standard method here...

each selection corresponds to a sequence of length 100 where three of the terms are 1 and the rest are 0. we are then looking for the number of such sequences where no two 1's are next to each other. call such sequences "good". given a good sequence, we may remove exactly one zero from between successive 1's in the sequence (these 0's are available because the sequence is good) for a total removal of two 0's. the result is a sequence of length 98 where three of the terms are 1 and the rest are 0's and on which there are no restrictions. conversely, given any sequence of length 98 where three terms are 1 and the rest 0's, on which there are no restrictions, we may insert a 0 in each of the two blocks between the three 1's (one or more of these blocks may be empty, of course) to get a good sequence. we have a one-to-one onto correspondence, so the answer is $ 98 \choose 3$.

[quote="dogcat"]There are 100 books on a shelf. We want to select 3 books which are not neighbors. How many selections are there?[/quote]
we can suppose that $ {(a\ books)\ (book\ A)\ (b\ books)\ (book\ B)\ (c\ books)\ (book\ C)\ (100 \minus{} (a \minus{} 1 \minus{} b \minus{} 1 \minus{} c \minus{} 1)\ books)}$ where $ a\ge 0$ etnd $ b,c > 0$
so we have to find the numbre of solution of $ (a \plus{} 1) \plus{} (b \plus{} 1) \plus{} (c \plus{} 1)\le 100$ where $ (a,b,c)\in\mathbb{N}\times\mathbb{N}^*\times \mathbb{N}^*$
$ \equal{} |\{(a,b,c)\in \mathbb{N}\times\mathbb{N}^*\times \mathbb{N}^*|\ a \plus{} b \plus{} c\le 97\}|$
$ \equal{} |\{(a,b,c)\in \mathbb{N}\times\mathbb{N}\times \mathbb{N}|\ a \plus{} b \plus{} c\le 95\}|$
$ \equal{} \sum_{k \equal{} 0}^{95}C_{k \plus{} 2}^{2}$
$ \equal{} \sum_{k \equal{} 0}^{95}\frac {(k \plus{} 2)(k \plus{} 1)}{2}$
$ \equal{} \frac {\sum_{k \equal{} 1}^{96}k^2 \plus{} \sum_{k \equal{} 1}^{96}k}{2}$


function
linear algebra
matrix
symmetry
superior algebra
superior algebra solved

[i]F[/i] is a field of numbers. A mapping $f:F\to F$ satisfy:
(1) for any matrices [b]A[/b], [b]B[/b] of size [i]n[/i], it holds f([b]A[/b]+[b]B[/b])=f([b]A[/b])+f([b]B[/b])
(2) for any $k \in F$,  it holds that f(k[b]A[/b]) = k f([b]A[/b]), where [b]A[/b] is a matrix of size [i]n[/i].
(3) for any matrices [b]A[/b], [b]B[/b] of size [i]n[/i], f([b]A[/b][b]B[/b])=f([b]B[/b][b]A[/b])
(4) f([b]E[/b])=n

Show that f([b]X[/b]) = tr([b]X[/b])

[quote="liyi"][i]F[/i] is a field of numbers. A mapping $f:F\to F$ satisfy:
[/quote]

I bet you ment A mapping $f:M_n(F)\to F$ satisfy:

$F(A)+F(B)=F(A+B) \Rightarrow F(0_n)=0$.

Let $E_{i,j}$ be the matrixe $(e_{i,j}^{k,l})_{k,l=\overline{1,n}}$ given by
$e_{i,j}^{k,l}=\{\begin{array}{cc} 1,&if\ (i,j)=(k,l) \\ 0,& otherwise\end{array}$.
Let $f_{i,j}=F(E_{i,j})$. It is easy to see that $E_{i,j}\cdot E_{k,l}\neq 0_n$ and $E_{k,l}\cdot E_{i,j}\neq 0_n$ iff i=j=k=l.
Let $L=\{ f_{i,j} | i\neq j\}$ It is easy to see that $ab=0 \forall a,b \in F$, not necessarily distinct. This quickly leads to L={0}. So $f_{i,j}=0$ $\forall i\neq j$.

$(E_{i,i}+E_{i,j})(E_{i,i}+E_{j,i})=2E_{i,i}$
$(E_{i,i}+E_{j,i})(E_{i,i}+E_{i,j})=E_{i,i}+E_{i,j}+E_{j,i}+E_{j,j}$
So $2\cdot f_{i,i}=2\cdot F(E_{i,i})=F(2\cdot E_{i,i})=F(E_{i,i}+E_{i,j}+E_{j,i}+E_{j,j})=f_{i,i}+f_{j,j}$
So $f_{i,i}=f_{j,j}$. And this is for all $i,j=\overline{1,n}$.

$F(E)=n \Rightarrow \sum_{k=0}^n f_{k,k}=n \Rightarrow f_{k,k}=1 \forall k=\overline{1,n}$

There is no problem now to get F(A)=Tr(A).

[quote="amfulger"] It is easy to see that $E_{i,j}\cdot E_{k,l}\neq 0_n$ and $E_{k,l}\cdot E_{i,j}\neq 0_n$ iff i=j=k=l.
Let $L=\{ f_{i,j} | i\neq j\}$ It is easy to see that $ab=0 \forall a,b \in F$, not necessarily distinct. This quickly leads to L={0}. So $f_{i,j}=0$ $\forall i\neq j$.
[/quote]

Replace this by:
It is easy to see that $E_{i,j}\cdot E_{k,l}\neq 0_n$ and $E_{k,l}\cdot E_{i,j}\neq 0_n$ iff j=k and i=l.
Let $L=\{ f_{i,j} | i<j\}$ It is easy to see that $ab=0 \forall a,b \in F$, not necessarily distinct. This quickly leads to L={0}. So, also using symmetry, $f_{i,j}=0$ $\forall i\neq j$.

Hope it's ok now.

[quote="amfulger"]
 It is easy to see that $ab=0 \forall a,b \in F$, not necessarily distinct. This quickly leads to L={0}. So, also using symmetry, $f_{i,j}=0$ $\forall i\neq j$.
[/quote]

I don't see it too easy, unless you assume that $\forall i,j,k,l:F(E_{i,j}\cdot E_{k,l}) = f_{i,j}\cdot f_{k,l}$. Which, in the case of the trace function, would be false.

[quote="amfulger"] Let $L=\{ f_{i,j} | i<j\}$ It is easy to see that $ab=0 \forall a,b \in F$, not necessarily distinct. [/quote]

Let $L=\{ f_{i,j} | i<j\}$ It is easy to see that $ab=0 \forall a,b \in L$, not necessarily distinct.

Sorry I ment L, not F there.

I didn't notice the typo, either. By the way, if it was $a,b\in L$ it would be trivially false :D.
So let me reformulate my question: Why is it so easy to see that $ab=0\forall a,b\in L$, not necessarily distinct?

I must regretfully admit that you are right, Fibonacci.

So let me correct the proof.
Let $i\neq j$.
$F(E_{i,i}\cdot E_{i,j})=F(E_{i,j})=f_{i,j}$
$F(E_{i,j}\cdot E_{i,i})=F(0_n)=0$
So $f_{i,j}=0$ for $i\neq j$.

Hope it's ok now.


Find out n natural, n>=2 for which the ineq is true for any a,b,c pozitive reals:
sqrt(a+sqrt(b+sqrt(c)))>=(abc)^(1/n).

bye bye!

Contest from Vaslui again. The answer is 14.

You can prove that 14 works by repeated applications (2, actually) of AM-GM in such a way that you make the exponents of a, b, c equal. For example, in the first step you apply AM-GM for b, (c^1/2)/2 and (c^1/2)/2. I think you can see what I mean.

To prove that 14 is the ONLY n you assume this works for n>14 (it's similar for n<14) and you take some sequences a_k, b_k, c_k which tend to 0 (or 00, I can't remember), in such a way that you have equality in the 2 applications of AM-GM. You'll eventually get a contradiction if you choose 0 or 00 in the right way. I really have no time for posting a soln right now, but this is what I did and I got 10 points for this (that's the max you can get for a prb).


limit
linear algebra
matrix
real analysis
real analysis unsolved

Let for any $E\subset R^{2}$ and $\alpha\in (0,2]$ consider
\[\parallel E\parallel _\alpha=\lim_{\delta\to 0}\inf\{\sum_{i=1}^{n}(diam(E_{i}))^\alpha\mid\ E\subset \bigcup E_{i},\ diam(E_{i})<\delta\}.\]
Suppose $T=M_{2}(R)$. Find
\[\parallel T\parallel _\alpha=\sup_{\parallel E\parallel _\alpha=1}\parallel TE\parallel _\alpha.\]

Matrix $2\times 2$. I though it is a common notation on this forum.

$T\in M_{2}(\mathbb R)$, then. The [url=http://en.wikipedia.org/wiki/Singular_value_decomposition]singular value decomposition[/url] reduces the problem to diagonal $T$ with nonnegative entries. Let $\lambda_{1}\ge \lambda_{2}$ be the diagonal elements. Naturally, $\parallel T\parallel_{\alpha}\le \lambda_{1}^{\alpha}$. This is attained for $0<\alpha\le 1$ by appropriate subsets of $\mathbb R e_{1}$. For $\alpha\in (1,2]$ one should expect $\parallel T\parallel_{\alpha}=\lambda_{1}\lambda_{2}^{\alpha-1}$, but I can't prove either inequality.  :(


calculus
integration
function
real analysis
real analysis unsolved

Suppose that $(E,A,\phi)$ is a measurable space of finite measure and $f_{n}$ are measurable and converge pointwise to some $f$. Also, assume that for some increasing function $H$ which goes to $\infty$ at $\infty$ we have $\sup_{n\geq 1}{\int_{E}{|f_{n}|H(|f_{n}|)d\phi}}<\infty$. Then $f_{n}$ converges to $f$ in $L^{1}$.

The condition with $H$ implies that $(f_{n})$ is a uniformly integrable family, that is, for any $\epsilon>0$ there is $M$ such that $\sup_{n}\int_{|f_{n}|>M}|f_{n}|<\epsilon$. Uniform integrability implies convergence in $L^{1}$ as follows: (1) truncate the sequence at level $M$ and apply the Lebesgue domination to the truncated sequence $f_{n}^{M}$; (2) control the contribution of $f_{n}-f_{n}^{M}$ with Fatou's lemma.


inequalities
geometry
circumcircle
trigonometry
geometry unsolved

Let $ABC$ be a triangle, let $H$ be the orthocentre and $L,M,N$ the midpoints of the sides $AB, BC, CA$ respectively. Prove that
\[HL^{2} + HM^{2} + HN^{2} < AL^{2} + BM^{2} + CN^{2}\]
if and only if $ABC$ is acute-angled.

It seems easy. Let $R$ be the circumadius of triangle $LMN$. Of course the point $H$ is it's circumcenter ( a well known fact). Also segments $AL$,$BM$,$CN$ are equal to the sides of triangle $LMN$ thus all we want to do, is to prove the following \[ LM^2+MN^2+NL^2>3R^2 \] and this is easy to prove using using formulas concerning triangles.
I think that even the stronger on holds  \[ LM^2+MN^2+NL^2>8R^2 \]

Well, that well known fact is wrong ... the orthocenter isn't the center of the Feuerbach circle. The well known fact is that the orthocenter of $LMN$ is the circumcenter of $ABC$ ...

[b][u]Lemma.[/u][/b] $\boxed {\ \mathrm {\ acute}\ \triangle ABC\Longrightarrow 8R^2<a^2+b^2+c^2\le 9R^2\ }\ .$

[u]Proof.[/u] $a^2+b^2+c^2-8R^2=4R^2\cdot(\sin ^2A+\sin ^2B+\sin^2C-2)=$
$2R^2[(1-\cos 2A)+(1-\cos 2B)+(1-\cos 2C)-4]=$
$-2R^2(1+\cos 2A+\cos 2B+\cos 2C)=-2R^2[2\cos^2A+2\cos(B+C)\cos(B-C)]=$
$-4R^2[\cos^2A-\cos A\cos (B-C)]=4R^2\cos A[\cos (B+C)+\cos (B-C)]=$
$8R^2\cos A\cos B\cos C>0\ .$ [u]Remark.[/u] $a^2+b^2+c^2=8R^2(1+\cos A\cos B\cos C)\le 9R^2$
because it is well-known that in any triangle $ABC$, $\cos A\cos B\cos C\le \frac 18\ .$

[b][u]The proof of the proposed problem.[/u][/b] It is well-knownn that $AH=2R\cos A$ a.s.o. Therefore,
$4\cdot HM^2=2\cdot(HB^2+HC^2)-a^2=8R^2(\cos^2B+\cos^2C)-a^2=$
$8R^2(2-\sin^2B+\sin^2C)-a^2=16R^2-2(b^2+c^2)-a^2\ .$ Thus,
$4\cdot\sum HM^2=48R^2-5(a^2+b^2+c^2)\ .$ From the above lemma results the relation
$\frac 13 (a^2+b^2+c^2)\le 4\cdot\sum HM^2<6(a^2+b^2+c^2)-5(a^2+b^2+c^2)=a^2+b^2+c^2$, i.e.
$\boxed {\ \frac 13 (AL^2+BM^2+CN^2)\le HM^2+HN^2+HL^2<AL^2+BM^2+CN^2\ }\ .$

The condition from problem can be write: $ a^{2}\plus{}b^{2}\plus{}c^{2} < 8R^{2}$
Hence, caculate the power of $ H$ with $ (O)$, we get the rusult


function
real analysis
real analysis unsolved

$X$ a separable metric space. (there exists a dense sequence).
$\mu$ a Borel measure such that $\mu(X)=1$.
Suppose that $f: X\to X$ is a continuous function such that $\mu(f^{-1}(A))=\mu(A)$ for any Borel set $A$.

A point $x \in X$ is "recurrent" if $x$ is an adherent point of the sequence $u_n=f^n(x)$.
Prove that the set of points that are not "recurrent" points is of null measure.

For any non-recurrent point $x$, there is a neighborhood $V$ of $x$ and a natural $n_0$ such that $f^{(n)}(x)\not\in V,\ \forall n\ge n_0\ (*)$. Given a countable basis $(V_m)_m,\ (*)$ still holds if we replace $V$ with some $V_m$ such that $x\in V_m\subset V$, so it suffices to prove that for each $m$, the set $N_m$ of non-recurrent points in $V_m$ has measure zero, since the set $N$ of non-recurrent points satisfies $N=\bigcup_mN_m\Rightarrow \mu(N)\le\sum_m\mu(N_m)=0$. Furthermore, we can restrict our attention to $n_0=1$, since any reasoning which will prove the desired assertion for $n_0=1$ can be repeated for any $n_0$ by replacing $f$ with $f^{(n_0)}$. This means that all we need to prove is that:

Given a Borel subset $A\subset X$, we have $\mu(A\cap\bigcap_{n\ge 1}f^{-n}(X\setminus A))=0$ (then we can aply this to $A=N_m$, since it’s easy to see that the $N_m$’s are Borel sets, and we’re done).

Let $M=A\cap\bigcap_{n\ge 1}f^{-n}(X\setminus A)$. This means that $M$ is a Borel subset of $X$ satisfying $M\subset f^{-n}(X\setminus M),\ \forall n\ge 1$. We can see from here that $M,f^{-1}(M),f^{-2}(M),\ldots$ are disjoint, and since they all have the same measure and $X$ has finite measure, we must have $\mu(M)=0$, as desired.


geometry
ratio
analytic geometry
geometry solved

Let $ A_1A_2A_3 $ be a triangle . $ B_i $ are the midpoints of $ A_iA_{i+1} $ , $ C_i $ are the midpoints of $ A_iB_i $ , $ D_i $ is the intersections of $ A_iC_{i+1} $ and $ B_iA_{i+2} $ . $ E_i $ are the intersections of $ A_iB_{i+1} $ and $ C_iA_{i+2} $ (i=1,2,3) . Find the ratio of $ S(D_1D_2D_3) $ and $ S(E_1E_2E_3) $ .


If n is a positive integer and $ n!+1 $ is not a prime find n such that
$ \frac{n^2+2n-576}{n!}+47=n! $

The cases $n \le 3$ are ruled out by the requirement that $n!+1$ be prime.
Since $n! - 47$ is an integer, it is necessary that $n! \mid n^2 +2n - 576$.  If $n \mid n^2+2n-576$ then $n \mid 576$; if $n-1 \mid n^2+2n-576$ then $n-1 \mid 573$; and if $n-2 \mid n^2+2n-576$ then $n-2\mid568$.  Together these imply $n=4$, which is the unique solution.


[color=blue]¿Alguien puede probar que cualquier número primo (positivo) de la forma $ 4k\plus{}1$ puede expresarse como la suma de dos cuadrados perfectos?[/color]

Primero hay que notar que (-1) es residuo cuadratico mod p=4k+1 ( se demuestra haciendo el test de elevar a la potencia (p-1)/2)
o sea, existe a≡-1(mod p).
Queremos mostrar que existen x,y tal que x^2≡-(y^2)(mod p) pero  x^2 +y^2<2p. Para eso consideremos los nros. de la forma ax-y con x e y enteros tales que 0≤x^2,y^2<p.  La cantidad de pares (x,y) es por lo tanto mayor que p, luego por Palomar existen dos nros. de esta forma q dejan el mismo resto mod p, es decir, existen b,c,d,e, tal que ab-c≡ad-e (mod p). Haciendo x=b-d e y=c-e tenemos que lo pedido pues ax≡y(mod p) y al elevar ambos miemros al cuadrado ya esta, pues x^2 +y^2<2p entonces se debe tener x^2 +y^2=p.

Falta completar la pregunta:
¿Qué enteros se pueden escribir como suma de ods cuadrados y porqué?

Buena pregunta por cierto..La respuesta es todo numero que cumple la propiedad:

 [hide]Al descomponerlo en factores primos, la potencia de cada primo $ p\equal{}3(mod 4)$ es par. [/hide]
Demostracion[hide]Queremos resolver la ecuacion $ x \equal{} a^2\plus{}b^2$ .
Sea d=mcd(a,b). Claramente $ d^2$ divide a x, luego $ x\equal{}Xd^2$.
Dividiendo la ecuacion por  $ d^2$ queda por resolver
 $ X \equal{} A^2\plus{}B^2$ con A,B,X coprimos dos a dos.
Supongamos por absurdo que X tiene un factor primo $ p\equal{}3(mod 4)$.Luego $ A^2\equal{}\minus{}B^2(mod p)$ Esto implica que $ A\equal{}B\equal{}0(mod p)$ que no puede ser pues A y B son coprimos, o bien que $ \minus{}1$ es residuo cuadratico mod  $ p\equal{}3(mod 4)$ que sabemos no puede ser. Analogamente $ 4\nmid X$ Luego es necesario que $ X\equal{}p_1^{k_1}....p_m^{k_m}$ o bien $ X\equal{}2p_1^{k_1}....p_m^{k_m}$ donde $ p_i\equal{}1(mod 4)$
Para mostrar la suficiencia, notamos que en la demostracion anterior solo era necesario que $ \minus{}1$ sea residuo cuadratico mod x para que x pueda expresarse como suma de dos cuadrados. Luego bastaria mostrar que $ \minus{}1$ es residuo cuadratico mod $ X\equal{}p_1^{k_1}....p_m^{k_m}$ y seguimos como en la demostracion anterior.
 Sea p un primo con $ p\equal{}1(mod 4)$ Mostramos por induccion en el exponente $ i$ que existe $ y_i$ tal que $ y_i^2\equal{}\minus{}1 (mod p^i)$.Sabemos que el caso base $ y\equal{}1$ es cierto.La induccion queda completa usando el lema de lifting del exponente ( Si $ a\equal{}b(mod p^k)$ entonces $ a^p\equal{}b^p(mod p^{k\plus{}1})$)
Entonces, siendo $ X\equal{}p_1^{k_1}....p_m^{k_m}$, para todo i existe $ y_i$ tal que $ y_i^2\equal{}\minus{}1 (mod p_i^{k_i})$. Por el teorema chino del resto concluimos que existe $ Y\equal{}y_i (mod p_i^{k_i})$ para todo i, lo que implica que $ Y^2\equal{}\minus{}1 (modX)$, osea,$ \minus{}1$ es residuo cuadratico mod $ X$ lo que completa la demostracion.[/hide]

[quote="maurizio"]Buena pregunta por cierto..La respuesta es todo numero que cumple la propiedad:

 [hide]Al descomponerlo en factores primos, la potencia de cada primo $ p \equal{} 3(mod 4)$ es par. [/hide]
Demostracion[hide]Queremos resolver la ecuacion $ x \equal{} a^2 \plus{} b^2$ .
Sea d=mcd(a,b). Claramente $ d^2$ divide a x, luego $ x \equal{} Xd^2$.
Dividiendo la ecuacion por  $ d^2$ queda por resolver
 $ X \equal{} A^2 \plus{} B^2$ con A,B,X coprimos dos a dos.
Supongamos por absurdo que X tiene un factor primo $ p \equal{} 3(mod 4)$.Luego $ A^2 \equal{} \minus{} B^2(mod p)$ Esto implica que $ A \equal{} B \equal{} 0(mod p)$ que no puede ser pues A y B son coprimos, o bien que $ \minus{} 1$ es residuo cuadratico mod  $ p \equal{} 3(mod 4)$ que sabemos no puede ser. Analogamente $ 4\nmid X$ Luego es necesario que $ X \equal{} p_1^{k_1}....p_m^{k_m}$ o bien $ X \equal{} 2p_1^{k_1}....p_m^{k_m}$ donde $ p_i \equal{} 1(mod 4)$
Para mostrar la suficiencia, notamos que en la demostracion anterior solo era necesario que $ \minus{} 1$ sea residuo cuadratico mod x para que x pueda expresarse como suma de dos cuadrados. Luego bastaria mostrar que $ \minus{} 1$ es residuo cuadratico mod $ X \equal{} p_1^{k_1}....p_m^{k_m}$ y seguimos como en la demostracion anterior.
 Sea p un primo con $ p \equal{} 1(mod 4)$ Mostramos por induccion en el exponente $ i$ que existe $ y_i$ tal que $ y_i^2 \equal{} \minus{} 1 (mod p^i)$.Sabemos que el caso base $ y \equal{} 1$ es cierto.La induccion queda completa usando el lema de lifting del exponente ( Si $ a \equal{} b(mod p^k)$ entonces $ a^p \equal{} b^p(mod p^{k \plus{} 1})$)
Entonces, siendo $ X \equal{} p_1^{k_1}....p_m^{k_m}$, para todo i existe $ y_i$ tal que $ y_i^2 \equal{} \minus{} 1 (mod p_i^{k_i})$. Por el teorema chino del resto concluimos que existe $ Y \equal{} y_i (mod p_i^{k_i})$ para todo i, lo que implica que $ Y^2 \equal{} \minus{} 1 (modX)$, osea,$ \minus{} 1$ es residuo cuadratico mod $ X$ lo que completa la demostracion.[/hide][/quote]

Bien, es cierto que basta mirar si $ \minus{}1$ es residuo cuadrático módulo cada primo. Está muy linda la idea de usar el lifting. Sin emargo, es más fácil abservar el siguiente lema:
[b]Lema:[/b]
Si [b]X[/b] y [b]Y[/b] se pueden escribir como suma de cuadrados [b]XY[/b] también.

Asñi como los primos $ P\equal{}4k\plus{}1$ se pueden escribir como suma de cuadrados $ X \equal{} p_1^{k_1}....p_m^{k_m}$ (con primos de esta forma) tambíen se puede escribir como suma de cuadrados.

Alguien se le ocurre como probar el lema??

[quote="skuge"][quote="maurizio"]Buena pregunta por cierto..La respuesta es todo numero que cumple la propiedad:

 [hide]Al descomponerlo en factores primos, la potencia de cada primo $ p \equal{} 3(mod 4)$ es par. [/hide]
Demostracion[hide]Queremos resolver la ecuacion $ x \equal{} a^2 \plus{} b^2$ .
Sea d=mcd(a,b). Claramente $ d^2$ divide a x, luego $ x \equal{} Xd^2$.
Dividiendo la ecuacion por  $ d^2$ queda por resolver
 $ X \equal{} A^2 \plus{} B^2$ con A,B,X coprimos dos a dos.
Supongamos por absurdo que X tiene un factor primo $ p \equal{} 3(mod 4)$.Luego $ A^2 \equal{} \minus{} B^2(mod p)$ Esto implica que $ A \equal{} B \equal{} 0(mod p)$ que no puede ser pues A y B son coprimos, o bien que $ \minus{} 1$ es residuo cuadratico mod  $ p \equal{} 3(mod 4)$ que sabemos no puede ser. Analogamente $ 4\nmid X$ Luego es necesario que $ X \equal{} p_1^{k_1}....p_m^{k_m}$ o bien $ X \equal{} 2p_1^{k_1}....p_m^{k_m}$ donde $ p_i \equal{} 1(mod 4)$
Para mostrar la suficiencia, notamos que en la demostracion anterior solo era necesario que $ \minus{} 1$ sea residuo cuadratico mod x para que x pueda expresarse como suma de dos cuadrados. Luego bastaria mostrar que $ \minus{} 1$ es residuo cuadratico mod $ X \equal{} p_1^{k_1}....p_m^{k_m}$ y seguimos como en la demostracion anterior.
 Sea p un primo con $ p \equal{} 1(mod 4)$ Mostramos por induccion en el exponente $ i$ que existe $ y_i$ tal que $ y_i^2 \equal{} \minus{} 1 (mod p^i)$.Sabemos que el caso base $ y \equal{} 1$ es cierto.La induccion queda completa usando el lema de lifting del exponente ( Si $ a \equal{} b(mod p^k)$ entonces $ a^p \equal{} b^p(mod p^{k \plus{} 1})$)
Entonces, siendo $ X \equal{} p_1^{k_1}....p_m^{k_m}$, para todo i existe $ y_i$ tal que $ y_i^2 \equal{} \minus{} 1 (mod p_i^{k_i})$. Por el teorema chino del resto concluimos que existe $ Y \equal{} y_i (mod p_i^{k_i})$ para todo i, lo que implica que $ Y^2 \equal{} \minus{} 1 (modX)$, osea,$ \minus{} 1$ es residuo cuadratico mod $ X$ lo que completa la demostracion.[/hide][/quote]

Bien, es cierto que basta mirar si $ \minus{} 1$ es residuo cuadrático módulo cada primo. Está muy linda la idea de usar el lifting. Sin emargo, es más fácil abservar el siguiente lema:
[b]Lema:[/b]
Si [b]X[/b] y [b]Y[/b] se pueden escribir como suma de cuadrados [b]XY[/b] también.

Asñi como los primos $ P \equal{} 4k \plus{} 1$ se pueden escribir como suma de cuadrados $ X \equal{} p_1^{k_1}....p_m^{k_m}$ (con primos de esta forma) tambíen se puede escribir como suma de cuadrados.

Alguien se le ocurre como probar el lema??[/quote]

$ X\equal{}a^2\plus{}b^2$, $ Y\equal{}c^2\plus{}d^2$,..$ a,b,c,d \in Z^{\plus{}}$
$ XY\equal{}(ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2$
$ 2\equal{}1^2\plus{}1^2$.
Disculpa, no entiendo eso de la idea de usar el lifting....explica profas..bye..

[quote="aev5peru"]
$ X \equal{} a^2 \plus{} b^2$, $ Y \equal{} c^2 \plus{} d^2$,..$ a,b,c,d \in Z^{ \plus{} }$
$ XY \equal{} (ac \plus{} bd)^2 \plus{} (ad \minus{} bc)^2$
$ 2 \equal{} 1^2 \plus{} 1^2$.
Disculpa, no entiendo eso de la idea de usar el lifting....explica profas..bye..[/quote]
Sí también se puede:
$ X \equal{} z\bar{z}$, $ Y \equal{} w\bar{w}$        $ z,w \in Z[i]$
$ XY \equal{} (zw)\bar{(zw)}$
(Como ni $ z$ ni $ w$ son enternos reales o complejos puros $ (zw)\bar{(zw)}$ es una suma de cuadrados perfectos en enteros reales):lol:

para los que no conoscan "Lifting the exponent":
[url=http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf]A nice and tricky lemma (lifting the exponent)[/url]

Bueno hablando de primos q se pueden expresar como la suma de dos cuadrados.
p=1(mod4), => existen $ x,y \in Z^\plus{}$ talque $ p\equal{}x^2\plus{}y^2$, mi duda es q si es conocido q estos $ x,y$ son unicos (porq lo son).

Amilcar, esses x e y são de fato únicos.
Não sei provar, mas são! :P

[quote="aev5peru"]Bueno hablando de primos q se pueden expresar como la suma de dos cuadrados.
p=1(mod4), => existen $ x,y \in Z^ \plus{}$ talque $ p \equal{} x^2 \plus{} y^2$, mi duda es q si es conocido q estos $ x,y$ son únicos (porq lo son).[/quote]
BUeno, conosco dos formas de probar este hecho.
-La primera es usando que si $ n\equal{}x^2\plus{}y^2$ , el número de soluciones $ (x,y)$ en los números enteros es $ 4(d_1\minus{}d_3)$, donde $ d_i$ es el número de divisores congruentes a $ i$ módulo 4.
-La segunda es suponer por contradicción: $ p\equal{}x^2\plus{}y^2\equal{}a^2\plus{}b^2$.
chaufas!


integration
calculus
logarithms
limit
function
derivative
real analysis

im trying to calculate the asymptotic estimate of

$ P_d(z)\equal{}\sum_n z^n n^{\minus{}d}$ as $ z\uparrow 1$,

even a link to a nice description on how to do things like this in general would be appreciated,
 but then again so would a solution if one exists

Start that sum at $ n\equal{}1$,so we don't have to worry about powers of zero.

Quick facts: $ P_d(z)$ is bounded for $ d>1$. The series converges uniformly on the closed disk, and the asymptotic estimate is simply the sum $ \zeta(d)$ of the series. For $ d\equal{}1$, this is $ \minus{}\ln(1\minus{}z)$. It's $ \frac{z}{1\minus{}z}$ for $ d\equal{}0$, $ \frac{z}{(1\minus{}z)^2}$ for $ d\equal{}\minus{}1$, and $ \frac{q_d(z)}{(1\minus{}z)^{1\minus{}d}}$ for integer values of $ d$, where $ q_d$ is a polynomial. Since $ P_{d\minus{}1}(z)\equal{}zP_d'(z)$, we have $ q_{d\minus{}1}(z)\equal{}(1\minus{}z)q_d'(z)\plus{}(1\minus{}d)q_d(z)$, and $ q_{d\minus{}1}(1)\equal{}(1\minus{}d)(\minus{}d)(\minus{}d\minus{}1)\cdots (1)\equal{}(1\minus{}d)!$.

Conjecture: $ P_d(z)\approx \begin{cases}\zeta(d)& d>1\\ \minus{}\ln(1\minus{}z)& d\equal{}1\\ \Gamma(1\minus{}d)\cdot(1\minus{}z)^{d\minus{}1}& d<1\end{cases}$

The first two lines are already established, so we focus on the third, with $ d<1$. Here's one way to do it:
As $ z\to 1^{\minus{}}$, the sum $ \sum_{n\equal{}1}^{\infty}\left(z^{n\minus{}1}\minus{}z^n\right)f(z^n)$ converges to $ \int_0^1 f(t)\,dt$; this is just the Riemann integral with a geometric partition. It would be nice to match the sum $ \sum_{n\equal{}1}^{\infty}z^n n^{\minus{}d}$ to this pattern. The $ z^n$ is easy to handle, and leaves us with merely a $ \frac1z\minus{}1$ factor to account for in the first term. To get $ n^{\minus{}d}$, we take $ f(t)\equal{}(\minus{}\ln t)^{\minus{}d}$ The sum becomes
$ I(d)\equal{}\int_0^1 (\minus{}\ln t)^{\minus{}d}\,dt$
$ \equal{}\lim_{z\to 1^\minus{}}\sum_{n\equal{}1}^{\infty}\left(\frac1z\minus{}1\right)\cdot z^n\cdot \left(\minus{}\ln z^n\right)^{\minus{}d}$
$ \equal{}\lim_{z\to 1^\minus{}}\left(\frac1z\minus{}1\right)\sum_{n\equal{}1}^{\infty}z^n\cdot n^{\minus{}d}\cdot (\minus{}\ln z)^{\minus{}d}$
$ \equal{}\lim_{z\to 1^\minus{}}\left(\frac{1\minus{}z}{z}\right)\cdot(\minus{}\ln z)^{\minus{}d}\cdot\sum_{n\equal{}1}^{\infty}z^n\cdot n^{\minus{}d}$
Now, $ \minus{}\ln z\approx 1\minus{}z$ for $ z$ near 1, so we get that $ P_d(z)\cdot (1\minus{}z)^{1\minus{}d}\approx I(d)$. Substituting $ t\equal{}e^{\minus{}x}, dt\equal{}\minus{}e^{\minus{}x}\,dx$, $ I(d)\equal{}\int_0^{\infty} x^{\minus{}d}e^{\minus{}x}\,dx\equal{}\Gamma(\minus{}d\plus{}1)$. This completes the proof of the estimate $ P_z(d)\approx I(d)\cdot (1\minus{}z)^{d\minus{}1}\equal{}\Gamma(1\minus{}d)(1\minus{}z)^{d\minus{}1}$ for $ d<1$.

That was a fairly tricky method. There are certainly others, but the key to anything which will get you the constant is to estimate the sum with an integral.

thank you very much, but i was actually trying to find an estimate of how $ P_d(z)$ approaches these limits.

I am actually working with a slightly different sum $ \hat{P}_d(z)\equal{}\sum_{n\ge 1} z^n L(n) n^{\minus{}d}$ where $ L(n)$ is a slowly varying function (i.e. $ \lim_{x \to \infty} \frac{L(cn)}{L(n)}\equal{}1$ for all c) we can just generalize to $ L(x) \to 1$ as $ x\to \infty$
and i have a nice relationship between $ z(\beta), \beta$ and $ \hat{P}_d(z(\beta))$,

so knowing how $ \hat{P}_d(z)\to \zeta(d)$ would give me something to examine how $ z(\beta) \uparrow 1$ as $ \beta \downarrow \beta_0$

That's pretty simple. Just differentiate term by term in $ z$:
$ P_d(z)\approx P_d(1) \plus{} (z \minus{} 1)\sum_{n \equal{} 1}^{\infty}nz^{n \minus{} 1}\cdot n^{ \minus{} d}$
$ \approx P_d(1) \plus{} \frac {z \minus{} 1}{z}\sum_{n \equal{} 1}^{\infty}z^n\cdot n^{1 \minus{} d}$
$ \approx P_d(1) \minus{} \frac {1 \minus{} z}{z}P_{d \minus{} 1}(z)$

The derivative basically just takes you up one power; this also applies to your variant.

OK, that formula's only really valid when the series for $ P_{d\minus{}1}$ converges uniformly. When $ d$ is too small for that, we use $ \frac1z\int_z^1 P_{d\minus{}1}(t)\,dt$. That integral converges and thus produces a meaningful estimate as long as $ d>1$. For larger $ d$, even more derivatives can be taken.


James got 75% of the questions on a test correct.  If he answered 11 of the first 13 questions correctly, and 2/3 of the remaining questions correctly, how many questions were on the test?

[quote="sharkman"]I understand what you did.

Why did you use x for 2 and 3?

I set it up this way:

11/13 + (2/3)x = 0.75 and got no where.

Your method is much easier.[/quote]
I put the number of questions that James got right on top, and the total number on the bottom. The problem states he got $ 11$ of the first $ 13$ right, so I put $ 11$ on top and $ 13$ on the bottom. He also got $ \tfrac23$ of the remaining questions right, so I added $ 2x$ on top, and $ 3x$ on the bottom, and set the whole thing equal to $ \tfrac34$. Then I solved for $ x$, and found the total amount of questions once I knew what $ x$ was.

post spam in the games and fun factory forum, not here, okay?


find all natural (x,y,z):
$ x^2\equal{}2(y\plus{}z)$
$ x^6\equal{}y^6\plus{}z^6\plus{}31(y^2\plus{}z^2)$

$ x^2 \equal{} 2(y \plus{} z)$
$ x^6 \equal{} y^6 \plus{} z^6 \plus{} 31(y^2 \plus{} z^2)$

$ 8(y\plus{}z)^3\equal{} y^6 \plus{} z^6 \plus{} 31(y^2 \plus{} z^2)$

WLOG $ z \ge y$
Suppose that $ z\ge 4$
$ z (z^3\minus{}64)\plus{}31>0 \iff z^4\plus{}31>64z \iff z^6\plus{}31z^2 >64z^3$
$ 64z^3 \ge 8(y\plus{}z)^3$ contradiction
then $ y \le z \le 3$

With only solution for $ (y,z)\equal{}(1,1)$ adn $ x\equal{}2$

is it ok?


trigonometry
inequalities
search
inequalities unsolved

As I know, this problems is belong to Darji's

[quote="kunny"]Let $a=\cos \alpha ,\ b=\cos \beta ,\ c=\cos \gamma \left (0<\alpha ,\beta , \gamma <\frac{\pi}{2}\right)$[/quote]
I think you are right only if $a,b,c\in (0,1)$ but there is no restriction.

Let $b+c=s,\ bc=t$, since $b>0,\ c>0\Longleftrightarrow s>0,\ t>0,\ s^{2}\geq 4t$.
Our inequality will become $a^{2}+2(s-t)a+s^{2}-4t+1\geq 0$.
Proof:$f(a): =a^{2}+2(s-t)a+s^{2}-4t+1\geq a^{2}-2(s-t)a+1$
$=\{a-(s-t)\}^{2}+1-(s-t)^{2}$. We prove that $1-(s-t)^{2}\geq 0$.
Let $P=\{(s,\ t)|s>0,\ t>0,\ t\leq \frac{1}{4}s^{2}\}, \ Q=\{(s,\ t)|s>0,\ t>0,-1\leq s-t\leq 1\}$.
Sketching the region of $P$ and $Q$ on $s-t$ plane gives $P\in{Q}\Longleftrightarrow f(a)\geq 0$ which completes the proof.

[quote="Mr.JN2P"]For $a, b,c >0$. Prove that
$a^{2}+b^{2}+c^{2}+2abc+1\geq 2(ab+bc+ca)$[/quote]

See darij's post on

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=575919256&t=19666

carij


[quote="Agr_94_Math"]Given real numbers $ a,b,c$ such that $ \frac {1}{2} \le a,b,c \le 1$
Prove that $ 2\le \sum \frac {a \plus{} b}{1 \plus{} c} \le 3$.[/quote]

Search for Romania 2006 .

$ \frac {a \plus{} b}{1 \plus{} c} < \equal{} \frac {2(a \plus{} b)}{3}, \frac {b \plus{} c}{1 \plus{} a} < \equal{} \frac {2(b \plus{} c)}{3}, \frac {a \plus{} c}{1 \plus{} b} < \equal{} \frac {2(a \plus{} c)}{3}$.on adding we get $ \sum_cyc \frac {a \plus{} b}{1 \minus{} c} < \equal{} \frac {4(a \plus{} b \plus{} c)}{3} < \equal{} 3.$
i am unable to prove the left hand side :(

http://www.mathlinks.ro/Forum/viewtopic.php?t=84221
also,
[quote="aadil"]$ \frac {a \plus{} b}{1 \plus{} c} < \equal{} \frac {2(a \plus{} b)}{3}, \frac {b \plus{} c}{1 \plus{} a} < \equal{} \frac {2(b \plus{} c)}{3}, \frac {a \plus{} c}{1 \plus{} b} < \equal{} \frac {2(a \plus{} c)}{3}$.on adding we get $ \sum_cyc \frac {a \plus{} b}{1 \minus{} c} < \equal{} \frac {4(a \plus{} b \plus{} c)}{3} < \equal{} 3.$
i am unable to prove the right hand side :([/quote]
am i too dum but how is  $ \frac {4(a \plus{} b \plus{} c)}{3} < \equal{} 3$ that is not true is it?wat abt $ a\equal{}b\equal{}c\equal{}1$


function
algebra
domain
algebra unsolved

I want to find the function solution of this equation:

$ xf(y)\plus{}yf(x)\equal{}(x^2\plus{}y^2)f(xy)$.

I found $ f(0)\equal{}0$, and $ f(x)\equal{}\frac{1}{x}$ if $ x\neq 0$. But I don't know if it is the

unique solution.

Thanks in advance for some hint.

Hint:
[hide]Fill in $ x \equal{} 1$ (i suppose it's in the domain) and look what happens[/hide]

Solution:
[hide] Filling in $ x \equal{} 1$ gives us $ yf(1) \equal{} y^2f(y)$, so all satisfying functions are of the form
\[ f(x) \equal{} \frac {c}{x}\]
for some $ c \in \mathbb{R}$; for all $ x \neq 0$. They do all satisfy the conditions. [/hide]


algebra
polynomial
linear algebra
matrix
invariant
linear algebra unsolved

Find the class of similitude of those real matrices $A$, $3nx3n$ such that $A^3=0$ and $rank A=2n$.

perhaps I'm wrong but cyclic spaces decomposition seems to work:

1/ it is easy to check that the minimal polynomial of $A$ is $x^3$
2/ we know that a matrix $M$ is similar to $A$ iff their invariant of similitude are the same (this is why it is called invariant of similitude isn't it  :D  ) So there exists a sequence of polynomial $P_1,..,P_k=x^3$ such that $P_i | P_{i+1}$ that are these invariants of similitude. So $P_i$ is either $x$, $x^2$ or $x^3$. Suppose that $P_1=P_2=..=P_a=x$, $P_{a+1}=P_2=..=P_{a+b}=x^2$, $P_{a+b+1}=P_2=..=P_{k}=x^3$. Then if we use the fact that $Rank(A)=2^n$ we can see that in fact $P_1=..=P_k=x^3$, if I did not made stupid mistakes.

  So I think that the similitude class of $A$ is exactly those matrices such that $A^3=0$ and $rank(A)=2n$. Is it ?

there exist $P\in GL_{3n}(R)$ such that

$P^{-1}AP=\left(\begin{array}{ccc}0&0&0\\I_n&0&0\\0&I_n&0\end{array}\right)$

Yes, but the proof is too complicated, Alekk. There is a simple proof in 4 lines.  ;)  I don't know if at X they will accept all the classical things you presented. Anyway, I like more your solution, since it is more natural.


real analysis
geometry
rectangle
algebra
function
domain
advanced fields

Given $n \in \mathbb{N}$, let $\lambda_{n}$ denote Lebesgue measure in $\mathbb{R}^{n}$. Show that $\lambda_{m+n}$ = $\lambda_{m} \times \lambda_{n}$.

It probably suffices to show $\lambda^{*}_{m+n}$ = $(\lambda_{m} \times \lambda_{n})^{*}$. (Equality of the outer measures)

Check any construction of the product measure: such a measure is entirely characterized by its values on measurable rectangles. In the case of the product of two Lebesgue measures, this product clearly corresponds to the lebesgue measure of the (cartesian) product space on the measurable rectangles whose sides are open intervals, and from here it is easy to show that they correspond on any measurable rectangles.

P.S: $B_{m} \times B_{n} = B_{m+n}$ but $L_{m} \times L_{n} \subsetneq L_{m+n}$ where $B$ and $L$ denote the Borel and Lebesgue $\sigma$-algebras respectively. In order to obtain $L_{m+n}$, you need to complete $L_{m} \times L_{n}$ first. That means what we did above is rigorously true only for Borel $\sigma$-algebras, otherwise the measures we talked about are different (since their domains are different).

I'm still having difficulty with this problem. Guess I need more help.


limit
function
real analysis
real analysis theorems

[b]If you read this thread and you're intimidated by all the discussion, please don't care. What I really want to know is: Is it obvious that we are ALLOWED TO REARRANGE THE TERMS IN A POSITIVE SERIES? If there is an easy proof/way to informally understand it please tell me. Even if there isn't one, please tell me so I can stop thinking about this![/b]

This deals with a Mandelbrot problem fron intermediate (http://www.mathlinks.ro/Forum/viewtopic.php?t=67347). We are to show that $\sum\limits_{j=1}^{\infty}\sum\limits_{k = 1}^{\infty}\frac{1}{(j+k)^3} = \sum_{n=1}^\infty \frac{n-1}{n^3}$. To quote scorpius, the method to see this is:

[quote="scorpius119"]Because the equation $j+k=n$ has $n-1$ solutions in positive integers $j$ and $k$, the term $\frac{1}{n^3}$ will appear $n-1$ times in the double sum, so the sum becomes $\sum_{n=1}^\infty \frac{n-1}{n^3}$. [/quote]

Now, always locking myself up in petty details as I am prone to do (I've spent the past two hours thinking about this. Being me is a curse), I see huge problems with this reasoning. When I establish equality between two series, I want the partial sums to equal each other (at least sometimes), so that the limits are equal to each other. But for this reasoning, the partial sums are NEVER actually equal to each other. So, while the reasoning is easy enough to understand intuitively, I don't see how to formalize the "rearrangement" of terms that is done... seeing as we can't cut off at a number of terms and establish the equality.

Do you understand my problem with this? If you read the original thread you will see that I've reduced the issue to just regrouping one series instead of the original "double series"*, but the same kind of problem with the rearrangement of terms appear.

Please help!

* I've shown that $\sum_{j=1}^m\sum_{k=1}^{\infty}\frac{1}{(j+k)^3} = \sum_{n=1}^{\infty}\frac{n-1}{n^3} - \sum_{k=m+2}^{\infty}\frac{k - m - 1}{k^3}$, but again, this is not true if you only look at a partial sum instead of infinity.

the fact is that series with positive terms commute, and can be rearranged...
it's a matter of absolute convergence, as far as i know..

Ok. A lot of strange things happening... Is this true then:
If both series are convergent, then $\sum_{k=1}^{\infty}f(k) + \sum_{j=1}^{\infty}f(j) = \sum_{n=1}^{\infty}(f(n) + g(n))$

It should be true from the definition of limits, but it seems like such strange things happen with series. So, is this result true, or I is there something I did not think of?

[i was just stating false and stupid things :p]

anyway.. your question is slightly different (the original one): is it always true that $\sum_i \sum_j a_{i,j} = \sum_j \sum_i a_{i,j}$ or $\sum_i \sum_j a_{i,j} = \sum_s \sum_k a_{k,s-k}$?
in general, answer is no...
i don't know the sharp hypotesis you need on it, though.. maybe someone (kent, fedja, moubinool) could give us some enlightenment :)

I am going nuts now. Nothing is what I want it to be... Ok, so what's wrong here:

$\sum_{k=1}^{\infty}(f(k) + g(k)) = \lim_{n\to\infty}\sum_{k=1}^{n}(f(k) + g(k)) = \lim_{n\to\infty}\left(\sum_{k=1}^{n}f(k) + \sum_{k=1}^{n}g(k)\right) = \lim_{n\to\infty}\sum_{k=1}^{n}f(k) + \lim_{n\to\infty}\sum_{k=1}^{n}g(k) = \sum_{k=1}^{\infty}f(k) + \sum_{k=1}^{\infty}g(k)$

From this we could make the appropriate changes in variable, and what I said should be true (we are working under the assumption that both series are convergent). Why is this not true? This is upsetting.  :P

er..
ok, i'll edit it :p
that works, my mistake :D

That's the beauty of the statement scorpius made.  It is a different procedure for counting.

$\sum\limits_{j=1}^{\infty}\sum\limits_{k = 1}^{\infty}\frac{1}{(j+k)^3} = \frac1{2^3}+\frac1{3^3}+\frac1{4^3}+...+\frac1{3^3}+\frac1{4^3}+...+\frac1{4^3}+...$

For this problem you are staring at infinite infinitely long series.  If you just look at the grouping's I wrote you can easily see the relevance of the single summation.

The problem is you can't easily look at a partial sum, unless you change the limit so that it might be like

$\sum_{j=1}^{m} \sum_{k=1}^{n-j} \frac1{(j+k)^3}$

Ignoring the ellipsis's above is this function for $m=3$ and $n=4$

If you want to more rigorously show that you are allowed to rearrange the terms as shown, although this is similar to the original method, then we can:

First let's make a chart for all $(j+k)$
2 3 4 5 6 7 8 9 10 11 ...
   3 4 5 6 7 8 9 10 11 ...
      4 5 6 7 8 9 10 11 ...
         5 6 7 8 9 10 11 ...
            6 7 8 9 10 11 ...
               7 8 9 10 11 ...
                  8 9 10 11 ...
                     9 10 11 ...
                        10 11 ...
                             11 ...

Now we can obviously see the first approach in action.

Let $S_j = \sum_{i=1}^{\infty} \frac1{(j+k)^3}$

Then $S_j - S_{j+1} = \frac1{(j+1)^3}$

Each successive $S_j$ is losing it's previous first term.  Then you can show that there are a given number of $j+k$ for the entire array of values, in fact $j+k-1$

So this can be rewritten as $\sum_{j+k=2}^{j+k = \infty} \frac{j+k-1}{(j+k)^3} = \sum_{n=2}^{\infty} \frac{n-1}{n^3} = \sum_{n=1}^{\infty} \frac{n-1}{n^3}$

Hopefully my explanation helps, but essentially you are just tracking the the sums of the two indices through the primary index.

[quote="Kalle"]$\sum_{k=1}^{\infty}(f(k) + g(k)) = \lim_{n\to\infty}\sum_{k=1}^{n}(f(k) + g(k)) = \lim_{n\to\infty}\left(\sum_{k=1}^{n}f(k) + \sum_{k=1}^{n}g(k)\right) = \lim_{n\to\infty}\sum_{k=1}^{n}f(k) + \lim_{n\to\infty}\sum_{k=1}^{n}g(k) = \sum_{k=1}^{\infty}f(k) + \sum_{k=1}^{\infty}g(k)$[/quote]

I think you meant that.

Where are you having your mistakes?

[quote="FMako"]
Where are you having your mistakes?[/quote]
Nah, apparently I'm not having any. It's just that ma_go said that this statement wasn't true at first, but he then changed his mind. Well, the only thing I really feel unsure about is the change of variable, I guess. If we have the sum of two limits with two different variables, it is always OK to make a substitution so that they depend on the same variable, right? So using the limit laws we can evaluate this as one limit?

And for the original problem, I'm glad you did all the work for me, but I still don't believe you :P I am okay with the fact that we have a rearrangement, I'm just not sure that we have shown the first series to have the same value as the second series. Isn't your more rigorous approach just a more rigorous way of saying that it is a rearrangement? Maybe I'm missing something.

[quote="FMako"]I didn't do a really good job of verifying it, mainly because it seems obvious.

I guess I don't understand what you don't understand.[/quote]
The value of a series DOES depend on order! See for example http://mathworld.wolfram.com/RiemannSeriesTheorem.html . I don't know if there is some obvious proof that rearrangements don't matter for positive series. I couldn't come up with it..


function
algebra
polynomial
calculus
derivative
calculus computations

42.  What is the greatest value of $ b$ for which any function $ f$ that satisfies the following properties must also satisfy $ f(1) < 5$?

(i) $ f$ is infinitely differentiable on the real numbers;
(ii) $ f(0)\equal{}1$, $ f'(0)\equal{}1$, and $ f''(0)\equal{}2$; and
(iii) $ |f'''(x)|<b$ for all $ x$ in $ [0,1]$.

(A) 1
(B) 2
(C) 6
(D) 12
(E) 24

This is a very direct application of a Taylor polynomial with remainder.

$ f(1)\equal{}f(0)\plus{}f'(0)\plus{}\frac{f''(0)}{2}\plus{}\frac{f'''(\xi)}{3!}$ for some $ \xi\in(0,1).$

Put in the specific information given:

$ 1\plus{}1\plus{}1\plus{}\frac{f''(\xi)}{6}<5$

$ f'''(\xi)<12.$

So the answer is (D), 12.

If you just wrote $ f(x)\equal{}1\plus{}x\plus{}x^2\plus{}\frac{bx^3}{6}$ and evaluated at $ x\equal{}1,$ you'd get there. (That may be playing a little fast and loose with the proof, but you weren't asked for the proof.) Speed is very important on the subject GRE, as you have less than 3 minutes per problem; this one should belong to the category of "settle it in less than a minute," as the form the information is given in should make you think of Taylor polynomials.

(The condition that $ f\in C^{\infty}$ is overkill, since you only need a specific finite number of derivatives - but it doesn't hurt anything. Stating it that way simply removes it from the list of things you have to think about.)


modular arithmetic
number theory proposed
number theory

For a positive integer $ n$ and $ n$ is a divisor of $ 1 \plus{} \displaystyle\sum_{1\equal{}1}^{n\minus{}1}\ i^{n\minus{}1}$
Prove that $ n$ is squarefree .

well just see that if $ p |n$ and $ pk \equal{} n$ then we have
$ 0 \equiv 1 \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}i^{n \minus{} 1} \equiv 1 \plus{} k(\sum_{i \equal{} 1}^{p \minus{} 1}i^{n \minus{} 1}) \pmod {p}\implies k(\sum_{i \equal{} 1}^{p \minus{} 1}i^{n \minus{} 1}) \equiv \minus{} 1 \pmod {p} \implies \gcd(p,k) \equal{} 1$.
so we are done as u saw no matter what the exponent is( we must just have $ n > 1$).!
about Giuga's conjecture:
 http://en.wikipedia.org/wiki/Giuga%27s_conjecture

this looks similar to the giuga's conjecture .
That is if $ \displaystyle\sum_{i \equal{} 1}^{n \minus{} 1}\ i^{n \minus{} 1} \plus{} 1 \equiv 0\bmod{n}$ then $ n$ is prime  :D


[quote="mr.danh"]On aline, take three distinct points A,B,C in this order. Draw the tangnts AD,AE to the circle $ (\omega)$ whose diameter is BC (D,E are the touching points). H is the orthogonal projection of D on CE. P is the midpoint of DH. CP meets again $ (\omega)$ at Q. Show that the circumcircle (ADQ) is tangent to the line AC.[/quote]
$ DE$ cut $ XY$ at the point X.
Because $ X,P$ is midpoints of $ DE$ and $ DH$ so $ XP\parallel{}HE$
So $ \angle{DXP} \equal{} \angle {DEC} \equal{} \angle{DBP}$
So $ B,X,P,D$ are cyclic .
Because $ DH\perp EC$ so $ XP\perp  DP$
It means that $ B,X,P,D$ lie on the circle diameter $ DX$ .Call it is $ (C)$
Because $ AX\perp DX$ so $ AX$ is a tangent of  $ (C)$
So $ \angle{BXA} \equal{} \angle{BDX} \equal{} \angle{AEB}$
It follows that $ A,B,E,C$ are cyclic on the circle with diameter $ AE$.
Because $ A,B,E,C$ are cyclic so $ \angle {XAQ} \equal{} \angle {BEX} \equal{} \angle {ADB}$
So $ AC$ is tangent of the circle $ (ADQ)$ at  point A.

[quote="mr.danh"]On aline, take three distinct points A,B,C in this order. Draw the tangnts AD,AE to the circle $ (\omega)$ whose diameter is BC (D,E are the touching points). H is the orthogonal projection of D on CE. P is the midpoint of DH. CP meets again $ (\omega)$ at Q. Show that the circumcircle (ADQ) is tangent to the line AC.[/quote]
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=123012#123012
Here is another solution:
[hide="Solution"]
Let $ \angle ACE \equal{} \beta$ and $ \angle QCB \equal{} \alpha$.  Furthermore, let $ DE$ and $ BC$ to meet at $ X$.  Now,\[ \angle QDX \equal{} \angle QDB \plus{} \angle BDE \equal{} \angle BCE \plus{} \angle QCB \equal{} \alpha \plus{} \beta\]Also, since the center of the circle lies on $ BC$, we have that $ D$ and $ E$ are symmetric about $ BC$, so $ DX \equal{} XE$.  This means that $ XP$ is a midline of $ \triangle DEH$, so $ XP\parallel EH$.  Then, $ \angle PXC \equal{} \angle XCE \equal{} \beta$, so $ \angle QPX \equal{} \angle PXC \plus{} \angle PCB \equal{} \alpha \plus{} \beta \equal{} \angle QDX$ and so $ QDPX$ is cyclic.  Now, $ \angle XPD \equal{} \angle EHD \equal{} 90$, so $ \angle QXD \equal{} \angle QPD \equal{} \angle DPX \minus{} \angle QPX \equal{} 90 \minus{} \alpha \minus{} \beta$.  $ D$ and $ E$ are symmetric about $ BC$, so $ \angle DXA \equal{} 90$, which means that $ \angle QXA \equal{} 90 \minus{} \angle QXD \equal{} \alpha \plus{} \beta$.  Also, $ \angle QEB \equal{} \angle QCB \equal{} \alpha$ and $ \angle BEA \equal{} \beta$.  Thus, $ \angle AEQ \equal{} \angle AEB \plus{} \angle BEQ \equal{} \alpha \plus{} \beta \equal{} \angle QXA$, so $ QXEA$ is cyclic.  Since $ D$ and $ E$ are symmetric about $ BC$, we notice that $ \beta \equal{} \angle AEB \equal{} \angle ACE \equal{} \angle BCD \equal{} \angle BEX$, so $ \angle AEX \equal{} \angle AEB \plus{} \angle BEX \equal{} 2\beta$.  Therefore, $ \angle XQA \equal{} 180 \minus{} \angle AEX \equal{} 180 \minus{} 2\beta$, so $ \angle QAX \equal{} 180 \minus{} \angle AQX \minus{} \angle QXA \equal{} \beta \minus{} \alpha$.  Yet, $ \angle QCD \equal{} \angle BCD \minus{} \angle BCQ \equal{} \beta \minus{} \alpha \equal{} \angle QAX$, so $ \angle ADQ \equal{} \angle QCD \equal{} \angle QAX$, so the circumcircle of $ \triangle AQD$ is tangent to $ AC$ at $ A$. [/hide]

Let $ T\equal{}DE\cdot BC$ then $ DT\equal{}TE$ this means $ TP\parallel CE$ $ \Rightarrow$ $ PT\perp DH$ because $ P$ is midpoint then, $ \angle{TDP}90\minus{}\angle{DEH}\equal{}90\minus{}\alpha\equal{}\angle{THP}$ because, $ \angle{PQD}\equal{}\alpha$ $ \Rightarrow$ $ QDPT$ is cyclic. Then, $ TQ\perp QD$ let $ DQ\cdot AB\equal{}X$ then, $ X$ is midpoint of $ AT$ $ \angle{ADX}\equal{}\angle{DCP}$ and, $ \angle{PCH}\equal{}\angle{QDT}$ and $ \triangle DHC \equiv \triangle{ATD}$. and because $ TQ\perp XD$ $ XQ\cdot XD\equal{}XT^2\equal{}XA^2$ $ \Rightarrow$ $ AX$ is tangent to the circumcircle of $ \triangle{AQD}$


modular arithmetic
number theory unsolved
number theory

Dtermine all x,y such that 2^x+3^y is a perfect square.

Suppose $2^x+3^y=z^2$. First we check small cases: for $x=0$ we have $z^2-1=(z+1)(z-1)=3^y$, then $z+1, z-1$ are powers of $3$ that differ by $2$, so $z-1=1, z+1=3, z=2, y=1$. Similarly if $y=0$ we have $(z+1)(z-1)=2^x$, so $z+1, z-1$ are powers of $2$ that differ by $2$. Then $z-1=2, z+1=4, z=3, 2^x=8, x=3$. So we've found $(0, 1)$ and $(3, 0)$. Now suppose $x, y \geq 1$.

Clearly since $3 \mid 3^y$, and $3 \nmid 2^x$ we see we must have $z^2 \equiv 2^x \equiv 1 \pmod{3}$, i.e. $x=2n$. Now since $x \geq 2$, $4 \mid 2^x, 4 \nmid 3^y$, so $z^2 \equiv 3^y \equiv 1 \pmod{4}$, i.e. $y=2k$.

So $2^{2n}+3^{2k}=z^2$. We have $3^{2k}=(z+2^n)(z-2^n)$. So $z-2^n=3^a, z+2^n=3^b$ are powers of 3 that differ by $2^{n+1}$, that is, $3^a+2^{n+1}=3^b$. So either $3 \mid 2^{n+1}$ or $a=0$. Since the first is absurd, we have $a=0$, that is $z-2^n=1, z+2^n=2^{n+1}+1=3^{2k}$. Then we have $2^{n+1}=(3^k+1)(3^k-1)$, so the factors are powers of 2 and differ by 2, which implies $3^k+1=4, 3^k-1=2, k=1, 2^{n+1}=8, n+1=3, n=2, 2n=x=4, y=2k=2$. So we have the solution $(4, 2)$ and there are no other solutions.


It's more powerful than God.
It's more evil than the devil.
The poor have it.
The rich need it.
If you eat it, you'll die.
What am I?

It is [hide]nothing[/hide]

but when I first found this teaser on braingle <-- place where you can ACTAULLY spam but I don't post, I didn't get the eating part, but now I do.

[hide]nothing is more powerful than God
nothing is more evil than the devil
the poor have nothing
the ricch need nothing
If you eat nothing, you will die[/hide]


The new topic is: Democracy is best served by strict separation of church and state.

This is excellent - though controversial.

Hmm...I don't believe it's such a great topic or that it's one of the most controversial issues. It isn't horrible of anything, but definitely could have been much better.

Why? I don't think it's all that bad, besides the fact that I can't seem to figure out a good value and VC for my AFF case.

History is too one-sided on this. How many successful clergy-controlled governments do you know of?

But that's not even the main issue. Where do you plan to find evidence? What arguments can you use? The argument is generally fairly one-sided towards the AFF, unless of course NEG can pull out an argument about the definition of "strict" and how a complete seperation of church and state isn't beneficial and etc.

I mean, just to name a few AFF arguments: freedom of religion, tolerance of different religions, political equality towards all religions, likeliness of discrimination, policy dictated by religious beliefs not necessarily accepted by populace, etc. And that's just what I came up with in the last 10 seconds.

So basically, NEG is running an uphill race here. And to be honest, I realli can't think of a good NEG argument off the bat except moral values.

But hey...we could discuss possible...uh...arguments here.

I agree. The negative has the challenge of proving grounds for which democracy is best served, and that's where the fundamental principles of religion might be handy. After all, where do we get all our moral pithy sayings? From religious beliefs, and religion often dictates what is moral. Thus, applying morality, religion might be essential for that matter. If not a theocracy, at least not utter, absolute, separation of church and state.  :?

[quote="Danbert"][quote="yif man12"][quote="SirErnest"]America was founded on religious beliefs.[/quote]

Funny, I thought the settlers came to Jamestown to plant 'baccy.[/quote]

That was a side effect.  :D
They came to get land for their king/queen/ruler.[/quote]

They came for a lot of reasons. But that has nothing to do with separation of church and state. :-)

And they  still believed in randomly finding large piles of gold back then.

See, the problem with that, is that it is way too ethnocentric, since the USA isn't even mentioned in the resolution.

Thus, to use it, you have to establish it as a democracy first, and even at that first step, you'll run into huge roadblocks if your opponents knows their stuff.

Then, you have to prove how that relates to the resolution... which is also difficult.

[quote="yif man12"][quote="MithsApprentice"]And they  still believed in randomly finding large piles of gold back then.[/quote]

I wouldn't mind randomly finding a large pile of gold.[/quote]

Neither would I!  :lol:

eh stacks of hundred dollar bills work better for me :-)

What is the formal procedure for a debate?

AFF CASE: 6 MIN
CX: 3 MIN
NEG CASE / 1NR: 7 MIN
CX: 3 MIN
1AR: 4 MIN
2NR: 6 MIN
2AR: 3 MIN

So, each must have a value and a value criterion by which to judge the value given. This is a value debate, not policy, so there isn't alot of burdens and specific examples and incidences. Those aren't qualified evidence. It is more important to argue philosophy and use philosophy as a backbone. Then, you can build your case from your contentions.

SO, a case would be.

Openning:

Value:
Value C:

Definitions:

Contention 1:

Sub A:

Sub B:

Contention 2:

Sub A:

Sub B:

Contention 3:

Sub A:

Sub B:

Policy Debate > Value Debate

Sorry, I was too tempted.  My brother did Value Debate (aka Lincoln-Douglas or LD), but I do Policy Debate (Cross-X or CX).  I've seen an LD round before and thought it was so very slow.  Then again, if anyone watches a CX round... I'd monitor their heart rate.

could just as easily be definitions then value...which is actually how most cases i know of are set up...

and there can be any number of contentions...although 2/3 is optimal...

Public Forum annhilates LD and Policy. What I hate about policy. Anything (peeing in the ocean, butterfly effect) could lead to global nuclear war and ultimate human genocide in policy. What I hate about LD: all the topics suck this year.

[quote="MithsApprentice"][quote="churchilljrhigh"]Anything (peeing in the ocean, butterfly effect) could lead to global nuclear war and ultimate human genocide in policy. [/quote]

Guess you've never heard of a factor we know as "significance".[/quote]

It doesn't matter how significant nuclear war is or how much the impact will be or how it outweigh all D/As.
When you run a Nuclearism Kritik...........that is if you glean my meaning.

[quote="Katok"]Tell me what you all think about using these philosophers for my cases:
Aff- Neitzche, Aristotole, Thomas Jefferson, James Madison

Neg- Spinoza, George Washington, James Madison, St. Thomas Aquinas, Kant's Universiality

I know what you'll say about using founding fathers for philosophers when the topic's not limited to the U.S. but they're ideas are no less substantial just because they were influences on the U.S.

Also, anyone know anything about the Hegelian Dialect? I printed it out, but as of now I'm way to lazy to actually read the thing.[/quote]

My two cents:
LD Philosophers
http://plato.stanford.edu/index.html
Policy Forums
http://www.cross-x.com

Thank you, you're wonderful. I'll check it out. Anyone else?

[quote="MithsApprentice"][quote="churchilljrhigh"]It doesn't matter how significant nuclear war is or how much the impact will be or how it outweigh all D/As.
When you run a Nuclearism Kritik...........that is if you glean my meaning.[/quote]

Actually I was referring to peeing in the ocean. Anyone suggesting that peeing in the ocean would ultimately lead to nuclear war needs to get their brain checked. It is if they can even find evidence to support their claim in the first place.

And honestly, I don't know what you mean. This significance factor as well as arguments of feasability usually known down outrageous arguments.[/quote]

The negative has a ton of cheap arguments to  make if their are some evident problems in the aff case (Extra T and Effects T). Yes, that's the whole point, the negative is saying that topicality is a voter issue (sure, neg is making the judge believe that the aff are crazy, not to sure if that will get you any courtesy points though).

I have no idea what that post was supposed to suggest. Care to clarify?

And fyi, policy debate actually requires some argumentation skills. It's not a game of who can make the stupidest remarks.

[quote="MithsApprentice"]It's not a game of who can make the stupidest remarks.[/quote]

Dang. Debate's not gonna be my future occupation.

[quote="History Genius"]If you argue on the neg side that religion in democracy promotes moral actions and decisions, how would you support it, especially if your opponent brings up examples such as abortion rights in the US Supreme Court (and please don't jump on my back about the topic not being about just the US:?)[/quote]

I would stay away from morality, since, unlike our previous cases, this really has nothing to do with it. In the topic states nothing about any moral obligation to protect the people, only to give the best service to [u] the democracy[/u]. We want the system that will best profit the democracy, not necessarily the people that live in it. Are you catching what I'm saying?

P.S. I love James and Cat!

[quote="History Genius"]If you argue on the neg side that religion in democracy promotes moral actions and decisions, how would you support it, especially if your opponent brings up examples such as abortion rights in the US Supreme Court (and please don't jump on my back about the topic not being about just the US:?)[/quote]

See, in my opinion, your value should at least be democracy related, since you need to prove that a non-strict separation upholds democracy. Then you would have to tie religion into your VC, since it will most likely be religion in your arguments which influences whether a democracy will be good or bad, or successful or not.

why do you guys talk so much about founding fathers, the united states, blalbala it's democracy that counts not united states. you can define democracy in a lot of different ways so the U.S constitution is not importent and btw u.s is a republic not democracy.

i'l try knowledge v and cultural diverstity for my neg

The United States is a democracy - it may have characteristics which define a republic, but it still runs on a democratic system.

hmm...that's a very detailed debate folks...i mean it depends on how u define democracy. the US is definitely not a pure democracy...but it's like half and half. some issues everyone votes on and other things we just vote for representatives to vote on

It's just because we have a electoral system which throws it out of whack. Otherwise...

Not only that -- in a "pure" democratic system, every citizen would have an opportunity to vote on every issue that required a societal decision.  (Not that this is necessarily a desirable system, but it is the logical extreme of a democracy.)  In our system, we have it broken down into all sorts of odd categorizations -- I mean, I have at least 3 major legislative bodies and 4 executive branches which are responsible for running things -- in New York, I hardly ever have the opportunity to vote directly on an issue.

What would you debate for a neg side anyway then? Couldn't you say that the reason why other countries have become so poor and war torn is because of neglient leaders and immoral dictators?

I dont get what you guys are talking about :blush:


inequalities
ratio
trigonometry
geometry
triangle inequality
inequalities unsolved

Prove that $\frac{1}{m_{a}}+\frac{1}{m_{b}}+\frac{1}{m_{c}}>\frac{5}{s}$,

where $m_{a}$, $m_{b}$, $m_{c}$ are the three medians of a triangle $ABC$, and $s=\frac{a+b+c}{2}$ is the semiperimeter of this triangle.

We have : $m_A < \frac{b+c}{2}$, $m_B < \frac{a+c}{2}$, $m_C < \frac{a+b}{2}$

Thus : $m_A+m_B+m_C < 2s$

Thus $\frac{1}{m_A}+\frac{1}{m_B}+\frac{1}{m_C}>= \frac{9}{m_A+m_B+m_C} > 9/2$

Then, I'm not sure your inequality is true.

I forgot the s, we have thus

$\frac{1}{m_A}+\frac{1}{m_B}+\frac{1}{m_C}>= \frac{9}{m_A+m_B+m_C} > \frac{4.5}{s}$

I agree with you.

$\displaystyle \frac{1}{m_a}+\frac{1}{m_b}+\frac{1}{m_c}>\frac{5}{s}>\frac{4.5}{s}$ ???

Labiliau, your estimation is correct but it is pretty weak. You should try to improve it :) (or disprove the inequality by W. Janous)

it seams right, because there is only one "extreme" situation, where it is equal, but never smaller...
i mean, when one side is zero ...

but i could prove it yet, but cauchy in the way it is used is definitly too weak and also the inequalities $2m_a<b+c$ etc. are too weak because they never get really close to equal at once...

peeta

I guess the way to go is not to put separate bounds on the $m$, but to make the substitutions $m_a=\frac{1}{4}\sqrt{4z^2+x^2+y^2+4xz+4yz-2xy}$ (and cycling), and $s=\frac{x+y+z}{2}$, with $x=a+b-c$, $y=c+a-b$ and $z=b+c-a$.

I tried this when the problem was first put up, but it got too ugly for me - I'm sure one of the inequality-crunching master-solvers on this forum will find a neat way to do it...

\[\displaystyle\sum_{cyc}\frac{4}{\sqrt{4x^2+y^2+z^2+4xy+4xz-2yz}}>\frac{10}{x+y+z}\]

Maybe not the most beautiful proof ever posted on ML, but still a proof:

[color=blue][b]Theorem 1.[/b] If $m_a$, $m_b$, $m_c$ are the medians and $s =\frac{a+b+c}{2}$ is the semiperimeter of a triangle ABC, then $\frac{1}{m_a}+\frac{1}{m_b}+\frac{1}{m_c}>\frac{5}{s}$.[/color]

[i]Proof of Theorem 1.[/i] We will verify a different inequality:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\frac{15}{2\left(m_a+m_b+m_c\right)}$.

This inequality will readily yield Theorem 1 by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59415]the medians duality principle[/url]: In fact, applying this inequality to the medians triangle of triangle ABC, which has sidelengths $m_a$, $m_b$, $m_c$ and medians $\frac34a$, $\frac34b$, $\frac34c$, we obtain $\frac{1}{m_a}+\frac{1}{m_b}+\frac{1}{m_c}>\frac{15}{2\left(\frac34a+\frac34b+\frac34c\right)}$, what simplifies to $\frac{1}{m_a}+\frac{1}{m_b}+\frac{1}{m_c}>\frac{5}{\left(\frac{a+b+c}{2}\right)}=\frac{5}{s}$, so that Theorem 1 is proven.

Thus, it remains to prove the inequality

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\frac{15}{2\left(m_a+m_b+m_c\right)}$.

This rewrites as

$\frac23\left(m_a+m_b+m_c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>5$.

Now, the centroid G of triangle ABC is the point of intersection of its medians $m_a$, $m_b$, $m_c$, and divides these medians in the ratio 2 : 1. Hence, $AG=\frac23m_a$, $BG=\frac23m_b$, $CG=\frac23m_c$, so that $\frac23\left(m_a+m_b+m_c\right)=\frac23m_a+\frac23m_b+\frac23m_c=AG+BG+CG$, and the inequality in question becomes

$\left(AG+BG+CG\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>5$.

Now, instead of directly attacking this inequality, we will show a more general assertion:

[color=blue][b]Theorem 2.[/b] If ABC is a triangle and P an arbitrary point in its plane, then

$\left(AP+BP+CP\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>5$.[/color]

[i]Proof of Theorem 2.[/i] We distinguish between two cases:

[i]Case 1:[/i] The triangle ABC has an angle > 120°. We can WLOG assume that A > 120°.

Then, it is well-known that the point Q in the plane of triangle ABC for which the sum AQ + BQ + CQ is minimal is the vertex A of the triangle ABC. Hence, for every point P in the plane, we have $AP+BP+CP\geq AA+BA+CA$. This simplifies to $AP+BP+CP\geq 0+c+b=b+c$. Hence, instead of proving the inequality $\left(AP+BP+CP\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>5$, it is enough to verify $\left(b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>5$. But this is easy: The triangle inequality for triangle ABC yields b + c > a, and the Cauchy-Schwarz inequality (or AM-HM) yields $\left(b+c\right)\left(\frac{1}{b}+\frac{1}{c}\right)\geq 4$; thus,

$\left(b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(b+c\right)\frac{1}{a}+\left(b+c\right)\left(\frac{1}{b}+\frac{1}{c}\right)>a\frac{1}{a}+4=1+4=5$.

Therefore, the proof of Theorem 2 in Case 1 is complete.

[i]Case 2:[/i] All angles of triangle ABC are $\leq 120^{\circ}$.

WLOG, we assume that A is the greatest angle of triangle ABC. Then, $A\leq 120^{\circ}$ by the condition of Case 2, but on the other hand, $A\geq 60^{\circ}$ (in fact, if the angle A would be < 60°, then the angles B and C would be < 60° as well - because A is the greatest angle of triangle ABC -, and thus we would have A + B + C < 60° + 60° + 60° = 180°, contradicting the sum of the angles in triangle ABC which states that A + B + C = 180°). From $A\leq 120^{\circ}$ and $A\geq 60^{\circ}$, it follows that $\sin A\geq\sin 60^{\circ}=\sin 120^{\circ}=\frac{\sqrt3}{2}$. On the other hand, the area $\Delta$ of triangle ABC equals $\Delta=\frac12bc\sin A$. Thus, $\Delta=\frac12bc\sin A\geq\frac12bc\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{4}bc$, so that $4\sqrt3\Delta\geq 4\sqrt3\frac{\sqrt3}{4}bc=3bc$.

Now, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=191721#p191721]http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 post #4[/url] Theorem 2, we have $AP+BP+CP\geq W$, where $W=\sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3\Delta\right)}$. Thus,

$AP+BP+CP\geq\sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3\Delta\right)}\geq\sqrt{\frac12\cdot\left(a^2+b^2+c^2+3bc\right)}$.

Hence, the inequality in question, $\left(AP+BP+CP\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>5$, will be evident once the following inequality is shown:

$\sqrt{\frac12\cdot\left(a^2+b^2+c^2+3bc\right)}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>5$.

Upon squaring, this inequality becomes

$\frac12\cdot\left(a^2+b^2+c^2+3bc\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2>25$,

and upon multiplication by 2, this becomes

$\left(a^2+b^2+c^2+3bc\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2>50$.

Now, denote by $h=\frac{2}{\frac{1}{b}+\frac{1}{c}}=\frac{2bc}{b+c}$ the harmonic mean of the numbers b and c. Then, by the QM-HM inequality, $\sqrt{\frac{b^2+c^2}{2}}\geq h$, so that $\frac{b^2+c^2}{2}\geq h^2$ and $b^2+c^2\geq 2h^2$, and by the GM-HM inequality, $\sqrt{bc}\geq h$, so that $bc\geq h^2$. Finally, $h=\frac{2}{\frac{1}{b}+\frac{1}{c}}$ yields $\frac{1}{b}+\frac{1}{c}=\frac{2}{h}$. Thus,

$\left(a^2+b^2+c^2+3bc\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\geq\left(a^2+2h^2+3h^2\right)\left(\frac{1}{a}+\frac{2}{h}\right)^2=\left(a^2+5h^2\right)\left(\frac{1}{a}+\frac{2}{h}\right)^2$,

and it remains to show that

$\left(a^2+5h^2\right)\left(\frac{1}{a}+\frac{2}{h}\right)^2>50$.

The following nice proof of this inequality is due to Peter Scholze (my own proof was pretty messy): By AM-GM, we have

$a^2+\frac53h^2+\frac53h^2+\frac53h^2\geq 4\sqrt[4]{a^2\cdot\frac53h^2\cdot\frac53h^2\cdot\frac53h^2}$
$\Longleftrightarrow\ \ \ \ \ a^2+5h^2\geq 4\sqrt[4]{\left(\frac53\right)^3a^2h^6}$

and

$\frac{1}{a}+\frac{2}{3h}+\frac{2}{3h}+\frac{2}{3h}\geq 4\sqrt[4]{\frac{1}{a}\cdot\frac{2}{3h}\cdot\frac{2}{3h}\cdot\frac{2}{3h}}$
$\Longleftrightarrow\ \ \ \ \ \frac{1}{a}+\frac{2}{h}\geq 4\sqrt[4]{\left(\frac23\right)^3\cdot\frac{1}{ah^3}}$;

hence,

$\left(a^2+5h^2\right)\left(\frac{1}{a}+\frac{2}{h}\right)^2\geq 4\sqrt[4]{\left(\frac53\right)^3a^2h^6}\cdot\left(4\sqrt[4]{\left(\frac23\right)^3\cdot\frac{1}{ah^3}}\right)^2$
$=64\sqrt[4]{\left(\frac53\right)^3\left(\frac23\right)^6}>50$

(in fact, $64\sqrt[4]{\left(\frac53\right)^3\left(\frac23\right)^6}\approx 51.101029$). And Theorem 2 is proven in Case 2. This completes the proof of Theorem 1.

  Darij

[quote="darij grinberg"]Maybe not the most beautiful proof ever posted on ML, but still a proof[/quote]
well, i think too, that its a proof ;) ... but it is that easy that i was able to understand it. i didnt expect that, so NICE! ;)
job job darij :)

[quote="manlio"]Prove that the triangle inequality
\[\frac{1}{m_{a}}+\frac{1}{m_{b}}+\frac{1}{m_{c}}>\frac{5}{s}\]
where $m_{a},\ m_{b},\ m_{c}$ are medians of a triangle and $s$ is its semi-perimeter.[/quote]
This is a problem was proposed by Walther Janous, see [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], Vol. 12 (1986), p. 177, Problem $1137^{*}$. A solution by Ji Chen, see Vol. 18 (1992) , pp. 208 - 209.

Can you post a complete solution ? You think all mathlinker can download Crux as you? [/color]

I have just merged this topic with another one, so you can find my proof of the inequality above, but I would love to know the one you proposed in CRUX, Ji Chen!

  darij


Prove that:
$ \sum_{n \equal{} 1}^{\infty}\frac {1}{F_nF_{n \plus{} 2}^2F_{n
\plus{} 3}} \plus{} \sum_{n \equal{} 1}^{\infty}\frac {1}{F_nF_{n \plus{} 1}^2F_{n \plus{} 3}} \equal{} \frac12$

For conbernience, let $ a\equal{}F_n$, $ b\equal{}F_{n\plus{}1}$, $ c\equal{}F_{n\plus{}2}$, and $ d\equal{}F_{n\plus{}3}$. Then $ a\plus{}b\equal{}c$ and $ b\plus{}c\equal{}d$.
$ LHS$
$ \equal{} \sum_{n\equal{}1}^\infty \frac{1}{ac^2d} \plus{} \sum_{n\equal{}1}^\infty \frac{1}{ab^2d}$
$ \equal{} \sum_{n\equal{}1}^\infty \left( \frac{1}{ac^2d} \plus{} \frac{1}{ab^2d} \right)$
$ \equal{} \sum_{n\equal{}1}^\infty \left( \frac{b}{abc^2d} \plus{} \frac{c}{ab^2cd} \right)$
$ \equal{} \sum_{n\equal{}1}^\infty \left( \frac{c\minus{}a}{abc^2d} \plus{} \frac{d\minus{}b}{ab^2cd} \right)$
$ \equal{} \sum_{n\equal{}1}^\infty \left( \frac{1}{abcd} \minus{} \frac{1}{bc^2d} \plus{} \frac{1}{ab^2c} \minus{} \frac{1}{abcd} \right)$
$ \equal{} \sum_{n\equal{}1}^\infty \left( \frac{1}{ab^2c} \minus{} \frac{1}{bc^2d} \right)$
$ \equal{} \sum_{n\equal{}1}^\infty \left( \frac{1}{F_n F_{n\plus{}1}^2 F_{n\plus{}2}} \minus{} \frac{1}{F_{n\plus{}1} F_{n\plus{}2}^2 F_{n\plus{}3}} \right) \ \ \leftarrow$ Telescoping sum
$ \equal{} \frac{1}{F_1 F_2^2 F_3}$
$ \equal{} \frac{1}{2}$


geometry
incenter
search
circumcircle
angle bisector
cyclic quadrilateral
geometry proposed

Let $\triangle ABC$ be a triangle,  $I$ its incentre and $\gamma _a$ the circle internally tangent to the sides $AB,AC$ and to the circumcircle.
Denote with $M,N$ the points of contact of $\gamma _a$ with $AB,AC$ respectively. Show that :

(1)  $I, M, N $ are collinear

(2)  $MN \perp AI$

discussed. a search for "casey's theorem" / "tranversal theorem" would probably pull up the topic as I remember that being used in one of the proofs. I think this might also be IMO78/4. Finally, a recent Belgrade problem was generalized and this was a special case; look for that too.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=46418 Soon will appear the solution.

Let us to present an elementary proof of the collinearity of $ I,\ M,\ N,$ which it is also enough for the proof of the result of $ AI\perp MN.$

$ \bullet$ We denote as $ M,\ N,$ the intersection points of $ AB,\ AC$ respectively, from the line through the incenter $ I$ of $ \bigtriangleup ABC$ and perpendicular to $ AI.$

The lines through the points $ M,\ N$ and perpendicular to $ AB,\ AC$ respectively, intersect each other at the point so be it $ O',$ lies on the angle bisector of $ \angle A$ and clearly, the circle $ (O')$ centered at this point with radius $ O'M \equal{} O'N,$ tangents to $ AB,\ AC,$ at points $ M,\ N,$ respectively.

We will prove that the circle $ (O'),$ also tangents to the circumcircle $ (O)$ of the given triangle $ \bigtriangleup ABC.$

$ \bullet$ Let be the points $ B'\equiv (O)\cap BI$ and $ C'\equiv (O)\cap CI.$

Let also be the points $ P\equiv OO'\cap B'N,$ $ P'\equiv OO'\cap C'M$ and we will prove trhat $ P'\equiv P.$

From $ O'N\parallel OB',$ applying the [b][size=100]Thales theorem[/size][/b], we have that $ \frac {PO'}{PO} \equal{} \frac {O'N}{OB'}$ $ ,(1)$

Similarly, from $ O'M\parallel OC'$ $ \Longrightarrow$ $ \frac {P'O'}{P'O} \equal{} \frac {O'M}{OC'}$ $ ,(2)$

From $ (1),$ $ (2)$ and because of $ \frac {O'N}{OB'} \equal{} \frac {O'M}{OC'}$ $ ,(3)$ $ \Longrightarrow$ $ \frac {PO'}{PO} \equal{} \frac {P'O'}{P'O}$ $ \Longrightarrow$ $ \frac {PO'}{PO \minus{} PO'} \equal{} \frac {P'O'}{P'O \minus{} P'O'}$ $ \Longrightarrow$ $ \frac {PO}{OO'} \equal{} \frac {P'O'}{OO'}$ $ \Longrightarrow$ $ PO' \equal{} P'O'$ $ \Longrightarrow$ $ P'\equiv P$ $ ,(4)$

$ \bullet$ We consider the non-convex hexagon $ ABB'PC'C$ and because of the collinearity of the points $ M\equiv AB\cap PC',$ $ I\equiv BB'\cap C'C,$ $ N\equiv B'P\cap AC,$ based on the [b][size=100]Pascal theorem[/size][/b], we conclude that this hexagon is cyclic.

So, the point $ P\equiv OO'\cap B'N\cap C'M,$ lies on the circumcircle of $ \bigtriangleup ABC.$

Because of now, $ OP \equal{} OB'$ and $ O'N\parallel OB',$ we conclude that $ O'P \equal{} O'N$ $ ,(5)$ and similarly we have $ O'P \equal{} O'M$ $ ,(6)$

From $ (5),$ $ (6)$ and because of the collinearity of the points $ O,\ O',\ P,$ we conclude that the circle $ (O'),$ which tangent to $ AB,\ AC$ at points $ M,\ N$ respectively, tangents also internally to the circle $ (O)$ at point $ P$ and the proof is completed.

Kostas Vittas.

Here is a nice generalization of the collinearity: http://www.mathlinks.ro/viewtopic.php?t=170192 (when cyclic quadrilateral $ ABCD$ degrades to triangle $ ABC$)


trigonometry
function
integration
calculus
calculus computations

Find

$\int{\sec^{-1}{x}}\,dx$

[hide="Hint"]
Use the substitution $x=\sec{y}$[/hide]

I used $x = \sec{k}$.
so $dx = \sec{k}. \tan{k}dk$

Subst. inside will get $\int{k\sec{k}\tan{k}}\,dk$

Then, use integration by parts,

$I = k\sec{k}-\int{\sec{k}}\,dk+C_{1}$
 $= k\sec{k}-ln|\sec{k}+\tan{k}|+C$

Subst. back will get,
$\int{\sec^{-1}{x}}\,dx = x\sec^{-1}{x}-ln|x+\sqrt{x^{2}-1}|+C$

[quote="maokid7"]would knowing that $\sec^{-1}(x)=\cos^{-1}\left(\frac{1}{x}\right)$ help at all?[/quote]

I don't think so. For me, it makes things more complicated.   :D

for me it was much simpler
[hide="here is how it is done"]
$\int\sec^{-1}(x)dx=\int\cos^{-1}\left(\frac{1}{x}\right)dx$
so integrating by-parts we use

$dv=dx$ $v=x$ $u=\cos^{-1}\left(\frac{1}{x}\right)$ and $du=\frac{1}{x\sqrt{x^{2}-1}}$

so

$\int\cos^{-1}\left(\frac{1}{x}\right)dx=x\cos^{-1}\left(\frac{1}{x}\right)-\int\frac{1}{\sqrt{x^{2}-1}}$

to evaluate the second integral we use $x=\cosh(\theta)$

so
$\int\cos^{-1}\left(\frac{1}{x}\right)dx=x\cos^{-1}\left(\frac{1}{x}\right)-\cosh^{-1}(x)=x\sec^{-1}(x)-\ln(|x+\sqrt{x^{2}-1}|)+C$
[/hide]

you're right, That's a great solution!  :D
didn't think of using hyperbolic functions. i only learnt them 2days ago  :P


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group of order $90$ with a non-normal subgroup,$H$,of order $5$.
a).Show that $G$ has a subgroup of order $15$.Show this subgroup is abelian.

I would not know how to do this without Sylow, you know Sylow theorems?

The number of subgroups of order five must be a divisor of ninety, and 1 mod 5
so it is 1 or 6
however is also is part of sylow theorems that they are all conjugated
so you said H is not normal, so the number is six

now by this conjugation, we get a permutation representation on G itself, transitively
so the stabilisator of H  is 90/6=15 by the orbit formula

now there is a much more general theorem that every group of order p*q with p<q primes and p not dividing q-1, is not only abelian, but even cyclic !!
the number of sylow p groups is 1 (q is not possible)
and of q it is also 1

so we have at most p-1+q-1+1 elements of order in {1,p,q}
thus there is an element of order p*q
thus the group is cyclic

i hope this helps, someone else will probably be able to do this exercise with less powerful theorems

thanks, it's ok-there was no reason why we mustn't use strong theorems
0f course it would be great if someone find easier solution


geometry
3D geometry
AMC
USA(J)MO
USAMO
algebra
polynomial

Here is my solution to 73/5.  I hope someone out there can help me improve it.  Any critique is welcome.

Problem:
Prove that the cube roots of three prime numbers can never be in an arithmetic progression (they are not necessarily consecutive terms).
Solution:
<<<This is a really drawn-out solution because it's for the USAMO and they take away points for tiny things>>>

Changelog:
> Set R was changed to set Q because of a stupid mistake.
> Roots of any NON-CUBES cannot be rational.
> Added proof of the non-rational statement.

Assume the contrary:
Let the three primes be $ x,y,z$.
Let (by the definition of an arithmetic progression)
$ \begin{cases} \sqrt [3]{y} \equal{} \sqrt [3]{x} \plus{} pr \\
\sqrt [3]{z} \equal{} \sqrt [3]{y} \plus{} qr \end{cases}$
Where $ p, q\in\mathbb{Z}$, $ p, q\ne 0$.
Then, multiplying by $ q$ in the first equation and $ p$ in the second, we have
$ \begin{cases} q\sqrt [3]{y} \equal{} q\sqrt [3]{x} \plus{} pqr \\
p\sqrt [3]{z} \equal{} p\sqrt [3]{y} \plus{} pqr \\
\end{cases}$
Then subtracting the second from the first, we have
$ q\sqrt [3]{y} \minus{} p\sqrt [3]{z} \equal{} q\sqrt [3]{x} \minus{} p\sqrt [3]{y}$
And then, adding $ p\sqrt [3]{y}$ to both sides we have
$ (p \plus{} q)\sqrt [3]{y} \minus{} p\sqrt [3]{z} \equal{} q\sqrt [3]{x}$
Now, let $ k \equal{} p \plus{} q$ for convenience.
$ k\sqrt [3]{y} \minus{} p\sqrt [3]{z} \equal{} q\sqrt [3]{x}$ (EQUATION 1)
We then cube.
$ k^3y \minus{} 3k^2p\sqrt [3]{y^2z} \plus{} 3kp^2\sqrt [3]{yz^2} \minus{} p^3z \equal{} q^3x$
Now, grouping some terms, we have
$ k^3y \minus{} p^3z \minus{} 3kp\sqrt [3]{yz}\left( k\sqrt [3]{y} \minus{} p\sqrt [3]{z}\right) \equal{} q^3x$
Substituting (1) and adding $ p^3z \minus{} k^3y$ we have
$ \minus{} 3kpq\sqrt [3]{xyz} \equal{} q^3x \plus{} p^3z \minus{} k^3y$
Dividing by $ \minus{} 3kpq$ we have
$ \sqrt [3]{xyz} \equal{} \dfrac{k^3y \minus{} q^3x \minus{} p^3z}{3kpq}$
Then, since the numerator and denominator in the RHS are both integer, $ \text{RHS} \in\mathbb{Q}$.  But this cannot be rational.  Assume $ \sqrt[3]{xyz}\equal{}\dfrac{g}{h}$.  Then $ xyz \equal{} \dfrac{g^3}{h^3}$, and since $ xyz$ is an integer, let $ n\equal{}1$.  Then $ xyz \equal{} m^3$, which is impossible since $ x,y,z$ are distinct primes.  Therefore, $ \sqrt [3]{xyz}\not\in\mathbb{Q}$.  Contradiction!

If I were writing this up, I'd probably start straight from equation 1; that's a well-known property of arithmetic sequences, I would think.

Roots of any kind cannot be rational is not quite a true statement. $ \sqrt{4}$? The cube root of the product of three distinct prime numbers on the other hand...

For the end, I think you mean Q (rationals), not R (reals). Also, for a bit more rigor, you can go like:
Assume $ \sqrt[3]{xyz}\equal{}\frac{m}{n}$, (a rational with all the usual m,n are integers stuff). Then $ xyz\equal{}\frac{m^3}{n^3}$, and since $ xyz$ is an integer, we can let $ n\equal{}1$. Then we have $ xyz\equal{}m^3$, which is impossible since $ x,y,z$ are distinct primes.

Another method you can use goes as follows:
Consider the polynomial $ f(t)\equal{}t^3\minus{}xyz$. Since the polynomial is monic, any rational root of the polynomial must be an integer. Thus, showing that there are no integer roots of the polynomial will prove that the polynomial has no rational roots. Brute force checking with all possible factors of $ xyz$ shows that this is true.

Since $ \sqrt[3]{xyz}$ [b]is[/b] a root of f(t), and f(t) has no rational roots, $ \sqrt[3]{xyz}$ must be irrational.

[hide="identity"]Let $ p<q<r$ be the three primes. It suffices to prove that $ \frac{r^{1/3}\minus{}q^{1/3}}{r^{1/3}\minus{}p^{1/3}}$ is not rational. For contradiction assume it is equal to rational $ \frac{x}{y}$ in lowest term. Rearranging gives $ (y\minus{}x)r^{1/3}\minus{}yq^{1/3}\plus{}xp^{1/3}\equal{}0$. Therefore, $ (y\minus{}x)^3r\minus{}y^3q\plus{}x^3p\equal{}\minus{}3xy(y\minus{}x)(pqr)^{1/3}$. Since $ p, q, r$ are primes, right hand side must be irrational, but left side is rational, so contradiction.
[/hide]

[quote="timwu"][hide="identity"]Let $ p < q < r$ be the three primes. It suffices to prove that $ \frac {r^{1/3} \minus{} q^{1/3}}{r^{1/3} \minus{} p^{1/3}}$ is not rational. For contradiction assume it is equal to rational $ \frac {x}{y}$ in lowest term. Rearranging gives $ (y \minus{} x)r^{1/3} \minus{} yq^{1/3} \plus{} xp^{1/3} \equal{} 0$. Therefore, $ (y \minus{} x)^3r \minus{} y^3q \plus{} x^3p \equal{} \minus{} 3xy(y \minus{} x)(pqr)^{1/3}$. Since $ p, q, r$ are primes, right hand side must be irrational, but left side is rational, so contradiction.
[/hide][/quote]

seems shorter  :D


calculus
integration
algorithm
induction
irrational number
absolute value

Show that a number $r$ can be expressed as an infinite repeating decimal iff $r$ is rational.

Some hints..

[hide="big hint"]Say your rational number is $\frac{s}{n}$. Then you can say $s=mn+r$ where $0\le{r}<n$. The integral part of $\frac{s}{n}$ is $m$. So we want to find the decimal representation of $\frac{r}{n}$. How do we use the division algorithm repeatedly to find equations that the digits of $\frac{r}{n}$ satisfy?

[hide="try this"]$\frac{10r}{n}=d_1+\frac{r_1}{n}$, and so on. The tenths digit will be $d_1$.[/hide][/hide]

[hide="next hint"]Use induction..[/hide]

Sorry if this doesn't fit here. I read about this in an old algebra book and thought it was pretty interesting, and very fundamental.

[quote="Altheman"]well, using the method for finding fractions for rational numbers would help with this, and by definition of an irrational number, it does not repeat, so that is pretty much it since we are only dealing with reals[/quote]Not sure where you're going there.. What is this method for finding fractions? And the fact that irrational numbers do not repeat (this is not their definition, but it can be proven) doesn't imply that rational numbers do repeat.

[quote="AntonioMainenti"][quote="Altheman"]remember if you have like $3.\bar{5} = x$
then $10x = 35.\bar{5}$
and subtracting gives $9x = 32$
so if it has n repeating digits, then multiply by $10^n$ and follow the same process[/quote]But we want to to show that a rational number can be expressed as a repeating decimal. Your method is for representing a number as a fraction once we know the repeating decimal..[/quote]Hmmm...maybe we can try to express every rational number in the form $\frac{a}{1-r}$?

well, i guess then one must show that the reverse of that procedure is always possible, which is what probability1.01 said

[quote="probability1.01"]Hint: this problem reduces (or can be reduced) to the following classic problem: show that for all n, $\displaystyle \sum_{i=0}^k 10^i$ is divisible by n for some k.[/quote]Are you sure it's not $10^k-1$? I see how it would reduce to that but not what you said. Please enlighten me.

[hide="how it reduces to this.."]This isn't the solution I'm looking for, which is more elementary.

Let $\frac{m}n$ be some rational number. Suppose for some $k$ that there is an integer $q$ such that $nq=(10^k-1)m$, and the that decimal representation of $q$ is $a_1\cdots{a}_s$. Then $\frac{m}n=\frac{a_1\cdots{a}_s}{10^k-1}=a_1\cdots{a}_s\left(\frac1{1-\frac1{10^k}}-1\right)=\sum_{i=1}^\infty\frac{a_1\cdots{a}_s}{10^{ik}}=.a_1\cdots{a}_s{a}_1\cdots{a}_s{a}_1\cdots{a}_s\cdots$

(edit: I just realized I made some serious mistake(s) here. I'll see if I can correct them tomorrow  :blush: :mad: )
(edit 2: Basically for what I said to make sense we just need to let $0<m<n$ and then we will have $s\le{k}$ [hide="proof"]$m\cdot10^{s-1}<n\cdot10^{s-1}\le{n}q=10^k\cdot{m}-m<10^k\cdot{m}\Longrightarrow{s}\le{k}$[/hide]. So there may be some $0$s between groups of $a_1\cdots{a}_s$ in the decimal representation.)

If you already have the infinite repeating decimal, you can easily represent it as a rational number (a fraction) by the method Altheman suggested.[/hide]

Well, no one seems to be trying so here's a simple demonstration..

Suppose we want to find the decimal expansion of $\frac{r}n$, where $0\le{r}<n$. From the division algorithm we get the equations
$10r=b_1n+r_1\\\\10r_1=b_2n+r_2\\\\\ldots\\\\10r_i=b_{i+1}n+r_{i+1},$

where $0\le{r_j}<n$ for all $j=1,2,\ldots$

It is easily shown that the $b_j$ are the digits in the decimal. Since $\frac{r}n=\frac{b_1}{10}+\frac{r_1}{10n}$ and in general $\frac{r_i}{10^in}=\frac{b_{i+1}}{10^{i+1}}+\frac{r_{i+1}}{10^{i+1}n}$ we have by induction that $\frac{r}{n}=\sum_i\frac{b_i}{10^i}$.

Since each $r_j$ must be one of $0,1,\ldots,n-1$, obviously the $r_j$ cannot all be different, so $r_p=r_{p+t}$ for some $p,t$. Let $p$ and $t$ be the least integers for which this is true. By definition we have $r_p=r_{p+t}\Longrightarrow{b}_{p+1}n+r_{p+1}=b_{p+t+1}n+r_{p+t+1}\Longrightarrow$
${n}(b_{p+1}-b_{p+t+1})=r_{p+t+1}-r_{p+1}$. But the RHS has an absolute value less than $n$, so $n$ can only divide it if it is zero. Thus $b_{p+1}=b_{p+t+1}$ (these digits are equal) and $r_{p+1}=r_{p+t+1}$. By induction, $b_k=b_{k+t}$ for all $k\ge{p}+1$.


[quote="azsxder"]On Section B of the S-III forms for the USAMO, it asks if you are "willing to attend the 2009 Mathematical Olympiad Summer Program".

I don't think I will make it to MOP, but just in case:

1. Are you allowed to leave MOP early to attend another camp (such as RSI)?

2. If you mark "no" and score high enough to qualify, does that mean you are not a MOP qualifier (which carries some weight)? And if you mark "yes" even though you know you can't make it, will that make them annoyed?[/quote]

1.If you leave earlier without any emergency, you will be responsible for all the fees.(it is free otherwise)
2.Just mark yes.

[quote="ZzZzZzZzZzZz"]Will you be responsible for fees if you mark "yes," but later it turns out that you can't go at all?[/quote]

I think if you tell them that you can't go shortly after scores out you'll be fine

[quote="soltixnj"]If the IMO takes place on 13-22 July, then how does MOP last 3 weeks? That mean's MOP would have to start right after school got out (or during the last week of school for some people).

And how does Titu Andreescu teach MOP and Awesome math at the same time?[/quote]

Yes, MOSP participants are very likely to miss at least the final exam. I was a junior last year so I need to take all exams earlier(a lot of trouble). However, seniors are in better position because school usually ends earlier.

Titu didn't come to MOSP last year.

MOP starts June 9 this year, which is before the end of many schools.

If you're not sure, then mark "yes."  Once USAMO grading finishes, you'll get a phone call or an e-mail invitation if your score is high enough, and there's no penalty for turning down the invitation - that's why they contact you right away instead of just assuming you'll come.

MOP really doesn't like people leaving for another camp, especially RSI, so I doubt they would let you do that.

Well considering that you are a junior, you will have to qualify for Blue MOP, Black MOP, or CGMO.  If you haven't qualified for MOP in the past it's definitely a stretch.  Nonetheless, say yes and if you qualify choose one or the other.

[quote="archimedes1"][quote="matt276eagles"]MOP really doesn't like people leaving for another camp, especially RSI, so I doubt they would let you do that.[/quote]

Why RSI especially?[/quote]
Probably because RSI steals a good deal of juniors, and they have a pretty strict policy too I hear.

Is there any sort of confirmation given that they received the S-III form?

They generally put up a "Forms Received" page after a week or two.


function
integration
real analysis
real analysis unsolved

I read a theorem in a book:
If f is an increasing and bound function in [a,b] and g is a function on [a,b] that exist $\int_{a}^{b}{g(x)dx}$
Prove that :
exist a real number c between a and b such that:
$\int_{a}^{b}{f(x)g(x)dx}= f(a). \int_{a}^{c}{g(x)dx}+f(b). \int_{c}^{b}{g(x)dx}$

Can anyone solve it?

I posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65758]here[/url] a similar problem (regarding a [i]monotonically decreasing function[/i] instead of a [i]monotonically increasing function[/i]).


A thief stole $ \frac{5}{7}$ of Brendon's money and spent $ \frac{5}{7}$ of the amount stolen. The thief was then caught, and the remaining money was returned to Brendon. The remaining amount was $ \$40$ less than the amount Brendon had after being robbed. How many dollars did Brendon have before the theft?

Let the amount of money Brendon had be $ \alpha$. This means the thief took $ \frac{5}{7}\alpha$ and spent $ \frac{25}{49}\alpha$. Thus what was returned to Brendon was $ \frac{5}{7}\alpha \minus{} \frac{25}{49}\alpha \equal{} \frac{10}{49}\alpha$. We have the following equation:

$ \frac{10}{49}\alpha \equal{} 40$

$ \alpha \equal{} \boxed{196}$

This is a bad problem
When it says 'after being robbed' it means after the theft but before the money is returned.
so amount after robbed: $ \frac27a$
remaining money: $ \frac57\cdot\frac27a$
$ (\frac27a\minus{}\frac57\cdot\frac27 )a\equal{}40$
$ \frac{14\minus{}10}{49}a\equal{}40$
$ a\equal{}490$


limit algebra function domain trigonometry topology calculus

I was helping my friend with this limit question. but i only know up through calc 2. so i was not sure if i was allowed to make some of the assumptions that i did.
basically i said that

$\lim_{{(x_{1},x_{2},\dots,x_{n})}\to{0}}\frac{\sin(x_{1}x_{2}\dots x_{n})}{x_{1}x_{2}\dots x_{n}}=1$

i was wondering if this assumption is correct, and how would you prove that it is or isn't?

Surprisingly, this depends on your exact definition of limit.

Do you want this to be the definition of limit in a metric space?

(1) We say that $\lim_{x\to p}f(x)=L$ if $f$ is defined on a punctured neighborhood of $p$ (that is, $\exists\,\eta>0$ such that $\{x|0<d(x,p)<\eta\}$ is contained in the domain of $f$) and $\forall\,\epsilon>0\,\exists\,\delta>0$ such that if $0<d(x,p)<\delta$ then $d(f(x),L)<\epsilon.$

Or would you rather this be the definition?

(2) We say that $\lim_{x\to p}f(x)=L$ if $p$ is a limit point of the domain of $f$ and $\forall\,\epsilon>0\,\exists\,\delta>0$ such that if $0<d(x,p)<\delta$ and $x$ is in the domain of $f,$ then $d(f(x),L)<\epsilon.$

(I was using generic distances in a metric space; you can substitute absolute values and Euclidean distances for familiarity.)

If you use definition 2, then your statement is correct. The origin is a limit point of the domain of this function, and if the the point if sufficiently close to the origin and in the domain of the function, then the value is close to 1.

But if you use definition 1, then your statement is incorrect because this function is not defined in a punctured neighborhood of the origin. It is not defined when any one of the variables is zero.

i understand number 2 since it is simlar to the nomral
$\epsilon$,$\delta$ definition of a limit.

however, could you please further explain number 1.

The statement $\lim_{x\to a}f(x)=L$ is supposed to mean that $f(x)$ is close to $L$ when $x$ is close (but not equal) to $a$. But how do we interpret this when $f(x)$ is not defined everywhere? Definition (1) [b]requires[/b] $f(x)$ to be defined for all $x$ sufficiently close (but not equal) to $a$. This rules out your function. Definition (2) is more forgiving: the points where $f$ is not defined are simply ignored.

A one-variable example: $\lim_{x\to0}x\sin\left(\tan (1/x)\right)$ is zero according to (2), but it does not exist according to (1).