why a Bezier curve is guaranteed to lie within the convex hull of its control points?
up vote
0
down vote
favorite
why if Bernstein basis polynomials are non-negative ($ B_{k,n}(x) geq 0 $) and also due to the Partition of Unity/sum up to one ($ sum_{k=0}^n B_{k,n}(x) = 1, for all x in [0,1] $) implies Bezier curve is guaranteed to lie within the convex hull of its control points $ CH=({p_{0},p_{1},p_{2},...,p_{n}})$
differential-geometry bezier-curve
add a comment |
up vote
0
down vote
favorite
why if Bernstein basis polynomials are non-negative ($ B_{k,n}(x) geq 0 $) and also due to the Partition of Unity/sum up to one ($ sum_{k=0}^n B_{k,n}(x) = 1, for all x in [0,1] $) implies Bezier curve is guaranteed to lie within the convex hull of its control points $ CH=({p_{0},p_{1},p_{2},...,p_{n}})$
differential-geometry bezier-curve
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
why if Bernstein basis polynomials are non-negative ($ B_{k,n}(x) geq 0 $) and also due to the Partition of Unity/sum up to one ($ sum_{k=0}^n B_{k,n}(x) = 1, for all x in [0,1] $) implies Bezier curve is guaranteed to lie within the convex hull of its control points $ CH=({p_{0},p_{1},p_{2},...,p_{n}})$
differential-geometry bezier-curve
why if Bernstein basis polynomials are non-negative ($ B_{k,n}(x) geq 0 $) and also due to the Partition of Unity/sum up to one ($ sum_{k=0}^n B_{k,n}(x) = 1, for all x in [0,1] $) implies Bezier curve is guaranteed to lie within the convex hull of its control points $ CH=({p_{0},p_{1},p_{2},...,p_{n}})$
differential-geometry bezier-curve
differential-geometry bezier-curve
edited Nov 17 at 11:34
asked Nov 17 at 2:13
GreenQuestioner
777
777
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Here are the facts:
A convex combination of the points $p_{0},p_{1},p_{2},...,p_{n}$ is a point of the form
$
sum_{k=0}^n lambda_k p_k
$.
where $lambda_k ge 0$ and $sum_{k=0}^n lambda_k = 1$.The convex hull of the points $p_{0},p_{1},p_{2},...,p_{n}$ is the set of all convex combination of $p_{0},p_{1},p_{2},...,p_{n}$.
A point on a Bézier curve is a convex combination of its control points and so is in their convex hull.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here are the facts:
A convex combination of the points $p_{0},p_{1},p_{2},...,p_{n}$ is a point of the form
$
sum_{k=0}^n lambda_k p_k
$.
where $lambda_k ge 0$ and $sum_{k=0}^n lambda_k = 1$.The convex hull of the points $p_{0},p_{1},p_{2},...,p_{n}$ is the set of all convex combination of $p_{0},p_{1},p_{2},...,p_{n}$.
A point on a Bézier curve is a convex combination of its control points and so is in their convex hull.
add a comment |
up vote
0
down vote
Here are the facts:
A convex combination of the points $p_{0},p_{1},p_{2},...,p_{n}$ is a point of the form
$
sum_{k=0}^n lambda_k p_k
$.
where $lambda_k ge 0$ and $sum_{k=0}^n lambda_k = 1$.The convex hull of the points $p_{0},p_{1},p_{2},...,p_{n}$ is the set of all convex combination of $p_{0},p_{1},p_{2},...,p_{n}$.
A point on a Bézier curve is a convex combination of its control points and so is in their convex hull.
add a comment |
up vote
0
down vote
up vote
0
down vote
Here are the facts:
A convex combination of the points $p_{0},p_{1},p_{2},...,p_{n}$ is a point of the form
$
sum_{k=0}^n lambda_k p_k
$.
where $lambda_k ge 0$ and $sum_{k=0}^n lambda_k = 1$.The convex hull of the points $p_{0},p_{1},p_{2},...,p_{n}$ is the set of all convex combination of $p_{0},p_{1},p_{2},...,p_{n}$.
A point on a Bézier curve is a convex combination of its control points and so is in their convex hull.
Here are the facts:
A convex combination of the points $p_{0},p_{1},p_{2},...,p_{n}$ is a point of the form
$
sum_{k=0}^n lambda_k p_k
$.
where $lambda_k ge 0$ and $sum_{k=0}^n lambda_k = 1$.The convex hull of the points $p_{0},p_{1},p_{2},...,p_{n}$ is the set of all convex combination of $p_{0},p_{1},p_{2},...,p_{n}$.
A point on a Bézier curve is a convex combination of its control points and so is in their convex hull.
answered Nov 17 at 11:40
lhf
161k9165384
161k9165384
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001888%2fwhy-a-bezier-curve-is-guaranteed-to-lie-within-the-convex-hull-of-its-control-po%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown