Recherche

/jsm> * <jsm>\left[\!| n b |\!\right]_{nat} = 2\times \left[\!| n |\!\right]_{nat} + \left[\!| b |\!\ri... narray} \left[\!| 1 0 1 |\!\right]_{nat} & = & 2\times \left[\!| 1 0 |\!\right]_{nat} + \left[\!| 1 |\!\right]_{nat} \\ & = & 2\times\left(2\times \left[\!| 1 |\!\right]_{nat} + \left[\!| 0 |\!\right]... ight) + \left[\!| 1 |\!\right]_{nat} \\ & = & 2\times\left(2\times \left[\!| 1 |\!\right]_{bin} + \left

t(n,m\right) |\!\right] = \left[\!| n |\!\right] \times \left[\!| m |\!\right]</jsm> * Premier cas : <... left[\!| mult\left(0,m\right) |\!\right] = 0 = 0 \times \left[\!| m |\!\right] = \left[\!| 0 |\!\right] \times \left[\!| m |\!\right]</jsm> * Second cas : <jsm>\left[\!| mult\left(1,m\right) |\!\right] = m = 1 \times \left[\!| m |\!\right] = \left[\!| 1 |\!\right] \times \left[\!| m |\!\right]</jsm> H

weighting of R, G and B: * <jsm>L_1 = 0.2126\times R + 0.7152 \times G + 0.0722 \times B </jsm> * ... interested in B satisfying * <jsm>\left(0.0722\times B + 0.05\right)/\left(0.05\right)\geq 3.0</jsm>, or * <jsm>0.0722\times B \geq 3.0\times 0.05 -0.05 = 2.0\times 0.05 = 0.1 </jsm> * <jsm>\Righta... en to achieve this, we need * <jsm>\left(0.0722\times B + 0.9278\times X +0.05\right)/\left(0.05\right)

''f_A'' de type noté <jsm>f_A : S_{1_A}\times S_{2_A}\times\ldots \times S_{n_A}\rightarrow S_{... \begin{eqnarray}f_{T_\Sigma} : & S_{1_{T_\Sigma}}\times S_{2_{T_\Sigma}} \times \cdots \times S_{n_{T_\Si

et non <jsm>x = a y^2 + b y +c</jsm> * <jsm>a \times 0^2 + b\times 0 +c = 0 \Rightarrow c=0 </jsm> * <jsm> h = a\times \left( 2h \right)^2 + b\times \left( 2h \right) = 4ah^2 + 2bh \quad \Rightarrow \... de la courbe, <jsm>dy = \frac{2x}{4h} \lt \frac{2\times 2h}{4h}= 1</jsm> car <jsm>x\leq 2h</jsm>. Ainsi,

ta y}{\Delta x}\left(x_c + 1\right)+b \\ & = & 2\times\left[\frac{\Delta y}{\Delta x}\left(x_c + 1\right) + b - y_c \right] -1 \\ d_c=\Delta x \times D & = & 2\times\left[\Delta y\left(x_c + 1\right) + b\times\Delta x - y_c \times\Delta x\right] -\Delta x \end{eqnarray} </jsmath> * Notons que pre

sm> * <jsm>\left[\!| n 0 |\!\right]_{nat} = 2 \times \left[\!| n |\!\right]_{nat}+\left[\!| 0 |\!\right]_{bin} = 2 \times \left[\!| n |\!\right]_{nat} </jsm> * <jsm>\left[\!| n 1 |\!\right]_{nat} = 2 \times \left[\!| n |\!\right]_{nat}+\left[\!| 1 |\!\right]_{bin} = 2 \times \left[\!| n |\!\right]_

m> * <jsm>\displaystyle = P_{B}\left(t\right)\times\left[1- \alpha_{T}\sum_{B\neq B'} 1\right] +\alph... m> * <jsm>\displaystyle = P_{B}\left(t\right)\times\left[1- \alpha_{T}\left(n-1\right)\right] +\alpha... m> * <jsm>\displaystyle = P_{B}\left(t\right)\times\left[1- \alpha_{T}\left(n-1\right)\right] +\alpha

- [[recipes:pita_bread|Pita Bread]] from the LA Times cookbook (p.374) - [[recipes:pain-de_campagne... _special|Joe's Special]] -- according to the L.A. Times cookbook. I stopped making this since it requires... ion. But I'm sure I made it at least a couple of times for Sunday lunch or brunch, and it is worth havin

es === Puisque <jsm>\left(n-1\right)^2= n^2 - 2 \times n + 1</jsm>, on constate que <jsm>n^2 - \left(n-1\right)^2 = n^2 - n^2 + 2\times n - 1=2n-1=2\left(n-1\right) + 1</jsm>. En effet, si l'on pense aux... +1), ou <jsm>\displaystyle \sum_{k=0}^n \left(1+2\times k\right)</jsm> pour un nombre ''n'' quelconque.

intra} \left(C_k\right) = \sum_{x_i\in C_k} p_i \times d\left(x_i, g_k \right)</jsm> mais je crois que l... yle I_{inter}\left(D\right) = \sum_{k=1}^{K} P_k \times d\left(g, g_k \right)</jsm>, où \$g_k\$ est le c... r serait K tel que <jsm>\displaystyle \frac{1}{T}\times \frac{M-K}{K-1} </jsm> est maximisé. Mais qu'est

iguration**__ de M est un élément de <jsmath> K \times \triangleright\Sigma^* \times\left(\Sigma^* \left

n * <jsm>\delta: \left(K \setminus H\right) \times \Sigma \mapsto K\times \left(\Sigma \cup \left\{

b then A else B '' * la répétition : ''do n times A'' * l'itération : ''while b do A'' ^ Rema... > \forall s\in S,\; \left[\!| \text{do } n \text{ times }A |\!\right]\left(s\right) = \underbrace{\left[\