Modified Hermite interpolation
$begingroup$
I asked similar questions here and here, but I tried to formulate this
one in a sharper way.
Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?
For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
$$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$
Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
$$left[
begin{matrix}
1&a_1 & a_1^2 & a_1^3\
1&a_2 & a_2^2 & a_2^3\
0&1 & 2a_3 & 3a_3^2\
0&1 & 2a_4 & 3a_4^2
end{matrix}
right]
left[
begin{matrix}
alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
end{matrix}
right] =
left[
begin{matrix}
f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
end{matrix}
right].
$$
I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.
In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.
calculus reference-request polynomials derivatives interpolation
$endgroup$
add a comment |
$begingroup$
I asked similar questions here and here, but I tried to formulate this
one in a sharper way.
Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?
For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
$$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$
Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
$$left[
begin{matrix}
1&a_1 & a_1^2 & a_1^3\
1&a_2 & a_2^2 & a_2^3\
0&1 & 2a_3 & 3a_3^2\
0&1 & 2a_4 & 3a_4^2
end{matrix}
right]
left[
begin{matrix}
alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
end{matrix}
right] =
left[
begin{matrix}
f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
end{matrix}
right].
$$
I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.
In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.
calculus reference-request polynomials derivatives interpolation
$endgroup$
add a comment |
$begingroup$
I asked similar questions here and here, but I tried to formulate this
one in a sharper way.
Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?
For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
$$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$
Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
$$left[
begin{matrix}
1&a_1 & a_1^2 & a_1^3\
1&a_2 & a_2^2 & a_2^3\
0&1 & 2a_3 & 3a_3^2\
0&1 & 2a_4 & 3a_4^2
end{matrix}
right]
left[
begin{matrix}
alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
end{matrix}
right] =
left[
begin{matrix}
f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
end{matrix}
right].
$$
I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.
In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.
calculus reference-request polynomials derivatives interpolation
$endgroup$
I asked similar questions here and here, but I tried to formulate this
one in a sharper way.
Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?
For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
$$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$
Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
$$left[
begin{matrix}
1&a_1 & a_1^2 & a_1^3\
1&a_2 & a_2^2 & a_2^3\
0&1 & 2a_3 & 3a_3^2\
0&1 & 2a_4 & 3a_4^2
end{matrix}
right]
left[
begin{matrix}
alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
end{matrix}
right] =
left[
begin{matrix}
f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
end{matrix}
right].
$$
I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.
In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.
calculus reference-request polynomials derivatives interpolation
calculus reference-request polynomials derivatives interpolation
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Sep 8 '14 at 21:25
Eric AuldEric Auld
13.3k433112
13.3k433112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
See my answer at your previous question.
If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
$$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
linear as a function of the $i$'th row tells you that the partial derivative
of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
from the Laplace expansion (aka minor expansion) along the $i$'th row:
$$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
where the $M_{ik}$ don't depend on $x_i$, so
$$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
which is the determinant of the matrix obtained from $V$ by replacing
each entry of row $i$ by its partial derivative wrt $x_i$.
$endgroup$
$begingroup$
Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
$endgroup$
– Eric Auld
Sep 16 '14 at 21:05
$begingroup$
Well, now you've taken care of the first question in the other post. :)
$endgroup$
– Eric Auld
Sep 17 '14 at 1:22
$begingroup$
I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
$endgroup$
– Eric Auld
Sep 18 '14 at 17:33
add a comment |
$begingroup$
Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:
Estimate the difference between $f$ and $p$ interpolating $f$
I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.
There is a unique polynomial interpolating $f$ and its derivatives
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See my answer at your previous question.
If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
$$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
linear as a function of the $i$'th row tells you that the partial derivative
of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
from the Laplace expansion (aka minor expansion) along the $i$'th row:
$$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
where the $M_{ik}$ don't depend on $x_i$, so
$$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
which is the determinant of the matrix obtained from $V$ by replacing
each entry of row $i$ by its partial derivative wrt $x_i$.
$endgroup$
$begingroup$
Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
$endgroup$
– Eric Auld
Sep 16 '14 at 21:05
$begingroup$
Well, now you've taken care of the first question in the other post. :)
$endgroup$
– Eric Auld
Sep 17 '14 at 1:22
$begingroup$
I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
$endgroup$
– Eric Auld
Sep 18 '14 at 17:33
add a comment |
$begingroup$
See my answer at your previous question.
If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
$$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
linear as a function of the $i$'th row tells you that the partial derivative
of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
from the Laplace expansion (aka minor expansion) along the $i$'th row:
$$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
where the $M_{ik}$ don't depend on $x_i$, so
$$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
which is the determinant of the matrix obtained from $V$ by replacing
each entry of row $i$ by its partial derivative wrt $x_i$.
$endgroup$
$begingroup$
Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
$endgroup$
– Eric Auld
Sep 16 '14 at 21:05
$begingroup$
Well, now you've taken care of the first question in the other post. :)
$endgroup$
– Eric Auld
Sep 17 '14 at 1:22
$begingroup$
I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
$endgroup$
– Eric Auld
Sep 18 '14 at 17:33
add a comment |
$begingroup$
See my answer at your previous question.
If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
$$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
linear as a function of the $i$'th row tells you that the partial derivative
of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
from the Laplace expansion (aka minor expansion) along the $i$'th row:
$$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
where the $M_{ik}$ don't depend on $x_i$, so
$$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
which is the determinant of the matrix obtained from $V$ by replacing
each entry of row $i$ by its partial derivative wrt $x_i$.
$endgroup$
See my answer at your previous question.
If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
$$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
linear as a function of the $i$'th row tells you that the partial derivative
of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
from the Laplace expansion (aka minor expansion) along the $i$'th row:
$$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
where the $M_{ik}$ don't depend on $x_i$, so
$$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
which is the determinant of the matrix obtained from $V$ by replacing
each entry of row $i$ by its partial derivative wrt $x_i$.
edited Dec 11 '18 at 4:12
Aloizio Macedo♦
23.8k24088
23.8k24088
answered Sep 16 '14 at 20:55
Robert IsraelRobert Israel
329k23218472
329k23218472
$begingroup$
Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
$endgroup$
– Eric Auld
Sep 16 '14 at 21:05
$begingroup$
Well, now you've taken care of the first question in the other post. :)
$endgroup$
– Eric Auld
Sep 17 '14 at 1:22
$begingroup$
I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
$endgroup$
– Eric Auld
Sep 18 '14 at 17:33
add a comment |
$begingroup$
Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
$endgroup$
– Eric Auld
Sep 16 '14 at 21:05
$begingroup$
Well, now you've taken care of the first question in the other post. :)
$endgroup$
– Eric Auld
Sep 17 '14 at 1:22
$begingroup$
I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
$endgroup$
– Eric Auld
Sep 18 '14 at 17:33
$begingroup$
Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
$endgroup$
– Eric Auld
Sep 16 '14 at 21:05
$begingroup$
Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
$endgroup$
– Eric Auld
Sep 16 '14 at 21:05
$begingroup$
Well, now you've taken care of the first question in the other post. :)
$endgroup$
– Eric Auld
Sep 17 '14 at 1:22
$begingroup$
Well, now you've taken care of the first question in the other post. :)
$endgroup$
– Eric Auld
Sep 17 '14 at 1:22
$begingroup$
I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
$endgroup$
– Eric Auld
Sep 18 '14 at 17:33
$begingroup$
I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
$endgroup$
– Eric Auld
Sep 18 '14 at 17:33
add a comment |
$begingroup$
Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:
Estimate the difference between $f$ and $p$ interpolating $f$
I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.
There is a unique polynomial interpolating $f$ and its derivatives
$endgroup$
add a comment |
$begingroup$
Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:
Estimate the difference between $f$ and $p$ interpolating $f$
I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.
There is a unique polynomial interpolating $f$ and its derivatives
$endgroup$
add a comment |
$begingroup$
Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:
Estimate the difference between $f$ and $p$ interpolating $f$
I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.
There is a unique polynomial interpolating $f$ and its derivatives
$endgroup$
Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:
Estimate the difference between $f$ and $p$ interpolating $f$
I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.
There is a unique polynomial interpolating $f$ and its derivatives
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Sep 18 '14 at 17:36
Eric AuldEric Auld
13.3k433112
13.3k433112
add a comment |
add a comment |
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