Amman, Jordan
e^{i\pi} + 1 = 0
e^{i x} = \cos x + i \sin x
\displaystyle \sum_{n} \frac{1}{n^s} = \prod_{p} {\frac{1}{1 - \frac{1}{p^s}}}
\displaystyle\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}
Following (1)
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135
57:29
Amman, Jordan
e^{i\pi} + 1 = 0
e^{i x} = \cos x + i \sin x
\displaystyle \sum_{n} \frac{1}{n^s} = \prod_{p} {\frac{1}{1 - \frac{1}{p^s}}}
\displaystyle\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}
Following (1)
See all135
57:29