cleaning more

doc/pub/DecisionTrees/html/._DecisionTrees-bs048.html

@@ -293,7 +293,7 @@ $$
 \frac{\partial {\cal C}}{\partial \gamma} =-2\sum_{i}\beta x_i(y_i-\beta(1+\gamma x_i))=0.
 $$
 
-We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma x_i) \) with \( \boldsymbol{e} \) being the unit vector)
+We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma \boldsymbol{x}) \) with \( \boldsymbol{e} \) being the unit vector)
 $$
 \gamma \boldsymbol{w}^T(\boldsymbol{y}-\beta\gamma \boldsymbol{w})=0,
 $$
@@ -305,7 +305,7 @@ $$
 
 <p>
 which leads to \( \gamma =(\boldsymbol{x}^T\boldsymbol{y}-\beta\boldsymbol{x}^T\boldsymbol{e})/(\beta\boldsymbol{x}^T\boldsymbol{x}) \).  Inserting
-for \( \beta \) gives us an equation for \( \gamma \).
+for \( \beta \) gives us an equation for \( \gamma \). This is a non-linear equation in the unknown \( \gamma \) and has to be solved numerically.
 
 <p>
 The solution to these two equations gives us in turn \( \beta_1 \) and \( \gamma_1 \) leading to the new expression for \( f_1(x) \) as

doc/pub/DecisionTrees/html/._DecisionTrees-bs056.html

@@ -269,7 +269,7 @@ MathJax.Hub.Config({
 <h2 id="___sec55" class="anchor">Gradient Boosting, algorithm </h2>
 
 <p>
-Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  square-error  function
+Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  squared-error  function
 $$
 C(\boldsymbol{y},\boldsymbol{f})=\sum_{i=0}^{n-1}(y_i-f(x_i))^2.
 $$

doc/pub/DecisionTrees/html/DecisionTrees-reveal.html

@@ -2098,7 +2098,7 @@ $$
 $$
 <p>&nbsp;<br>
 
-We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma x_i) \) with \( \boldsymbol{e} \) being the unit vector)
+We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma \boldsymbol{x}) \) with \( \boldsymbol{e} \) being the unit vector)
 <p>&nbsp;<br>
 $$
 \gamma \boldsymbol{w}^T(\boldsymbol{y}-\beta\gamma \boldsymbol{w})=0,
@@ -2114,7 +2114,7 @@ $$
 
 <p>
 which leads to \( \gamma =(\boldsymbol{x}^T\boldsymbol{y}-\beta\boldsymbol{x}^T\boldsymbol{e})/(\beta\boldsymbol{x}^T\boldsymbol{x}) \).  Inserting
-for \( \beta \) gives us an equation for \( \gamma \).
+for \( \beta \) gives us an equation for \( \gamma \). This is a non-linear equation in the unknown \( \gamma \) and has to be solved numerically.
 
 <p>
 The solution to these two equations gives us in turn \( \beta_1 \) and \( \gamma_1 \) leading to the new expression for \( f_1(x) \) as
@@ -2376,7 +2376,7 @@ See discussion during lecture November 8.
 <h2 id="___sec55">Gradient Boosting, algorithm </h2>
 
 <p>
-Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  square-error  function
+Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  squared-error  function
 <p>&nbsp;<br>
 $$
 C(\boldsymbol{y},\boldsymbol{f})=\sum_{i=0}^{n-1}(y_i-f(x_i))^2.

doc/pub/DecisionTrees/html/DecisionTrees-solarized.html

@@ -2094,7 +2094,7 @@ $$
 \frac{\partial {\cal C}}{\partial \gamma} =-2\sum_{i}\beta x_i(y_i-\beta(1+\gamma x_i))=0.
 $$
 
-We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma x_i) \) with \( \boldsymbol{e} \) being the unit vector)
+We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma \boldsymbol{x}) \) with \( \boldsymbol{e} \) being the unit vector)
 $$
 \gamma \boldsymbol{w}^T(\boldsymbol{y}-\beta\gamma \boldsymbol{w})=0,
 $$
@@ -2106,7 +2106,7 @@ $$
 
 <p>
 which leads to \( \gamma =(\boldsymbol{x}^T\boldsymbol{y}-\beta\boldsymbol{x}^T\boldsymbol{e})/(\beta\boldsymbol{x}^T\boldsymbol{x}) \).  Inserting
-for \( \beta \) gives us an equation for \( \gamma \).
+for \( \beta \) gives us an equation for \( \gamma \). This is a non-linear equation in the unknown \( \gamma \) and has to be solved numerically.
 
 <p>
 The solution to these two equations gives us in turn \( \beta_1 \) and \( \gamma_1 \) leading to the new expression for \( f_1(x) \) as
@@ -2337,7 +2337,7 @@ See discussion during lecture November 8.
 <h2 id="___sec55">Gradient Boosting, algorithm </h2>
 
 <p>
-Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  square-error  function
+Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  squared-error  function
 $$
 C(\boldsymbol{y},\boldsymbol{f})=\sum_{i=0}^{n-1}(y_i-f(x_i))^2.
 $$

doc/pub/DecisionTrees/html/DecisionTrees.html

@@ -2099,7 +2099,7 @@ $$
 \frac{\partial {\cal C}}{\partial \gamma} =-2\sum_{i}\beta x_i(y_i-\beta(1+\gamma x_i))=0.
 $$
 
-We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma x_i) \) with \( \boldsymbol{e} \) being the unit vector)
+We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma \boldsymbol{x}) \) with \( \boldsymbol{e} \) being the unit vector)
 $$
 \gamma \boldsymbol{w}^T(\boldsymbol{y}-\beta\gamma \boldsymbol{w})=0,
 $$
@@ -2111,7 +2111,7 @@ $$
 
 <p>
 which leads to \( \gamma =(\boldsymbol{x}^T\boldsymbol{y}-\beta\boldsymbol{x}^T\boldsymbol{e})/(\beta\boldsymbol{x}^T\boldsymbol{x}) \).  Inserting
-for \( \beta \) gives us an equation for \( \gamma \).
+for \( \beta \) gives us an equation for \( \gamma \). This is a non-linear equation in the unknown \( \gamma \) and has to be solved numerically.
 
 <p>
 The solution to these two equations gives us in turn \( \beta_1 \) and \( \gamma_1 \) leading to the new expression for \( f_1(x) \) as
@@ -2342,7 +2342,7 @@ See discussion during lecture November 8.
 <h2 id="___sec55">Gradient Boosting, algorithm </h2>
 
 <p>
-Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  square-error  function
+Suppose we have a cost function \( C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i)) \) where \( y_i \) is our target and \( f(x_i) \) the function which is meant to model \( y_i \). The above cost function could be our standard  squared-error  function
 $$
 C(\boldsymbol{y},\boldsymbol{f})=\sum_{i=0}^{n-1}(y_i-f(x_i))^2.
 $$

doc/pub/DecisionTrees/ipynb/DecisionTrees.ipynb

@@ -2172,7 +2172,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "We can then rewrite these equations as (defining $\\boldsymbol{w}=\\boldsymbol{e}+\\gamma x_i)$ with $\\boldsymbol{e}$ being the unit vector)"
+    "We can then rewrite these equations as (defining $\\boldsymbol{w}=\\boldsymbol{e}+\\gamma \\boldsymbol{x})$ with $\\boldsymbol{e}$ being the unit vector)"
    ]
   },
   {
@@ -2205,7 +2205,7 @@
    "metadata": {},
    "source": [
     "which leads to $\\gamma =(\\boldsymbol{x}^T\\boldsymbol{y}-\\beta\\boldsymbol{x}^T\\boldsymbol{e})/(\\beta\\boldsymbol{x}^T\\boldsymbol{x})$.  Inserting\n",
-    "for $\\beta$ gives us an equation for $\\gamma$.\n",
+    "for $\\beta$ gives us an equation for $\\gamma$. This is a non-linear equation in the unknown $\\gamma$ and has to be solved numerically. \n",
     "\n",
     "The solution to these two equations gives us in turn $\\beta_1$ and $\\gamma_1$ leading to the new expression for $f_1(x)$ as\n",
     "$f_1(x) = \\beta_1(1+\\gamma_1x)$. Doing this $M$ times results in our final estimate for the function $f$. \n",
@@ -2569,7 +2569,7 @@
     "\n",
     "## Gradient Boosting, algorithm\n",
     "\n",
-    "Suppose we have a cost function $C(f)=\\sum_{i=0}^{n-1}L(y_i, f(x_i))$ where $y_i$ is our target and $f(x_i)$ the function which is meant to model $y_i$. The above cost function could be our standard  square-error  function"
+    "Suppose we have a cost function $C(f)=\\sum_{i=0}^{n-1}L(y_i, f(x_i))$ where $y_i$ is our target and $f(x_i)$ the function which is meant to model $y_i$. The above cost function could be our standard  squared-error  function"
    ]
   },
   {

doc/src/DecisionTrees/DecisionTrees.do.txt

@@ -1727,7 +1727,7 @@ and
 \frac{\partial {\cal C}}{\partial \gamma} =-2\sum_{i}\beta x_i(y_i-\beta(1+\gamma x_i))=0.
 \]
 !et
-We can then rewrite these equations as (defining $\bm{w}=\bm{e}+\gamma x_i)$ with $\bm{e}$ being the unit vector)
+We can then rewrite these equations as (defining $\bm{w}=\bm{e}+\gamma \bm{x})$ with $\bm{e}$ being the unit vector)
 !bt
 \[
 \gamma \bm{w}^T(\bm{y}-\beta\gamma \bm{w})=0,
@@ -1741,7 +1741,7 @@ which gives us $\beta = \bm{w}^T\bm{y}/(\bm{w}^T\bm{w})$. Similarly we have
 !et
 
 which leads to $\gamma =(\bm{x}^T\bm{y}-\beta\bm{x}^T\bm{e})/(\beta\bm{x}^T\bm{x})$.  Inserting
-for $\beta$ gives us an equation for $\gamma$.
+for $\beta$ gives us an equation for $\gamma$. This is a non-linear equation in the unknown $\gamma$ and has to be solved numerically. 
 
 The solution to these two equations gives us in turn $\beta_1$ and $\gamma_1$ leading to the new expression for $f_1(x)$ as
 $f_1(x) = \beta_1(1+\gamma_1x)$. Doing this $M$ times results in our final estimate for the function $f$. 
@@ -1953,7 +1953,7 @@ See discussion during lecture November 8.
 !split
 ===== Gradient Boosting, algorithm =====
 
-Suppose we have a cost function $C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i))$ where $y_i$ is our target and $f(x_i)$ the function which is meant to model $y_i$. The above cost function could be our standard  square-error  function
+Suppose we have a cost function $C(f)=\sum_{i=0}^{n-1}L(y_i, f(x_i))$ where $y_i$ is our target and $f(x_i)$ the function which is meant to model $y_i$. The above cost function could be our standard  squared-error  function
 !bt
 \[
 C(\bm{y},\bm{f})=\sum_{i=0}^{n-1}(y_i-f(x_i))^2.