Joy of Mathematics

An equation!

June 21, 2024June 21, 2024 Tolaso

Let $f(x) = \left(x^2 +1 \right) \ln x \; , \; x>0$ .

Prove that $f$ is strictly increasing on $(0, +\infty)$ .
If $x \in \left(0, \frac{\pi}{2} \right)$ , solve the equation
$\cos^2 x \cdot f \left ( \tan x \right ) + \sin^2 x \cdot f \left ( \cot^{22} x \right ) =0$

Solution

It is $\displaystyle f'(x) = x \left ( \ln x^2 + 1 + \frac{1}{x^2} \right )$ . However, recalling that
$\ln x \geq 1 - \frac{1}{x} \quad \text{foreach} \quad x>0$

we conclude that $f$ is strictly increasing.
We note that
$\begin{align*} \cos^2 x \cdot f \left ( \tan x \right ) + \sin^2 x \cdot f \left ( \cot^{22} x \right ) =0 &\Leftrightarrow \frac{\cos^2 x}{\sin^2 x} \cdot f \left ( \tan x \right ) + f \left ( \cot^{22}x \right ) = 0 \\ &\Leftrightarrow \cot^2 x \cdot f \left ( \frac{1}{\cot x} \right ) + f \left ( \cot^{22}x \right ) =0 \\ &\!\!\!\!\!\!\!\!\!\overset{u = \cot x}{\Leftarrow \! =\! =\! =\! =\! \Rightarrow } u^2 f \left ( \frac{1}{u} \right ) + f \left ( u^2 \right ) = 0 \end{align*}$

On the other hand, $f \left ( \frac{1}{x} \right ) = - \frac{f(x)}{x^2}$ and hence:

$\begin{align*} f \left ( u^{22} \right ) = f(u) &\Leftrightarrow u^{22} = u \\ &\Leftrightarrow u = 1 \\ &\Leftrightarrow x = \frac{\pi}{4} \end{align*}$

A logarithm

August 18, 2023 Tolaso

Let $a \in (0, 1)$ . If $\log_a \left ( 1 - a \right ) = 5$ , prove that $\log_a \left ( 1 + a \right ) = -2$ .

An identity

July 23, 2023 Tolaso

If $\alpha + \beta + \gamma =0$ , prove that

$2 \left ( \alpha^5 + \beta^5 + \gamma^5 \right ) = 5 \alpha \beta \gamma \left ( \alpha^2 + \beta^2 + \gamma^2 \right )$

A finite product

April 10, 2023 Tolaso

Evaluate the product

$\Pi = \prod_{k=1}^{n}\frac{(2k)^4+(2k)^2+1}{(2k-1)^4+(2k-1)^2+1}$

Solution

We are making use of the identity

$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) = (x^2 + x + 1)((x-1)^2 + (x-1) + 1)$

Hence,

$\begin{align*} \prod_{k=1}^{n}\frac{(2k)^4+(2k)^2+1}{(2k-1)^4+(2k-1)^2+1} &= \prod_{k=1}^{n} \frac{\cancel{\left ( 4k^2- 2k+1 \right )} \left ( 4k^2+2k+1 \right )}{\left ( 4k^2-6k+3 \right ) \cancel{\left ( 4k^2-2k+1 \right )}} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n} \left ( 4k^2-6k+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=2}^{n} \left ( 4k^2-6k+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n-1} \left ( 4(k+1)^2-6(k+1)+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n-1} \left ( 4k^2 + 2k+1 \right )} \\ &= 4n^2 +2n+1 \end{align*}$

A convergent sequence

April 9, 2023 Tolaso

Let $\left \{ a_n \right \}_{n \in \mathbb{N}}$ be a strictly increasing sequence of positive integers. Prove that the series $\displaystyle \mathcal{S} = \sum_{i=0}^{\infty} \frac{1}{\left [ a_i, a_{i+1} \right ]}$ , where $\left [ \cdot, \cdot \right ]$ denotes the least common multiple, converges.

Solution

We have successively:

$\begin{align*} \sum_{i=0}^{\infty} \frac{1}{\left [ a_i, a_{i+1} \right ]} &= \sum_{i=0}^{\infty} \frac{\left ( a_i,a_{i+1} \right )}{a_i a_{i+1}} \\ &\leq \sum_{i=0}^{\infty} \frac{a_{i+1} - a_i}{a_i a_{i+1}} \\ &= \sum_{i=0}^{\infty} \left ( \frac{1}{a_i} - \frac{1}{a_{i+1}} \right ) \\ &= \frac{1}{a_0} - \lim_{N \rightarrow +\infty} \frac{1}{a_N} \end{align*}$