From 472a8079c1f685b7b1179e28843557068a566352 Mon Sep 17 00:00:00 2001 From: nunzip Date: Tue, 20 Nov 2018 15:04:22 +0000 Subject: Add dots and correct grammar to part 2 --- report/paper.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/report/paper.md b/report/paper.md index f5b1934..44d3d70 100755 --- a/report/paper.md +++ b/report/paper.md @@ -230,13 +230,13 @@ affect recognition the most are: glasses, hair, sex and brightness of the pictur To combine both method it is possible to perform LDA in a generative subspace created by PCA. In order to maximize class separation and minimize the distance between elements of the same class it is necessary to -maximize the function J(W) (generalized Rayleigh quotient): $J(W) = \frac{W\textsuperscript{T}S\textsubscript{B}W}{W\textsuperscript{T}S\textsubscript{W}W}$ +maximize the function J(W) (generalized Rayleigh quotient): $J(W) = \frac{W\textsuperscript{T}S\textsubscript{B}W}{W\textsuperscript{T}S\textsubscript{W}W}$. With S\textsubscript{B} being the scatter matrix between classes, S\textsubscript{W} being the within-class scatter matrix and W being the set of projection vectors. $\mu$ represents the mean of each class. -It can be proven that when we have a singular S\textsubscript{W} we obtain [@lecture-notes]: $W\textsubscript{opt} = arg\underset{W}max\frac{|W\textsuperscript{T}S\textsubscript{B}W|}{|W\textsuperscript{T}S\textsubscript{W}W|} = S\textsubscript{W}\textsuperscript{-1}(\mu\textsubscript{1} - \mu\textsubscript{2})$ +It can be proven that when we have a singular S\textsubscript{W} we obtain [@lecture-notes]: $W\textsubscript{opt} = arg\underset{W}max\frac{|W\textsuperscript{T}S\textsubscript{B}W|}{|W\textsuperscript{T}S\textsubscript{W}W|} = S\textsubscript{W}\textsuperscript{-1}(\mu\textsubscript{1} - \mu\textsubscript{2})$. However S\textsubscript{W} is often singular since the rank of S\textsubscript{W} is at most N-c and usually N is smaller than D. In such case it is possible to use @@ -258,7 +258,7 @@ small number) are H\textsubscript{pca}(*e*)= Through linear interpolation, for $0\leq t \leq 1$: $F\textsubscript{t}(e)=\frac{1-t}{2} H\textsubscript{pca}(e)+\frac{t}{2}H\textsubscript{lda}(e)= -\frac{1-t}{2}+\frac{t}{2}\frac{}{ + \epsilon}$ +\frac{1-t}{2}+\frac{t}{2}\frac{}{ + \epsilon}$. The objective is to find a unit vector *e\textsubscript{t}* in **R**\textsuperscript{n} (with n being the number of samples) such that: $e\textsubscript{t}=arg\underset{et}min F\textsubscript{t}(e)$. @@ -268,7 +268,7 @@ We can model the Lagrange optimization problem under the constraint of ||*e*|| To minimize we take the derivative with respect to *e* and equate L to zero: $\frac {\partial L(e\lambda)}{\partial e}=\frac{\partial F\textsubscript{t}(e)}{\partial e} -+\frac{\partial\lambda(||e||\textsuperscript{2}-1)}{\partial e}=0$ ++\frac{\partial\lambda(||e||\textsuperscript{2}-1)}{\partial e}=0$. Being $\nabla F\textsubscript{t}(e)= (1-t)Se+\frac{t}{ +\epsilon}S\textsubscript{B}e-t\frac{}{(