Root finding for a convex 3D function
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Say I have a convex 2D function $f(x,y)$ ; x-absciss y-ordinate
Getting zeros of this function is ok for me with the Newton method.
Now say this time I still have a convex function but in 3D $f(x,y,z)$.
You have to imagine a function such as $f(x,y)=x^2-2$ And increasing in the z-dimension (depth).
I'd like to find the value of $z$ such as the whole function is positive.
I was thinking about using Newton and dichotomy but the function is not monotonic... Any clue ?
Edit:
I know the range of z where to search, say [5, 15]
I could do a Newton search on different points: 5,6,7,...,15 and then look at the optimal value but this would not be optimal.
roots
add a comment |Â
up vote
0
down vote
favorite
Say I have a convex 2D function $f(x,y)$ ; x-absciss y-ordinate
Getting zeros of this function is ok for me with the Newton method.
Now say this time I still have a convex function but in 3D $f(x,y,z)$.
You have to imagine a function such as $f(x,y)=x^2-2$ And increasing in the z-dimension (depth).
I'd like to find the value of $z$ such as the whole function is positive.
I was thinking about using Newton and dichotomy but the function is not monotonic... Any clue ?
Edit:
I know the range of z where to search, say [5, 15]
I could do a Newton search on different points: 5,6,7,...,15 and then look at the optimal value but this would not be optimal.
roots
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Say I have a convex 2D function $f(x,y)$ ; x-absciss y-ordinate
Getting zeros of this function is ok for me with the Newton method.
Now say this time I still have a convex function but in 3D $f(x,y,z)$.
You have to imagine a function such as $f(x,y)=x^2-2$ And increasing in the z-dimension (depth).
I'd like to find the value of $z$ such as the whole function is positive.
I was thinking about using Newton and dichotomy but the function is not monotonic... Any clue ?
Edit:
I know the range of z where to search, say [5, 15]
I could do a Newton search on different points: 5,6,7,...,15 and then look at the optimal value but this would not be optimal.
roots
Say I have a convex 2D function $f(x,y)$ ; x-absciss y-ordinate
Getting zeros of this function is ok for me with the Newton method.
Now say this time I still have a convex function but in 3D $f(x,y,z)$.
You have to imagine a function such as $f(x,y)=x^2-2$ And increasing in the z-dimension (depth).
I'd like to find the value of $z$ such as the whole function is positive.
I was thinking about using Newton and dichotomy but the function is not monotonic... Any clue ?
Edit:
I know the range of z where to search, say [5, 15]
I could do a Newton search on different points: 5,6,7,...,15 and then look at the optimal value but this would not be optimal.
roots
edited Aug 21 at 9:04
asked Aug 21 at 8:55
Cedric_W
12
12
add a comment |Â
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%2f2889636%2froot-finding-for-a-convex-3d-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