From 9671ec07b1d8cb6c9225e33920786e18b8cbd1ef Mon Sep 17 00:00:00 2001 From: Morten Hjorth-Jensen Date: Wed, 30 Aug 2023 08:03:11 +0200 Subject: [PATCH] small typo --- doc/pub/week35/html/._week35-bs010.html | 6 +- doc/pub/week35/html/week35-reveal.html | 6 +- doc/pub/week35/html/week35-solarized.html | 6 +- doc/pub/week35/html/week35.html | 6 +- doc/pub/week35/ipynb/ipynb-week35-src.tar.gz | Bin 191 -> 192 bytes doc/pub/week35/ipynb/week35.ipynb | 858 ++++++++++--------- doc/src/week35/week35.do.txt | 6 +- 7 files changed, 450 insertions(+), 438 deletions(-) diff --git a/doc/pub/week35/html/._week35-bs010.html b/doc/pub/week35/html/._week35-bs010.html index 408f18d6c..eb91dd405 100644 --- a/doc/pub/week35/html/._week35-bs010.html +++ b/doc/pub/week35/html/._week35-bs010.html @@ -380,7 +380,7 @@ $$

with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors. In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first -\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then +\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then, using the definition of the Jacobian,

$$ \alpha = \boldsymbol{z}^T\boldsymbol{x}, @@ -388,9 +388,11 @@ $$

which means that (using our previous example) we have

$$ -\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A}. +\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}. $$ +

Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one if 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

$$ \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T. diff --git a/doc/pub/week35/html/week35-reveal.html b/doc/pub/week35/html/week35-reveal.html index cb595855f..42d1a5b03 100644 --- a/doc/pub/week35/html/week35-reveal.html +++ b/doc/pub/week35/html/week35-reveal.html @@ -530,7 +530,7 @@ $$

with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors. In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first -\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then +\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then, using the definition of the Jacobian,

 
$$ @@ -541,10 +541,12 @@ $$

which means that (using our previous example) we have

 
$$ -\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A}. +\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}. $$

 
+

Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one if 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

 
$$ diff --git a/doc/pub/week35/html/week35-solarized.html b/doc/pub/week35/html/week35-solarized.html index 9308cb802..1e0c3bb46 100644 --- a/doc/pub/week35/html/week35-solarized.html +++ b/doc/pub/week35/html/week35-solarized.html @@ -602,7 +602,7 @@ $$

with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors. In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first -\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then +\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then, using the definition of the Jacobian,

$$ \alpha = \boldsymbol{z}^T\boldsymbol{x}, @@ -610,9 +610,11 @@ $$

which means that (using our previous example) we have

$$ -\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A}. +\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}. $$ +

Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one if 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

$$ \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T. diff --git a/doc/pub/week35/html/week35.html b/doc/pub/week35/html/week35.html index 1d054cbb3..e8dff76fd 100644 --- a/doc/pub/week35/html/week35.html +++ b/doc/pub/week35/html/week35.html @@ -679,7 +679,7 @@ $$

with \( \boldsymbol{y} \) a vector of length \( m \), \( \boldsymbol{A} \) an \( m\times n \) matrix and \( \boldsymbol{x} \) a vector of length \( n \). We assume also that \( \boldsymbol{A} \) does not depend on any of the two vectors. In order to find the derivative of \( \alpha \) with respect to the two vectors, we define an intermediate vector \( \boldsymbol{z} \). We define first -\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then +\( \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A} \), a vector of length \( n \). We have then, using the definition of the Jacobian,

$$ \alpha = \boldsymbol{z}^T\boldsymbol{x}, @@ -687,9 +687,11 @@ $$

which means that (using our previous example) we have

$$ -\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}^T=\boldsymbol{y}^T\boldsymbol{A}. +\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}. $$ +

Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one if 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

$$ \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T. diff --git a/doc/pub/week35/ipynb/ipynb-week35-src.tar.gz b/doc/pub/week35/ipynb/ipynb-week35-src.tar.gz index dd8d64c6588fe792eb9a716d93611da239d4d988..9b781eaaf09b2a2eb4a52b6f0d3b1edffe57e130 100644 GIT binary patch literal 192 zcmV;x06+g9iwFRa+wNol1MSaC3c@fD2H>uHia9|^%%f|;E?fvAULd8ZjkQTlQna_X z573q3rihSl^E1pa%p9`yW|t-Y?xV#ZggB)zX3E)=Xf@Xpj2U1`Q^FYID+CA=MtB5h zxszTx=Y<_kX{xi-PU?4a!&q5<*fYNZ&-@d|N;%l>dtVt1+IX28Q^QTlny^H*z0RR9 u(v2;!^4cjYLFgVtQ7EsB7MHlS=43_VlK9up1VIo4UwZ(cmlGia9|^P13FfyKo_hc!88oZLCddlA^u6 z{Q+GmZi)!`HjglmFmuS(n_U*Uy9*XW2uT=&G1G)kNmlcF62}~3#!^anN`@kdDW^05 zvfN59opr(tt2EUal|}WgH`L1V!$0#W@XSAPtfYnQzIT;YptQqW>l$u|HRX|PdzC|> t(T*c9c