Mixing
Confusion in Second Derivative of a Vector Valued Function
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I have a slight confusion in Newton Raphson method applied in Higher Dimensions. I have a vector valued function $f(x)$ which maps from $mathbbC^mrightarrow mathbbC^m$ whose root I have to find. I resorted to Newton-Raphson first order method to do this, namely: $$bfx_n+1=bfx_n-fracf(x)dotf(x)$$ However, I started facing problems since sometimes the linear approximation of the function $f(x)$ was quite poor and hence tended to diverge or oscillate. To fix this, I have to introduce some sort of scaling for the Newton step and HouseHolder's second order iterate seems like a suitable candidate. Its given by:$$bfx_n+1=bfx_n-fracf(x)dotf(x)left(1+fracf(x)ddotf(x)2dotf(x)^2right)$$ I understand that $dotf(x)$ is the $bfJacobian$. However, I am quite unsure of what $ddotf(x)$ means for a vector valued function(it would be the $bfHessian$ for a multivariate function mapping to $mathbbR$) and hence I am unable to compute it. Could someone please clarify?
multivariable-calculus newton-raphson jacobian
asked Jul 23 at 12:51
Aakusti
1163
add a comment |Â
up vote
0
down vote
favorite
I have a slight confusion in Newton Raphson method applied in Higher Dimensions. I have a vector valued function $f(x)$ which maps from $mathbbC^mrightarrow mathbbC^m$ whose root I have to find. I resorted to Newton-Raphson first order method to do this, namely: $$bfx_n+1=bfx_n-fracf(x)dotf(x)$$ However, I started facing problems since sometimes the linear approximation of the function $f(x)$ was quite poor and hence tended to diverge or oscillate. To fix this, I have to introduce some sort of scaling for the Newton step and HouseHolder's second order iterate seems like a suitable candidate. Its given by:$$bfx_n+1=bfx_n-fracf(x)dotf(x)left(1+fracf(x)ddotf(x)2dotf(x)^2right)$$ I understand that $dotf(x)$ is the $bfJacobian$. However, I am quite unsure of what $ddotf(x)$ means for a vector valued function(it would be the $bfHessian$ for a multivariate function mapping to $mathbbR$) and hence I am unable to compute it. Could someone please clarify?
multivariable-calculus newton-raphson jacobian
asked Jul 23 at 12:51
Aakusti
1163
$dot f$ is replaced by the jacobian matrix for the multivariate case, not only the jacobian (which is the determinant of the jacobian matrix when it has a meaning). So first thing first, try to think how you would rewrite the first equation in a multivariate case. Then, if $f(x)$ is a vector and $dot f(x)$ is a matrix, what $ddot f(x)$ could be ?
– nicomezi
Jul 23 at 13:09
@nicomezi my bad, it has to be tensor
– Aakusti
Jul 23 at 19:36
Would you add details defining your specific function $f$? It is entirely possible, that a sufficiently good initial guess can be constructed and Newton's method will succeed. It is also vital to investigate if you are operating near a point where the Jacobian is singular.
– Carl Christian
Jul 24 at 17:23
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a slight confusion in Newton Raphson method applied in Higher Dimensions. I have a vector valued function $f(x)$ which maps from $mathbbC^mrightarrow mathbbC^m$ whose root I have to find. I resorted to Newton-Raphson first order method to do this, namely: $$bfx_n+1=bfx_n-fracf(x)dotf(x)$$ However, I started facing problems since sometimes the linear approximation of the function $f(x)$ was quite poor and hence tended to diverge or oscillate. To fix this, I have to introduce some sort of scaling for the Newton step and HouseHolder's second order iterate seems like a suitable candidate. Its given by:$$bfx_n+1=bfx_n-fracf(x)dotf(x)left(1+fracf(x)ddotf(x)2dotf(x)^2right)$$ I understand that $dotf(x)$ is the $bfJacobian$. However, I am quite unsure of what $ddotf(x)$ means for a vector valued function(it would be the $bfHessian$ for a multivariate function mapping to $mathbbR$) and hence I am unable to compute it. Could someone please clarify?
multivariable-calculus newton-raphson jacobian
asked Jul 23 at 12:51
Aakusti
1163
I have a slight confusion in Newton Raphson method applied in Higher Dimensions. I have a vector valued function $f(x)$ which maps from $mathbbC^mrightarrow mathbbC^m$ whose root I have to find. I resorted to Newton-Raphson first order method to do this, namely: $$bfx_n+1=bfx_n-fracf(x)dotf(x)$$ However, I started facing problems since sometimes the linear approximation of the function $f(x)$ was quite poor and hence tended to diverge or oscillate. To fix this, I have to introduce some sort of scaling for the Newton step and HouseHolder's second order iterate seems like a suitable candidate. Its given by:$$bfx_n+1=bfx_n-fracf(x)dotf(x)left(1+fracf(x)ddotf(x)2dotf(x)^2right)$$ I understand that $dotf(x)$ is the $bfJacobian$. However, I am quite unsure of what $ddotf(x)$ means for a vector valued function(it would be the $bfHessian$ for a multivariate function mapping to $mathbbR$) and hence I am unable to compute it. Could someone please clarify?
multivariable-calculus newton-raphson jacobian
asked Jul 23 at 12:51
Aakusti
1163
asked Jul 23 at 12:51
Aakusti
1163
asked Jul 23 at 12:51
Aakusti
1163
asked Jul 23 at 12:51
Aakusti
1163
1163
$dot f$ is replaced by the jacobian matrix for the multivariate case, not only the jacobian (which is the determinant of the jacobian matrix when it has a meaning). So first thing first, try to think how you would rewrite the first equation in a multivariate case. Then, if $f(x)$ is a vector and $dot f(x)$ is a matrix, what $ddot f(x)$ could be ?
– nicomezi
Jul 23 at 13:09
@nicomezi my bad, it has to be tensor
– Aakusti
Jul 23 at 19:36
Would you add details defining your specific function $f$? It is entirely possible, that a sufficiently good initial guess can be constructed and Newton's method will succeed. It is also vital to investigate if you are operating near a point where the Jacobian is singular.
– Carl Christian
Jul 24 at 17:23
add a comment |Â
$dot f$ is replaced by the jacobian matrix for the multivariate case, not only the jacobian (which is the determinant of the jacobian matrix when it has a meaning). So first thing first, try to think how you would rewrite the first equation in a multivariate case. Then, if $f(x)$ is a vector and $dot f(x)$ is a matrix, what $ddot f(x)$ could be ?
– nicomezi
Jul 23 at 13:09
@nicomezi my bad, it has to be tensor
– Aakusti
Jul 23 at 19:36
Would you add details defining your specific function $f$? It is entirely possible, that a sufficiently good initial guess can be constructed and Newton's method will succeed. It is also vital to investigate if you are operating near a point where the Jacobian is singular.
– Carl Christian
Jul 24 at 17:23
$dot f$ is replaced by the jacobian matrix for the multivariate case, not only the jacobian (which is the determinant of the jacobian matrix when it has a meaning). So first thing first, try to think how you would rewrite the first equation in a multivariate case. Then, if $f(x)$ is a vector and $dot f(x)$ is a matrix, what $ddot f(x)$ could be ?
– nicomezi
Jul 23 at 13:09
$dot f$ is replaced by the jacobian matrix for the multivariate case, not only the jacobian (which is the determinant of the jacobian matrix when it has a meaning). So first thing first, try to think how you would rewrite the first equation in a multivariate case. Then, if $f(x)$ is a vector and $dot f(x)$ is a matrix, what $ddot f(x)$ could be ?
– nicomezi
Jul 23 at 13:09
@nicomezi my bad, it has to be tensor
– Aakusti
Jul 23 at 19:36
@nicomezi my bad, it has to be tensor
– Aakusti
Jul 23 at 19:36
Would you add details defining your specific function $f$? It is entirely possible, that a sufficiently good initial guess can be constructed and Newton's method will succeed. It is also vital to investigate if you are operating near a point where the Jacobian is singular.
– Carl Christian
Jul 24 at 17:23
Would you add details defining your specific function $f$? It is entirely possible, that a sufficiently good initial guess can be constructed and Newton's method will succeed. It is also vital to investigate if you are operating near a point where the Jacobian is singular.
– Carl Christian
Jul 24 at 17:23
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860324%2fconfusion-in-second-derivative-of-a-vector-valued-function%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
$dot f$ is replaced by the jacobian matrix for the multivariate case, not only the jacobian (which is the determinant of the jacobian matrix when it has a meaning). So first thing first, try to think how you would rewrite the first equation in a multivariate case. Then, if $f(x)$ is a vector and $dot f(x)$ is a matrix, what $ddot f(x)$ could be ?
– nicomezi
Jul 23 at 13:09
@nicomezi my bad, it has to be tensor
– Aakusti
Jul 23 at 19:36
Would you add details defining your specific function $f$? It is entirely possible, that a sufficiently good initial guess can be constructed and Newton's method will succeed. It is also vital to investigate if you are operating near a point where the Jacobian is singular.
– Carl Christian
Jul 24 at 17:23