Team:ETH Zurich/modeling/reactions

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(Difference between revisions)
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$$c = \pm\sqrt{a^2 + b^2}$$
$$c = \pm\sqrt{a^2 + b^2}$$
$$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$
$$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$
-
$$\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$
+
$$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$
$$f(x) = \int_{-\infty}^\infty
$$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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Revision as of 14:21, 23 September 2014

iGEM ETH Zurich 2014