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new 6807cca [MINOR][DOC] Fix latex format and header
6807cca is described below
commit 6807cca3c64682c88464aefb972e995a246871de
Author: Janardhan Pulivarthi <[email protected]>
AuthorDate: Tue May 4 13:41:18 2021 +0530
[MINOR][DOC] Fix latex format and header
---
docs/_includes/header.html | 1 +
docs/site/dml-vs-r-guide.md | 10 ++++++----
2 files changed, 7 insertions(+), 4 deletions(-)
diff --git a/docs/_includes/header.html b/docs/_includes/header.html
index 874d0cf..82506e1 100644
--- a/docs/_includes/header.html
+++ b/docs/_includes/header.html
@@ -47,6 +47,7 @@ limitations under the License.
<li><b>Language Guides:</b></li>
<li><a href=".{% if page.path contains 'site' %}/..{%
endif %}/site/dml-language-reference.html">DML Language Reference</a></li>
<li><a href=".{% if page.path contains 'site' %}/..{%
endif %}/site/builtins-reference.html">Built-in Functions Reference</a></li>
+ <li><a href=".{% if page.path contains 'site' %}/..{%
endif %}/site/dml-vs-r-guide.html">DML vs R guide</a></li>
<li class="divider"></li>
<li><b>ML Algorithms:</b></li>
<li><a href=".{% if page.path contains 'site' %}/..{%
endif %}/site/algorithms-reference.html">Algorithms Reference</a></li>
diff --git a/docs/site/dml-vs-r-guide.md b/docs/site/dml-vs-r-guide.md
index 4b8664d..dbf62e6 100644
--- a/docs/site/dml-vs-r-guide.md
+++ b/docs/site/dml-vs-r-guide.md
@@ -217,6 +217,7 @@ Given lower triangular matrix L, we compute its inverse X
which is also lower tr
both matrices in the middle into 4 blocks (in a 2x2 fashion), and multiplying
them together to get
the identity matrix:
+$$
\begin{equation}
L \text{ %*% } X = \left(\begin{matrix} L_1 & 0 \\ L_2 & L_3
\end{matrix}\right)
\text{ %*% } \left(\begin{matrix} X_1 & 0 \\ X_2 & X_3 \end{matrix}\right)
@@ -224,26 +225,27 @@ L \text{ %*% } X = \left(\begin{matrix} L_1 & 0 \\ L_2 &
L_3 \end{matrix}\right)
= \left(\begin{matrix} I & 0 \\ 0 & I \end{matrix}\right)
\nonumber
\end{equation}
+$$
If we multiply blockwise, we get three equations:
-$
+$$
\begin{equation}
L1 \text{ %*% } X1 = 1\\
L3 \text{ %*% } X3 = 1\\
L2 \text{ %*% } X1 + L3 \text{ %*% } X2 = 0\\
\end{equation}
-$
+$$
Solving these equation gives the following formulas for X:
-$
+$$
\begin{equation}
X1 = inv(L1) \\
X3 = inv(L3) \\
X2 = - X3 \text{ %*% } L2 \text{ %*% } X1 \\
\end{equation}
-$
+$$
If we already recursively inverted L1 and L3, we can invert L2. This suggests
an algorithm
that starts at the diagonal and iterates away from the diagonal, involving
bigger and bigger
