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Convergent sequence

January 12, 2019 / Leave a comment

Let $\{x_n\}_{n \in \mathbb{N}}$ be a sequence of positive real numbers such that

(1) $\begin{equation*} x_{n+m} \leq x_n + x_m \quad , \quad m, n \in \mathbb{N} \end{equation*}$

Prove that $\left\{\dfrac{x_n}{n} \right\}_{n \in \mathbb{N}}$ converges.

Solution

Fix $m$ and let $n \geq m$ . Then, there exist $k, r$ such that $n=km+r$ where $0\leq r <m$ . Thus,

$\frac {x_n}{n} = \frac {x_ {km+r}}{n}\leq \frac {kx_ {m}}{n} + \frac {x_{r}}{n}$

Letting $n \rightarrow +\infty$ it follows that

$\limsup \frac {x_n}{n} \leq \frac {x_ {m}}{m} + 0$

Since this holds forall $m$ it follows that $\displaystyle \limsup \frac {x_n}{n} \leq \liminf \frac {x_ {m}}{m}$ and the result follows.

Root inequality

December 15, 2018 / Leave a comment

Let $a, b, c, d$ be positive real numbers satisfying the following equality

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} =4$

Prove that

$\displaystyle \sqrt[3]{\frac{a^3+b^3}{2}} + \sqrt[3]{\frac{b^3+c^3}{2}} + \sqrt[3]{\frac{c^3+d^3}{2}} + \sqrt[3]{\frac{d^3+a^3}{2}} \leq 2 \left ( a+b+c+d \right ) -4$

Solution

We begin by stating a lemma:

Lemma: Let $a, b$ be positive real numbers, then:

$\frac{a+b}{2} \leq \sqrt[3]{\frac{a^3+b^3}{2}} \leq \frac{a^2+b^2}{a+b}$

Now, making use of the lemma we have that:

$\begin{align*} \sum \sqrt[3]{\frac{a^3+b^3}{2}} &\leq \sum \frac{a^2+b^2}{a+b} \\ &=\sum \left ( a+b \right )-\sum \frac{2ab}{a+b} \\ &= 2\left ( a+b+c+d \right )- 2\sum \frac{1}{\frac{1}{a}+\frac{1}{b}} \end{align*}$

Making use of the Cauchy – Schwartz inequality we have that

$\begin{align*} \sum \frac{1}{\frac{1}{a}+\frac{1}{b}} & \geq \frac{\left ( 1+1+1+1 \right )^2}{2\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d} \right )} \\ &=\frac{8}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d}} \\ &= 2 \end{align*}$

The inequality now follows.

Trigonometric identity

December 14, 2018 / Leave a comment

Let $a, b, c$ be positive real numbers such that $a+b+c=\pi$ . Prove the following trigonometric identities:

$\displaystyle \sin a + \sin b + \sin c = 4 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2}$
$\displaystyle \cos a + \cos b + \cos c = 1+4 \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2}$

Solution

We have successively:
$\begin{align*} \sin a + \sin b +\sin c &= 2 \sin \frac{a+b}{2} \cos \frac{a-b}{2} + \sin c \\ &=2 \sin \left ( \frac{\pi}{2} - \frac{c}{2} \right ) \cos \frac{a-b}{2} + 2 \sin \frac{c}{2} \cos \frac{c}{2} \\ &=2 \cos \frac{c}{2} \cos \frac{a-b}{2} + 2 \sin \frac{c}{2} \cos \frac{c}{2} \\ &=2 \cos \frac{c}{2} \left ( \cos \frac{a-b}{2} + \sin \frac{c}{2} \right ) \\ &= 2 \cos \frac{c}{2} \left ( \cos \frac{a-b}{2}+ \cos \frac{a+b}{2} \right ) \\ &= 2 \cos \frac{c}{2} \cdot 2 \cos \frac{a}{2} \cos \frac{b}{2} \\ &= 4 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2} \end{align*}$
Similarly, we have successively:
$\begin{align*} \cos a + \cos b + \cos c &=2 \cos \frac{a+b}{2} \cos \frac{a- b}{2} + \cos c \\ &=2 \cos \left ( \frac{\pi}{2} - \frac{c}{2} \right ) \cos \frac{a-b}{2} + \cos c \\ &=2 \sin \frac{c}{2} \cos \frac{a-b}{2} + \left ( 1 - 2 \sin^2 \frac{c}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \left (\cos \frac{a-b}{2} - \sin \frac{c}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \left [ \cos \frac{a-b}{2} - \sin \left ( \frac{\pi}{2} -\frac{a+b}{2} \right ) \right ] \\ &=1 + 2 \sin \frac{c}{2} \left ( \cos \frac{a-b}{2} - \cos \frac{a+b}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \cdot 2 \sin \frac{a}{2} \sin \frac{b}{2} \\ &=1 + 4 \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2} \end{align*}$

Number of solutions

December 10, 2018 / Leave a comment

Find the number of solutions of the equation:

$x_1+x_2+\cdots + x_{100} = 2017$

in the positive integers.

Solution

We set $y_n=x_n-1 \;\;, \;\; 1\leq n \leq 100$ . It suffices to find the number of solutions of the equation

(1) $\begin{equation*} y_1+y_2+ \cdots + y_n =1917 \end{equation*}$

in the non negative numbers. We represent each sum $z_1+\cdots+z_m$ of non negative integers with a sequence of $z_1$ dots ( $\bullet$ ) followed by a vertical bar ( $\big \mid$ ), after $z_2$ dots another one vertical bar etc, till we place the last $z_m$ dots ( without the vertical bar at the end.) For example the sum $5+2+0+1=8$ can be represented as

$\bullet \bullet \bullet \bullet \bullet \mid \bullet \bullet \mid \mid \bullet$

We note that every solution of $(1)$ matches a sequence that has $1917$ dots in total and $99$ vertical bars. Conversely, every such sequence matches a solution of $(1)$ .

Thus, in total there are

$\binom{1917+99}{99} = \binom{2016}{99}$

solutions.

Comment: In general the equation

$x_1+x_2+\cdots+x_m=n$

has $\displaystyle \binom{n-1}{m-1}$ solutions in the positive integers and $\displaystyle \binom{n+m-1}{m-1}$ solutions in the non negative integers.