A booklet on the Basic AI Technics

The result from the latest successful Travis build can be found the 'releases' pane of this page.

$${\left(\sum _{k=1}^n{a}_k{b}_k\right)}^2\le \left(\sum _{k=1}^n{a}_k^2\right)\left(\sum _{k=1}^n{b}_k^2\right)$$

$${\left(\sum _{k=1}^n{a}_k{b}_k\right)}^2\le \left(\sum _{k=1}^n{a}_k^2\right)\left(\sum _{k=1}^n{b}_k^2\right)$$

Ne marche pas

$$ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) $$caption=This is a super cool formula&anchor=eq1

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$$caption=This is a super cool formula&anchor=eq1
\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)
$$

$$ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) $$caption=This is a super cool formula

$$ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) $$anchor=eq1

$$[T=t_1,\dots ,t_m][\mathrm{\forall }{t}_k\in T{t}_k\subseteq B]$$