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A factorization

July 30, 2022 / Leave a comment

Let $ABC$ be a right triangle at $A$ . Factor $a^3 + b^3 + c^3$ .

Solution

Since $a^2 = b^2 + c^2$ we have successively:

$\begin{align*} a^3 + b^3 + c^3 &= a \cdot a^2 + b^3 + c^3 \\ &= a \left ( b^2 + c^2 \right ) + b^3 + c^3 \\ &= ab^2 + ac^2 + b^3 + c^3 \\ &= b^2 \left ( a + b \right ) + c^2 \left ( a + c \right ) \\ &= \left ( a^2 - c^2 \right ) \left ( a + b \right ) + \left ( a^2 - b^2 \right ) \left ( a + c \right ) \\ &= \left ( a + c \right ) \left ( a + b \right ) \left ( 2a - b - c \right ) \end{align*}$

Equality

July 24, 2022 / Leave a comment

Prove that in any triangle $ABC$ it holds that

$16\sum m_a^2 m_b^2 = 9 \left ( s^2 + r^2 + 4Rr \right )^2 - 144 s^2 Rr$

Solution

We have successively:

$\begin{align*} 16 \sum m_a^2 m_b^2 &= 16 \sum \frac{2 \left ( b^2 + c^2 \right ) - a^2}{4} \cdot \frac{2 \left ( a^2 + c^2 \right ) - b^2}{4} \\ &= \sum \left ( 2b^2 + 2c^2 - a^2 \right ) \left ( 2a^2 + 2c^2 - b^2 \right ) \\ &= \sum \left ( 5a^2 b^2 + 2b^2 c^2 + 2a^2 c^2 - 2a^4 - 2b^4 + 4c^4 \right ) \\ &= 9 \left ( a^2 b^2 + b^2 c^2 + c^2 a^2 \right ) \\ &= 9 \left ( ab + bc + ca \right )^2 - 8abc \left ( a + b +c \right ) \\ &= 9 \left ( s^2 + r^2 + 4Rr \right )^2 - 144 Rrs^2 \end{align*}$

On a Lambert W integral

July 18, 2022 / Leave a comment

Prove that

$\int_{0}^{\infty} W_0 \left ( x e^{-x} \right )\, \mathrm{d}x = \int_{0}^{\infty} \frac{W_0 \left ( xe^{-x} \right )}{x} \, \mathrm{d}x = \frac{\pi^2}{12}$

A Fibonacci series

July 17, 2022 / Leave a comment

Let $\mathcal{F}_n$ denote the $n$ – th Fibonacci number. Prove that

$\sum_{n=2}^{\infty}\arctan \left(\frac{\mathcal{F}_{n-1}}{\mathcal{F}_n \mathcal{F}_{n+1}+1}\right)\arctan \left(\frac{\mathcal{F}_{n+2}}{\mathcal{F}_n \mathcal{F}_{n+1}-1}\right)=\frac{\pi^2}{16}$

Solution

Let $S$ denote the given sum. Then,

$\begin{align*} \mathcal{S} & =\sum_{n=2}^{\infty}\arctan \left(\frac{\mathcal{F}_{n+1}-\mathcal{F}_n}{\mathcal{F}_n\mathcal{F}_{n+1}+1}\right)\arctan \left(\frac{\mathcal{F}_{n+1}+\mathcal{F}_n}{\mathcal{F}_n\mathcal{F}_{n+1}-1}\right) \\ &=\sum_{n=2}^{\infty}(\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n)(\pi-\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n) \\ &= \pi \sum_{n=2}^{\infty}(\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n)- \sum_{n=2}^{\infty}\left(\arctan^2 \mathcal{F}_{n+1} -\arctan^2 \mathcal{F}_n\right) \\ &= \pi \lim_{n \rightarrow +\infty}(\arctan \mathcal{F}_n-\arctan \mathcal{F}_2)- \lim_{n \rightarrow + \infty} \left( \arctan^2 \mathcal{F}_n -\arctan^2 \mathcal{F}_2 \right) \\ &= \pi \left ( \frac{\pi}{2} - \frac{\pi}{4} \right ) - \left ( \frac{\pi^2}{4} - \frac{\pi^2}{16} \right ) \\ &= \frac{\pi^2}{16} \end{align*}$

A floor series

July 17, 2022 / Leave a comment

Let $x \in [0, 1)$ . Prove that

$\displaystyle x = \sum_{n=1}^{\infty} \frac{\left \lfloor 2^n x \right \rfloor - 2 \left \lfloor 2^{n-1} x \right \rfloor}{2^n}$ .
$\displaystyle x = \sum_{n \geq 1 , 2 \nmid \left \lfloor 2^n x \right \rfloor} \frac{1}{2^n}$ .
$\displaystyle \sum_{n=1}^{\infty} \frac{1 - \mathrm{sgn} \left ( \tan 2^n \right )}{2^{n+2}} = \frac{1}{\pi}$ where $\mathrm{sgn}$ denotes the sign function.

Solution

We have successively:
$\begin{align*} \sum_{n=1}^{\infty} \frac{\left \lfloor 2^n x \right \rfloor - 2 \left \lfloor 2^{n-1} x \right \rfloor}{2^n} &= \lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \frac{\left \lfloor 2^n x \right \rfloor - 2 \left \lfloor 2^{n-1} x \right \rfloor}{2^n} \\ &=\lim_{N \rightarrow +\infty} \sum_{n=1}^{\infty} \left (\frac{\left \lfloor 2^n x \right \rfloor}{2^n} - \frac{ \left \lfloor 2^{n-1} x \right \rfloor}{2^{n-1}} \right ) \\ &=\lim_{N \rightarrow +\infty} \frac{\left \lfloor 2^N x \right \rfloor}{2^N} \end{align*}$

However,

$x-\frac{1}{2^N} = \frac{2^Nx-1}{2^N} < \frac{\lfloor 2^Nx \rfloor}{2^N} \leq \frac{2^Nx}{2^N}=x$

and the result follows.
Let $a_n=\lfloor 2^nx \rfloor$ . Then, $a_n \leq 2^n x < a_n+1$ and $a_{n-1} \leq 2^{n-1}x < a_{n-1}+1$ . It follows that $-1 < a_n-2a_{n-1} < 2$ . Hence,
$\lfloor 2^nx \rfloor -2\lfloor 2^{n-1}x \rfloor =a_n-2a_{n-1} \in \{0,1\}$

since $a_n$ is integer for all $n$ . The result follows from the previous question as well as the fact that $a_n-2a_{n-1}=1$ if-f $n$ is odd; $2 \nmid a_n$ .
It follows from the previous question since $1-\mathrm{sgn}(\tan 2^n)= \left\{\begin{matrix} 2 & , & 2 \nmid \left \lfloor \frac{2^{n+1}}{\pi} \right \rfloor \\ 0 & , & \text{otherwise} \end{matrix}\right.$ .

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