From 8e4948fb2a0e4167c7ef00f7947accbc9a1749d0 Mon Sep 17 00:00:00 2001
From: Morten Hjorth-Jensen
which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have
$$ -\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}. +\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}. $$Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.
-Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
+Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
$$ \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T. $$ diff --git a/doc/pub/week35/html/._week35-bs014.html b/doc/pub/week35/html/._week35-bs014.html index 3e251c1f6..9ebd409ce 100644 --- a/doc/pub/week35/html/._week35-bs014.html +++ b/doc/pub/week35/html/._week35-bs014.html @@ -364,9 +364,14 @@ MathJax.Hub.Config({We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)
$$ -\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b}, +\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T, $$ +$$ +\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T, +$$ + + $$ \frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T, $$ diff --git a/doc/pub/week35/html/week35-reveal.html b/doc/pub/week35/html/week35-reveal.html index fb3474d30..78deaf60f 100644 --- a/doc/pub/week35/html/week35-reveal.html +++ b/doc/pub/week35/html/week35-reveal.html @@ -537,16 +537,16 @@ $$ $$
-
which means that (using our previous example) we have
+which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have
$$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
$$
Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.
-Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
+Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
$$
\frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T.
@@ -715,7 +715,13 @@ $$
We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)
$$
-\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
+$$
+
+
+
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
$$
diff --git a/doc/pub/week35/html/week35-solarized.html b/doc/pub/week35/html/week35-solarized.html
index eabfc92d3..9310fb771 100644
--- a/doc/pub/week35/html/week35-solarized.html
+++ b/doc/pub/week35/html/week35-solarized.html
@@ -602,14 +602,14 @@ $$
\alpha = \boldsymbol{z}^T\boldsymbol{x},
$$
-
which means that (using our previous example) we have
+which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have
$$ -\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}. +\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}. $$Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.
-Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
+Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
$$ \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T. $$ @@ -739,9 +739,14 @@ $$We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)
$$ -\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b}, +\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T, $$ +$$ +\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T, +$$ + + $$ \frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T, $$ diff --git a/doc/pub/week35/html/week35.html b/doc/pub/week35/html/week35.html index ce93085af..64d36e6e1 100644 --- a/doc/pub/week35/html/week35.html +++ b/doc/pub/week35/html/week35.html @@ -679,14 +679,14 @@ $$ \alpha = \boldsymbol{z}^T\boldsymbol{x}, $$ -which means that (using our previous example) we have
+which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have
$$ -\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}. +\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}. $$Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.
-Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
+Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that
$$ \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T. $$ @@ -816,9 +816,14 @@ $$We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)
$$ -\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b}, +\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T, $$ +$$ +\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T, +$$ + + $$ \frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T, $$ diff --git a/doc/pub/week35/ipynb/ipynb-week35-src.tar.gz b/doc/pub/week35/ipynb/ipynb-week35-src.tar.gz index 2a9642d24f53d26324aafb355f3d552d03943356..49dc5e02d46b13322cd939ab8275d24fddca5228 100644 GIT binary patch literal 192 zcmV;x06+g9iwFQ#K+R?V1MSaC3c@fD2H>uHia9|^(j;9Ax^N+gc!89rHr6IJNzvZk zK0sHBn<7HK&Cf8yFmu?f*1JvO?><@#LWoleV`iL9iO5_}FlK-$=P98G6AlQcjARj@ z