Newton's Method for a step size to move in the direction of the gradient
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I am reading this article that talks about Newton's method that can give us an ideal step size to move in the direction of the gradient.
I do not understand what $epsilon$ is in the following part of the article:
I am confused on what $g$ is and what this math expression represents.
Can someone help me to understand it?
Thank you!
calculus gradient-descent hessian-matrix
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
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add a comment |
$begingroup$
I am reading this article that talks about Newton's method that can give us an ideal step size to move in the direction of the gradient.
I do not understand what $epsilon$ is in the following part of the article:
I am confused on what $g$ is and what this math expression represents.
Can someone help me to understand it?
Thank you!
calculus gradient-descent hessian-matrix
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
$endgroup$
$begingroup$
You may also be interested in Steffensen's method, another Newton variant that has shrinking step size as we approach the root.
$endgroup$
– J.G.
Dec 24 '18 at 23:37
$begingroup$
If you approximate the cost near $x$ by $f(x+h) approx langle g, h rangle >+ { 1over 2} langle h, H h rangle$ ($g$ is the gradient of $f$, $H$ is the Hessian) and you consider a step in the direction $-g$ then the step size that extremises the approximate cost is given by the above formula. (That is, consider the approximation with parameter $h=-lambda g$.)
$endgroup$
– copper.hat
Dec 25 '18 at 1:40
add a comment |
$begingroup$
I am reading this article that talks about Newton's method that can give us an ideal step size to move in the direction of the gradient.
I do not understand what $epsilon$ is in the following part of the article:
I am confused on what $g$ is and what this math expression represents.
Can someone help me to understand it?
Thank you!
calculus gradient-descent hessian-matrix
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
$endgroup$
I am reading this article that talks about Newton's method that can give us an ideal step size to move in the direction of the gradient.
I do not understand what $epsilon$ is in the following part of the article:
I am confused on what $g$ is and what this math expression represents.
Can someone help me to understand it?
Thank you!
calculus gradient-descent hessian-matrix
calculus gradient-descent hessian-matrix
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
asked Dec 24 '18 at 23:29
YohanRothYohanRoth
6471715
6471715
$begingroup$
You may also be interested in Steffensen's method, another Newton variant that has shrinking step size as we approach the root.
$endgroup$
– J.G.
Dec 24 '18 at 23:37
$begingroup$
If you approximate the cost near $x$ by $f(x+h) approx langle g, h rangle >+ { 1over 2} langle h, H h rangle$ ($g$ is the gradient of $f$, $H$ is the Hessian) and you consider a step in the direction $-g$ then the step size that extremises the approximate cost is given by the above formula. (That is, consider the approximation with parameter $h=-lambda g$.)
$endgroup$
– copper.hat
Dec 25 '18 at 1:40
add a comment |
$begingroup$
You may also be interested in Steffensen's method, another Newton variant that has shrinking step size as we approach the root.
$endgroup$
– J.G.
Dec 24 '18 at 23:37
$begingroup$
If you approximate the cost near $x$ by $f(x+h) approx langle g, h rangle >+ { 1over 2} langle h, H h rangle$ ($g$ is the gradient of $f$, $H$ is the Hessian) and you consider a step in the direction $-g$ then the step size that extremises the approximate cost is given by the above formula. (That is, consider the approximation with parameter $h=-lambda g$.)
$endgroup$
– copper.hat
Dec 25 '18 at 1:40
$begingroup$
You may also be interested in Steffensen's method, another Newton variant that has shrinking step size as we approach the root.
$endgroup$
– J.G.
Dec 24 '18 at 23:37
$begingroup$
You may also be interested in Steffensen's method, another Newton variant that has shrinking step size as we approach the root.
$endgroup$
– J.G.
Dec 24 '18 at 23:37
$begingroup$
If you approximate the cost near $x$ by $f(x+h) approx langle g, h rangle >+ { 1over 2} langle h, H h rangle$ ($g$ is the gradient of $f$, $H$ is the Hessian) and you consider a step in the direction $-g$ then the step size that extremises the approximate cost is given by the above formula. (That is, consider the approximation with parameter $h=-lambda g$.)
$endgroup$
– copper.hat
Dec 25 '18 at 1:40
$begingroup$
If you approximate the cost near $x$ by $f(x+h) approx langle g, h rangle >+ { 1over 2} langle h, H h rangle$ ($g$ is the gradient of $f$, $H$ is the Hessian) and you consider a step in the direction $-g$ then the step size that extremises the approximate cost is given by the above formula. (That is, consider the approximation with parameter $h=-lambda g$.)
$endgroup$
– copper.hat
Dec 25 '18 at 1:40
add a comment |
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$begingroup$
You may also be interested in Steffensen's method, another Newton variant that has shrinking step size as we approach the root.
$endgroup$
– J.G.
Dec 24 '18 at 23:37
$begingroup$
If you approximate the cost near $x$ by $f(x+h) approx langle g, h rangle >+ { 1over 2} langle h, H h rangle$ ($g$ is the gradient of $f$, $H$ is the Hessian) and you consider a step in the direction $-g$ then the step size that extremises the approximate cost is given by the above formula. (That is, consider the approximation with parameter $h=-lambda g$.)
$endgroup$
– copper.hat
Dec 25 '18 at 1:40