Team:ETH Zurich/modeling/reactions
From 2014.igem.org
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- | + | $$c = \pm\sqrt{a^2 + b^2}$$ | |
- | + | $$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$ | |
- | + | $$\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$ | |
- | + | $$f(x) = \int_{-\infty}^\infty | |
\hat f(\xi)\,e^{2 \pi i \xi x} | \hat f(\xi)\,e^{2 \pi i \xi x} | ||
\,d\xi | \,d\xi | ||
- | + | $$ | |
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</html> | </html> | ||
{{:Team:ETH Zurich/tpl/foot}} | {{:Team:ETH Zurich/tpl/foot}} |
Revision as of 19:53, 18 September 2014
Reactions
$$c = \pm\sqrt{a^2 + b^2}$$ $$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$ $$\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$ $$f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi $$