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funny picture

Fri, 04 May 2012 12:28:34 -0700

What do you think ?

$e\cdot n! - \lfloor e\cdot n! \rfloor < 1/n$

Thu, 31 Mar 2011 10:11:15 -0700

Show tat $\displaystyle{e\cdot n! - \lfloor e\cdot n! \rfloor < \frac{1}{n}}$

Find positive integers by the 7/2 rule

Mon, 07 Mar 2011 16:31:16 -0700

Find all positive integers with the property that, when the first digit is moved to the end, the resulting number is $\frac{7}{2}$ times the original one.

Limit of a recursion sequence

Sat, 26 Feb 2011 22:26:44 -0700

Let $ y_{0}\geq 2,\; y_{n}=y_{n-1}^{2}-2, \quad \displaystyle{s_{n}=\frac{1}{y_{0}}+\frac{1}{y_{0}y_{1}}+\cdots + \frac{1}{y_{0}y_{1}\cdots y_{n}} }$
prove that : $\displaystyle {\lim_{n\to\infty} s_n=\frac{1}{2}\left(y_0-\sqrt{y_0^2-4}\right)} $

Solve $a!b! = a!+b!+c^2,\quad a,b,c\in\mathbb{N}^+$

Fri, 18 Feb 2011 16:32:20 -0700

Find positive integer solutions $(a,b,c)$ such that $a! b! = a! + b! + c^2$

Colored convex polyhedron

Fri, 28 Jan 2011 09:26:28 -0700

Let be given a convex polyhedron whose all faces are triangular. The vertices of the polyhedron are colored using three colors. Prove that the number of faces with vertices in all the three colors is even.

Points of tangency and concyclic

Sun, 09 Jan 2011 18:50:53 -0700

[attachment=37]See the picture，given $\triangle ABC$ and its excircle $\odot I_a$，let $\ell_1$、$\ell_2$ the common tangents of $\odot O(ABC)$ and $\odot I_a$ that cut $BC$ at $D$、$E$; let $P$、$Q$ on $\odot O$ be the tangent poins，and $I_b$、$I_c$ be the other escenters of $\triangle ABC$ .

Prove that：
(1): $P$、$Q$、$I_b$、$I_c$ are on a circle.
(2): $\angle BAD=\angle CAE$ .

Power of $(\sqrt{2} +1)$ and floor function

Fri, 07 Jan 2011 18:24:15 -0700

Prove that $\lfloor (\sqrt{2} + 1)^{2n} \rfloor + 1 - (\sqrt{2} + 1)^{2n} \geq \frac{1}{2} $ and $ (\sqrt{2} + 1)^{2n-1} - \lfloor (\sqrt{2} + 1)^{2n-1} \rfloor \leq \frac{1}{2} , \forall n \in \mathbb{N} $

Inscribed circle centre problem

Thu, 06 Jan 2011 21:26:42 -0700

[attachment=38]Given three circles $(O_1), (O_2), (O)$, (see the figure) with $(O_1)$ contact to $(O_2)$ at point $T$, $(O_1)$ contact to $(O)$ at point $E$, $(O_2)$ contact to $(O)$ at point $F$, $BC$ tangent to $(O_1)$ and $(O_2)$ at points $P, Q$ respectively, $AT$ tangent to $(O_1)$ and $(O_2)$ at point $T$, And $A, B, C$ lie on the circle $(O)$.

Prove that $T$ is the inscribed circle centre
of $\triangle ABC$.

http://www.artofproblemsolving.com/Forum...er+problem

Partition and Equivalent Class

Thu, 06 Jan 2011 10:38:51 -0700

(1) If $f: X \to Y$ is a mapping, then $f_{=}=\{f^{-1}(y) \mid y \in f(X)\}$ is a partition of $X$

Prove the equality: $\bigcup$ and $\bigcap$

Tue, 04 Jan 2011 11:45:54 -0700

Prove that
\[ \bigcup_{i \in I} \bigcap_{j \in J} A_{i,j}=\bigcap_{f \in J^I} \bigcup_{i \in I} A_{i,f(i)} \]
where $f \in J^I$ is a function such that $f:I \rightarrow J$.

http://www.artofproblemsolving.com/Forum...&p=2137628

Ash:Abstract Algebra The Basic Graduate Year

Thu, 30 Dec 2010 13:37:36 -0700

0.1 Elementary Number Theory
$\quad\quad \gcd(a,b) =: \min\{d \mid d\in \mathbb{N}^+,\; (d \mid a)\wedge (d\mid b)\}\left(=\min\{sa+tb \mid s,t \in \mathbb{Z},\; sa+tb > 0\} \right )$

$\quad\quad \textbf{0.1.1} \quad d=\gcd(a,b) \Rightarrow \exists s,t \in \mathbb{Z}\; (d=sa+tb)$

$\quad\quad \textbf{0.1.2} \quad p\in \mathbb{N}^+$ is called a prime if $(p > 1) \wedge (\forall d\in \mathbb{N}^+\;(d \mid p \Rightarrow d\in \{1,p\}) )$
$\quad\quad\quad\quad $We have that $(a_1,\cdots,a_n \in \mathbb{Z},\; p \text{ is prime},\; p \mid a_1\cdots a_n)\Rightarrow (\exists k\in \{1,\cdots ,n\}\;\; (p\mid a_k))$
$\quad\quad\quad\quad \textbf{0.1.2.1}\quad (p \nmid c)\Rightarrow \gcd(p,c)=1$ $\left(\gcd(p,c)=d \Rightarrow (d \mid p \nmid c) \Rightarrow (d \in \{1,p\}) \Rightarrow (d=1)\right)$
$\quad\quad\quad\quad \textbf{0.1.2.2}\quad (c \mid ab)\wedge (\gcd(c,a)=1)\Rightarrow (c \mid b) $ $c \mid scb+tab =b\gcd(c,a)=b$

$\quad\quad \textbf{0.1.3 Unique Factorization Th.} 1

Taylor's Theorem (Taylor polynomial)

Mon, 27 Dec 2010 14:16:01 -0700

Consider $\displaystyle{\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n dt}$. Apply integral by parts to get
$\displaystyle{-\frac{f^{(n)}(a)}{n!}(x-a)^n+\int_a^x\frac{f^{(n)}(t)}{(n-1)!}(x-t)^{(n-1)}dt=\cdots = -\sum_{k=1}^n \frac{f^{(k)}(a)}{k!}(x-a)^k+\int_a^x f^{\;'}(t)dt}$
Therefore $\quad (1)\quad\displaystyle{f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k+\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n dt}$
Since $(x-t)^n$ will not change sign when $t \in [a,x]$, we have
$\displaystyle{\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n dt = \frac{f^{(n+1)}(\xi )}{n!}\int_a^x (x-t)^n dt = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}}$ and so
$\quad\quad\quad\quad\quad (2)\quad \displaystyle{f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}}$
These expressions form the Taylor's theorem for single variable functions

Decimal number presentation

Sun, 26 Dec 2010 16:43:54 -0700

(1) For $n = \sum_0^{k-1} a_j 10^j, \quad 10 > a_j \in \mathbb{N}, \quad j=\overline{0,k-1}$ define $S(n) = \sum_0^{k-1} a_j$
$\quad\quad$ Prove that $\forall m \in \mathbb{N} \quad \exists n \in \mathbb{N}:\; n+S(n)\in \{m,m+1\}$

$\sum y_k \sum \frac{x_j^2}{y_j} - \left(\sum x_k \right )^2 = \sum_{1\le i
Sun, 26 Dec 2010 16:19:10 -0700
(1) $\displaystyle{\sum_{k=1}^n y_k \sum_{j=1}^n \frac{x_j^2}{y_j} - \left(\sum_{k=1}^n x_k \right )^2 = \sum_{1\le i

Non-linear Recursion Problems

Fri, 24 Dec 2010 15:21:05 -0700
(1) Find $\{a_n\}$ such that $\displaystyle{a_1 = \frac{1}{2}, \; \sum_{k=1}^n a_k = n^2 a_n}$

Misc little: Find $x$ such that $x-\lfloor x \rfloor , \lfloor x \rfloor, x $ form a geometric progression

Thu, 23 Dec 2010 16:06:13 -0700
(1) Find $x$ such that $x-\lfloor x \rfloor , \lfloor x \rfloor, x $ form a geometric progression

Polynomials$\cdots$: Find $m,n \in \mathbb{N}^+$ s.t. $(1+x+\cdots + x^m) \mid (1+x^n +x^{2n} + \cdots + x^{mn})$

Thu, 23 Dec 2010 15:52:27 -0700
(1) Find $m,n \in \mathbb{N}^+$ s.t. $(1+x+\cdots + x^m) \mid (1+x^n +x^{2n} + \cdots + x^{mn})$

Liouville's theorem (On diophantine approximation)

Thu, 23 Dec 2010 15:45:31 -0700
Theorem: If $\alpha$ is an irrational number which is the root of a polynomial $P$ of degree $n > 0$
with integer coefficients, then $\exists \;A \in \mathbb{R}, \; (A>0)$ such that $\displaystyle{\left|\alpha - \frac{p}{q}\right | > A/q^n \quad\forall p,q \in \mathbb{Z}, \; (q > 0)}$

Proof: Let $M = \displaystyle{\max_{[\alpha -1, \alpha +1]} |P'|}$ and take $A \in (0, \min(1,1/M,|\alpha - \alpha_1| , \cdots, |\alpha-\alpha_m|))$
where $\alpha_1, \cdots, \alpha_m$ are distinct roots of $P$ which differ from $\alpha$.

Assume the claim is not true: $\displaystyle{\left|\alpha - \frac{p}{q}\right| \le A/q^n < \min(1,1/M,|\alpha - \alpha_1| , \cdots, |\alpha-\alpha_m|)}$ ,
for some integers $p$ and $q \ (>0)$, then $\displaystyle{\frac{p}{q} \in [\alpha-1,\alpha+1]}$ and $\displaystyle{\frac{p}{q} \not\in \{ \alpha_1,\cdots,\alpha_m\}}$
and so $\displaystyle{\frac{p}{q}}$ is not a root of $P$ and there is no root between $\alpha$ and $\displaystyle{\frac{p}{q}}$.

By the mean value theorem, there is a $\theta$ between $\alpha$ and $p/q$ such that
$\displaystyle{0 \ne -P\left(\frac{p}{q}\right) = P(\alpha)-P\left(\frac{p}{q}\right) = \left(\alpha-\frac{p}{q}\right)P'(\theta)}$ hence
$\displaystyle{\left|\alpha - \frac{p}{q}\right| = \frac{|P(\frac{p}{q})|}{|P'(\theta)|}>A\left|P\left(\frac{p}{q}\right)\right|=A\left|\sum_{i=0}^n c_i p^i q^{-i}\right| = \frac{A|\sum_{i=0}^n c_i p^i q^{n-i}|}{q^n} \ge A/q^n \ge \left|\alpha - \frac{p}{q}\right|}$
A contradiction! and this completed the proof.

Lagrange's identity $\displaystyle{ (a \cdot b)^2 =|| a||^2 || b||^2 - \sum_{1\le i
Thu, 23 Dec 2010 11:21:41 -0700
By induction we have $\displaystyle{\left(\sum_{k=1}^n x_k\right) = \sum_{k=1}^n x_k^2 + \sum_{1\le i < j \le n} 2x_i x_j}$. This implies that

$\begin{align} \left \| \mathbf{a}\right \|^2 \left \| \mathbf{b}\right \|^2 &=\sum_{k=1}^n a_k^2 \sum_{k=1}^n b_k^2 = \sum_{1\le k,j \le n} a_k^2 b_j^2 = \sum_{k=1}^n a_k^2 b_k^2 + \sum_{1\le k & = \sum_{k=1}^n (a_k b_k)^2+ \sum_{1\le k < j\le n} \left(2a_k b_k a_j b_j+\left( a_k^2 b_j^2+a_j^2 b_k^2 -2a_k b_k a_j b_j \right )\right )\\
& = \left(\mathbf{a} \cdot \mathbf{b} \right )^2 + \sum_{1 \le k