Taylor expansion on PRML(5.28)












0












$begingroup$


I'm reading PRML and I have a question,don't solve by myself.

So, please anyone help me.



PRML





Q:Why following third time is not multiplied with $E(boldsymbol{hat{w}})$



$cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} E(boldsymbol{hat{w}}) boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}})$



$E(boldsymbol{w}) simeq E(boldsymbol{hat{w}}) + ( boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{b} + cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}}) cdots(5.28) $



$ boldsymbol{b} equiv nabla E ,|_{w=hat{w}} , cdots(5.29)$





Taylor series



$f(boldsymbol{x}) approx f(boldsymbol{a}) + Df(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a})
+ frac{1}{2} (boldsymbol{x}-boldsymbol{a})^T Hf(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a}).$










share|cite|improve this question











$endgroup$












  • $begingroup$
    A) For all we know $E(hat w)$ might be zero, rendering that term useless. B) Why should it be multiplied? The only effect would be a rescaling of the matrix $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:00












  • $begingroup$
    Are you familiar with the usual (=single variable) Taylor series? With several variable Taylor series the vector $mathbf{b}$ consists of the first order partial derivatives, the matrix $H$ contains the second order partial derivatives etc.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:01












  • $begingroup$
    Thank you for your comment.
    $endgroup$
    – Eiji
    Dec 22 '18 at 8:45












  • $begingroup$
    That $Hf(a)$ is NOT a matrix $H$ multiplied by $f(a)$. It is the Hessian of the function $f$ evaluated at the point $a$. It is probably better to typeset it differently. Like $H(f)(a)$ or $H_f(a)$. Possibly $(Hf)(a)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:48












  • $begingroup$
    Anyway, it looks like the author of the text abbreviated $H(E)$ to just $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:51
















0












$begingroup$


I'm reading PRML and I have a question,don't solve by myself.

So, please anyone help me.



PRML





Q:Why following third time is not multiplied with $E(boldsymbol{hat{w}})$



$cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} E(boldsymbol{hat{w}}) boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}})$



$E(boldsymbol{w}) simeq E(boldsymbol{hat{w}}) + ( boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{b} + cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}}) cdots(5.28) $



$ boldsymbol{b} equiv nabla E ,|_{w=hat{w}} , cdots(5.29)$





Taylor series



$f(boldsymbol{x}) approx f(boldsymbol{a}) + Df(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a})
+ frac{1}{2} (boldsymbol{x}-boldsymbol{a})^T Hf(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a}).$










share|cite|improve this question











$endgroup$












  • $begingroup$
    A) For all we know $E(hat w)$ might be zero, rendering that term useless. B) Why should it be multiplied? The only effect would be a rescaling of the matrix $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:00












  • $begingroup$
    Are you familiar with the usual (=single variable) Taylor series? With several variable Taylor series the vector $mathbf{b}$ consists of the first order partial derivatives, the matrix $H$ contains the second order partial derivatives etc.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:01












  • $begingroup$
    Thank you for your comment.
    $endgroup$
    – Eiji
    Dec 22 '18 at 8:45












  • $begingroup$
    That $Hf(a)$ is NOT a matrix $H$ multiplied by $f(a)$. It is the Hessian of the function $f$ evaluated at the point $a$. It is probably better to typeset it differently. Like $H(f)(a)$ or $H_f(a)$. Possibly $(Hf)(a)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:48












  • $begingroup$
    Anyway, it looks like the author of the text abbreviated $H(E)$ to just $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:51














0












0








0





$begingroup$


I'm reading PRML and I have a question,don't solve by myself.

So, please anyone help me.



PRML





Q:Why following third time is not multiplied with $E(boldsymbol{hat{w}})$



$cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} E(boldsymbol{hat{w}}) boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}})$



$E(boldsymbol{w}) simeq E(boldsymbol{hat{w}}) + ( boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{b} + cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}}) cdots(5.28) $



$ boldsymbol{b} equiv nabla E ,|_{w=hat{w}} , cdots(5.29)$





Taylor series



$f(boldsymbol{x}) approx f(boldsymbol{a}) + Df(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a})
+ frac{1}{2} (boldsymbol{x}-boldsymbol{a})^T Hf(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a}).$










share|cite|improve this question











$endgroup$




I'm reading PRML and I have a question,don't solve by myself.

So, please anyone help me.



PRML





Q:Why following third time is not multiplied with $E(boldsymbol{hat{w}})$



$cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} E(boldsymbol{hat{w}}) boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}})$



$E(boldsymbol{w}) simeq E(boldsymbol{hat{w}}) + ( boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{b} + cfrac{1}{2} (boldsymbol{w} - boldsymbol{hat{w}} )^mathrm{T} boldsymbol{H}(boldsymbol{w} - boldsymbol{hat{w}}) cdots(5.28) $



$ boldsymbol{b} equiv nabla E ,|_{w=hat{w}} , cdots(5.29)$





Taylor series



$f(boldsymbol{x}) approx f(boldsymbol{a}) + Df(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a})
+ frac{1}{2} (boldsymbol{x}-boldsymbol{a})^T Hf(boldsymbol{a}) (boldsymbol{x}-boldsymbol{a}).$







optimization machine-learning hessian-matrix






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 8:36







Eiji

















asked Dec 22 '18 at 7:29









EijiEiji

11




11












  • $begingroup$
    A) For all we know $E(hat w)$ might be zero, rendering that term useless. B) Why should it be multiplied? The only effect would be a rescaling of the matrix $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:00












  • $begingroup$
    Are you familiar with the usual (=single variable) Taylor series? With several variable Taylor series the vector $mathbf{b}$ consists of the first order partial derivatives, the matrix $H$ contains the second order partial derivatives etc.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:01












  • $begingroup$
    Thank you for your comment.
    $endgroup$
    – Eiji
    Dec 22 '18 at 8:45












  • $begingroup$
    That $Hf(a)$ is NOT a matrix $H$ multiplied by $f(a)$. It is the Hessian of the function $f$ evaluated at the point $a$. It is probably better to typeset it differently. Like $H(f)(a)$ or $H_f(a)$. Possibly $(Hf)(a)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:48












  • $begingroup$
    Anyway, it looks like the author of the text abbreviated $H(E)$ to just $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:51


















  • $begingroup$
    A) For all we know $E(hat w)$ might be zero, rendering that term useless. B) Why should it be multiplied? The only effect would be a rescaling of the matrix $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:00












  • $begingroup$
    Are you familiar with the usual (=single variable) Taylor series? With several variable Taylor series the vector $mathbf{b}$ consists of the first order partial derivatives, the matrix $H$ contains the second order partial derivatives etc.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:01












  • $begingroup$
    Thank you for your comment.
    $endgroup$
    – Eiji
    Dec 22 '18 at 8:45












  • $begingroup$
    That $Hf(a)$ is NOT a matrix $H$ multiplied by $f(a)$. It is the Hessian of the function $f$ evaluated at the point $a$. It is probably better to typeset it differently. Like $H(f)(a)$ or $H_f(a)$. Possibly $(Hf)(a)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:48












  • $begingroup$
    Anyway, it looks like the author of the text abbreviated $H(E)$ to just $H$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 8:51
















$begingroup$
A) For all we know $E(hat w)$ might be zero, rendering that term useless. B) Why should it be multiplied? The only effect would be a rescaling of the matrix $H$.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:00






$begingroup$
A) For all we know $E(hat w)$ might be zero, rendering that term useless. B) Why should it be multiplied? The only effect would be a rescaling of the matrix $H$.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:00














$begingroup$
Are you familiar with the usual (=single variable) Taylor series? With several variable Taylor series the vector $mathbf{b}$ consists of the first order partial derivatives, the matrix $H$ contains the second order partial derivatives etc.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:01






$begingroup$
Are you familiar with the usual (=single variable) Taylor series? With several variable Taylor series the vector $mathbf{b}$ consists of the first order partial derivatives, the matrix $H$ contains the second order partial derivatives etc.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:01














$begingroup$
Thank you for your comment.
$endgroup$
– Eiji
Dec 22 '18 at 8:45






$begingroup$
Thank you for your comment.
$endgroup$
– Eiji
Dec 22 '18 at 8:45














$begingroup$
That $Hf(a)$ is NOT a matrix $H$ multiplied by $f(a)$. It is the Hessian of the function $f$ evaluated at the point $a$. It is probably better to typeset it differently. Like $H(f)(a)$ or $H_f(a)$. Possibly $(Hf)(a)$.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:48






$begingroup$
That $Hf(a)$ is NOT a matrix $H$ multiplied by $f(a)$. It is the Hessian of the function $f$ evaluated at the point $a$. It is probably better to typeset it differently. Like $H(f)(a)$ or $H_f(a)$. Possibly $(Hf)(a)$.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:48














$begingroup$
Anyway, it looks like the author of the text abbreviated $H(E)$ to just $H$.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:51




$begingroup$
Anyway, it looks like the author of the text abbreviated $H(E)$ to just $H$.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 8:51










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