@Rui
You think your math is advanced? Try this:
1-1
\begin{align*}&\quad 1 - 1\\ &= \frac{2}{2} - \frac{3}{3}\\ &= \frac{2 \times 3 - 3 \times 2}{6}\\ &= \frac{1}{6}(\sqrt4 \times \sqrt9 - \lfloor \pi \rfloor \lceil e \rceil)\\ &= \frac16 \left( \sqrt{ \lim_{(x,y) \to (1,1)} \frac{(x+1)^2 + (y+1)^2}{x^2+y^2} } \times \sqrt{-1 + \sum_{n=1}^4 n} - \left \lfloor 4\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \right \rfloor\left \lceil \lim_{\theta \to 0}(1 + \theta)^{1/\theta} \right \rceil\right)\\ &= \frac{1}{\int_0^\infty x^3 e^{-x}\,dx}\left( \sqrt{\lim_{(x,y)\to (1,1)}\frac{(x+1+iy+i)(x+1-iy-i)}{\left(nx,my)\cdot (n^{-1}x,m^{-1}y\right)}} \sqrt{e^{2001i\pi} + \sum_{n=1}^N \frac{n^2}{n} - \sum_{n=5}^N \sqrt[3]{n^3}} - \left \lfloor 16777216^{1/144^{1/2}} \sum_{k=0}^\infty \int_0^1 \int_{2k+1}^\infty \frac{2v}{u^2} \,du\,dv\right \rfloor \left \lceil \lim_{\epsilon \to 0} \frac{1-\cos x}{x^2} \lim_{\theta \to 0} \lim_{\phi \to 2\theta} \lim_{\rho \to \frac\pi2}2^{2\rho/\pi} \theta^{-1/\phi} \left( \sqrt{\frac\phi2} + \sqrt{\frac2\phi}\right) \sin \rho \right \rceil \right)\\ &= 0\end{align*}
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