[Olena-patches] 4467: Remove trailing spaces.

* green/doc/formulae/formulae.tex : Correct file --- trunk/milena/sandbox/ChangeLog | 6 ++++++ .../milena/sandbox/green/doc/formulae/formulae.tex | 8 ++++---- 2 files changed, 10 insertions(+), 4 deletions(-) diff --git a/trunk/milena/sandbox/ChangeLog b/trunk/milena/sandbox/ChangeLog index 923875c..8bb5ec1 100644 --- a/trunk/milena/sandbox/ChangeLog +++ b/trunk/milena/sandbox/ChangeLog @@ -1,5 +1,11 @@ 2009-09-11 Yann Jacquelet <jacquelet(a)lrde.epita.fr> + Remove trailing spaces. + + * green/doc/formulae/formulae.tex : Correct file. + +2009-09-11 Yann Jacquelet <jacquelet(a)lrde.epita.fr> + Correct english writing in the documentation file. * green/doc/formulae/formulae.tex : Correct file. diff --git a/trunk/milena/sandbox/green/doc/formulae/formulae.tex b/trunk/milena/sandbox/green/doc/formulae/formulae.tex index 655cb1a..43bfbf5 100644 --- a/trunk/milena/sandbox/green/doc/formulae/formulae.tex +++ b/trunk/milena/sandbox/green/doc/formulae/formulae.tex @@ -1084,7 +1084,7 @@ $ %%================================================================= \subsection{MCO for eigenvalues} -In 3d, it's difficult to extract the cubic roots from the characteristic +In 3d, it's difficult to extract the cubic roots from the characteristic polynomia. The difficulties disappear when we find one of the three roots. A planar regression allows us to reach the equation of the plane. From the equation, we can determine its normal vector $\bm{w}$. It satisfies the @@ -1096,17 +1096,17 @@ the two others eigenvalues (just solve $\lambda^2 - (trace(A) - \lambda_3) $\lambda_2$ let us find $\bm{u}$ and $\bm{v}$ by the equations $A\bm{u} = \lambda_1\bm{u}$ and $A\bm{v} = \lambda_2\bm{v}$. -Let's center the points by subtracting their center of mass. Now, we have +Let's center the points by subtracting their center of mass. Now, we have three equivalent ways to estimate the coefficients of the plane: \begin{itemize} \item if not $c = 0$, then $\frac{a}{c}x + \frac{b}{c}y + z = 0$, \item if not $b = 0$, then $\frac{a}{b}x + y + \frac{c}{b}z = 0$, \item if not $a = 0$, then $x + \frac{b}{a}y + \frac{c}{a}z = 0$. \end{itemize} -As we cannot decide which way is the best, may be we have to test the three +As we cannot decide which way is the best, may be we have to test the three ones. -Let's choose the linear model of the major inertia plane +Let's choose the linear model of the major inertia plane ($ax + by + cz + d = 0$). With not $c = 0$. \begin{tabular}{lcl} -- 1.5.6.5