Recent content by dromarand

## \sqrt{3} = 1+\frac{\sqrt{2}}{1+\frac{\sqrt{2+\sqrt{2}}}{1+\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\ddots}}}##

Please this: ##2^{p-1}\left(2^{p}-1\right)=\sum_{k=0}^{n=\left\lfloor\frac{p-1}{4}\right\rfloor}\binom{2p+2}{4k+p-4n-1}##

##\frac{2}{ \pi } = 1\cdot \frac{1}{2\cdot \frac{2}{3\cdot \frac{3}{\ddots} } }##

## \prod_{p \ prime}{} \frac{p^{2}+1}{p^{2}-1} = \frac{5}{2} ## ## \prod_{p \ prime}{} \frac{p^{4}+1}{p^{4}-1} = \frac{7}{6} ##

##\pi = 6 - \sum_{n=1}^\infty \frac{1}{\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)} + \frac{1}{2\left(n+\frac{1}{4}\right)\left(n-\frac{1}{4}\right)}## ##\phi = \sqrt{2+\frac{1}{\sqrt{2+\frac{1}{\sqrt{2+...}}}}}##

\pi = 6 - \sum_{n=1}^{ \infty } \frac{1}{\left( n+ \frac{1}{2} \right)\left( n- \frac{1}{2} \right) } + \frac{1}{2\left( n+ \frac{1}{4} \right)\left( n- \frac{1}{4} \right) } \phi = \sqrt{2+ \frac{1}{ \sqrt{2+ \frac{1}{ \sqrt{2+...} } } } }

The ideal model of the discussed experiment assumes that the first falling snowflakes (constant vertical precipitation velocity, constant density) touch the ground at the moment of starting the measurement of the dozer movement (horizontal, uniform rectilinear motion). Which dozer, under the...