}\right]'\equal{}\frac {x^{2}}{1 \plus{} cos^{2}x}\ .$

Good job. However, the original poster wanted $ P \equal{} \int \frac{x}{1\plus{} \cos^2x}\,dx$.

I'm sorry I don't know where I take second degree from ... indeed I think my solution is wrong. What if $ cos^{2}x \equal{} \frac {1 \plus{} cos2x}{2}$ and we substitude $ t \equal{} 3 \plus{} cos2x$ ? Is there any progress?

suweijie, have you a solution?

As Extremal showed, I think there are more interesting primitives to "play with".

You may be right, but sometimes with such integrals there is some 'trick' but not every time. I'm interested to see the suweijie's solution.


geometry
geometric transformation
homothety

[b]Problem.[/b] [i]Two circles are internally tangent at $T$. Let a chord $AB$ of the larger circle  be tangent to the smaller circle at $P$.  Prove that $PT$ bisects $\angle ATB$.[/i]

let a parallel to AB touch the larger circle at P', [opposite half plane with respect to T], by homothety, APP' is a line, then since those lines are parallel, $P'$ bisects arc $AB$, the result follows from inscribed angles

Center of smaller circle: $M$
Diameter of smaller circle: $TC$
Center of larger circle: $O$
Diameter of larger circle: $TD$
Extend $TP$ to meet the larger circle at $Q$. By definition, since $TD$ is a diameter and $Q$ is a point on circle $O$, $\angle DQT = 90$. Similarly, $TC$ is a diameter and $P$ is a point of circle $M$, so $\angle CPT = 90$.
This makes $\triangle DQT \sim \triangle CPT \Longrightarrow \frac{QT}{PT}=\frac{DT}{CT}$.
Since $\frac{QO}{PM}= \frac{\frac{DT}{2}}{\frac{CT}{2}}= \frac{DT}{CT}$, $\triangle OQT \sim \triangle MPT \Longrightarrow QO \parallel PM$. So $PM \perp AB$ (because tangency) $\Longrightarrow QO \perp AB$
Radius perpendicular to a chord bisects it, so $AQ = QB \Longrightarrow \angle AOQ = \angle QOB \Longrightarrow\angle ATQ = \angle QTB$ QED

the result you proved can be summarized as follows: two internally tangent circles are homothetic with respect to the point of tangency


quadratics
algebra
polynomial
number theory unsolved
number theory

Suppose that a is an irrational number.
a. If there is a real no. b such that both  (a+b) and ab are rational, show that a is a quadratic surd, i.e. of the form (r+s^1/2) or (r-s^1/2), where s is not the square of any rational no.

Could you please help me with this?
I am new to problem solving, and i need this as early as possible.

$ x^2 \minus{} (a \plus{} b)x \plus{} ab$ is the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=134466]minimal polynomial[/url] for $ a$ over the rationals.  The result follows by the quadratic formula.


A positive integer is called perfectly balanced if adding and subtracting the same positive integer results in two perfect squares. (What a boring name...) For example, 53 is a perfectly balanced integer since 53-28=25 and 53+28=81 . Suppose that all perfectly balanced integers are ordered in increasing order. What is the third perfectly balanced number?

Please show your work.

[hide]
Halfway between 0 and 1, there is not an integer.
Halfway between 0 and 4, there is 2.
Halfway between 1 and 4 there is not one
Halfway between 1 and 9 there is 5.
Halfway between 4 and 9 or 1 and 9 there is not one.
Halfway between 0 and 16 there is 8.

So 8 is the third.
Answer=8[/hide]

[hide]i did it by trial and error because i didn't think it would take too long.
0,4, which is 2
1,9, which is 5
0,16, which is 8
it goes a lot faster if you realize perfectly balanced numbers can only be between two even squares or two odd squares.[/hide]

[quote="mathnerd314"][quote="Captain Sugar"]between two even primes or two odd primes.[/quote]

Don't you mean even or odd squares?

666th post!!! :o :o[/quote]

yeah, oops.  I'll fix it. :)

[quote="Ignite168"][quote="pgpatel"]Is 0 a perfect square?[/quote]
Considering $0^2 = 0$, yes.   :P[/quote] Still... Anyways is 0 even.

Since 0 is evenly divisible by any number, it is divisible by 2 and so it is even.