geometry
geometric transformation
reflection
incenter
geometry unsolved

Let $ Y_{a}$ be the reflection of $ X_{a}$ into $ M_{a}$. This point is the tangency point of the $ A$-excircle with the sideline $ BC$. Thus $ AY_{a}$ passes through the Nagel point of $ ABC$. So, $ AY_{a}$ passes also through the antipode of $ X_{a}$ in the incircle. In this case, let the perpendicular at $ X_{a}$ to $ X_{a}M_{a}$ cut $ AY_{a}$ at $ X_{a}'$. Let $ I$ be the midpoint of $ X_{a}X_{a}'$. This is the incenter of $ ABC$. Draw the incircle $ \Gamma$ of $ ABC$ (circle through $ X_{a}$ centered at $ I$) and let the tangents from $ A$ to $ \Gamma$ meet the line $ X_{a}M_{a}$ at $ B$, $ C$.


geometry
rectangle
analytic geometry
percent

A rectangle with length 8 units and width 5 units is centered at the origin of a coordinate plane, and its longer side is parallel to the x-axis. What percent of the rectangular region is to the right of the line with equation $ x \equal{} \minus{}2$?

[geogebra]023e68529c789717a5a7b02a1e280131d88eeaf3[/geogebra]

From that picture, we can easily see that $ \frac{3}{4}$ of the rectangle is to the right of the line $ x \equal{} \minus{}2$. $ \frac{3}{4} \equal{} \boxed{75\%}$


geometry
inradius
trigonometry
area of a triangle
Heron\u0027s formula
geometry solved

A triangle has sides 18', 24', and 26'. WHat is the length of the inradius to the nearest tenth of a foot..

I know the formula to find the inradius is S/(2(a+b+c)) which S is the area and a,b,c is the lengths of the sides...

How ever after I used Law of cosine to find the area which is 209 square foot..but then when I applied the formula above ..I got the inradius equal 5.1
but the choices were
a) 6.7 ft.      b)7.9ft       c)6.0ft           D)8.0ft            E)6.1ft

please help..thank you

Since you know the sides $a = 18\ \text{ft},\ b = 24\ \text{ft},\ c = 26\ \text{ft}$ and not any angles, it is better to use Heron's formula for the triangle area:

$s = \frac{a + b + c}{2} = 34\ \text{ft}$

$|\triangle ABC| = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{34 \cdot 16 \cdot 10 \cdot 8} = 16 \sqrt{170} \doteq 208.6\ \text{ft}^2$

$r = \frac{|\triangle ABC|}{s} = \frac{16 \sqrt{170}}{34} = \frac{ 8 \sqrt{170}}{17} = 6.136\ \text{ft}$


linear algebra
matrix
vector
linear algebra unsolved

This is the same as requiring that (1,-1,0,1)', (1,2,1,1)', (0,1,-1,0)', (1,2,1,2)' form a basis of R^4, which they form, since the determinant of the matrix whose columns (or rows) are these vectors is not zero.


limit
real analysis
real analysis theorems

$ (t_n)$ - bounded sequence of reals

$ f_j\rightrightarrows f$

$ f,f_j$ continuous

can we write
\[ \lim_{j\to\infty}\lim_{k\to\infty}\frac{1}{k}\sum_{n\equal{}1}^kf_j(t_n)
\equal{}
\lim_{k\to\infty}\lim_{j\to\infty}\frac{1}{k}\sum_{n\equal{}1}^kf_j(t_n)\]
if yes, why ?

I assume that $ f_j\rightrightarrows f$ means "$ f_j$ converge uniformly to $ f$". Denote $ F(j,k)\equal{}\frac {1}{k}\sum_{n \equal{} 1}^kf_j(t_n)$ and note that $ \lim_{j\to\infty}F(j,k)$ is uniform. This is enough to swap it with the limit as $ k\to\infty$, provided that the latter exists for every fixed $ j$.

[quote="fedja"]I assume that $ f_j\rightrightarrows f$ means "$ f_j$ converge uniformly to $ f$".[/quote]
yes

[quote="fedja"] $ \lim_{j\to\infty}F(j,k)$ is uniform. [/quote]
Could you explain what do you mean by this?

For every $ \varepsilon>0$ there exists $ J$ such that for every $ k$, one has $ |F(j,k)\minus{}F(k)|<\varepsilon$ whenever $ j>J$ (with $ F(k)\equal{}\frac 1k\sum_{n\equal{}1}^k f(t_n)$, of course). The key point is that $ J$ depends only on $ \varepsilon$ but not on $ k$.


How do you define the "center" of a tetrahedron? Is it hard to find?

Depends on what you mean by center (How do you define the "center" of a triangle?).

Well for a tetrahedron with congruent isoceles faces, it would be the intersection of the segments connecting each vertex to the centroid of the opposite face... i dont know if these segments will intersect for tetrahedrons without isoceles congruent faces.

I've attached what I'd consider the centre for a unit tetrahedron.


inequalities
real analysis
trigonometry
integration
function
calculus
real analysis unsolved

Can anyone please help with the following problems from Rudin Chapter 11:

[hide="Rudin. Chapter 11: #17"]
Suppose $ E \subset( \minus{} \pi, \pi), m(E) > 0, \delta > 0.$
Use the Bessel inequality to prove that there are at most finitely many integers $ n$ such that $ \sin nx \geq \delta$ for all $ x \in E.$

By the way, I think the statement of the problem is confusing and can be simplified to saying that we want to prove

The set
\[ \{ n \in \mathbb{Z} \quad : \quad \sin nx \geq \delta \text{ for all } x \in E\}
\]
is finite.

Also, Bessel inequality is given in a theorem on page 188 of Rudin, which states:
If $ \{\phi_n\}$ is orthonormal on $ [a,b]$, and if
\[ f(x) \sim \sum_{n \equal{} 1}^{\infty} c_n \phi_n (x),
\]
then
\[ \sum_{n \equal{} 1}^{\infty} |c_n|^2 \leq \int_a^b |f(x)|^2 dx.
\]
[/hide]

[hide="Rudin. Chapter 11: #18"]
Suppose $ f \in \mathcal{L}^2(\mu), g \in \mathcal{L}^2(\mu).$ Prove that
\[ \left| \int f \bar{g} d\mu \right|^2 \equal{} \int |f|^2 d\mu \int |g|^2 d\mu
\]
if and only if there is a constant $ c$ such that $ g(x) \equal{} cf(x)$ almost everywhere.
[/hide]

#17: Apply Bessel inequality to the characteristic function of the set E.
#18: The statement is false. Take $ f(x)\equal{}0$, $ g(x)\equal{}1$ :D. If $ f(x)\not\equiv0$, then put

\[ c\equal{}\frac{\int \overline fg\,d\mu}{\int|f|^2d\mu}\]

and consider

\[ \int|g\minus{}cf|^2d\mu.\]

To RIP,

For # 17, that's what I thought, but I didn't know what to do after.

So we have the coefficients of the Fourier series given by
\[ c_m \equal{} \frac {1}{2\pi} \int_{ \minus{} \pi}^{\pi} K_E(x) \ e^{ \minus{} i m x} dx \equal{} \frac {1}{2\pi} \int_{ \minus{} \pi}^{\pi} e^{ \minus{} i m x} dx
\]
Here I don't know if I'm supposed to take the integral over the interval $ [ \minus{} \pi,\pi],$
or just $ E.$

Could you explain a little more? I really appreciate it!

Also for #18, are you saying that the $ \Leftarrow$ part is wrong?

#17: use the following orthonormal system: $ \pi^{\minus{}1/2}\sin nx$, $ n\in\mathbb Z_{>0}$. By the way,
\[ \int\limits_{\minus{}\pi}^{\pi}K_E(x)g(x)\,dx\equal{}\int\limits_Eg(x)\,dx.\]

#18: Quite the contrary. The $ \Leftarrow$ part is trivially true, but the $ \Rightarrow$ one is false as it is stated. It should changed to something like $ cf(x)\equal{}c'g(x)$ a.e., for some constants $ c,c'$ such that $ |c|\plus{}|c'|>0$.


In a pentagon $ ABCDE$ with $ BC\parallel{}AD$ and $ BD\parallel{}AE$, $ M$ and $ N$ are the midpoitns of $ CD, DE$ respectively. Let $ O\equal{}AM\cap BN$. Prove that the areas of quadrilateral $ MDNO$ and triangle $ ABO$ are equal.

[quote="discredit"] [color=darkred]Let $ ABCDE$ be a pentagon with $ BC\parallel{}AD$ , $ BD\parallel{}AE$ . Denote the midpoints  $ M$ , $ N$

of $ CD, DE$ respectively and $ O \equal{} AM\cap BN$. Prove that $ [MDNO]\equal{}[ABO]$ .[/color] [/quote]

[b][u]Proof.[/u][/b] $ [MDNO]\equal{}[ABO]\ \Longleftrightarrow\ [MDNO]\plus{}[AONE]\equal{}$ $ [ABO]\plus{}[AONE]\ \Longleftrightarrow\ [AMDE]\equal{}[ABNE]$ .

$ [AMDE]\ \equal{}[ADE]\plus{}[AMD]\stackrel {(MC\equal{}MD)}{\ \equal{}\equal{}\equal{}\equal{}\equal{}\ }[ADE]\plus{}\frac 12\cdot [ACD]\stackrel{(BC\parallel AD)}{\ \equal{}\equal{}\equal{}\equal{}\equal{}\ }$ $ [ADE]\plus{}\frac 12\cdot [ABD]\stackrel{(AE\parallel BD)}{\ \equal{}\equal{}\equal{}\equal{}\equal{}\ }$

$ [ADE]\plus{}\frac 12\cdot [BED]\ \stackrel {\left(\begin{array}{c}
AE\parallel BD\\\
NE\equal{}ND\end{array}\right)}{\ \equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\ }\ [ABE]\plus{}[BNE]\equal{}[ABNE]$ . Thus, $ [AMDE]\ \equal{}[ABNE]$ .


I have a very interesting math riddle here:
Why did the acute angle always get math problems wrong?

And another:
What kind of a figure would always win a beauty contest?

This should be in Puzzles and Brainteasers  :wink: .

1:[hide]Acute angles are never right?[/hide]

archimedes, you were right!

My math teacher proposed this joke to me, and i loved it. It was funny.

Why didn't you just put these in the math joke topic?


function
topology
calculus
calculus computations

My book defines a contraction of $ M$ is a mapping $ f: M\rightarrow M$ such that for some constant $ k<1$ and for all $ x,y \in M$,

$ d(f(x), f(y)) \le kd(x,y)$

I've read in other books that a weak contraction is something similar, except $ d(f(x), f(y)) < d(x,y)$. So it looks like a weak contraction is technically a contraction right? ( for example,  $ d(f(x), f(y)) < kd(x,y)< 1 \cdot d(x,y)$ for $ k <1$)

Secondly, does compactness have any application to showing something is a contraction?

What you wrote is right - what you said is wrong. Technically a contraction is also a weak contraction, not the other way around.

As for your other question, compactness is paramount in many applications, but here?
Think of real functions defined on $ [0, 1]$. Even if we assume $ f$ to be of $ C^1$ class, so $ f'$ is bounded, that only yields Lipschitz, not necessarily contraction.

There is a real difference between strong and weak contractions. For a (strong) contraction on a complete metric space, we have the Banach theorem that there is a unique fixed point. For weak contractions, we don't.
As an example, consider $ f(x)\equal{}\sqrt{x^2\plus{}1}$ on $ [0,\infty)$. That's a weak contraction, with no fixed point.

Let $ M \equal{} [1,\infty)$ and let $ f(x) \equal{} x \plus{} \frac1x.$

This is a weak contraction but not a contraction. And it makes a big difference. A contraction on a complete space must have a fixed point. $ M$ is a complete space, but this $ f$ does not have a fixed point.

[Or what jmerry said. Simultaneous posts.]

The point here is that the constant $ k$ appearing in the definition of a contraction is a [b]uniform[/b] bound.  As you can see, if the bound isn't uniform you don't get the contraction principle.

As for your second question, perhaps you were thinking along the following lines: suppose $ f$ is a weak contraction on a compact metric space $ M$.  Let $ k(x, y) \equal{} \frac{d(f(x), f(y))}{d(x, y)}$, which is defined on $ M \times M$ away from the diagonal.  Since $ k$ is continuous, it attains a maximum on compact subsets of $ M \times M$.

Unfortunately, $ M \times M$ minus the diagonal isn't closed, hence not compact, so one can't conclude that $ \sup_{x \neq y} k(x, y) < 1$.

Now, there is one thing a compact space gets you: a weak contraction must have a unique fixed point- because $ d(x,f(x))$ must have a minimum, and this minimum must be zero to avoid $ d(f(x),f(f(x)))$ being smaller.


MATHCOUNTS
geometry
3D geometry
pyramid
calculus

Explain why the volume of a pyramid is 1/3bh.

this is pretty much a calculus problem...


Find all functions R to R such that
f(f(m)+n)=2m+f(f(n)-m)

$ P(m,\minus{}f(m)): f(f(\minus{}f(m))\minus{}m)\equal{}\minus{}2m\plus{}f(0)$. So $ f$ is surjective
Let $ a\equal{}f(0)$.
$ P(0,n): f(f(n))\equal{}f(a\plus{}n)$ (F1)
$ P(m,f(n)): f(f(m)\plus{}f(n))\equal{}2m\plus{}f(f(f(n))\minus{}m)\equal{}2m\plus{}f(f(a\plus{}n)\minus{}m)$, where the last step uses (F1)
$ P(m,a\plus{}n): f(f(m)\plus{}n\plus{}a)\equal{}2m\plus{}f(f(a\plus{}n)\minus{}m)$.
Equating the RHS of above two, we get
$ f(f(m)\plus{}f(n))\equal{}f(f(m)\plus{}n\plus{}a)$
Since $ f$ is surjective, we can get
$ f(m\plus{}f(n))\equal{}f(m\plus{}n\plus{}a)$ (F2)
Also, $ f(m\plus{}f(n))\equal{}2n\plus{}f(f(m)\minus{}n)\equal{}2n\plus{}f(m\minus{}n\plus{}a)$, where the last step uses (F2)
Equating the LHS of the above two,
$ f(m\plus{}n\plus{}a)\minus{}f(m\minus{}n\plus{}a)\equal{}2n$.
Hence, we have
$ f(u)\minus{}f(v)\equal{}u\minus{}v,\forall u,v$.
or
$ f(u)\equal{}u\plus{}c$, where $ c$ is constant. It is easy to check that is the solution.


linear algebra
matrix
linear algebra unsolved

[color=darkblue][size=150]Is there exists $ A,B \in M_n(\mathbb{R})$

such that: $ rank(AB\minus{}BA)\equal{}n$[/size][/color]?

Here's an example with $ n\equal{}2:$

Let $ A\equal{}\begin{bmatrix}1&1\\0&0\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}0&0\\1&0\end{bmatrix}.$

Then $ AB\minus{}BA\equal{}\begin{bmatrix}1&0\\\minus{}1&\minus{}1\end{bmatrix},$ which does have full rank.

Let $ A,B\in\mathcal{M}_2(K)$ s.t. $ A$ has 2 distinct eigenvalues and $ A,B$ has no common eigenvectors (It's the generic case if $ K$ is algebraically closed).
Then $ AB\minus{}BA$ is invertible.
Proof: we may assume $ A\equal{}diag(\alpha,\beta),B\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix}$ where $ \alpha\not\equal{}\beta$ and $ bc\not\equal{}0$. $ \cdots$


induction
combinatorics unsolved
combinatorics

For all positive integer n let$p(n)$be the number of ways express n into the sum of positive integers.In each of such way,we consider the distinct number in the sum.These number are summed ,call the sum $q(n)$e.g: $6=1+1+1+1+1+1=1+1+1+1+2=1+1+1+3=1+1+4=1+5=1+2+3=2+4=3+3=6,p(6)=9,q(6)=16$
Prove that$a,q(n)=1+p(1)+..+p(n-1)$
               $b,q(n)<\sqrt{2n}.p(n)$

For a) we can count q(n) like that:For a i,count the numbers of partition which include i,which is p(n-i)
For b),we have the identity:$np(n)=\sum_{i=1}^{n} d(i)p(n-i)$,d(i) is the numbers of divisors of i

Thanks for your solution.My solution use induction of system condition,but too difficult

To spdf:For b,you are incorrect.My solution:Consider$n=x_1+x_2+..+x_k$has l  distinct values$\rightarrow n\geq 1+2+..+l\rightarrow l<\sqrt{2n}\rightarrow q(n)<p(n)\sqrt{2n}$


[quote="FightTheTide"]Let $ f(n) \equal{} \lfloor{\frac {n}{1}}\rfloor \plus{} \lfloor{\frac {n}{2}}\rfloor \plus{} \lfloor{\frac {n}{3}}\rfloor \plus{} \lfloor{\frac {n}{4}}\rfloor \plus{} \lfloor{\frac {n}{5}}\rfloor \plus{} ...$

Compute $ \displaystyle\sum^{10}_{k \equal{} 1}f(k!) \minus{} \displaystyle\sum^{10}_{k \equal{} 1}f(k! \minus{} 1)$[/quote]

Note that $ \lfloor{\frac {n}{k}}\rfloor$ is the number of multiples of k in the interval [1, n]. Then
$ \lfloor{\frac {n}{k}}\rfloor \minus{} \lfloor{\frac {n \minus{} 1}{k}}\rfloor$ is 1 if k divides n and 0 otherwise. So, the sum becomes the sum of the number of divisors of k! with k from 1 to 10. This could be computed by brute force now, is there a better way though?

EDIT: Brute forced and got 647, might have done something wrong there though.


vector
inequalities
advanced fields
advanced fields solved

Prove that if M in Cnxn is hermitian positive definite then
$\phi(x) = \sqrt{x^{*}Mx}$ is a norm.

Now I proved everythin except $\phi(x+y)\leq \phi(x)+\phi(y)$
$x^*$ is the hermitian conjugate of course
hermitian positive definite means that for any vector x, $\langle x^{*}Mx \rangle \ge 0$ (by definition) and has some other properties...

Is it ok (by definitions) if I prove that
$\langle x,y \rangle = x^{*}My$ is an inner product and then get it's square... $\sqrt{\langle x,x \rangle} =$ ||x||

thanx.

[quote="alexrait1"]Prove that if M in Cnxn is hermitian positive definite then
phi(x) = sqrt(x*Mx) is a norm.

Now I proved everythin except phi(x+y)<=phi(x)+phi(y)
x* is the hermitian conjugate of course
hermitian positive definite means that for any vector x, x*Mx>0 (by definition) and has some other properties...

Is it ok (by definitions) if I prove that
<x,y> = x*My is an inner product and then get it's square... sqrt(<x,x>) = ||x||

thanx.[/quote]

But why do you want to prove triangular inequality, if you can just prove that the inner product that you define is actually an inner product, what is kind of trivial. Let's see, the properties to be an inner product are:

1. Linearity on the first component component (trivial, comes from the linearity of the multiplication by $M$ and the linearity of the conjugation)
2. $\langle x,y\rangle=\overline{\langle y,x\rangle}$. (This one requires a little of work, but it is easy)
3. $\langle x,x\rangle\geq0$ and $\langle x,x\rangle=0\Leftrightarrow x=0$. That is the definition to be positive definite!

If it is an inner product, the norm property is for free!


Evaluate $\int^a_0\int^b_0e^{\max(b^2x^2,a^2y^2)}dydx$, where $a$ and $b$ are positive.


calculus
integration
analytic geometry
graphing lines
slope
function
absolute value

Let $ f(x)\equal{}x^2\plus{}3$ and $ y\equal{}g(x)$ be the equation of the line with the slope $ a$, which pass through the point $ (0,\ f(0))$ .
Find the maximum and minimum values of $ I(a)\equal{}3\int_{\minus{}1}^1 |f(x)\minus{}g(x)|\ dx$.

since its symmetric, we only need to consider $ a\ge 0$

$ a\in [0,1]$:

$ I(a)\equal{}3\int_{\minus{}1}^1 |x^2\minus{}xa| dx\equal{}3\int_{\minus{}1}^0 (x^2\minus{}ax) dx \minus{}3\int_0^a (x^2\minus{}ax) dx \plus{} 3\int_a^1 (x^2\minus{}ax) dx$

$ \equal{}\left[ \frac{2x^3\minus{}3ax^2}{2}\right]_{\minus{}1}^0\minus{}\left[ \frac{2x^3\minus{}3ax^2}{2}\right]_0^a\plus{}\left[ \frac{2x^3\minus{}3ax^2}{2}\right]_a^1$

$ \equal{} \frac{2\plus{}3a}{2}\plus{}a^3\plus{}\frac{2\minus{}3a}{2}\equal{}2\plus{}a^3$

$ a >1$

$ I(a)\equal{}3\int_{\minus{}1}^1 |x^2\minus{}xa| dx\equal{}3\int_{\minus{}1}^0 (x^2\minus{}ax) dx \minus{}3\int_0^1 (x^2\minus{}ax) dx$

$ \equal{}\left[ \frac{2x^3\minus{}3ax^2}{2}\right]_{\minus{}1}^0\plus{}\left[ \frac{2x^3\minus{}3ax^2}{2}\right]_{1}^0\equal{}\frac{2\plus{}3a}{2}\plus{}\frac{\minus{}2\plus{}3a}{2}\equal{}3a$

so the min is at $ a\equal{}0$ with $ I(a)\equal{}2$, and the max is as $ |a|\to \infty$ have $ I(a) \to \infty$

Please check the equation of $ g(x)$. :lol:

Note that $ g(x)\equal{}ax\plus{}3$ , and define $ h(a)\equal{}\int^{1}_{\minus{}1}|f(x)\minus{}g(x)|dx\equal{}\int^{1}_{\minus{}1}|x(x\minus{}a)|dx$

case 1: $ a>1$ ;
$ h(a)\equal{}\int^{0}_{\minus{}1}x(x\minus{}a) dx\plus{}\int^1_{0}x(a\minus{}x)dx\equal{}a$

case 2: $ 0<a\leq 1$ ;
$ h(a)\equal{}\int^{0}_{\minus{}1}x(x\minus{}a)dx\plus{}\int^a_{0}x(a\minus{}x)dx\plus{}\int^1_{a}x(x\minus{}a)dx\equal{}\frac{2(1\plus{}a^3)}{3}$

case 3: $ \minus{}1<a\leq 0$ ;
$ h(a)\equal{}\int^{a}_{\minus{}1}x(x\minus{}a)dx\plus{}\int^{0}_{a}x(a\minus{}x)dx\plus{}\int^{1}_{0}x(x\minus{}a)dx\\\equal{}\frac{2(a^3\plus{}1)}{3}\plus{}\frac{a\minus{}2a^3\minus{}1}{2}\equal{}\frac{\minus{}2a^3\plus{}3a\plus{}1}{6}$

case 4: $ a\leq \minus{}1$ ;
$ h(a)\equal{}\int^{0}_{\minus{}1}x(x\minus{}a)dx\plus{}\int^{1}_{0}x(a\minus{}x)dx\equal{}a$

then $ a\rightarrow\infty$ , $ I(a)\rightarrow\infty$
$ a\rightarrow\minus{}\infty$ , $ I(a)\rightarrow\minus{}\infty$

well like the guy above me shows, $ g(x)\equal{}ax\plus{}3$

but im pretty sure you'll never get a negative value for $ I(a)$ since its integration of positive function (absolute value)

maybe i scrubbed the integration in case |a|<1?


:? I was just wandering if I was the only one who plays runescape
[hide]<http://www.runescape.com>[/hide]

uhm... you posted this at a wrong forum....


integration
trigonometry
calculus
calculus computations

Calculate:
\[ \int_{0}^{\frac{\pi}{2}}\cos (\pi\sin^{2}\theta)\;d\theta\]

hello
Answer is $ 0$ as $ f(\frac {\pi}{4} \plus{} x) \equal{}\minus{} f(\frac {\pi}{4} \minus{} x)$
thank u


un problema que quiero ver varias soluciones  :D

en una cuadricula de $nxn$ se escriben los numeros del a al $n^{2}$ en el orden habitual (de izquierda a derecha y de arriba hacia a abajo)
por ejemplo para $n=3$:

   $1-2-3$
   $4-5-6$
   $7-8-9$

llamamos camino en la cuadricula a una sucesion de pasos de un cuadro a otro desde el cuadro que tiene el numero $1$ hasta el que tiene el numero $n^{2}$ , de tal manera que en cada paso el movimiento sea hacia la derecha o hacia abajo . si $C$ es un camino, denotamos por $L_{(}C)$ a la suma de los numeros por los que pasa el camino $C$.

(i) sea $M$ la mayor $L_{(}C)$ que se pueda obtener de entre todos los caminos $C$ en una cuadricula fija de tamaño $nxn$  y sea $m$ la menor $L_{(}C)$ . pruebe que $M-m$ es un cubo perfecto .

(ii) pruebe que en ninguna cuadricula hay un camino $C$ tal que $L_{(}C)=1996$

Oooooooh!!!! Un nacional de Mexico! Este problema pertenece al nacional que se celebro aqui en Merida Yucatan (donde fue la IMO 2005).
Es un bonito problema!!
 :D

[quote="mhuarancca"]que paso con las soluciones al problema  :lol:[/quote]
bueno creo que la primera parte es asi,
yo considerare $n\geq 2$, ya que para $1$ , no se como seria.
ahora demotraremos que $L(C)$ maximo es la suma de todos los numeros de la 1era columna en la cuadricula, mas los ultimos numeros de la ultima fila, osea
$L(C)=(1+n+1+2n+1+...+1+n^{2}-n)+(n^{2}-n+2+n^{2}-n+3+...+n^{2})$
$L(C)=n+n^{2}(n-1)/2+(n^{2}-n)(n-1)+n(n+1)/2-1...(i)$
dem.
como apartir de la casilla $K$ (arbitraria), podemos ir solo para derecha o abajo, entonces notemos que si vamos para abajo, ese numero sera mayor que si vamos para la derecha. entonces si iniciamo en el numero $1$, tendremos que bajar hasta la ultima y de ahi solo nos keda la posibildad de ir hacia la derecha por lo tanto $L(C)$ es maximo,
---
ahora analogamente $L(C)$ minimo es cuando va primero de $1$, hacia la derecha hasta la casilla $n$, y de ahi hacia abajo:
$L(C)'= (1+2+3+..+n)+(2n+3n+..+n^{2})$
$L(C)'= n(n+1)/2+n^{2}(n+1)/2-n..(ii)$
ahora restando (i) y (ii) obtenemos lo que nos piden que si bien mis calculos no fallan sale:
$L(C)-L(C)'=(n-1)^{3}$--> demotrando asi que es un cubo.
ahora para la segunda parte no se aun no la intento, supongo que debe ser acotando, o si no analisar en algun modulo conventientemente tal que cumpla con todas y no con este numero.
bueno chaufas espero que posteen la 2da parte,

De la primera parte, obtenemos:
$\frac{3n^{3}}{2}\geq M\geq L(C)\geq m\geq \frac{n^{3}}{2}$
Supongamos que exista $C$ con $L(C)=1996$, entonces $11\leq n\leq15$.
Sea $1=x_{1},x_{2},\dots,x_{2n-1}=n^{2}$ la secuencia que define $C$.
Se puede ver que $x_{k}\cong k mod(n-1)$ y entonces
$1996=L(C)=\sum_{k=1}^{2n-1}x_{k}\cong\sum_{k=1}^{2n-1}k mod(n-1)\cong 1 mod(n-1)$
Lo cual es falso para $11\leq n\leq15$.


$\bigstar$ Find the greatest of the expression:
$N=\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}-\frac{abcd}{(ab+cd)^{2}}$
when $a,b,c,d$ are distinct real numbers satisfying the condition: $ac=bd$ and $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}$

I think above inequality very interesting. And the equaly hold on many case. Now, let try it ... I am waiting a nice solution from you :!:

[quote="kunny"]$ac=bd\Longrightarrow \frac{a}{d}=\frac{b}{c}$ or $\frac{a}{b}=\frac{d}{c}$, so Is $a^{2}=d^{2}$ or $c^{2}=d^{2}$?[/quote]
I misunderstand you, kunny  :roll: What does your mean ?
I am sure above inequality is true  :P Because it was solved by me. And I have another solution. This is toanhocmuonmau's solution.
@my brother: How to solve it, hungkhtn?    I am sure you will solve it easily  :wink:

O.K.

$N=2\left(\frac{c}{a}+\frac{a}{c}\right)-\frac{abcd}{(ad+bc)^{2}}$.

$\left|\frac{c}{a}+\frac{a}{c}\right|\geq 2,\ (ad+bc)^{2}\geq 4abcd$.


this is almost like 2 different questions...

$k:=2x-y$.  Then $(1+4^k)(5^{1-k})=1+2^{k+1}$.

So $(1+2^{2k})(2^2+2^0)^{1-k}=2^0+2^{k+1}$.  For $k$=1, equality holds.

$f(x):=(1+2^{2k})(2^2+2^0)^{1-k}-2^0+2^{k+1}$
$f ' (x)$ = some mother that's always negative, so that means the functions monotinically decreasing, and therefore $k=1$ is only solution.  So 2x-y=1.

for the 2nd one see that $(0,-1)$ is a solution and then use $x=(1+y)/2$, then differentiate and show that the derivative is always positive and hence the functions monotonically increasing for all $y$, so that $(0, -1)$ is the only solution.

Hmm... it looks like it'd be rather difficult to find $f'(x)$

if by difficult you mean computation-intensive, i guess thats true but once you do it out its obvious it's negative.  i didnt do it by hand, i used ti-89 but it can be done for sure.

on another note, i think it's kind of a weird system, like they could be two completely different questions.  The first equation could be a question on its own, just solve for $x$ in terms of $y$.

Actually I'd like to know what's the result after differentiating (the one you say it's obviously negative). Do you still have it?

What of complex solutions? The questions dosnt say 'real values', and your (wonderful it must be said!) approach breaks down....

well if nothing is said it usually implies real solutions.

i will think about it later but if we're looking at complex solutions i think we should first work at the 2nd equation because the $\ln{y^2+2x}$ limits the possibilities.

if it's complex no's i think its pretty complicated.  its safe to assume real numbers.


On a recent AIME problem I did, the solution to a geometry problem used "mass points", and said it was a powerful tool. I tried searching it up but didn't quite understand. Can someone explain to me how it works and how it is used? A simple example of usage would be appreciated!

Some Googling turned up this, the handout from the very lecture where I learned about mass-point geometry: http://mathcircle.berkeley.edu/mpgeo.pdf

It's nice way to learn, as it has a lot of exercises.

Well, it is quite powerful tool. I did a paper on the subject, but it is in Bosnian, so it wouldn't be of much use, would it...

As LydianRain pointed out, here's a great resource:

[url]http://mathcircle.berkeley.edu/mpgeo.pdf[/url]

I enjoyed the text, although it could be little bit confusing for someone new to the subject. Nevertheless that shouldn't stop you from analyzing it. I recommend you to see also:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=261353#261353[/url]

and search ML for mass point geometry. Maybe tomorrow I'll have more time to write about mass points: very interesting subject indeed. BTW, after getting the basics on MP Geometry, you can proceed to barycentric coordinates...


In any triangle ABC  and  for any  [tex] x,y,z >0 [/tex]  show that

[tex]\Sigma\frac{xa^2(b+c-a)^2}{y+z} \geq8S^2  [/tex]
Thanks!

My sol:
Put [tex]
m = a + b - c
[/tex]
[tex]n=b+c-a[/tex]
[tex]p=c+a-b[/tex]
then the question becomes proving
[tex]
\frac{x}{{y + z}}n^2 (m + p)^2  + \frac{y}{{x + z}}p^2 (m + n)^2  + \frac{z}{{x + y}}m^2 (n + p)^2  \ge 2mnp(m + n + p)
[/tex]

Add [tex]
n^2 (m + p)^2  + p^2 (m + n)^2  + m^2 (n + p)^2
[/tex]
 in both sides,
[tex]
 \Leftrightarrow \left( {x + y + z} \right)\left[ {\frac{1}{{y + z}}n^2 (m + p)^2  + \frac{1}{{x + z}}p^2 (m + n)^2  + \frac{1}{{x + y}}m^2 (n + p)^2 } \right] \ge 2mnp(m + n + p) + n^2 (m + p)^2  + p^2 (m + n)^2  + m^2 (n + p)^2
[/tex]

[tex]
L.H.S. \ge \frac{1}{2}\left[ {n(m + p) + p(m + n) + m(n + p)} \right]^2
[/tex]by cauchy
[tex]
\begin{array}{l}
  = 2(mn + np + mp)^2  \\
  = (mn + np)^2  + (np + mp)^2  + (mn + mp)^2  + 2\left[ {\left( {mn} \right)\left( {np} \right) + \left( {mn} \right)\left( {mp} \right) + \left( {mp} \right)\left( {np} \right)} \right] \\
  = n^2 (m + p)^2  + p^2 (n + m)^2  + m^2 (n + p)^2  + 2mnp(m + n + p) \\
 \end{array}
[/tex]
proven.

Edit: let me try to tex my solution :)

Here is my solution:

Denoting $s=\frac12 \left( a+b+c\right) $, we have $b+c-a=2\left( s-a\right) $, and thus $\left( b+c-a\right) ^{2}=4\left( s-a\right) ^{2}$. Hence your inequality

$\frac{xa^{2}\left( b+c-a\right) ^{2}}{y+z}+\frac{yb^{2}\left( c+a-b\right) ^{2}}{z+x}+\frac{zc^{2}\left( a+b-c\right) ^{2}}{x+y}\geq 8S^{2}$

rewrites as

$\frac{xa^{2}\cdot 4\left( s-a\right) ^{2}}{y+z}+\frac{yb^{2}\cdot 4\left( s-b\right) ^{2}}{z+x}+\frac{zc^{2}\cdot 4\left( s-c\right) ^{2}}{x+y}\geq 8S^{2}$,

i. e. as

$\frac{xa^{2}\left( s-a\right) ^{2}}{y+z}+\frac{yb^{2}\left( s-b\right) ^{2}}{z+x}+\frac{zc^{2}\left( s-c\right) ^{2}}{x+y}\geq 2S^{2}$.

Well, if we denote y + z = X, z + x = Y and x + y = Z, then $x=\frac{1}{2}\left( Y+Z-X\right) $, $y=\frac{1}{2}\left( Z+X-Y\right) $, $z=\frac{1}{2}\left( X+Y-Z\right) $. Hence the inequality we have to prove turns out to be equivalent to

$\frac{\frac{1}{2}\left( Y+Z-X\right) a^{2}\left( s-a\right) ^{2}}{X}+\frac{\frac{1}{2}\left( Z+X-Y\right) b^{2}\left( s-b\right) ^{2}}{Y}+\frac{\frac{1}{2}\left( X+Y-Z\right) c^{2}\left( s-c\right) ^{2}}{Z}\geq 2S^{2}$,

i. e. as

$\frac{\left( Y+Z-X\right) a^{2}\left( s-a\right) ^{2}}{X}+\frac{\left( Z+X-Y\right) b^{2}\left( s-b\right) ^{2}}{Y}+\frac{\left( X+Y-Z\right) c^{2}\left( s-c\right) ^{2}}{Z}\geq 4S^{2}$.

This can be rewritten as

$\frac{Y+Z-X}{X}a^{2}\left( s-a\right) ^{2}+\frac{Z+X-Y}{Y}b^{2}\left( s-b\right) ^{2}+\frac{X+Y-Z}{Z}c^{2}\left( s-c\right) ^{2}\geq 4S^{2}$,

or, equivalently, as

$\left( \frac{Z}{Y}b^{2}\left( s-b\right) ^{2}+\frac{Y}{Z}c^{2}\left( s-c\right) ^{2}\right) +\left( \frac{X}{Z}c^{2}\left( s-c\right) ^{2}+\frac{Z}{X}a^{2}\left( s-a\right) ^{2}\right) $
$+\left( \frac{Y}{X}a^{2}\left( s-a\right) ^{2}+\frac{X}{Y}b^{2}\left( s-b\right) ^{2}\right) -a^{2}\left( s-a\right) ^{2}-b^{2}\left( s-b\right) ^{2}-c^{2}\left( s-c\right) ^{2}\geq 4S^{2}$.

So we must prove this inequality. Now, AM-GM yields

$\frac{Z}{Y}b^{2}\left( s-b\right) ^{2}+\frac{Y}{Z}c^{2}\left( s-c\right) ^{2}\geq 2\sqrt{\frac{Z}{Y}b^{2}\left( s-b\right) ^{2}\cdot \frac{Y}{Z}c^{2}\left( s-c\right) ^{2}}$
$=2b\left( s-b\right) c\left( s-c\right) $,

and similarly,

$\frac{X}{Z}c^{2}\left( s-c\right) ^{2}+\frac{Z}{X}a^{2}\left( s-a\right) ^{2}\geq 2c\left( s-c\right) a\left( s-a\right) $;
$\frac{Y}{X}a^{2}\left( s-a\right) ^{2}+\frac{X}{Y}b^{2}\left( s-b\right) ^{2}\geq 2a\left( s-a\right) b\left( s-b\right) $,

so that

$\left( \frac{Z}{Y}b^{2}\left( s-b\right) ^{2}+\frac{Y}{Z}c^{2}\left( s-c\right) ^{2}\right) +\left( \frac{X}{Z}c^{2}\left( s-c\right) ^{2}+\frac{Z}{X}a^{2}\left( s-a\right) ^{2}\right) $
$+\left( \frac{Y}{X}a^{2}\left( s-a\right) ^{2}+\frac{X}{Y}b^{2}\left( s-b\right) ^{2}\right) -a^{2}\left( s-a\right) ^{2}-b^{2}\left( s-b\right) ^{2}-c^{2}\left( s-c\right) ^{2}$
$\geq 2b\left( s-b\right) c\left( s-c\right) +2c\left( s-c\right) a\left( s-a\right) +2a\left( s-a\right) b\left( s-b\right) $
$-a^{2}\left( s-a\right) ^{2}-b^{2}\left( s-b\right) ^{2}-c^{2}\left( s-c\right) ^{2}$.

But a calculation shows that

$2b\left( s-b\right) c\left( s-c\right) +2c\left( s-c\right) a\left( s-a\right) +2a\left( s-a\right) b\left( s-b\right) $
$-a^{2}\left( s-a\right) ^{2}-b^{2}\left( s-b\right) ^{2}-c^{2}\left( s-c\right) ^{2}=4S^{2}$.

Thus, we get

$\left( \frac{Z}{Y}b^{2}\left( s-b\right) ^{2}+\frac{Y}{Z}c^{2}\left( s-c\right) ^{2}\right) +\left( \frac{X}{Z}c^{2}\left( s-c\right) ^{2}+\frac{Z}{X}a^{2}\left( s-a\right) ^{2}\right) $
$+\left( \frac{Y}{X}a^{2}\left( s-a\right) ^{2}+\frac{X}{Y}b^{2}\left( s-b\right) ^{2}\right) -a^{2}\left( s-a\right) ^{2}-b^{2}\left( s-b\right) ^{2}-c^{2}\left( s-c\right) ^{2}\geq 4S^{2}$,

completing our proof.

BTW, if one takes the triangle ABC to be equilateral (so a = b = c), then your inequality becomes the well-known Nesbitt inequality. So it's a nice generalization of Nesbitt...

  Darij


abstract algebra
linear algebra
matrix
Ring Theory
group theory
superior algebra
superior algebra unsolved

Let $ A$ be an additive Abelian group generated by elements $ x, y, z, w$ satisfying the relations
$ 3x \plus{} y \plus{} w \equal{} 0$
$ x \minus{} y \plus{} z \plus{} 3w \equal{} 0$
$ \minus{}2x \plus{} 2y \plus{} 3z \equal{} 0$
What is $ A$ as a product of cyclic groups?

First of all, what exactly does this question mean? I am guessing it means that $ A$ should be such that every other abelian group generated by four elements satisfying the relations is the homomorphic image of $ A$.

Now, based on this, how do you proceed? By googling, I know the torsion coefficients are given by the Smith normal form, but I don't know why. How do you prove this? Also, how do I know how large the free part of the group is?

As a service, the Smith normal form of
$ \begin{bmatrix}3 & 1 & 0 & 1 \\ 1 & \minus{}1 & 1 & 3 \\ \minus{}2 & 2 & 3 & 0\end{bmatrix}$
is
$ \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0\end{bmatrix}$

For proof and everything, see Artin's chapter on modules. But I think that my explanation should give some intuition at least.

When we obtain the normal form, we add multiples of rows and columns. It is easy to see that adding rows doesn't change the relation. I.e.,
$ 3x\plus{}y\plus{}w\equal{}0\\
x\minus{}y\plus{}z\plus{}3w\equal{}0\\
\minus{}2x\plus{}2y\plus{}3z\equal{}0$ is equivalent to

$ 0\plus{}4y\minus{}3z\minus{}8w\equal{}0\\
x\minus{}y\plus{}z\plus{}3w\equal{}0\\
\minus{}2x\plus{}2y\plus{}3z\equal{}0$

But when we perform row operations, it is not as obvious why it works. What we do when we perform row operations is change variables (in this setting). Set $ \alpha\equal{}x\minus{}y\plus{}z\plus{}3w\equal{}0$. Every finite sum of $ x,y,z,w$ can be written uniquely as a sum of $ \alpha, y,z,3w$ (basically because the row/column operations have determinant $ \pm 1$).

Hence we can rewrite the relation we have as

$ 0\plus{}4y\minus{}3z\minus{}8w\equal{}0\\
\alpha\plus{}0\plus{}0\plus{}0\equal{}0\\
\minus{}2x\plus{}2y\plus{}3z\equal{}0$

And then we continue. If I made no mistakes, this group is generated by $ x\minus{}y\plus{}z\plus{}3w,z\minus{}8w\minus{}10z, 10y\plus{}19w,w$ with relations $ x\minus{}y\plus{}z\plus{}3w\equal{}z\minus{}8w\minus{}10z\equal{}0$, $ 2(10y\plus{}19w)\equal{}0$ and is hence $ \mathbb{Z}\times \mathbb{Z}/(2)$.

Normally, we might not be interested in the explicit generators, so we can just carelessly simplif the matrix and then see that the group is $ \mathbb{Z}\times \mathbb{Z}/(2)$.

I have the feeling I am terribly confusing. Am I?

I should probably add that the underlying idea is to view this group as the free module $ \mathbb{Z}^4$ quotient out by the image of  $ \begin{bmatrix}3 & 1 & 0 & 1 \\
1 & \minus{}1 & 1 & 3 \\
\minus{}2 & 2 & 3 & 0\end{bmatrix}^T$

When we've written the matrix on normal form, how can we directly see how big the free part should be (i.e. the betti number), without working the new generators as you did? As I wrote to you in a PM, I thought that the Betti number would always be the difference between the number of generators and the number of coefficients appearing in the Smith normal form. I am reading Jacobson, and following the proof of the Structure theorem for finitely generated modules over a PID, this seems true.

I found an example online (written by a professor) that contradicts this idea, but I think it may be wrong. Can someone else please give their opinion?


Euler
vector
advanced fields
advanced fields unsolved

Let $ X$ be a projective scheme over a field $ k$, and let $ F$ be a coherent sheaf on $ X$. We define the Euler characteristic of $ F$ by: $ \chi(F) \equal{} \sum ( \minus{} 1)^i dim _k H^i (X,F)$.
If $ 0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow 0$ is a short exact sequence of coherent sheaves on $ X$, show that:
$ \chi(F) \equal{} \chi(F') \plus{} \chi(F'')$

[hide]the only interesting thing I get from hypothesis is that $ dim _k H^i (X,F)$ are finite-dimesional (theorem 5.2 of III in Hartshorne) and I have no idea what to do next; certainly, one needs to look at long exact sequence associated with cohomology though I cannot take out anything from it)  :maybe:  [/hide]

[hide="Hint"]If $ 0\rightarrow V_1\rightarrow \ldots\rightarrow V_n\rightarrow 0$ is an exact sequence of finite dimensional vector spaces, then $ \sum_m(\minus{}1)^m\dim V_m\equal{}0$.[/hide]

Thank you, your hint solves the problem (as cohomology groups vanishes from certain point and are finite-dimensional vector spaces so we can use the fact you've given) so simple and cute :)


function
complex analysis
complex analysis unsolved

Show that there is no function $ f$ analytic on $ \mathbb{C} \minus{} \left\{0\right\}$  such that \[ \left|f(z)\right| \geq \frac{1}{\sqrt{\left|z\right|}}\] for all $ z \neq 0$.

Just consider $ 1/f$, which has removable singularity at the origin and grows too slowly at $ \infty$ to be non-constant.

Yes, I should have said to reserve it to beginners  :wink:


probability
calculus
integration
logarithms
limit
calculus computations

Recently in HSB there was a problem about the probability of three numbers chosen from the interval (0,2) having a product less than 1.

An extension to you all in the calculus forum:
Derive a general expression (will probably require multiple integrals) for the probability of $ n$ numbers evenly distributed in (0,2) to be less than 1.

That is, for $ x_1 \ldots x_n \in (0,2)$, find the probability that $ (\prod_{i \equal{} 1}^{n} x_i) < 1$

Let $ Y_i\equal{}\ln X_i$ $ Y_i$ has distribution on $ (\minus{}\infty,\ln 2)$ that belongs to the exponential family. In particular,

$ E(Y_i)\equal{}E(\ln(X_i))\equal{}\frac12\int_0^2\ln x\,dx\equal{}\ln 2\minus{}1$

Note that that is negative; this will be important.

$ \ln\left(\prod_{i\equal{}1}^nX_i\right)\equal{}\sum_{i\equal{}1}^nY_i.$

$ P\left(\prod_{i\equal{}1}^nX_i<1\right)\equal{}P\left(\sum_{i\equal{}1}^nY_i<0\right)\equal{} P\left(\frac1n\sum_{i\equal{}1}^nY_i<0\right)$

Since $ 0$ is greater than the common means of the i.i.d. $ Y_i,$ the Weak Law of Large Numbers will show that

$ \lim_{n\to\infty}P\left(\frac1n\sum_{i\equal{}1}^nY_i<0\right)\equal{}1$ and hence $ \lim_{n\to\infty}P\left(\prod_{i\equal{}1}^nX_i<1\right)\equal{}1.$

That isn't what you asked for - you asked for a formula in terms of $ n$ for what this limit should be - but we know one property such a formula should have.

In summary, could it be said that the sequence of probabilities will approach 1 as more numbers are added? This makes intuitive sense, I suppose. That's a nice proof of it, though.

The sum of identical independent exponential distributions will be a gamma distribution. The formula you seek is a partial probability of a gamma distribution - that is, the cumulative distribution (or its complement) of a gamma distribution evaluated at a specific point.


For some reason my ti-83+ has a weird display problem.  I didn't do anything funny to it or something like that but was merely using it when it was low on batteries.  When the batteries got really low, my screen started showing weird stuff like floating symbols and shaded boxes in place of numbers.  Not a function displayed correctly nor have I seen a complete word show up.  When i replaced the batteries, it continued acting weird. After a while of messed up display the whole screen would go white and would display nothing at all..

The only thing i remembered doing wrong is placing my calculator in my jacket and then placing the jacket on top of a moderately warm heater. It was working for around 48 hours from that point so i don't think that is the problem.

Things i have tried-
1. Cleared All Ram
2. Imported another OS

Nothing works, does anyone knows the cause of this problem or how to fix this problem?

Thanks!

Yes, my friend has a similar problem, but it might be caused due to the heater... :wink: .

Describe the effects of wierd symptoms.

If you can read numbers and stuff off of it I see no reason why you should get another calcluator lol  :lol:

Holy moly!! A calculator will get messed up because of the some heat? This suck!! Anyways I CAN"T see the numbers. Not a single word displays correctly...

Howd you clear the ram and get a new OS then?

My sister has her own calculator so i transfered OS by holding Delete key and using the battery trick.

I cleared my ram because my calculator is functioning as in you can close it and open it, but display is messed up.

Extensive heat can cause the pixels to change, but it should not mess with the actual OS.

When you turn your calculator on, are you able to type stuff in?  Also, are the same characters shown each time you press a key?  Can you run programs or applications?

I'd think a RAM clear would do the trick, but I'm not sure...

EDIT:  try [url=http://www.cemetech.net/forum/viewtopic.php?p=60067#60067]here[/url] for more help.

Its a calculator based forum.

Try this: turn calc off, while holding the delete button, take out one battery, and then replace it, all while holding delete.  Then turn it on and it should say something.  If that doesn't work try the same thing but with holding the clear button.


Μου ήρθε μία σκέψη γιατί πολύ μου άρεσε το προβληματάκι και δεν το είχα ξαναδεί... κατά καιρούς, έχουν μπει στο forum μερικά πολύ ωραία προβληματάκια πανεπιστημιακής ανάλυσης. Μερικά είναι well known αλλά ορισμένα σίγουρα δεν είναι και (προσωπικά τουλάχιστον) τα βρίσκω πολύ ενδιαφέροντα facts/pathologies/counterexamples.

Τί λέτε? Σε ό,τι απομένει απο το spare time μας (spare time... yeah, right  :rotfl:), τα μαζεύουμε να τα κάνουμε μια συλλογή a la counterexamples in analysis? Πολλά από όσα έχουν μπει είναι αρκετά πιο ενδιαφέροντα από τα counterexamples... και σε κάθε περίπτωση θα είναι fun. Δεν έχω στο μυαλό μου μεγάλα σχέδια, βιβλία και τέτοια πράγματα, αλλά μου φαίνεται πολύ ωραίο για side project "ελεύθερου χρόνου". Πίεση χρόνου δε θα έχουμε, πλάκα θα έχει (επίσης, ίσως μάθω επιτέλους λίγη ανάλυση in the process!) και όταν τελειώσει μπορεί να έχει βγει κάτι όμορφο.

So... what do you guys think?

Cheerio,

Durandal 1707

[quote="Durandal"]Anyway, off topic μου ήρθε μία σκέψη γιατί πολύ μου άρεσε το προβληματάκι και δεν το είχα ξαναδεί... κατά καιρούς, έχουν μπει στο forum μερικά πολύ ωραία προβληματάκια πανεπιστημιακής ανάλυσης. Μερικά είναι well known αλλά ορισμένα σίγουρα δεν είναι και (προσωπικά τουλάχιστον) τα βρίσκω πολύ ενδιαφέροντα facts/pathologies/counterexamples.

Τί λέτε? Σε ό,τι απομένει απο το spare time μας (spare time... yeah, right  :rotfl:), τα μαζεύουμε να τα κάνουμε μια συλλογή a la counterexamples in analysis? Πολλά από όσα έχουν μπει είναι αρκετά πιο ενδιαφέροντα από τα counterexamples... και σε κάθε περίπτωση θα είναι fun. Δεν έχω στο μυαλό μου μεγάλα σχέδια, βιβλία και τέτοια πράγματα, αλλά μου φαίνεται πολύ ωραίο για side project "ελεύθερου χρόνου". Πίεση χρόνου δε θα έχουμε, πλάκα θα έχει (επίσης, ίσως μάθω επιτέλους λίγη ανάλυση in the process!) και όταν τελειώσει μπορεί να έχει βγει κάτι όμορφο.

So... what do you guys think?

Cheerio,

Durandal 1707[/quote]

panagioti polu kali idea kai sigoura tha vgei kati omorfo kai xrhsimo! tha ithela na voithiso an kai se kamia periptosi den eimai sto epipedo to diko sas (durandal, cmad, demetres...klp..r_boris,xp, stergiu...). opote lets start!

Για την ιδέα του Παναγιώτη αυτό που θα μπορούσαμε να κάνουμε είναι ένα sticky με "Παραδείγματα/Αντιπαραδείγματα Ανάλυσης" και σιγά σιγά να παιρνάμε όλες αυτές τις παθολογιές που έχουμε συζητήσει. Όποιος βλέπει μια παλια ενδιαφέρουσα συζήτηση που δεν έχει περαστεί ακόμα μπορεί σιγά σιγά να την περνάει. Μερικές φορές θα χρειαστεί να αφήνει αρκετά πράγματα πίσω (αρκετές φορές απαντούσαμε τελείως λάθος μέχρι να καταλάβουμε τι ζητάει η άσκηση) και άλλες φορές θα χρειαστεί να βάζει λίγα περισσότερα πράγματα (π.χ. να επεκτείνει λίγο την μονολεκτική μου απάντηση εδώ.)  Στο μέλλον συζητάμε τα προβλήματα σε καινούργια topics και όσα τα λύνουμε τα παιρνάμε στο sticky.

Επίσης, για να μην χαλάμε την ροή, καλό θα ήταν να βάζαμε ένα link που να λέει που συζητήθηκε το θέμα ώστε αν κάποιος έχει αποριές να ρωτάει στο αντίστοιχο topic αντί στο sticky.

Δεν χρειάζεται να κρατάμε κάποια λογική σειρά. Δηλαδή μπορεί αύριο να περάσουμε την κατασκευή για το θεώρημα του Zahorski, και σε μια βδομάδα να δούμε ότι πιο παλιά υπήρξε μια ωραία συζήτηση για τις συναρτήσεις Weirstrass ή άλλες παρόμοιες συναρτήσεις που αναγκαστικά θα περαστούν μετά.

Δεν ξέρω αν αυτό ακριβώς σκεφτόταν ο Παναγιώτης. Οποιαδήποτε σχόλια/παρατηρήσεις σε αυτά που έγραψα είναι καλοδεχούμενα.  :)

Βρίσκω ωραία την ιδέα του Durandal, αλλά

[quote="Durandal"](spare time... yeah, right  :rotfl:)[/quote]

Seconded... Ο μόνος λόγος που μπαίνω τις τελευταίες μέρες είναι για "ξελαμπικάρισμα" από το τρελά φορτωμένο πρόγραμμά μου. Θα προσπαθήσω να βοηθήσω όμως, στο βαθμό που μου επιτρέπουν οι γνώσεις και ο χρόνος μου.

Είναι πολύ εύκολο να δημιουργηθεί ένα sticky και σε αυτό να βάλουμε όποια ποστς από κάθε άλλο τόπικ που θέλουμε (υπαρχει η δυνατότητα tou splitting) . Επομένως θα το ανοίξω εγώ σε πρώτη φάση και απλά αν μπορείτε στείλτε μου με pm το πρόβλημα και θα το εισάγωγω στο sticky .

[quote="Durandal"]
Η ιδέα που προτάθηκε για stickies από τον Dem είναι cool, ιδίως για την οργάνωση του υλικού [i]στο forum[/i]. Ιδανικά, θα μπορούσαμε να βάλουμε και διαφορετικά stickies για κάθε field (π.χ. άλγεβρα ή τοπολογία), αλλά δεν είμαι σίγουρος αν αυτό θα δούλευε αρκετά καλά.

Τώρα, εγώ είχα στο μυαλό μου κάτι διαφορετικό που να απευθύνεται και σε ανθρώπους εκτός του forum. Μαζεύουμε (συνεργατικά ή το ξεκινάω κι εγώ ο ίδιος αλλά σιγάααα σιγά γιατί δεν έχω και πολύ χρόνο) όλα τα προβλήματα και τα TeXάρουμε properly. Everyone should feel free to add in/add on. Μετά τα κάνουμε compile σε ένα μεγάλο standalone pdf το οποίο μπορεί γίνεται updated περιοδικά στο forum αλλά και μπορεί να γίνει hosted ή distributed και με άλλους τρόπους.

Eventually, αν οργανώσουμε properly τη δομή του, π.χ. ακολουθώντας το μοντέλο των Counterexamples in XXX θα προκύψει έτσι μία πολύ ωραία συλλογή προβλημάτων και ενδιαφέροντων facts.

Τα παραπάνω είναι πολύ επιγραμματικά και off the top of my head γιατί κι εγώ δεν έχω κάτσει να σκεφτώ πώς θα μπορούσε να γίνει, αλλά ας ξεκινήσουμε τη συζήτηση εδώ και ας δούμε πού θα μας πάει :)

Cheerio,

Durandal 1707

Edit: να τονίσω ότι αυτό απευθύνεται [b][i]μόνο[/i][/b] για τις στιγμές που θα έχουμε όρεξη να ξελαμπικάρουμε λίγο όπως λέει και ο cmad με το να γράψουμε καθαρά ένα προβληματάκι ή ίσως λίγη θεωρία αλλά δεν υπάρχει κάτι καινούριο στο forum. Aφού μπαίνουν που μπαίνουν προβλήματα ούτως ή άλλως και η πραγματική δουλεια έχει γίνει (διατύπωση + λύση), σε βάθος (όχι πολύ) χρόνου, θα μπορούμε να έχουμε κάτι πραγματικά όμορφο... γι'αυτό μου κόλλησε η ιδέα :)[/quote]


geometry
ratio
angle bisector
geometry proposed

Let $ABCD$ ba a quadrilateral, and let $E=AB\cap CD,\ F=AD\cap BC,\ P=AC\cap BD$. The projection of $P$ on $EF$ is $Q$. Show that the projections of $Q$ on $AB,BC,CD,DA$ lie on a circle.

[You might notice that, after playing around with it for a while, we find it to be equivalent to another problem on the board :)]

My solution which is based on this lemma
   Lemma
     QP is the bisector of $\measuredangle{AQC}$
   This lemma can be proved easily. Let U, V, W, X be respectively the projections of Q on AD, BC, AB, CD. We will prove $\measuredangle{UWX} = \measuredangle{UVX}$.
    It is sufficient to prove  $\measuredangle{UWX}= 180 + \measuredangle{FED} - \measuredangle{FAQ}$.
  It is a nice one though.

Wow, Treegoner! The problem posted by Grobber is equivalent to the problem with the op02 project number China 7 (this is a problem of the Chinese IMO TST 2002). That China 7 problem asked to prove that < BQC = < AQD; this is easily seen to be equivalent to the problem Grobber posted.

I had tried that China 7 problem for some days, then I gave it up thinking it must be super-hard since it is Chinese. Now I see your five lines solution, Treegoner. Big work!

One little comment: That lemma you use follows from Theorem 1 at http://www.mathlinks.ro/Forum/viewtopic.php?t=6088 , applied to triangle DEF and the point B in its plane.

Cheers!

  Darij

Thanks for the replies. In order to show that $QP$ is the bisector of $\angle AQC$ (or $\angle BQD$), I used [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17]this[/url] problem.

This lemma also appeared in the INMO. I forget what year it was. :D

In Darij's prove isn't it enough to notice that after B'A''\A''C'=B'X\C'X, from Appolonius's theorem, RA'' is the bisector of C'RB'?

Yeah, it is enough, but I just didn't want to use Apollonius' theorem. By the way, did you know that many of all geometry textbooks where the Apollonius theorem is mentioned give incomplete proofs of this theorem? They all prove that, for two points A and B in the plane and an arbitrary positive number k, every point M with AM : BM = k lies on the circle with diameter PQ, where P and Q are the two points dividing the segment AB in the ratio k. But 50% of them don't show that conversely, every point M on the circle with diameter PQ satisfies AM : BM = k, and they don't even admit that this must be proven!

  Darij

I don't think it is difficult to prove $\angle AQP= \angle PQC$
If intersection point of $AC$ and $EF$ we call it $T$
$(TP,AC)=-1$
and $TQ$ is perpendicular to $QP$ so QP is angle bisector of $\angle AQC$

I don't think that's too difficult.

[quote="Omid Hatami"]I don't think it is difficult to prove $\angle AQP= \angle PQC$
If intersection point of $AC$ and $EF$ we call it $T$
$(TP,AC)=-1$
and $TQ$ is perpendicular to $QP$ so QP is angle bisector of $\angle AQC$[/quote]

Hmm, the last conclusion (that two mutually perpendicular harmonic rays with respect to an angle must be the angle bisectors of this angle) again must be proved. Don't forget that I was speaking about textbooks, and in most textbooks, the notion of Apollonius circles comes long before the definition of cross-ratios!

  Darij

we can prove all the things I used easily.

Grobber, your problem is not completely true.
  They may lie on a line.
   For example let ABCD  be a quadrilateral exscribed by the circle with diameter AC. :D

[quote="treegoner"]Grobber, your problem is not completely true.
  They may lie on a line.[/quote]

Okay, it is known that a line is a degenerate circle...

  Darij

[quote="treegoner"]
    Let U, V, W, X be respectively the projections of Q on AD, BC, AB, CD. We will prove $\measuredangle{UWX}= \measuredangle{UVX}$.
    It is sufficient to prove  $\measuredangle{UWX}= 180+\measuredangle{FED}-\measuredangle{FAQ}$.
 [/quote]

Can you explain here better?

To maggot,

I hope you are not asking me to explain the lemma I used because you can either find it in some post in ML or find its solution in Kalva (it was Indian Olympiad) :)

About the second line, it is just based on angle chasing. To prove a quadrilateral cyclic, often you need to either prove some two angles equal or sum of two angles equals $180^{0}$. With this direction, you will see why.  :)

[quote="treegoner"]

About the second line, it is just based on angle chasing. To prove a quadrilateral cyclic, often you need to either prove some two angles equal or sum of two angles equals $180^{0}$. With this direction, you will see why.  :)[/quote]

Yes, I know the direction well. However, it's the angle chasing I have problem about! Can you show how to "chase"?


I'm not 100% sure if this belongs here. Anway,

Given $x,y \in R$ and $\bar{x} = fl(x), \bar{y} = fl(y)$ (floating point numbers) then the relative roundoff errors for each number are:

$\frac{|x - \bar{x}|}{|x|} \leq \epsilon$ and  $\frac{|y - \bar{y}|}{|y|} \leq \epsilon$

Where $\epsilon = 2^{-53}$ is the machine precision.

Show that the relative error for the sum $\overline{(x+y)} = fl(\bar{x} + \bar{y})$ does not have a bound.

-----

Note: I think it's best to use: $\bar{x} = x + \epsilon_{x}$ and $\bar{y} = y + \epsilon_{y}$ where $\epsilon_{y}$ and $\epsilon_{x}$ are the absolute roundoffs.

Also, I think the key lies in knowing that addition is not associative in floating point arythmatic.

A bound for the product is:

$\frac{|xy - \overline{xy}|}{|xy|} \leq K\epsilon$

for some constant K.

Suppose $x=10^{20}$ and ${y=-10^{20}+1}$

Then $x+y=1$ but $\overline{\overline{x}+\overline{y}}=0.$

That's a pretty nasty relative error.


I have recently received a PM from an unknown person.

[quote="gilado"][i]Message Removed by Administrator, so as not to spread spam[/i].[/quote]

This, as I am considered, is complete spam, an exploitation of the Private Messaging feature. Could someone please tell me how I should respond if I receive a private message like this.

Thanks.

Sorry for any problems.

who sent you this? some new spammer? or someone over 100 posts?

I spelled the name wrong.

http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=25512

Registered a few seconds ago and started to spam my inbox. :roll:

EDIT: I stopped the spam by adding him/her to the foes list

Also, the problem is that that person has spammed white_horse's email+pm too. I think that he/she should be banned.

I am glad I don't give my email out in my profile. :(
They should be banned!.

Meh, we're discussing it in the mod forum. Just report it and delete it.

[quote="Valentin Vornicu"][quote="junggi"][quote="anirudh"]
EDIT: I stopped the spam by adding him/her to the foes list[/quote]
Does that actually work? :maybe:[/quote]Not yet. It will do something similar to that in the future.[/quote]

Oh. Its just that after putting on teh foes list, the spam stopped. That's why.

I got spammed by this person too, but I pmed Mr. Vornicu about it instead.


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope
perpendicular bisector

Given the following points:

$ K\equal{}(\minus{}4,\minus{}3),
L\equal{}(\minus{}3,4),
M\equal{}(\minus{}6,3),
X\equal{}(0,\minus{}5),
Y\equal{}(6,\minus{}3),
Z\equal{}(5,0)$

(a) describe a transformation that transforms triangle KLM onto triangle XZY
(b) where does this transformation send the point $ (\minus{}5,0)$?

Thanks.

[hide="part a"]
drawing and labeling the sides of the triangles shows you that the triangles have opposite orientations. Thus the transformation is a reflection with, possibly, a translation.

In order to see if the transformation is simply a reflection, we will find the midpoints of corresponding vertices. The midpoint of a preimage and the image always lies on the line of reflection. This is, however, not a sufficient condition. We must also have that the line of reflection is perpendicular to each line segment connecting corresponding vertices. (the perpendicular bisector of a preimage and the image is the line of reflection)

$ K \equal{} ( \minus{} 4, \minus{} 3), L \equal{} ( \minus{} 3,4), M \equal{} ( \minus{} 6,3), X \equal{} (0, \minus{} 5), Y \equal{} (6, \minus{} 3), Z \equal{} (5,0)$

The midpoint of $ KX$ is $ ( \minus{} 2, \minus{} 4)$
The midpoint of $ LZ$ is $ (1,2)$
The midpoint of $ MY$ is $ (0,0)$
These points are colinnear, and the line passing through them has slope $ 2$
The slopes of lines $ KX, LZ, MY$ are all $ \minus{} \frac {1}{2}$, so both conditions are satisfied.

The transformation is a reflection over the line $ y \equal{} 2x$
[/hide]

[hide="part b"]
It is not challenging to reflect $ ( \minus{} 5,0)$ over the line of reflection, but we can notice something else. The point that we are trying to reflect is the midpoint of $ KM$ Therefore, we only need to find the midpoint of $ XY$, the corresponding segment. This is simply the midpoint of $ (0, \minus{} 5); (6, \minus{} 3)$ which is $ \boxed{(3, \minus{} 4)}$
[/hide]

Brilliant, especially on part (b).

Thank you.


Define the sequence $ (x_n)_{n \geq 0}$ by :
1) $ x_0 \equal{} 0$ if and only if $ n \equal{} 0$ and
2) $ x_{n \plus{} 1} \equal{} x_{[\frac {n \plus{} 3}{2}]}^{2} \plus{} ( \minus{} 1)^nx_{[\frac {n}{2}]}^2$ for all $ n \geq 1$
  Find $ x_n$ is closed

you mean "proobe that $ \{x_n\ | \ n\ge 0\}$ is closed?"

or do you mean "find $ x_n$ in closed form"?

It's the latter, and it surely has nothing to do with calculus.  (Just compute the first five or six terms to figure out what's going on.)


What, if any, books do you use to work on technique?

For which instrument?  I use the [i]Pumping Nylon[/i] series for classical guitar, which is great, as far as I'm concerned.

I don't use books for building technique (piano this is...) I mainly just let the technique come to me as I learn harder songs.

At the moment for piano, it's nothing, I haven't done any real "technique" work for a long time, I probably should.  Many years ago when I had Chinese teachers I did Hanon and Czerny, but since then, all I've done is play through all the scales, but that was a year or two ago...

Do you mean like books that attempt to teach you how to play with correct technique?
Etudes don't exactly teach you technique...However those books that you may be talking about might not be helpful because sometimes it's kinda hard to explain in words.

I mean anything other than a piece.
For example, I also do:
scales and arpeggios
double stops
positions
mazas
kreutzer

hm...

well i have to play scales and arpeggios. i also have some hanon exercises and czerny etudes to do

I mainly do exercises from Hanon, Czerny, and Dohnanyi. I basically just do Dohnanyi for independence of the fingers.

I just do hanon. My teacher makes me do all these scales~ it makes my fingers hurt at times.

I can't understand that what does it mean by the technique books? Can you explain it a bit?

Well for violin I use a Tune a Day, Kreutzer, and Kayser.

technique books (scales/etudes):

Ones I've finished so far:
Kayser
Kreutzer
Fiorillo
Hrimaly
Shraedik
Dancla
Sevcik

Working on:
Bach 6 Sonatas& Partitas
Paganini 24 Caprices
Ysaye Sonatas

Sooo much stuff for violin (but I bet theres more for piano)

[quote="smartazn"]
Sooo much stuff for violin (but I bet theres more for piano)[/quote]

I'm not really familiar with violin repertoire, but I can certainly confirm that there exists a ton of etudes and exercises for piano.  Of course, it depends on how obscure a composer is picked, and what exactly is defined as a technique exercise.

Piano: Hanon 60 exercises.
Cello: An Organized Method of String Playing.

I use some weird scales book for piano.
For viola, I play Suzuki and Building Technique With Beautiful Music.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ A,B\in \mathcal{M}_{3}(\mathbb{C})$ ,prove that :
\[ \det(A \plus{} B) \minus{} \det(A \minus{} B) \equal{} 2\det(B) \plus{} 2Tr(A^*\cdot B)
\]

if M is a matrix in M(3,C) then its polynomial characteristic is

$ det(M\minus{}xI) \equal{} \minus{}x^{3}\plus{}tr(M)x^{2}\minus{}tr(M*)x\plus{}det(M)$

use this formula, and suppose $ B$ is inversible

if B not invertible use density of $ GL(3,C)$ in $ M(3,C)$

$ det(A\plus{}xB)\equal{}detBx^3\plus{}mx^2\plus{}nx\plus{}detA$
$ det(A\plus{}B)\equal{}detB\plus{}m\plus{}n\plus{}detA$
$ det(A\minus{}B)\equal{}\minus{}detB\plus{}m\minus{}n\plus{}detA$
$ det(A\plus{}B)\minus{}det(A\minus{}B)\equal{}2detB\plus{}2n$
How to prove that $ n\equal{}Tr(A^*\cdot B)$ ?

use coefficient of $ x$ is $ \minus{}tr(M^{*})$

$ det(M \minus{} xI) \equal{} \minus{} x^{3} \plus{} tr(M)x^{2} \minus{} tr(M*)x \plus{} det(M)$
and $ det(A \plus{} xB) \equal{} detBx^3 \plus{} mx^2 \plus{} nx \plus{} detA$
[quote="Moubinool"]use coefficient of $ x$ is $ \minus{} tr(M^{*})$[/quote] .
$ det(M \minus{} xI)$  is not  $ det(A \plus{} xB)$
About coefficients of $ det(A \plus{} xB)$ we don't know nothing ( or I don't know )

If $ A$ is invertible,then it is right.
Then consider $ A \plus{} \lambda I$ instead of $ A$,
Take the limit on the both side.(Because the $ \lambda$ such that \det ($ A \plus{} \lambda I) \equal{} 0$ is finite.)
Then get the result.


Chào các bác,em là lính mới  :P  :P  :P
 Chúng ta sẽ sưu tập các đề toán hay thành một tuyển tập bằng tiếng Việt các bài toán trên Mathlinks gồm đề bài,lời giải(Một Topic)
 Như thế sẽ tiện cho viêc theo rõi các bài vì em thấy nó đồ sộ quá,hơn nữa lại là tiênga Anh
 Các bác có thâm niên xin cho ý kiến

 Nếu bác nào phản đối thì đùng đánh em nhé! :D  :D  :D

Thats a good idea!! but we have been doing it.!!

chao` bac'  :D

[quote="Viet Math"]Chúng ta sẽ sưu tập các đề toán hay thành một tuyển tập bằng tiếng Việt các bài toán trên Mathlinks gồm đề bài,lời giải(Một Topic)[/quote]
y' kie^n' nay` ta.m xai` ddu*o*.c
nhu*ng cha('c dda^u co' tha(`ng na`o sie^ng na(ng nhu* va^.y
hay la` em tu*. la`m mo^.t mi`nh ddi  :)

đồng ý áh,nhưng hình như ko có ai siêng hết :D

Cảm ơn các bác nhiều!
 Thôi không nói nhiều nữa ta sẽ đi vào cụ thể :
 Việc xây dựng các bác tính sẽ làm ra sao bởi nếu không cẩn thận thì đây sẽ là một Diễn đàn toán học thu nhỏ mất
 Phải có qui định riêng :
 +Chỉ post bài toán và lời giải (Miễn bàn luận)
 +Mỗi bài post phải đảm bảo có đúng một bài toán và phải là toán trên MathLinks ,ghi rõ nguồn của đề và lời giải,có thể có nhiều lời giải
 +Ta sẽ chia ra 2 loại : Các bài toán đã có lời giải và các bài toán chưa có lời giải (Cái này dùng để anh em chiến đấu tiếp)
 +Các bài sẽ được đánh số thứ tự cho dễ tìm
 +Chia ra các phần : Bất đẳng thức
                               Số học
                               Đại số
                               Hình học
                               ............
  Mong các bác cho ý kiến  :?

Minh nghi y kien nay ko thiet thuc cho lam:Boi vi cac bai toan tren mathlinks thi moi ngay deu tang len-->Viec chon loc la rat kho khan(Corat nhieu bai toan hay ma kho co the suu tap het).Toi nghi viec ma chung ta co the lam la:Tham gia forum nay va doc chung ..ban co the hoc duoc rat nhieu dieu tu 1 bt du no ko hay cho lam... :D


integration
logarithms
calculus
LaTeX
calculus computations

-gt + u * ln (m0 / (m0 - at)) dt from 0 to t
where  g, u, mo, and a are constants.

I appreciate any help!

Do you mean this?

$\int_{0}^{t}-gs+u\ln\left(\frac{m_{0}}{m_{0}-as}\right)\,ds$

(And you're talking about rockets?)

Note that the name of the variable that is a limit of integration cannot be the same as the name of the variable of integration; I changed your $t$'s inside into the dummy variable $s.$

Write $\ln\left(\frac{m_{0}}{m_{0}-as}\right) =\ln m_{0}-\ln(m_{0}-as)$

The constant causes no trouble, and you can integrate by parts for the logarithm, or use that $\int \ln u\,du=u\ln u-u+C.$

My answer

[hide] utln(m0) -au[[(mo-at)ln(mo-at)-(mo-at)]-[moln(mo)+mo]]-0.5gt^2

sorry if that is painful to read due to latex failing on my computer. I am not sure if that is right and my answer is partiall aided by the helpful natural log integral formula provided by kent merryfield :lol: And also, due to the fact that i haven't even taken calculus yet, it's probably wrong lol but i thought i would give it a shot anyways.[/hide]


real analysis
algorithm
combinatorics proposed
combinatorics

Given $I_{1}, I_{2}, \cdots, I_{n}$ a finite collection of open intervals of finite length, show that there is a subcollection of pairwise disjoint intervals $J_{1}, J_{2}, \cdots, J_{m}$  such that
\[m( \bigcup_{i}I_{i}) \le 2 \cdot m( \bigcup_{j}J_{j}) \]
($m$ is the Lebesgue measure.)

This is the lemma used to prove Vitali's Covering Theorem. Note that the bounding constant is 2 instead of the usual 3. The case of 3 can be [url=http://everything2.com/index.pl?node_id=1773590]proved by greedy algorithm[/url]

Nothing changes in the problem if we eliminate an interval which is covered by the other intervals (that is, $I_{i}\subset\bigcup_{j\ne i}I_{j}$), so we may as well assume the collection $\{I_{i}\}$ is minimal, in the sense that no proper subcollection has the same union.

Notice that minimality implies that no three of our intervals intersect, and also that all the endpoints of our intervals are distinct. Now, if we label the intervals $I_{1},\ldots,I_{n}$ in the increasing order of their left endpoints, $I_{i}$ cannot intersect $I_{j}$ if $|i-j|\ge 2$. For our family of $J$'s we can now take either the odd-labeled intervals $I$ or the even-labeled ones.


The digits 1,2,3 and 4 are randomly arranged to form a four-digit number. What is the probability that the resulting number is divisible by 4? Express your answer as a common fraction.

There are $ 4 \times 3$ numbers for the last two digits. Because of the divisibility rule of 4, we need look at only the last two digits. The last 2 digits divisible by 4 are 12, 24, and 34. The answer is $ \boxed{\frac {1}{4}}$.

There are $ 4\cdot 3\equal{}12$ possible pairs for the last two digits.  Of these, only 12, 24, and 32 work, so $ \frac{3}{12}\equal{}\boxed{\frac{1}{4}}$


A HAHAHAHAHA :rotfl: HAHA :rotfl: HAH :rotfl: HAH :rotfl: HHAAA :rotfl: A :rotfl: HA

i decree this deserves a national holiday dedicated it to it

That is probably the best <1 minute video that have ever seen :rotfl:  :rotfl:  :rotfl:

I bet it's true!


function
calculus
derivative
real analysis
real analysis unsolved

Let function $f(x) \in C^{2}(-\infty;+\infty)$ and for all x,y
let $f(x) \ne f'(y)$ and $f'(x) \ne f''(y)$. Prove that either
$f(x)$ or $f(-x)$ is positive decreasing convex downwards function

Here's my solution.
Given that $f(x) \in C^{2}(-\infty,+\infty)$,it is obviously observed that there are only four possiblities of the relations among $f,f'$ and $f''$.If $f'(x)<f''(y)$,then $f''(y)\geq 0$ for all $y \in R$,on the grounds that if a $y_{0}$ exists that suffices $f''(y_{0})<0$, we have $f'(x)<f''(y_{0})$ for all $x \in R$,which means $f(x)$ is unbounded and leads to a contradiction.And then an existing $x_{0}$ for which $f'(x_{0})>0$ would similarly render $f'(x)$ unbounded.For the positiveness of $f(x)$,the proof is substantially the same.Thus the desired result is obtained.In the case where $f'(x)>f''(y)$,an almost identical proof seems valid,but I don't propose to show you how.

If the proof is flawed,let me know.


abstract algebra
group theory
probability and stats

Let $ Z$ be a finite Abelian group with the operation $ \plus{}$.

Let $ A$ and $ B$ be subsets (need not be subgroups) of $ Z$.  Randomly (uniformly) pick $ x$ from $ Z$.

Show that
\[ E(1\minus{}\frac{|A\cup (B\plus{}x)|}{|Z|})\equal{}(1\minus{}\frac{|A|}{|Z|})(1\minus{}\frac{|B|}{|Z|}).\]

Got the proof after some thought.

Claim (which I shall not prove): Consider the singleton set $ \{ b\}$. Then $ P(|A\backslash \{b\plus{}x\}|\equal{}0)\equal{}\frac{|A|}{|Z|}$ and $ P(|A\backslash \{b\plus{}x\}|\equal{}1)\equal{}1\minus{}\frac{|A|}{|Z|}$. Hence, $ E(|A\backslash \{b\plus{}x\}|)\equal{}1\minus{}\frac{|A|}{|Z|}$.

Now, let $ B\equal{}\{b_1, b_2, \ldots, b_n\}$. Then $ A\cup (B\plus{}x)$ is a disjoint union of $ A$, $ A\backslash \{ b_i\plus{}x \}$'s. Hence, $ |A\cup (B\plus{}x)|$ is the sum of  $ |A|$, $ |A\backslash \{ b_i\plus{}x \}|$'s.  By the linearity property of expectations, we sum the expectations over the latter components and with some computation obtain the answer.


Please, try to prove that for every real numbers a,b,c>0, inequality:

a^2 + b^2 + c^2 + 2abc +1 > 2(ab + bc + ca) is true.
when equality holds?

Happy New Year, everyone!  :D

[quote="Klaas"]Please, try to prove that for every real numbers a,b,c>0, inequality:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{}1 \geq 2(ab \plus{} bc \plus{} ca)$ is true.
when equality holds?

Happy New Year, everyone!  :D[/quote]
Yes,indeed,inequality is very simple:
$ a^2\plus{}b^2\plus{}c^2\plus{}abc\plus{}abc\plus{}1\geq a^2\plus{}b^2\plus{}c^2\plus{}3\sqrt[3] {a^2b^2c^2}\geq \sum a^{\frac{2}{3}}b^{\frac{2}{3}}(a^{\frac{2}{3}}\plus{}b^{\frac{2}{3}})$
The last one is from Schur inequality,and the first is just am-gm.
Now we will prove that
$ \sum a^{\frac{2}{3}}b^{\frac{2}{3}}(a^{\frac{2}{3}}\plus{}b^{\frac{2}{3}})\geq 2(ab\plus{}bc\plus{}ca)\Longleftrightarrow$
$ \sum a^{\frac{2}{3}}b^{\frac{2}{3}}(a^{\frac{1}{3}}\minus{}b^{\frac{1}{3}})^2\geq 0$
Thus we are done.
[color=red]Happy New Year![/color]

Other solution:
We can rewrite this inequality that:
$ (a \minus{} 1)^2 \plus{} (b \minus{} 1)^2 \plus{} (c \minus{} 1)^2 \plus{} 2(a \minus{} 1)(b \minus{} 1)(c \minus{} 1)\ge 0$
We can suppose that  $ (a \minus{} 1)(b \minus{} 1)\ge 0$ then
$ (a \minus{} 1)^2 \plus{} (b \minus{} 1)^2 \plus{} (c \minus{} 1)^2 \plus{} 2(a \minus{} 1)(b \minus{} 1)(c \minus{} 1)\ge (c \minus{} 1)^2 \plus{} 2(a \minus{} 1)(b \minus{} 1)c\ge 0$
Done!


For a given opened set X. It is true that ri(X) = X?

Do you mean something like $X^\circ$ as in `interior'? If so, yes, the interior of an open set is the set itself.

What do you call "relative interior" here ?

Oh! I should have stated interior of X instead of relative interior of X. In a lecture note I found in the web uses them interchangebly although the authors noted their difference (though in quite an vague manner).

Is int(X) = X, given X is an opened set?

Can someone explain what is the difference between relative interior and interior of a set?

I also encountered relative boundary and boundary of a set. What does the word 'relative' signify?

"relative"="in the subspace topology". This is useless when talking about an interior, since every set is open in itself (thus, for any $X$, the relative interior of $X$ is $X$). It's possible the source has another definition in mind, but it would be very tricky to make that definition interesting.


It is IMO 1978..already discussed on AoPS... http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367388#367388

[quote="Cyber-Shreyas"][quote="madness"]This is some inmo or rmo sum im not quite sure  :D  Anyways i knw the method so i wont comment[/quote]

//NV No 1[/quote]

Now i know u source da shre and i told ya that place  :P
http://www.kalva.demon.co.uk/imo/imo78.html

Why do I need to go all the way to Kalva when I have this http://www.mathlinks.ro/resources.php?c=1&cid=16

[quote="madness"]Bcos Kalva is the original source  :P[/quote]

The original source of many of the problems you see in your IIT Books are probably some nice organic book you haven't heard of. Then according to your logic, you should go and buy the original source and not the IIT Obj books you buy.

And by original source do you mean that the owner of Kalva set the problem??

No i meant AOPS might have dibbied the problems  :P  And also i told you the site recently so..........  :P   :D


Let $ a, b, c$ be positive real numbers. Prove o disprove that

$ \frac{5a \plus{} 3c}{2a \plus{} b} \plus{} \frac{5b \plus{} 3a}{2b \plus{} c} \plus{} \frac{5c \plus{} 3b}{2c \plus{} a} \ge 8$ .

________
 :(

[quote="Ligouras"]Let $ a, b, c$ be positive real numbers. Prove o disprove that

$ \frac {5a \plus{} 3c}{2a \plus{} b} \plus{} \frac {5b \plus{} 3a}{2b \plus{} c} \plus{} \frac {5c \plus{} 3b}{2c \plus{} a} \ge 8$ .

________
 :([/quote]
This inequality is equivalent to:
$ \frac {1}{3} \left(\frac {(a \minus{} b)^{2}}{(2b \plus{} c)(2a \plus{} b)} \plus{} \frac {(a \minus{} b)(a \minus{} c)}{(2b \plus{} c)(2a \plus{} b)} \plus{} \frac {(b \minus{} c)^{2}}{(2c \plus{} a)(2a \plus{} b) } \plus{} \frac {(a \minus{} b)(b \minus{} c)}{(2c \plus{} a)(2a \plus{} b)} \right) \plus{}$
$ \plus{} \frac {8}{3} \left (\frac {(a \minus{} b)^2}{(2b \plus{} c)(2c \plus{} a)} \minus{} \frac {(a \minus{} b)(b \minus{} c)}{(2b \plus{} c)(2c \plus{} a)} \plus{} \frac {2(a \minus{} b)(b \minus{} c)}{(2b \plus{} c)(2a \plus{} b)} \plus{} \frac {(b \minus{} c)^2}{(2b \plus{} c)(2a \plus{} b)} \right) \geq 0$


modular arithmetic
number theory unsolved
number theory

solve the system in Z:

$ a^2\plus{}5b^2\equal{}c^2$
$ a^2\minus{}5b^2\equal{}d^2$

we can demostrate that the unique solution of $ (a,b,c,d)\in Z^4$ is $ (a,0,a,a)$. (in fac if sme variables are zeros that necessarily $ b = 0$.wlog $ a,b,c,d > 0$.
$ a^2 + 5b^2 = c^2$
$ a^2 - 5b^2 = d^2$
by summing we obtain $ 2a^2 = c^2 + d^2 \implies c \equiv d \pmod 2$. if $ c \equiv d \equiv 0 \pmod 2$ then also $ a\equiv b \equiv 0 \pmod 2$ and we can semplify unless we obtain some odd variables. so we must have $ c\equiv d \equiv a \equiv 1 \pmod 2 \implies$$ c^2 \equiv d^2 \equiv a^2 \equiv 1 \pmod 8 \implies b^2 \equiv 0 \pmod 8 \implies b\equiv 0 \pmod 4$.
so we have the equivalent system:
$ a^2 + 80b^2 = c^2$(*)
$ a^2 - 80b^2 = d^2$(**)
we have $ a^2 = \frac {c^2 + d^2}{2}$ so there must exist $ (e,f) \in N^2$ such that:
i)$ e > f > 0$
ii)$ e\equiv 0 \pmod 2 \text{ and } f \equiv 1\pmod 2$
iii)$ c = e + f$
iv)$ d = e - f$.
so we have $ a^2 = e^2 + f^2$ and there must exist $ (m,n)\in N^2$ such that $ a = m^2 + n^2, e = 2mn, f = m^2 - n^2, m > n > 0, m\equiv 0 \pmod 2, n\equiv 1 \pmod2$ or $ m\equiv 1 \pmod 2, n \equiv 0 \pmod 2$

by (**) it is true that $ b^2 = \frac {(a + d)(a - d)}{80} = \frac {(m^2 + n^2 + 2mn - m^2 + n^2)(m^2 + n^2 - 2mn + m^2 - n^2)}{80}$ and so:
$ \displaystyle b^2 = \frac {(2n^2 + 2mn)(2m^2 - 2mn)}{80} = nm\frac {m^2 - n^2}{20}$
now the problem is $ 20b^2 = mn(m + n)(m - n)$ :?:

one solution is (m,n)=(4,5): is it the unique?


Duke college ARML trigonometry function probability LaTeX

I have heard about the Duke Math Meet, and was wondering, what exactly is it.  Would it be worth going to (I live in Winston-Salem), and what types of problems are there?

[quote="Iversonfan2007"]it's pretty much like arml...except so much cooler cuz it's duke. yea arnav, unc sucks[/quote]

what screw you

sheesh

you don't even live in the state

wtf

- DUKE TOO

[quote="jb05"]They don't have a date yet... but everybody should go! They haven't posted anything except dates and dead links on their site for many years, probably because it's run by the problem solving class (not faculty). If you're in middle school, you need to find a team of 6 first.

Also, I think the individual problems are harder than ARML.[/quote]

what no

I can gets 8s at Duke

apparently not at ARML

UNC PWNS J00 STFU N00B

You didn't get an 8 last year... that was definitely harder than ARML!

dude

either last year was the year that just passed

or it is way too long ago to remember

how can you remember stuff from more long ago than last year

sheesh

this is not cool

what did you get last year at duke arnav?

If last year means the year that just passed, I'm pretty sure an 8. And I thought an 8 before that too? Maybe not. Huh. IT IS TOO LONG AGO FOR REMEMBERING PURPOSES

i thought that guy said no one got that ginormous trig problem right.

ok. then maybe I didn't get an 8. I DO NOT REMEMBER THESE THINGS!

No, Arnav. You missed: Express $\sum_{i=0}^{17}\frac{tan(17*2^{i})}{cos(17*2^{i+1})}$ in the form $u*\phi(v*w^{x})+y*\psi(z)$, where $\phi$ and $\psi$ can each be one of the six trigonometric functions, and each of u, v, w, x, y, and z are integers such that v and z are positive, x is greater than or equal to 2, and w is a positive prime number that does not divide v.

That was possible the most amusing competition problem I've seen.

!*)&*#!@#!@#&*^(*!@#%^&@%^@&!!!!!!!!!!!!!!!!!!!

 :rotfl:  :rotfl:

It must have a simple solution, being that disgusting and all.

EDIT: synchronized laughing smileys, lol

East, TJ, hmm yeah ok I guess there's room for y'all

Is Vestavia Hills going?  They have a chance of placing too, if they are as good as they have been in previous years.

Do you guys know a website of the past competitions? I'm going to start practicing for next year.

Vestavia Hills is this team that always comes to SC contests and pwns them.  Look at Division 1 (highest division) Furman Results from the last few years (I don't think they went in 2005 though):

[url=http://math.furman.edu/tournament/tourney01.html]2001[/url]
[url=http://math.furman.edu/tournament/tourney02.html]2002[/url]
[url=http://math.furman.edu/tournament/tourney03.html]2003[/url]
[url=http://math.furman.edu/tournament/tourney04.html]2004[/url]

Oh my gosh, they came in 2nd in 2002!

But they're a really good team from like Alabama or something.  They'll be top 5 contenders.

Our team will be like top 30 contenders  :D

Like we have no contest experience except for AMC/AIME and like one MathCounts guy, plus me, lol

SC... you guys should do pretty well

[quote="mysmartmouth"]Maybe, but Asif might be competition if Seva doesn't have a good day.[/quote]

are you in love with asif?

Wow... there are too many randomly posting SCers in here

It's because our forum is much better and cooler than theirs  :lol:

Obviously

Ever since our forum was started, the SC forum has all but died (half the posters there like left)

yeah it's because of that contest thing without the prizes, whatever it's called

Um... jb05, where do you get your problems? Do you still have them because you took the tests? Or are they online?


function
linear algebra
matrix
algebra
polynomial
calculus
derivative

It is not hard to verify that if $ (f_n)$ be a sequence of real-valued functions that converges uniformly to $ f$ and for each $ \lambda_i$ (as an eigenvalue) where $ 1\leq i \leq n$  ,$ f_n(\lambda_i)$ converges to $ f(\lambda_i)$ ,then for each matrix $ A$ that all $ \lambda_i$ are its eigenvalues, $ f_n (A)$ converges to $ f(A)$.

Now my question is :  if $ f_n$ converges pointwise to $ f$, will it conclude in the same way ?

[hide]I don't think so and I wanted to make a counter-example with this sequence of functions $ (f_n): (0,1) \longrightarrow \mathbb{R}$ where $ f_n(x)\equal{}x^n$ that $ f_n$ converges pointwise to $ 0$ and does not converge uniformly to $ 0$. But now... ??? :maybe: [/hide]

what is $ f(A)$ when $ f: \mathbb{R} \to \mathbb{R}$ ?

As an example, for $ f_n(x)\equal{}x^n$, we define $ f_n(A)\equal{}A^n$.

OK, I think that I got that. But what if the function is not a polynomial (or an entire function ?). You should be more explicit about the way you define the object you are working with ! or maybe you assume that you are only working with polynomials ?

Dear [b]alekk[/b], you're right   :oops:  . I will explain it with more details. I'm sure it's true but now I forget the details because one of the PhD. students in Sharif University explained it to me about three weeks ago.  :blush:

Correct me if I am wrong, but if we consider values of functions $ f_{n}$ only for a finite set of arguments, there is no difference between uniform and pointwise convergence, because for any $ \varepsilon>0$ we can find $ n_{0}$ such that $ \left|f_{n}\left(\lambda_{i}\right) \minus{}f\left(\lambda_{i}\right)\right|<\varepsilon$ for each $ n>n_{0}$ and each $ i$, for instance the maximum of $ n_{0}$ corresponding to all $ i$. Thus for $ A\equal{}S^{\minus{}1}DS$, where $ D\equal{}\mathrm{diag}\left(\lambda_{1},\ldots,\lambda_{k}\right)$, we have $ f_{n}\left(A\right)\equal{} S^{\minus{}1}\mathrm{diag}\left(f_{n}\left(\lambda_{1}\right),\ldots,f_{n}\left(\lambda_{k}\right)\right)S\longrightarrow$ $ S^{\minus{}1}\mathrm{diag}\left(f\left(\lambda_{1}\right),\ldots,f\left(\lambda_{k}\right)\right)S\equal{}f\left(A\right)$. There is nothing wrong with the example $ f_{n}\left(x\right)\equal{}x^{n}$ with $ f_{n}: \left(0;1\right)\longrightarrow \mathbb{R}$, because if all eigenvalues of $ A$ belong to the interval $ \left(0;1\right)$, the sequence $ f_{n}\left(A\right)$ converges to $ 0$.

[quote="Rzeszut"]Correct me if I am wrong, but if we consider values of functions $ f_{n}$ only for a finite set of arguments, there is no difference between uniform and pointwise convergence, because for any $ \varepsilon > 0$ we can find $ n_{0}$ such that $ \left|f_{n}\left(\lambda_{i}\right) \minus{} f\left(\lambda_{i}\right)\right| < \varepsilon$ for each $ n > n_{0}$ and each $ i$, for instance the maximum of $ n_{0}$ corresponding to all $ i$. Thus for $ A \equal{} S^{ \minus{} 1}DS$, where $ D \equal{} \mathrm{diag}\left(\lambda_{1},\ldots,\lambda_{k}\right)$, we have $ f_{n}\left(A\right) \equal{} S^{ \minus{} 1}\mathrm{diag}\left(f_{n}\left(\lambda_{1}\right),\ldots,f_{n}\left(\lambda_{k}\right)\right)S\longrightarrow$ $ S^{ \minus{} 1}\mathrm{diag}\left(f\left(\lambda_{1}\right),\ldots,f\left(\lambda_{k}\right)\right)S \equal{} f\left(A\right)$...[/quote]

I think your reasoning is essentially correct if you assume $ A$ is diagonalizable, but I believe the statement originally posted holds for non-diagonalizable $ A$, as well.  In that case you'd have to consider the messier Jordan decomposition, in which case $ f(A)$ would depend not only on the function evaluated at the eigenvalues, but also on the local behavior of $ f$ near the eigenvalues (i.e. derivatives at the eigenvalues).  I imagine that's where the uniform convergence comes into play, and where the difficulty in weakening to pointwise convergence is.

As I promised, for more information you can see , [b]Gantmacher , ch.5[/b].


[quote="Jan"]Solve the following system, when $x,y,z \in \mathbb{R}$:
\[\{ \begin{array}{c}x^{2}+2y^{2}+3z^{2}=36\\ 3x^{2}+2y^{2}+z^{2}=84\\ xy+yz+xz =-7 \end{array}\]
[/quote]
So this is basically trickery with the first and second equations. Adding them together, then subtracting one from two, then subtracting two from one yields all one needs.

[quote="Jan"]PenguinIntegral, your post doesn't help anyone.
Please, post your miraculous solution, and surprise us.

I'm convinced it won't be as easy as you think...  :wink:[/quote]
Sorry, that was a bit too pretentious. :oops:

The substitution I had in mind doesn't pan out too well...

[quote="cincodemayo5590"][hide="A start:"]
Well, adding $(1)$ and $(2)$ gives $x^{2}+y^{2}+z^{2}= 30$. Then, $(x+y+z)^{2}= 30+2(-7) = 16 \implies x+y+z = \pm 4$. Then, if we could only find $xyz$, we could get a polynomial going... :P  [/hide][/quote]
Yeah, this is what I had going until I realized I couldn't get xyz.

Maybe subtracting the first from the second or the second from the first for a substitution?

Also, I can't really see the third equation being useful because it doesn't have many terms in common with the first or second, except in situations like the above.

[quote="Jan"][quote="boxedexe"]$x^{2}+y^{2}+z^{2}=40$[/quote]
No.  :wink:

Finding $xyz$ would indeed solve the problem, but...[/quote]
Oops. I inadvertently divided by 3 instead of 4.  No wonder I acquired unreasonable equations.

[quote="Kurt Gödel"]Funny. Jan and I were also substracting and adding as many equations as possible, but we didn't find anything. After a while, we tried something else, and it worked.  :)[/quote]
What? Some of these *might* suggest geometrical interpretation, I think...

[quote="tim1234133"]It's easy enough to 'guess' the answer, and check that it works, but I suppose that isn't the idea you're looking for  :)[/quote]

Mm, yeah, at least I've found a solution set now, but that's quite unsatisfying  :|

[quote="tim1234133"]It's easy enough to 'guess' the answer, and check that it works, but I suppose that isn't the idea you're looking for  :)[/quote]

If you could prove that that answer is unique, that would be OK.

As derived in a previous post, $x+y+z = \pm 4$.  First, we take the case $x+y+z = 4$.  This implies $x+z = 4-y$.

Subtracting the first equation from the second and dividing by 2, we get $x^{2}-z^{2}= (x+z)(x-z) = 24$.  Therefore,
\[x-z = \frac{24}{x+z}= \frac{24}{4-y}. \]
From the first and second equation, $x^{2}+y^{2}+z^{2}= 30$.  Therefore,
\[2x^{2}+2y^{2}+2z^{2}= (x+z)^{2}+(x-z)^{2}+2y^{2}= (4-y)^{2}+\left( \frac{24}{4-y}\right)^{2}+2y^{2}= 60. \]
Expanding and simplifying, this gives us
\[3y^{4}-32y^{3}+68y^{2}+224y-128 = 0. \]
This factors as
\[(y+2)(3y^{3}-28y^{2}+144y-64) = 0. \]
From $y =-2$, we obtain the solution $(x,y,z) = (5,-2,1)$.  The cubic has one real root, approximately $0.510426$.  This gives us another solution $(x,y,z) \approx (5.183602, 0.510426,-1.694028)$.

The case $x+y+z =-4$ gives us the negatives of the solutions above.

[quote="Jan"]Rust: Ok, but the system is intended to be solved without a calculator. I don't think we can easiley solve that quartic equation. A more clever solution is needed.  :wink:
Besides, the quartic equation you give is not correct, you must have made an error somewhere.

Svejk: Indeed, that's easy to figure out... Now finish the solution if you can please!  :wink:[/quote]
This is what Jan Vonk wrote in another thread currently in the Olympiad Algebra section.

Thanks for the clear solution nsato.

I'm stunned to see that the system has such weird solution apart from the nice ones. It is certainly meant to be solved without a calculator, that's clear, so I'm very confused.
I sent an email to the writer of the document where this system came from, to ask for the 'official' solution. I hope it is not a typo in the document. I'll let you know.


I have a series of equations, and use
\begin{align}
...
E(V_{r}^{\ast}-\pi_{r}^{\ast})&=E(V_{r}^{\ast}-\pi_{r}^{\ast})\gtreqless 0\\
...
\end{align}

Dots represents other equations.  The ouput displays other equations correctly but the equation with \gtreqless. the output does not show the relation sign. I also have \usepackage{amsmath}

How to solve this? Thank you!

Please look at your log which will tell you that \gtreqless is undefined and hence not shown. You need the extra symbols provided by amssymb so add \usepackage{amssymb} to the preamble.

Stevem, thank you so much.
It did show undefined control sequence. And I did not understand it.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Supply a group $ G$ and a subgroup $ H$ for which $ x\sim y$ if $ xyx^{\minus{}1}y^{\minus{}1}\in H$ is not an equivalence relation on $ G$.

I know $ G$ is not abelian since if it was, then $ xyx^{\minus{}1}y^{\minus{}1}\equal{}e \in H$ for all possible $ H$, but after trying some groups that weren't abelian, I still can't find one.

You did not try very many nonabelian groups. :)  If you set $ H\equal{}\{e\}$, then every single nonabelian group is an example...

In fact, $ \sim$ is an equivalence relation if and only if $ H$ contains every commutator.


inequalities
logarithms
function
algebra
floor function

I don't think this problem is suitable for Pre-Olympiad level(I meant it's easier than the level of this forum), but I need to have someone point out my mistake in my solution, which apparently doesn't come up with the same answer as the official one:
Give that $ [log(2)(log(2)n] > [log(2)\sqrt {n}]$($ [x]$ is the greatest integer function). Find the maximum value of positive integer of n?

My solution:
Let's solve for the "normal" inequality, that is: $ log(2)log(2)n > log(2).sqrt{n}$.
This inequality is equivalent to: $ 2^(sqrt{n}) < n$. The solution gives me $ n \equal{} 63$, which is wrong according to this inequality(but it's true according to the calculator). So, my solution is wrong in that substitution. But I can't find why it's wrong? Can someone help me out with it?

Well, the 1st step is: log(2)n>2^(log(2).sqrt{n})=sqrt{n}. Then raise 2 to both sides, we get the inequality I have above! The problem seems to lie in the getting rid of the greatest integer function! :(


A solution is prepared by dissolving 100.0 grams of ammonium sulfate in a sufficient volume of water to make 250.0 mL of solution. What is the osmotic pressure of this solution at 0 degrees Celsius?

So you just use $\Pi=iMRT$ right?

I'm getting an answer of around 203.5 atm, which sounds unreasonable...

Perhaps you have used an unappropiate value of R, or make some error in calculating the molar mass of amonium sulfate, because I get a value of $\Pi = 67.87 atm$:

amonium sulfate = $(NH_{4})_{2}SO_{4}$, with M = 132,1442 g/mol.
concentration of solution = 0.003028 mol/cm3.
R = 82.058 atm cm3 mol-1 K-1.

It turns out my answer was correct.  :ninja:

I still think it's pretty counterintuitive, though.

[quote="Rnwang2"]It turns out my answer was correct.[/quote]

Then show your calculations, I want to see what I did wrong.

Yes, you are right, I forget the i factor. Thanks


calculus
derivative
calculus computations

Define $ f(x) \equal{} \frac{x^{n}}{1\minus{}x}$ and $ g(x) \equal{} e^{ax}\sin(bx)$, where $ a$ and $ b$ are positive numbers.

Find the nth derivatives of $ f(x)$ and $ g(x)$.

For the first one use the power series $ \frac {1}{1 \minus{} x} \equal{} 1 \plus{} x \plus{} x^2 \plus{} \dots$. For the second one I'd use either Euler's formula $ \sin(bx) \equal{} (e^{ibx} \minus{} e^{ibx})/(2i)$ or the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171293]generalized product rule[/url].

Thanks mlok. I had the same solution for number one using geometric series.
For the second part, your suggested substitution works (I tested it), but I am interested in a solution that uses tools from calculus I or simply no complex numbers.


inequalities
function
logarithms
calculus
induction
algebra
functional equation

Let $ f: \left[\alpha,\beta\right]\to\mathbb{R}$ be a continuous function such that for any two reals $ a$ and $ b$ from the interval $ \left[\alpha ,\beta\right]$, we have

$ f\left(a\right)f\left(b\right) \ge f\left(\sqrt{ab}\right)^{2}$.

Then prove that for any $ n$ reals $ a_{1}$, $ a_{2}$, ..., $ a_{n}$ from the interval $ \left[\alpha,\beta\right]$, we have

$ f\left(a_{1}\right)f\left(a_{2}\right) ... f\left(a_{n}\right) \ge f\left(\sqrt[n]{a_{1}a_{2}...a_{n}}\right)^{n}$.

Use jensen on the convex function $g(x)=\ln f(e^x)$

lomos_lupin, I think you mean to write $f(a)f(b) \ge \left(f(\sqrt{ab})\right)^2$.

Isn't this the classic "it does not suffice to show $f(x)+f(y) \ge 2f\left(\frac{x+y}{2}\right)$ for all $x,y$ to claim that $f$ is convex" problem/"almost solution"? Someone who is well-versed in Calculus, please tell me if I am wrong, but don't we need $\lambda f(x) + (1-\lambda)f(y) \ge f(\lambda x + (1-\lambda)y)$ for all $x,y$ and $0 \le \lambda \le 1$ to claim $f$ convex? What if $f$ is continuous?

What did i wrote?
I have just ommited two $(,)$?

I wrote $(f\sqrt{xy})^2$ and you changed it to $(f(\sqrt{xy}))^2$?

And i dont not understant you at all,

What do you mean by?
[quote="Mildorf"]
Isn't this the classic "it does not suffice to show $f(x)+f(y) \ge 2f\left(\frac{x+y}{2}\right)$ for all $x,y$ to claim that $f$ is convex"?[/quote]

First as blah has said it is enough for f to be convex by $f(x)+f(y) \ge 2f\left(\frac{x+y}{2}\right)$

second, How this and the problem i post are related? :?

[quote="blahblahblah"]The answer to your question is basically this:

$f$ convex on $(a,b)$ $\iff f\left(\frac {x+y}{2}\right)\leq \frac {f(x)+f(y)}{2}$ for all $x,y\in (a,b)$ and $f$ is continuous.

There are discontinuous functions satisfying the latter condition which are not convex, and there are discontinuous convex functions on closed, or half-closed intervals.[/quote]

Ok, that is what I thought. (Edited to be more precise) That also means siuhochung's Jensen solution will work if and only if we add in that $f$ is continuous over $[\alpha,\beta]$. That said, it may be possible to prove without Jensen even without $f$ continuous.

[quote="lomos_lupin"]Let $f$ be a function such that for

$\all a,b \in [\alpha ,\beta] ,f(x)f(y)\ge(f\sqrt{xy})^2$

For  $a_1,...,a_n \in [\alpha,\beta] $ prove

$f(a_1)f(a_2)...f(a_n) \ge f(\sqrt[n]{a_1a_2...a_n})^n$[/quote]

What I meant to point out is that $a$ and $b$ became $x$ and $y$ in that given.

:rotfl: Sorry i thought you mean the $(,)$,I edited the post.

Why is $g=\log\circ f\circ \exp$ convex?

[quote="blahblahblah"]The answer to your question is basically this:

$f$ convex on $(a,b)$ $\iff f\left(\frac {x+y}{2}\right)\leq \frac {f(x)+f(y)}{2}$ for all $x,y\in (a,b)$ and $f$ is continuous.

There are discontinuous functions satisfying the latter condition which are not convex, and there are discontinuous convex functions on closed, or half-closed intervals.[/quote]

i thought that (well, i remember seeing smth like this on the forum not long ago)  $f\left(\frac {x+y}{2}\right)\leq \frac {f(x)+f(y)}{2}$ implies $f$ continouos function (at least for closed intervals), hence convex.

anyway, could you show me some examples? i can't construct them by myself  :blush:  (i.e. for both * There are discontinuous functions satisfying the latter condition which are not convex * (1) and * there are discontinuous convex functions on closed, or half-closed intervals *(2), but especially for 1)

i would like add smth else: i think it is the definition of the semi-convex function on an interval $I$ (the property you stated above). this property surely implies jensen's inequality for semi-convex functions, right? (by the backwards inductions procedure) . I mean I see no reason why it should fail in some cases, but I may be wrong.

[b]1.[/b] The function f needs not be continous, but it must have only nonnegative values, and $\alpha$ and $\beta$ also must be nonnegative. So the problem should read:

[color=blue][b]Problem.[/b] Let $\alpha$ and $\beta$ be two nonnegative numbers, and let f be a function defined on the interval $\left[\alpha;\;\beta\right]$ and having all of its values nonnegative. Assume that for any two numbers a and b from the interval $\left[\alpha;\;\beta\right]$, the inequality $f\left(a\right)f\left(b\right)\geq\left(f\left(\sqrt{ab}\right)\right)^2$ holds. Then, prove that for any n numbers $a_1$, $a_2$, ..., $a_n$ from the interval $\left[\alpha;\;\beta\right]$, the inequality $f\left(a_1\right)f\left(a_2\right)...f\left(a_n\right)\geq\left(f\left(\sqrt[n]{a_1a_2...a_n}\right)\right)^n$ holds.[/color]

[b]2.[/b] The solution of this problem can be done either analogously to the proof of the (semi-)Jensen inequality for semi-convex functions, with the only difference that all arithmetic means have to be replaced by geometric means, or using Siuhochung's argument from post #2 with "semi-convex" instead of "convex". It is nowhere necessary that the function f is continuous.

[b]3.[/b] [quote="perfect_radio"][quote="blahblahblah"]There are discontinuous functions satisfying the latter condition which are not convex, and there are discontinuous convex functions on closed, or half-closed intervals.[/quote]

[...] anyway, could you show me some examples? i can't construct them by myself  :blush:  (i.e. for both * There are discontinuous functions satisfying the latter condition which are not convex * (1) and * there are discontinuous convex functions on closed, or half-closed intervals *(2), but especially for 1)[/quote]

For [b](1)[/b], take any of the famous discontinuous solutions of the Cauchy functional equation f(x + y) = f(x) + f(y). It is trivially semi-convex (or, in Blahblahblah's terms, it satisfies the "latter condition"), since it satisfies $\frac{f\left(x\right)+f\left(y\right)}{2}=\frac{f\left(x+y\right)}{2}=f\left(\frac{x+y}{2}\right)$, but it is not convex since it is nowhere continuous.

(Note that I am here assuming the axiom of choice to hold, but we are doing calculus now, so I think this is justified.)

For [b](2)[/b], take e. g. the function f(x) on the interval [0; 1] which equals x for $0<x\leq 1$ and equals 1 for x = 0.

  Darij

To prove an answer to lomos, the inequality is easily proven as follows

Since $f(x_1)f(x_2)f(x_3)f(x_4)\geq f(\sqrt{x_1x_2})^2f(\sqrt{x_3x_4})^2\geq f(\sqrt{x_1x_2x_3x_4})^4$, and so on, it is easily proven for powers of $2$.

If it is proven for $N$, then

$f(x_1)\ldots f(x_{N-1})f(\sqrt[n-1]{x_1\ldots x_{N-1}})\geq f(\sqrt[n-1]{x_1\ldots x_{N-1}})^N$, which proves it for $N-1$, and the result follows.

[quote="lomos_lupin"]Nice and smooth ,Thanks to all.

PS.What is the name of this way of induction?(its too interesting,its like walking backward :lol: )[/quote]
backward induction


geometry
circumcircle
rotation
geometry proposed

[color=darkblue]Points $ X$, $ Y$ and $ Z$ are situated on the sides $ (BC)$, $ (CA)$ and $ (AB)$ of the triangles $ ABC$, such that triangles $ XYZ$ and $ ABC$ are similiar. Prove that circumcircle of $ AYZ$ passes through a fixed point.[/color]

Sorry Ahiles, I do not understnad the question.
What do u mean by the fixed point?

I think that he means not the, but a fixed point.  :wink:

Ok but see when you want to prove the circumcircle of $ \triangle ABC$ passes through a fixed point, whcih point do you take  for the 4 to be concyclic?
Like in a circle which is a circumcircle for a trinagle, I can always find a point other than the 3 on the circle and specify its position with respect to the triangle rite?
Please explain the question propelry Ahiles.

The triangle $ \triangle XYZ$ is not unique.

Triangle $ ABC$ is fixed, but points $ X,Y,Z$ vary...

Since $ \angle YAZ \equal{} \angle ZXY$, the circumcircles of triangles $ \triangle AZY$ and $ \triangle ZXY$ are of equal size. Likewise, the circumcircles of triangles $ \triangle BXZ$, $ \triangle CYX$ and $ \triangle XYZ$ are of equal size. Moreover, the circumcircles of triangles $ \triangle AZY$, $ \triangle BXZ$ and $ \triangle CYX$ have a point in common. Let this point be $ S$. A short angle chaise shows that $ \angle BXS \equal{} 180 \minus{} \angle SXC \equal{} \angle CYS$. Therefore, $ SB\equal{}SC$. Likewise, $ SA\equal{}SB$ and $ SC\equal{}SA$. Thus, $ SA\equal{}SB\equal{}SC$ and it follows that $ S$ is the circumcenter of triangle $ \triangle ABC$. Hence, the circumcircle of triangle $ \triangle AYZ$ passes through the circumcenter of triangle $ \triangle ABC$ which is a fixed point.

ok nice problem!!
lemma 1:
the orthocenter of $XYZ$ is the circumcenter of triangle $ABC$:
proof:
suppose $O$the orthocenter of $XYZ$we know that $HYCX$is cyclic so $\angle OCX = \angle OYX = 90 - \angle YXZ = 90 - \angle BAC$
so $CO$is isogonal with Height of $C$to $AB$ and $BO$ is isogonal with height of $B$ to $AC$ so $O$ is circumcenter of $ABC$!
proof of problem:
we know that $H$ is on the circle of $AYZ$ so $AYZ$ is go from cicumcenter of $ABC$ :blush:

If you know how to construct the points $X, Y, Z$, it's very easy!

Take two identical pieces of paper, of triangular shape, perfectly overlapped, pick them with a needle on the circumcenter, rotate the upper one with an angle $\alpha$; the intersections of the equal sides are the three points $X, Y, Z$; obviously, $\widehat{(AB, A'B')}= \widehat{AOA'}=\alpha$, a.s.o., done.

Best regards,
sunken rock